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What allows us to use imaginary numbers?
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What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Does anyone know about how mathematical operations' ranges and domains were expanded to include imaginaries?
EDIT: An interesting comment notes a first use of complex numbers where,
those values would cancel in the end.
But can I refute that with, "from an inconsistency, anything is provable"?
A corollary question: Could I define a new number z which is 1/0 and simply begin using it? Seems ludicrous.
complex-numbers
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show 24 more comments
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Does anyone know about how mathematical operations' ranges and domains were expanded to include imaginaries?
EDIT: An interesting comment notes a first use of complex numbers where,
those values would cancel in the end.
But can I refute that with, "from an inconsistency, anything is provable"?
A corollary question: Could I define a new number z which is 1/0 and simply begin using it? Seems ludicrous.
complex-numbers
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29
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Would it bother you less simply to think of $a + bi$ as shorthand for the matrix $$begin{pmatrix}a & -b \ b & aend{pmatrix}?$$
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– Bungo
2 days ago
40
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You may be surprised to learn that the real numbers are not really older: when $i$ was first considered, negative numbers were not generally accepted as numbers either.
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– Robert Israel
2 days ago
20
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Interestingly, one can introduce $sqrt{-1}$ so that all operations 'behaves the usual way', while introducing $1/0$ will definitely break some familiar laws. Nevertheless, it can be done..
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– Berci
2 days ago
5
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One particular turning point was the search for a cubic formula similar to the familiar quadratic. (See Wikipedia's "Cubic Function" entry.) The mathematician Tartaglia had a method for solving certain cubics; the method required taking square roots of negatives; this didn't really bother him, since the those values would cancel in the end. Eventually, it was found that, for cubics in general, those values didn't always cancel, and so mathematics needed to come to grips with them. Then Gauss came along and put everything together.
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– Blue
2 days ago
9
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Also, in the beginning there were only rational numbers. The discovery that $sqrt{2}$ was irrational apparently cost the discoverer his life.
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– Lee Mosher
2 days ago
|
show 24 more comments
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Does anyone know about how mathematical operations' ranges and domains were expanded to include imaginaries?
EDIT: An interesting comment notes a first use of complex numbers where,
those values would cancel in the end.
But can I refute that with, "from an inconsistency, anything is provable"?
A corollary question: Could I define a new number z which is 1/0 and simply begin using it? Seems ludicrous.
complex-numbers
$endgroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Does anyone know about how mathematical operations' ranges and domains were expanded to include imaginaries?
EDIT: An interesting comment notes a first use of complex numbers where,
those values would cancel in the end.
But can I refute that with, "from an inconsistency, anything is provable"?
A corollary question: Could I define a new number z which is 1/0 and simply begin using it? Seems ludicrous.
complex-numbers
complex-numbers
edited 2 days ago
Rodrigo de Azevedo
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asked 2 days ago
DogweatherDogweather
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29
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Would it bother you less simply to think of $a + bi$ as shorthand for the matrix $$begin{pmatrix}a & -b \ b & aend{pmatrix}?$$
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– Bungo
2 days ago
40
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You may be surprised to learn that the real numbers are not really older: when $i$ was first considered, negative numbers were not generally accepted as numbers either.
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– Robert Israel
2 days ago
20
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Interestingly, one can introduce $sqrt{-1}$ so that all operations 'behaves the usual way', while introducing $1/0$ will definitely break some familiar laws. Nevertheless, it can be done..
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– Berci
2 days ago
5
$begingroup$
One particular turning point was the search for a cubic formula similar to the familiar quadratic. (See Wikipedia's "Cubic Function" entry.) The mathematician Tartaglia had a method for solving certain cubics; the method required taking square roots of negatives; this didn't really bother him, since the those values would cancel in the end. Eventually, it was found that, for cubics in general, those values didn't always cancel, and so mathematics needed to come to grips with them. Then Gauss came along and put everything together.
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– Blue
2 days ago
9
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Also, in the beginning there were only rational numbers. The discovery that $sqrt{2}$ was irrational apparently cost the discoverer his life.
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– Lee Mosher
2 days ago
|
show 24 more comments
29
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Would it bother you less simply to think of $a + bi$ as shorthand for the matrix $$begin{pmatrix}a & -b \ b & aend{pmatrix}?$$
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– Bungo
2 days ago
40
$begingroup$
You may be surprised to learn that the real numbers are not really older: when $i$ was first considered, negative numbers were not generally accepted as numbers either.
$endgroup$
– Robert Israel
2 days ago
20
$begingroup$
Interestingly, one can introduce $sqrt{-1}$ so that all operations 'behaves the usual way', while introducing $1/0$ will definitely break some familiar laws. Nevertheless, it can be done..
$endgroup$
– Berci
2 days ago
5
$begingroup$
One particular turning point was the search for a cubic formula similar to the familiar quadratic. (See Wikipedia's "Cubic Function" entry.) The mathematician Tartaglia had a method for solving certain cubics; the method required taking square roots of negatives; this didn't really bother him, since the those values would cancel in the end. Eventually, it was found that, for cubics in general, those values didn't always cancel, and so mathematics needed to come to grips with them. Then Gauss came along and put everything together.
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– Blue
2 days ago
9
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Also, in the beginning there were only rational numbers. The discovery that $sqrt{2}$ was irrational apparently cost the discoverer his life.
$endgroup$
– Lee Mosher
2 days ago
29
29
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Would it bother you less simply to think of $a + bi$ as shorthand for the matrix $$begin{pmatrix}a & -b \ b & aend{pmatrix}?$$
$endgroup$
– Bungo
2 days ago
$begingroup$
Would it bother you less simply to think of $a + bi$ as shorthand for the matrix $$begin{pmatrix}a & -b \ b & aend{pmatrix}?$$
$endgroup$
– Bungo
2 days ago
40
40
$begingroup$
You may be surprised to learn that the real numbers are not really older: when $i$ was first considered, negative numbers were not generally accepted as numbers either.
$endgroup$
– Robert Israel
2 days ago
$begingroup$
You may be surprised to learn that the real numbers are not really older: when $i$ was first considered, negative numbers were not generally accepted as numbers either.
$endgroup$
– Robert Israel
2 days ago
20
20
$begingroup$
Interestingly, one can introduce $sqrt{-1}$ so that all operations 'behaves the usual way', while introducing $1/0$ will definitely break some familiar laws. Nevertheless, it can be done..
$endgroup$
– Berci
2 days ago
$begingroup$
Interestingly, one can introduce $sqrt{-1}$ so that all operations 'behaves the usual way', while introducing $1/0$ will definitely break some familiar laws. Nevertheless, it can be done..
$endgroup$
– Berci
2 days ago
5
5
$begingroup$
One particular turning point was the search for a cubic formula similar to the familiar quadratic. (See Wikipedia's "Cubic Function" entry.) The mathematician Tartaglia had a method for solving certain cubics; the method required taking square roots of negatives; this didn't really bother him, since the those values would cancel in the end. Eventually, it was found that, for cubics in general, those values didn't always cancel, and so mathematics needed to come to grips with them. Then Gauss came along and put everything together.
$endgroup$
– Blue
2 days ago
$begingroup$
One particular turning point was the search for a cubic formula similar to the familiar quadratic. (See Wikipedia's "Cubic Function" entry.) The mathematician Tartaglia had a method for solving certain cubics; the method required taking square roots of negatives; this didn't really bother him, since the those values would cancel in the end. Eventually, it was found that, for cubics in general, those values didn't always cancel, and so mathematics needed to come to grips with them. Then Gauss came along and put everything together.
$endgroup$
– Blue
2 days ago
9
9
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Also, in the beginning there were only rational numbers. The discovery that $sqrt{2}$ was irrational apparently cost the discoverer his life.
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– Lee Mosher
2 days ago
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Also, in the beginning there were only rational numbers. The discovery that $sqrt{2}$ was irrational apparently cost the discoverer his life.
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– Lee Mosher
2 days ago
|
show 24 more comments
11 Answers
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We can do anything we want!
Specifically, we can define anything we want (as long as our definitions don't contradict each other). So if we want to allow ourselves to use imaginary numbers, all we have to do is write something like the following:
Define a complex number as an ordered pair of the form $(a, b)$, where $a$ and $b$ are real numbers.
Define $i$ as the complex number $(0, 1)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) + (c, d)$ as $(a + c, b + d)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) cdot (c, d)$ as $(ac - bd, ad + bc)$.
And define subtraction and division in similar ways.
Is that it? Are we done? No, there's still more that we want to do. There are a lot of useful theorems about real numbers that also apply to the complex numbers, but we don't know that they apply to the complex numbers until we prove them. For example, one very useful theorem about real numbers is:
Theorem: If $a$ and $b$ are real numbers, then $a + b = b + a$.
The analogous theorem about the complex numbers is:
Theorem (not yet proven): If $a$ and $b$ are complex numbers, then $a + b = b + a$.
This theorem is, in fact, true, but we didn't know that it was true until somebody proved it.
Once we've proved all the theorems that we want to prove, then we can say that we're "done."
(Do we have to prove these theorems? No, we don't have to if we don't want to. But without these theorems, complex numbers aren't very useful.)
As for your corollary question:
Could I define a new number $z$ which is $1/0$ and simply begin using it? Seems ludicrous.
Yes, you absolutely can! All you have to do is write:
Assume that there is a value $z$. Define $1/0$ as $z$.
And that's perfectly valid; this definition doesn't contradict any other definitions. This is completely legal, acceptable and proper.
Is that it? Are we done? Probably not; there's more we'd like to do. For example, what do you suppose $z cdot 0$ is? There are a couple of theorems here we might like to use, but we can't. Let's take a look at them:
Theorem: If $x$ is a real number, then $x cdot 0 = 0$.
Theorem: If $x$ and $y$ are real numbers, and $y ne 0$, then $(x / y) cdot y = x$.
Do you see why we can't use these theorems?
Does the first theorem tell us that $z cdot 0 = 0$? No, because we don't know that $z$ is a real number. So the first theorem doesn't apply.
How about the second theorem? We know that $z = 1/0$. Does the second theorem tell us that $(1 / 0) cdot 0 = 1$ (and therefore $z cdot 0 = 1$)? No, because the second theorem is only applicable when the denominator is not $0$, and here, the denominator is $0$. So the second theorem doesn't apply, either.
If we want, we can add more definitions and maybe make some of these theorems "work" for $z = 1/0$, just like we have a lot of theorems that "work" for the complex numbers. But when we do this, we encounter a lot of problems. Rather than dealing with these problems, most mathematical writers simply refuse to define $1/0$. (That's what the sentence "$1/0$ is undefined" means: the expression $1/0$ is an undefined expression, because we have refused to define it.)
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As an addition, I'd point out that you can define any operator you please, but if one wishes to call an operation "addition" or "multiplication," there are some behaviors that are informally expected of your operator, such as associative properties. The operators for complex numbers satisfy all of them. Of interest is that they satisfy all of the expectations of division. There are only a small number of real algebras which can admit a division operator that acts as one might expect division to act.
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– Cort Ammon
2 days ago
17
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Your answer is missing a very big piece, namely that one can prove that the complex numbers as defined in your post have multiplicative inverses except for zero. This with the other obvious ring properties make them form a field, which is the very reason we still consider them numbers.
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– user21820
2 days ago
6
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@user21820: Semi-rhetorical question: Are you suggesting that the integers (i.e. elements of $mathbb{Z},$ which is not a field), do not constitute numbers? :)
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– Will R
2 days ago
4
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@WillR Playing devil's advocate, the integers can be embedded into the rationals, reals, or complex numbers. Via this embedding, the integers can be regarded as a subset of a field of numbers, and therefore numbers themselves. That being said, I'm not quite sure what user21820 is getting at.
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– Xander Henderson
2 days ago
3
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@XanderHenderson: Bearing in mind the obvious point about "field of numbers" being undefined, I presume you mean that $mathbb{Z}$ can be seen as a subset of a field, and that's what makes it a system of numbers. But the same logic applies to any integral domain, e.g., $mathbb{Z}[x].$ What makes integers numbers, while polynomials (or power series!) aren't? I know of no satisfactory answer to this. My point being, user21820's comment seems misguided, in my opinion. I see no simple explanation as to why we consider complex numbers to be "numbers"; "they form a field" certainly doesn't cut it.
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– Will R
2 days ago
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show 6 more comments
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What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
It is set theory that allows us to give a rigorous foundation for complex numbers. In particular, as explained here the axiom of pairing plays a crucial role, It allows us to construct the product set $,Bbb R^2,$ and then reduce complex arithmetic to arithmetic on pairs of reals - as Hamilton did when he gave the first rigorous construction of $,Bbb C,,$ representing $,a + b,i $ by the pair $,(a,b),$ with operations
$$begin{align}
(a!+!bi) + (c!+!di) &= , a!+!c!+! (b!+!d)i\[.2em]
rightsquigarrow, (a, b) + (c, d) &= (a!+!c, b+d)\[.4em]
(a!+!bi)times (c!+!di) &= ac!-!bd!+!(ad!+!bc)i\[.2em]
rightsquigarrow, (a, b) times (c, d), &= (ac!-!bd, ad!+!bc) end{align}qquadqquad$$
This reduces the consistency of $,Bbb C,$ to the consistency of $,Bbb R,$ i.e. any contradiction derived in $,Bbb C,$ would yield a contradiction on such pairs of reals, so a contradiction in $,Bbb R.$
Further, a major accomplishment of the set-theoretical construction of $,Bbb C,$ (and
algebraic structures) is that it eliminates imprecise syntax and semantics in informal approaches. The imprecise term $, a + b, i, $
is replaced it by its rigorous set-theoretic representation
$,(a,b),$ - which eliminates many ambiguities, e.g. doubts about the meaning of symbols $,i,$ and $,+,$ and $,=,$ in complex arithmetic. Such questions were rampant in the early development of complex numbers, and without set theory or any other rigorous foundation it was difficult to provide convincing precise answers. For example below is how Cauchy tried to explain them.
In analysis, we call a symbolic expression any combination of symbols or algebraic signs which means nothing by itself but which one attributes a value different from the one it should naturally be [...] Similarly, we call symbolic equations those that, taken literally and interpreted according to conventions generally established, are inaccurate or have no meaning, but from which can be deduced accurate results, by changing and altering, according to fixed rules, the equations or symbols within [...] Among the symbolic expressions and equations whose theory is of considerable importance in analysis, one distinguishes especially those that have been called imaginary. -- Cauchy, Cours d'analyse,1821, S.7.1
It's no surprise that Cauchy's peers were not persuaded by such handwaving, e.g. Hankel replied
If one were to give a critique of this reasoning, we can not
actually see where to start. There must be something "which
means nothing," or "which is assigned a different value than
it should naturally be" something that has "no sense" or is
"incorrect", coupled with another similar kind, producing
something real. There must be "algebraic signs" - are these
signs for quantities or what? as a sign must designate something
- combined with each other in a way that has "a meaning." I do
not think I'm exaggerating in calling this an unintelligible
play on words, ill-becoming of mathematics, which is proud
and rightly proud of the clarity and evidence of its concepts. $quad$-- Hankel
Hamilton's elimination of such "meaningless" symbols - in favor of pairs of reals - served as a major step forward in placing complex numbers on a foundation more amenable to his contemporaries. Although there was not yet any theory of sets in which to rigorously axiomatize the notion of pairs, they were far easier to accept naively - esp. given the already known closely associated geometric interpretation of complex numbers.
See said answer for further discussion of this and related topics (above is excerpted from there).
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The axiom of pairing is not really important. For one thing, it can be proven from the other axioms of set theory. For another, even in the first-order theory of the real numbers where you have no sets at all, you can still reason about pairs of real numbers without needing a single object in the syntax to represent them (you just always represent them with two separate variables).
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– Eric Wofsey
yesterday
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You could make a reasonable case that being able to talk about $mathbb{R}^2$ as a single object is crucial, but again this doesn't actually really have to do with the axiom of pairing. The concept of pairs is obviously very important, but that does not translate into importance for the axiom of pairing in the technical development of set theory.
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– Eric Wofsey
yesterday
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@Eric Of course it goes without saying that judgments about such foundational matters and the role they played in historical development of algebra will be subjective. Please read the full linked article for more on my viewpoint (above is only an excerpt).
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– Bill Dubuque
yesterday
add a comment |
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The "necessary and sufficient" axioms to define the complex numbers are
$$(a,b)+(a',b')=(a+a',b+b')$$
$$(a,b)cdot(a',b')=(aa'-bb',ab'+a'b).$$
(Subtraction and division can be defined as the inverses of addition and multiplication, as usual.)
In particular,
$$(a,b)+(0,0)=(a,b)$$ so that $(0,0)$ is the zero and
$$(a,b)cdot(1,0)=(a,b)$$ so that $(1,0)$ is the unity.
As you can check, $(a,b)$ can also be represented as the expression $a+ib$, where $i$ is a reserved symbol, with the usual computation rules on polynomials (with $i$ seen as the variable).
Using this notation,
$$(0,1)cdot(0,1)=(-1,0)$$
translates to the famous
$$i^2=-1.$$
As you can check, the "pair" representation and the "$i$" representation are completely interchangeable. $i$ has a simple geometric interpretation: in a 2D plane, multiplication by $i$ corresponds to a rotation around the origin by a quarter turn.
Note that there are absolutely no undefined operations here.
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It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
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– Bill Dubuque
2 days ago
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@BillDubuque: take it informally.
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– Yves Daoust
2 days ago
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At the risk of sounding like a postmodernist: all numbers are imaginary.
Long ago, someone abstracted: what is the thing that this collection of sheep has in common with the number of fingers on my left hand, and called that thing "five." No inconsistencies were introduced and there were great simplifications to be made.
Someone asked how to divide two pies among three people and the abstraction of fractions was born. Someone thought about debt and the abstraction of negative numbers was born. Someone realized that positive and negative fractions didn't describe the intuitive nature of a continuum and the abstraction of the reals was born.
And eventually someone abstracted solutions to $x^2 + 1 = 0$; they're not any more imaginary than any of the other abstractions, they're all products of human imagination. The name "imaginary numbers" is unfortunate.
You say: why can't I abstract a solution to $0*z = 1$, i.e. $1/0$? The problem is that your abstraction will be incompatible with your other abstractions, i.e. you will break arithmetic. But there are areas of geometry (e.g. Mobius transformations of the plane) where there is a consistent way to do a little arithmetic with an idea of $1/0 = infty$ (although one has to be careful to remain consistent).
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We can always use the usual rules for doing arithmetic with real numbers on complex numbers too, provided we always substitute $-1$ for $i^2$ whenever we encounter it. It follows that once we allow the imaginary unit, we have a consistent algebra which obeys the usual laws.
Thus, we discover that if we extend all the usual laws of addition and multiplication on binomials of the form $a+ib,$ with $i^2=-1,$ everything goes on smoothly. One way to formally do this is by using ordered pairs, as William Hamilton first did, but the idea I think you need is that if we allow ourselves to calculate with complex numbers as we did with real numbers, only remembering to replace $i^2$ with $-1,$ then we have a consistent algebra. Mathematicians usually call the system so defined a field. There are many other fields apart from the ones formed by the real or complex numbers with addition and multiplication as usually defined, but that's another story. The gist is just that we can define two operations over the complex numbers similar in behavior to the usual addition and multiplication over the real numbers -- and with a certain lightness we may think of these as extensions of the usual addition and multiplication. Thus, the $+$ and $×$ of the complex system is technically different from that of the reals, but of course for symbolic economy (and also because they behave very similarly), we retain the same symbols. In general we use these symbols for the operations in any field, too, whether the members of the field are numbers or people.
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Why "adding a new element" $i$ with $i^2 = -1$ to $Bbb R$ "works", the technical cause.
What means "works": we want a field extension of $Bbb R$. Intuitively, a set greater than $Bbb R$ (this is the easy part) that is also a field: we have two operations $+$ and $cdot$ extending the operations sum and product of $Bbb R$ and verifying the same properties (see https://en.wikipedia.org/wiki/Field_(mathematics)). This is the hard part.
The "addition of a new element" $i$ with $i^2 = -1$ makes perfect sense in algebra: we are taking the quotient
$$Bbb R[x]/(x^2 + 1)$$
where $Bbb R[x]$ is the polynomial ring in one indeterminate and $(x^2 + 1)$ is the ideal generated by $x^2 + 1$.
The essential fact: a quotient like this will be a field iff the ideal is maximal. You can check easily that $(x^2 + 1)$ is maximal because $x^2 + 1$ is a polynomial of degree 2 without roots in $Bbb R$. Also important: the quotient contains (an isomorphic copy of) $Bbb R$ strictly.
Two examples where "adding a new element" does not "work":
$Bbb R[x]/(x)$ is a field because $(x)$ is a maximal ideal, but the quotient is isomorphic to $Bbb R$.
$Bbb R[x]/(x^2)$ isn't a field because $(x^2)$ isn't a maximal ideal.
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add a comment |
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What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
The definition of the complex numbers? There are really no new axioms involved since complex numbers are defined in terms of real ones and their behavior is completely derivable from definitions.
In the beginning, when there were just reals, these operations were defined for them.
See, there is your mistake. "there were just reals". Reals are imaginary constructs with no inherent existence. They are the consequence of choosing convenient definitions (and some axioms).
Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
That is a common misconception and actually untenable. You cannot "define" i as "the square root of -1" since -1 has two square roots in the complex numbers. Instead, complex numbers are defined via their arithmetic properties and then the real numbers are embedded into the complex numbers. Since that embedding retains basically all operations on the real numbers (logarithms and exponentiation get a lot more iffy than they are in the reals, though, even though exp(x) remains perfectly well defined), one can continue using the same symbols and operators without introducing trouble, just like embedding whole numbers into rational numbers and rational numbers into reals worked without necessitating different operators for writing down relations.
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add a comment |
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In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Not quite: in the beginning were the natural numbers, and addition over the natural numbers, and everything was good.
But then people wanted to "undo" addition, and subtraction turned out to be really useful, and related to many different real-world examples (taking things away from a collection). And suddenly a problem arose: you could add any two natural numbers and get a third natural number, but when you subtracted, the result might be undefined. What natural number is five minus seven?
Some people were okay with this being undefined, but others weren't, and the ones who weren't started wondering: what if we define a new quantity that's equal to five minus seven? And so the negative numbers were born! (And for quite a long time, mainstream mathematicians derided them and didn't consider them real numbers, because can you show me negative two apples, or people, or coins? No you can't.)
Similarly, division led to the introduction of rational numbers ("one divided by two is defined to be a new type of number, which I'll just write as 1/2"), and square roots led to the introduction of irrational numbers, and eventually we ended up with the real numbers that we all know and love today. "Imaginary" numbers are just another extension of the concept of "number", giving it new capabilities.
So then, would you like to define 1/0? Go for it! Look at the projectively extended real numbers, which do exactly that: augmenting the real numbers with a single "point at infinity". In the projectively-extended reals, 1/0=∞, and there's no problem there. As Tanner Swett put it, you can do anything you want! The real question in mathematics is: does this lead to anything interesting? And the only way to answer that is, try it and see!
P.S. The summary in this answer isn't meant to be chronologically accurate—the Greeks were dealing with "real numbers" in the form of lengths long before anyone came up with the Peano axioms. Instead, take this as a summary of how axiomatic definitions of "numbers" have evolved over time.
P.P.S. There's a legend that one ancient mathematician was either murdered or struck down by the gods for the hubris of constructing an irrational number. I haven't heard of this happening with the projectively-extended reals, but watch your back, just in case.
P.P.P.S. If you're curious about the complex numbers in particular, this website has a better explanation than I could ever give. Geometrically, $i$ can be considered as a 90-degree rotation in the plane; in geometric algebra, it would be called a unit bivector. But that's a can of worms too big for an answer here to encompass.
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add a comment |
$begingroup$
What allow us to use anything in mathematics are the axioms. Real numbers form a complete ordered field, if you take the set of real numbers and "add" $i$ to it (considering its usual behavior), you lose this order (that is: They won't be a complete ordered field anymore). You can still define some order but it will no longer be compatible with the ordering of the reals. (It is a good exercise to verify what axioms this "addition" of $i$ violates. See this or this).
In the previous case, the natural ordering was broken but mathematicians found a way to cope with it by using a new system in which the broken axioms were removed. These axioms give us a plethora of stuff we can do in the real numbers, it was natural to think: "In this new system, how can we do something that is at least remotely similar to what we could do with the ordering in the real numbers?"
There is no "moral ground" to define and use things in mathematics, you can define anything you want. There is - for example - a number system in which division by zero is meaningful. In the same way, you may run into trouble (such as the aforementioned loss of ordering) when you compare some two systems.
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add a comment |
$begingroup$
I think you already have a good interpretation of complex numbers algebraic view. So I propose here the geometric approach of this numbers, following Hamilton, Clifford and Grassmann geometric view of algebra.
First, we need to separate the meaning about the number (1) and the number (-1). You can do this for trying to count negative numbers as one does with positive numbers, e.g. one cup, two notes, three boxes, minus four houses???
The later seems meaningless because we have not added a good definition about (-). For this, we need to think in the number not only as a quantification, but as a quantification AND a direction. So (-1) is not just a number as (4), but a number that one can interpret as a quantity in a certain direction.
So the geometric view of (+) and (-) is that of symbols whose when fixed with number defines a direction in a certain quantity, e.g. If (+3) means 3 steps of unity in the certain direction, say right, then (-3) means 3 steps in the opposite direction. This could be demonstrated with the axiomatic approaches that my colleagues presented above, but I want to propose here just a geometric view of all this.
In the Euclidean Geometry, we can assert more than the opposite notion. If we suppose a segment $AB$, with a certain magnitude, when applying (-1) in the segment, we rotate it $pi$ rad. So $AB=-BA$.
This notion can be extended to "when applying positive number to a segment, it contracts (when between 0 and 1), maintain or stretch this segment. But when negative number are applied, besides the homothety, it rotates through $pi$ rad."
It's just a geometric view, I'm not proposing here any rigorous justifications.
But this is just in the $mathbb{R}$ line. Let's extend this thought to $mathbb{R}^2$ plane.
We already know that $(-1)$ means rotates in $pi$ rad, but as we now have this two lines orthogonally positioned, we also need to define an application (a number) which could rotate the segment $pi/2$ rad. Let's suppose a new type of number, whose faces i don't see, but algebraically i can give him a name (a letter), say $i$. So as we can see (and you could prove it using the axioms of above answers), when applying twice $i$ to this segment, it rotates to $pi$ rad and we can assert that $i^2=-1$. In this view, the meaning $i=sqrt{-1}$ is just a consequence of $i^2=-1$.
Now, as Kline say here, Hamilton pointed out that
A complex number $a+bi$ is not a genuine sum in the sense that $2 + 3$ is. The
use of the plus sign is a historical accident and $bi$ cannot be added to $a$. The
complex number $a + bi$ is no more than an ordered couple $(a, b)$ of real
numbers.
i suppose this answer your saying "In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined"
This allows us to use the knowledge of vectors. Indeed, Hamilton knew that a complex number is nothing more that a ratio between two segments. If $AB$ and $AC$ are two segments conveniently positioned at the origin $A$, a complex number $z=a+bi=(a,b)=e^{itheta}$, with $theta$ the angle between the complex number $z$ and the origin, is the operation necessary to rotate and stretch the segment $AB$ to give $AC$, or $ABcdot z=ACto z=frac{AC}{AB}$, which allows us to use complex numbers as a type of rotational numbers, as the negative numbers are a directional system of numbers.
And Hamilton gave latter a full usefulness of the imaginary numbers to also rotate in $mathbb{R}^3$, where the Quaternions, a four dimensional algebra, was born. And you could ask "why a four dimensional system of numbers is required to rotate three dimensional directed segments (vectors)?"
I'm not allowed to answer this question here, but i can tell you to read about the Clifford Algebras, the Geometric Algebra, which are an excellent extension of Hamilton's Quaternions with the concept of Grassmann's exterior algebra. In the geometric algebra, real numbers, complex numbers, quaternions, biquaternions, and other n-dimensional systems of numbers are synthesized in a rigorous way giving to all this abstract algebraic views a good geometrical interpretation.
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add a comment |
$begingroup$
Well, coming from the algebraic side, consider field extensions of the form
$${Bbb Q}(sqrt n)={a+bsqrt nmid a,bin{Bbb Q}},$$
where $nne0,1$ is a square-free integer.
The addition is ''componentwise'',
$$(a+bsqrt n) + (c+dsqrt n) = (a+c) + (b+d)sqrt n$$
and the multiplication is given by noting that $sqrt n^2 = n$,
$$(a+bsqrt n) cdot (c+dsqrt n) = (ac + bdn) + (ad + bc)sqrt n.$$
Here the case $n=-1$ comes up naturally and writing $i=sqrt{-1}$, we obtain
${Bbb Q}(i)={a+bimid a,bin{Bbb Q}}$ and also ${Bbb C} = {Bbb R}(i)={a+bimid a,bin{Bbb R}}$, the field of complex numbers.
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add a comment |
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$begingroup$
We can do anything we want!
Specifically, we can define anything we want (as long as our definitions don't contradict each other). So if we want to allow ourselves to use imaginary numbers, all we have to do is write something like the following:
Define a complex number as an ordered pair of the form $(a, b)$, where $a$ and $b$ are real numbers.
Define $i$ as the complex number $(0, 1)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) + (c, d)$ as $(a + c, b + d)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) cdot (c, d)$ as $(ac - bd, ad + bc)$.
And define subtraction and division in similar ways.
Is that it? Are we done? No, there's still more that we want to do. There are a lot of useful theorems about real numbers that also apply to the complex numbers, but we don't know that they apply to the complex numbers until we prove them. For example, one very useful theorem about real numbers is:
Theorem: If $a$ and $b$ are real numbers, then $a + b = b + a$.
The analogous theorem about the complex numbers is:
Theorem (not yet proven): If $a$ and $b$ are complex numbers, then $a + b = b + a$.
This theorem is, in fact, true, but we didn't know that it was true until somebody proved it.
Once we've proved all the theorems that we want to prove, then we can say that we're "done."
(Do we have to prove these theorems? No, we don't have to if we don't want to. But without these theorems, complex numbers aren't very useful.)
As for your corollary question:
Could I define a new number $z$ which is $1/0$ and simply begin using it? Seems ludicrous.
Yes, you absolutely can! All you have to do is write:
Assume that there is a value $z$. Define $1/0$ as $z$.
And that's perfectly valid; this definition doesn't contradict any other definitions. This is completely legal, acceptable and proper.
Is that it? Are we done? Probably not; there's more we'd like to do. For example, what do you suppose $z cdot 0$ is? There are a couple of theorems here we might like to use, but we can't. Let's take a look at them:
Theorem: If $x$ is a real number, then $x cdot 0 = 0$.
Theorem: If $x$ and $y$ are real numbers, and $y ne 0$, then $(x / y) cdot y = x$.
Do you see why we can't use these theorems?
Does the first theorem tell us that $z cdot 0 = 0$? No, because we don't know that $z$ is a real number. So the first theorem doesn't apply.
How about the second theorem? We know that $z = 1/0$. Does the second theorem tell us that $(1 / 0) cdot 0 = 1$ (and therefore $z cdot 0 = 1$)? No, because the second theorem is only applicable when the denominator is not $0$, and here, the denominator is $0$. So the second theorem doesn't apply, either.
If we want, we can add more definitions and maybe make some of these theorems "work" for $z = 1/0$, just like we have a lot of theorems that "work" for the complex numbers. But when we do this, we encounter a lot of problems. Rather than dealing with these problems, most mathematical writers simply refuse to define $1/0$. (That's what the sentence "$1/0$ is undefined" means: the expression $1/0$ is an undefined expression, because we have refused to define it.)
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5
$begingroup$
As an addition, I'd point out that you can define any operator you please, but if one wishes to call an operation "addition" or "multiplication," there are some behaviors that are informally expected of your operator, such as associative properties. The operators for complex numbers satisfy all of them. Of interest is that they satisfy all of the expectations of division. There are only a small number of real algebras which can admit a division operator that acts as one might expect division to act.
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– Cort Ammon
2 days ago
17
$begingroup$
Your answer is missing a very big piece, namely that one can prove that the complex numbers as defined in your post have multiplicative inverses except for zero. This with the other obvious ring properties make them form a field, which is the very reason we still consider them numbers.
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– user21820
2 days ago
6
$begingroup$
@user21820: Semi-rhetorical question: Are you suggesting that the integers (i.e. elements of $mathbb{Z},$ which is not a field), do not constitute numbers? :)
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– Will R
2 days ago
4
$begingroup$
@WillR Playing devil's advocate, the integers can be embedded into the rationals, reals, or complex numbers. Via this embedding, the integers can be regarded as a subset of a field of numbers, and therefore numbers themselves. That being said, I'm not quite sure what user21820 is getting at.
$endgroup$
– Xander Henderson
2 days ago
3
$begingroup$
@XanderHenderson: Bearing in mind the obvious point about "field of numbers" being undefined, I presume you mean that $mathbb{Z}$ can be seen as a subset of a field, and that's what makes it a system of numbers. But the same logic applies to any integral domain, e.g., $mathbb{Z}[x].$ What makes integers numbers, while polynomials (or power series!) aren't? I know of no satisfactory answer to this. My point being, user21820's comment seems misguided, in my opinion. I see no simple explanation as to why we consider complex numbers to be "numbers"; "they form a field" certainly doesn't cut it.
$endgroup$
– Will R
2 days ago
|
show 6 more comments
$begingroup$
We can do anything we want!
Specifically, we can define anything we want (as long as our definitions don't contradict each other). So if we want to allow ourselves to use imaginary numbers, all we have to do is write something like the following:
Define a complex number as an ordered pair of the form $(a, b)$, where $a$ and $b$ are real numbers.
Define $i$ as the complex number $(0, 1)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) + (c, d)$ as $(a + c, b + d)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) cdot (c, d)$ as $(ac - bd, ad + bc)$.
And define subtraction and division in similar ways.
Is that it? Are we done? No, there's still more that we want to do. There are a lot of useful theorems about real numbers that also apply to the complex numbers, but we don't know that they apply to the complex numbers until we prove them. For example, one very useful theorem about real numbers is:
Theorem: If $a$ and $b$ are real numbers, then $a + b = b + a$.
The analogous theorem about the complex numbers is:
Theorem (not yet proven): If $a$ and $b$ are complex numbers, then $a + b = b + a$.
This theorem is, in fact, true, but we didn't know that it was true until somebody proved it.
Once we've proved all the theorems that we want to prove, then we can say that we're "done."
(Do we have to prove these theorems? No, we don't have to if we don't want to. But without these theorems, complex numbers aren't very useful.)
As for your corollary question:
Could I define a new number $z$ which is $1/0$ and simply begin using it? Seems ludicrous.
Yes, you absolutely can! All you have to do is write:
Assume that there is a value $z$. Define $1/0$ as $z$.
And that's perfectly valid; this definition doesn't contradict any other definitions. This is completely legal, acceptable and proper.
Is that it? Are we done? Probably not; there's more we'd like to do. For example, what do you suppose $z cdot 0$ is? There are a couple of theorems here we might like to use, but we can't. Let's take a look at them:
Theorem: If $x$ is a real number, then $x cdot 0 = 0$.
Theorem: If $x$ and $y$ are real numbers, and $y ne 0$, then $(x / y) cdot y = x$.
Do you see why we can't use these theorems?
Does the first theorem tell us that $z cdot 0 = 0$? No, because we don't know that $z$ is a real number. So the first theorem doesn't apply.
How about the second theorem? We know that $z = 1/0$. Does the second theorem tell us that $(1 / 0) cdot 0 = 1$ (and therefore $z cdot 0 = 1$)? No, because the second theorem is only applicable when the denominator is not $0$, and here, the denominator is $0$. So the second theorem doesn't apply, either.
If we want, we can add more definitions and maybe make some of these theorems "work" for $z = 1/0$, just like we have a lot of theorems that "work" for the complex numbers. But when we do this, we encounter a lot of problems. Rather than dealing with these problems, most mathematical writers simply refuse to define $1/0$. (That's what the sentence "$1/0$ is undefined" means: the expression $1/0$ is an undefined expression, because we have refused to define it.)
$endgroup$
5
$begingroup$
As an addition, I'd point out that you can define any operator you please, but if one wishes to call an operation "addition" or "multiplication," there are some behaviors that are informally expected of your operator, such as associative properties. The operators for complex numbers satisfy all of them. Of interest is that they satisfy all of the expectations of division. There are only a small number of real algebras which can admit a division operator that acts as one might expect division to act.
$endgroup$
– Cort Ammon
2 days ago
17
$begingroup$
Your answer is missing a very big piece, namely that one can prove that the complex numbers as defined in your post have multiplicative inverses except for zero. This with the other obvious ring properties make them form a field, which is the very reason we still consider them numbers.
$endgroup$
– user21820
2 days ago
6
$begingroup$
@user21820: Semi-rhetorical question: Are you suggesting that the integers (i.e. elements of $mathbb{Z},$ which is not a field), do not constitute numbers? :)
$endgroup$
– Will R
2 days ago
4
$begingroup$
@WillR Playing devil's advocate, the integers can be embedded into the rationals, reals, or complex numbers. Via this embedding, the integers can be regarded as a subset of a field of numbers, and therefore numbers themselves. That being said, I'm not quite sure what user21820 is getting at.
$endgroup$
– Xander Henderson
2 days ago
3
$begingroup$
@XanderHenderson: Bearing in mind the obvious point about "field of numbers" being undefined, I presume you mean that $mathbb{Z}$ can be seen as a subset of a field, and that's what makes it a system of numbers. But the same logic applies to any integral domain, e.g., $mathbb{Z}[x].$ What makes integers numbers, while polynomials (or power series!) aren't? I know of no satisfactory answer to this. My point being, user21820's comment seems misguided, in my opinion. I see no simple explanation as to why we consider complex numbers to be "numbers"; "they form a field" certainly doesn't cut it.
$endgroup$
– Will R
2 days ago
|
show 6 more comments
$begingroup$
We can do anything we want!
Specifically, we can define anything we want (as long as our definitions don't contradict each other). So if we want to allow ourselves to use imaginary numbers, all we have to do is write something like the following:
Define a complex number as an ordered pair of the form $(a, b)$, where $a$ and $b$ are real numbers.
Define $i$ as the complex number $(0, 1)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) + (c, d)$ as $(a + c, b + d)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) cdot (c, d)$ as $(ac - bd, ad + bc)$.
And define subtraction and division in similar ways.
Is that it? Are we done? No, there's still more that we want to do. There are a lot of useful theorems about real numbers that also apply to the complex numbers, but we don't know that they apply to the complex numbers until we prove them. For example, one very useful theorem about real numbers is:
Theorem: If $a$ and $b$ are real numbers, then $a + b = b + a$.
The analogous theorem about the complex numbers is:
Theorem (not yet proven): If $a$ and $b$ are complex numbers, then $a + b = b + a$.
This theorem is, in fact, true, but we didn't know that it was true until somebody proved it.
Once we've proved all the theorems that we want to prove, then we can say that we're "done."
(Do we have to prove these theorems? No, we don't have to if we don't want to. But without these theorems, complex numbers aren't very useful.)
As for your corollary question:
Could I define a new number $z$ which is $1/0$ and simply begin using it? Seems ludicrous.
Yes, you absolutely can! All you have to do is write:
Assume that there is a value $z$. Define $1/0$ as $z$.
And that's perfectly valid; this definition doesn't contradict any other definitions. This is completely legal, acceptable and proper.
Is that it? Are we done? Probably not; there's more we'd like to do. For example, what do you suppose $z cdot 0$ is? There are a couple of theorems here we might like to use, but we can't. Let's take a look at them:
Theorem: If $x$ is a real number, then $x cdot 0 = 0$.
Theorem: If $x$ and $y$ are real numbers, and $y ne 0$, then $(x / y) cdot y = x$.
Do you see why we can't use these theorems?
Does the first theorem tell us that $z cdot 0 = 0$? No, because we don't know that $z$ is a real number. So the first theorem doesn't apply.
How about the second theorem? We know that $z = 1/0$. Does the second theorem tell us that $(1 / 0) cdot 0 = 1$ (and therefore $z cdot 0 = 1$)? No, because the second theorem is only applicable when the denominator is not $0$, and here, the denominator is $0$. So the second theorem doesn't apply, either.
If we want, we can add more definitions and maybe make some of these theorems "work" for $z = 1/0$, just like we have a lot of theorems that "work" for the complex numbers. But when we do this, we encounter a lot of problems. Rather than dealing with these problems, most mathematical writers simply refuse to define $1/0$. (That's what the sentence "$1/0$ is undefined" means: the expression $1/0$ is an undefined expression, because we have refused to define it.)
$endgroup$
We can do anything we want!
Specifically, we can define anything we want (as long as our definitions don't contradict each other). So if we want to allow ourselves to use imaginary numbers, all we have to do is write something like the following:
Define a complex number as an ordered pair of the form $(a, b)$, where $a$ and $b$ are real numbers.
Define $i$ as the complex number $(0, 1)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) + (c, d)$ as $(a + c, b + d)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) cdot (c, d)$ as $(ac - bd, ad + bc)$.
And define subtraction and division in similar ways.
Is that it? Are we done? No, there's still more that we want to do. There are a lot of useful theorems about real numbers that also apply to the complex numbers, but we don't know that they apply to the complex numbers until we prove them. For example, one very useful theorem about real numbers is:
Theorem: If $a$ and $b$ are real numbers, then $a + b = b + a$.
The analogous theorem about the complex numbers is:
Theorem (not yet proven): If $a$ and $b$ are complex numbers, then $a + b = b + a$.
This theorem is, in fact, true, but we didn't know that it was true until somebody proved it.
Once we've proved all the theorems that we want to prove, then we can say that we're "done."
(Do we have to prove these theorems? No, we don't have to if we don't want to. But without these theorems, complex numbers aren't very useful.)
As for your corollary question:
Could I define a new number $z$ which is $1/0$ and simply begin using it? Seems ludicrous.
Yes, you absolutely can! All you have to do is write:
Assume that there is a value $z$. Define $1/0$ as $z$.
And that's perfectly valid; this definition doesn't contradict any other definitions. This is completely legal, acceptable and proper.
Is that it? Are we done? Probably not; there's more we'd like to do. For example, what do you suppose $z cdot 0$ is? There are a couple of theorems here we might like to use, but we can't. Let's take a look at them:
Theorem: If $x$ is a real number, then $x cdot 0 = 0$.
Theorem: If $x$ and $y$ are real numbers, and $y ne 0$, then $(x / y) cdot y = x$.
Do you see why we can't use these theorems?
Does the first theorem tell us that $z cdot 0 = 0$? No, because we don't know that $z$ is a real number. So the first theorem doesn't apply.
How about the second theorem? We know that $z = 1/0$. Does the second theorem tell us that $(1 / 0) cdot 0 = 1$ (and therefore $z cdot 0 = 1$)? No, because the second theorem is only applicable when the denominator is not $0$, and here, the denominator is $0$. So the second theorem doesn't apply, either.
If we want, we can add more definitions and maybe make some of these theorems "work" for $z = 1/0$, just like we have a lot of theorems that "work" for the complex numbers. But when we do this, we encounter a lot of problems. Rather than dealing with these problems, most mathematical writers simply refuse to define $1/0$. (That's what the sentence "$1/0$ is undefined" means: the expression $1/0$ is an undefined expression, because we have refused to define it.)
answered 2 days ago
Tanner SwettTanner Swett
5,73621 silver badges41 bronze badges
5,73621 silver badges41 bronze badges
5
$begingroup$
As an addition, I'd point out that you can define any operator you please, but if one wishes to call an operation "addition" or "multiplication," there are some behaviors that are informally expected of your operator, such as associative properties. The operators for complex numbers satisfy all of them. Of interest is that they satisfy all of the expectations of division. There are only a small number of real algebras which can admit a division operator that acts as one might expect division to act.
$endgroup$
– Cort Ammon
2 days ago
17
$begingroup$
Your answer is missing a very big piece, namely that one can prove that the complex numbers as defined in your post have multiplicative inverses except for zero. This with the other obvious ring properties make them form a field, which is the very reason we still consider them numbers.
$endgroup$
– user21820
2 days ago
6
$begingroup$
@user21820: Semi-rhetorical question: Are you suggesting that the integers (i.e. elements of $mathbb{Z},$ which is not a field), do not constitute numbers? :)
$endgroup$
– Will R
2 days ago
4
$begingroup$
@WillR Playing devil's advocate, the integers can be embedded into the rationals, reals, or complex numbers. Via this embedding, the integers can be regarded as a subset of a field of numbers, and therefore numbers themselves. That being said, I'm not quite sure what user21820 is getting at.
$endgroup$
– Xander Henderson
2 days ago
3
$begingroup$
@XanderHenderson: Bearing in mind the obvious point about "field of numbers" being undefined, I presume you mean that $mathbb{Z}$ can be seen as a subset of a field, and that's what makes it a system of numbers. But the same logic applies to any integral domain, e.g., $mathbb{Z}[x].$ What makes integers numbers, while polynomials (or power series!) aren't? I know of no satisfactory answer to this. My point being, user21820's comment seems misguided, in my opinion. I see no simple explanation as to why we consider complex numbers to be "numbers"; "they form a field" certainly doesn't cut it.
$endgroup$
– Will R
2 days ago
|
show 6 more comments
5
$begingroup$
As an addition, I'd point out that you can define any operator you please, but if one wishes to call an operation "addition" or "multiplication," there are some behaviors that are informally expected of your operator, such as associative properties. The operators for complex numbers satisfy all of them. Of interest is that they satisfy all of the expectations of division. There are only a small number of real algebras which can admit a division operator that acts as one might expect division to act.
$endgroup$
– Cort Ammon
2 days ago
17
$begingroup$
Your answer is missing a very big piece, namely that one can prove that the complex numbers as defined in your post have multiplicative inverses except for zero. This with the other obvious ring properties make them form a field, which is the very reason we still consider them numbers.
$endgroup$
– user21820
2 days ago
6
$begingroup$
@user21820: Semi-rhetorical question: Are you suggesting that the integers (i.e. elements of $mathbb{Z},$ which is not a field), do not constitute numbers? :)
$endgroup$
– Will R
2 days ago
4
$begingroup$
@WillR Playing devil's advocate, the integers can be embedded into the rationals, reals, or complex numbers. Via this embedding, the integers can be regarded as a subset of a field of numbers, and therefore numbers themselves. That being said, I'm not quite sure what user21820 is getting at.
$endgroup$
– Xander Henderson
2 days ago
3
$begingroup$
@XanderHenderson: Bearing in mind the obvious point about "field of numbers" being undefined, I presume you mean that $mathbb{Z}$ can be seen as a subset of a field, and that's what makes it a system of numbers. But the same logic applies to any integral domain, e.g., $mathbb{Z}[x].$ What makes integers numbers, while polynomials (or power series!) aren't? I know of no satisfactory answer to this. My point being, user21820's comment seems misguided, in my opinion. I see no simple explanation as to why we consider complex numbers to be "numbers"; "they form a field" certainly doesn't cut it.
$endgroup$
– Will R
2 days ago
5
5
$begingroup$
As an addition, I'd point out that you can define any operator you please, but if one wishes to call an operation "addition" or "multiplication," there are some behaviors that are informally expected of your operator, such as associative properties. The operators for complex numbers satisfy all of them. Of interest is that they satisfy all of the expectations of division. There are only a small number of real algebras which can admit a division operator that acts as one might expect division to act.
$endgroup$
– Cort Ammon
2 days ago
$begingroup$
As an addition, I'd point out that you can define any operator you please, but if one wishes to call an operation "addition" or "multiplication," there are some behaviors that are informally expected of your operator, such as associative properties. The operators for complex numbers satisfy all of them. Of interest is that they satisfy all of the expectations of division. There are only a small number of real algebras which can admit a division operator that acts as one might expect division to act.
$endgroup$
– Cort Ammon
2 days ago
17
17
$begingroup$
Your answer is missing a very big piece, namely that one can prove that the complex numbers as defined in your post have multiplicative inverses except for zero. This with the other obvious ring properties make them form a field, which is the very reason we still consider them numbers.
$endgroup$
– user21820
2 days ago
$begingroup$
Your answer is missing a very big piece, namely that one can prove that the complex numbers as defined in your post have multiplicative inverses except for zero. This with the other obvious ring properties make them form a field, which is the very reason we still consider them numbers.
$endgroup$
– user21820
2 days ago
6
6
$begingroup$
@user21820: Semi-rhetorical question: Are you suggesting that the integers (i.e. elements of $mathbb{Z},$ which is not a field), do not constitute numbers? :)
$endgroup$
– Will R
2 days ago
$begingroup$
@user21820: Semi-rhetorical question: Are you suggesting that the integers (i.e. elements of $mathbb{Z},$ which is not a field), do not constitute numbers? :)
$endgroup$
– Will R
2 days ago
4
4
$begingroup$
@WillR Playing devil's advocate, the integers can be embedded into the rationals, reals, or complex numbers. Via this embedding, the integers can be regarded as a subset of a field of numbers, and therefore numbers themselves. That being said, I'm not quite sure what user21820 is getting at.
$endgroup$
– Xander Henderson
2 days ago
$begingroup$
@WillR Playing devil's advocate, the integers can be embedded into the rationals, reals, or complex numbers. Via this embedding, the integers can be regarded as a subset of a field of numbers, and therefore numbers themselves. That being said, I'm not quite sure what user21820 is getting at.
$endgroup$
– Xander Henderson
2 days ago
3
3
$begingroup$
@XanderHenderson: Bearing in mind the obvious point about "field of numbers" being undefined, I presume you mean that $mathbb{Z}$ can be seen as a subset of a field, and that's what makes it a system of numbers. But the same logic applies to any integral domain, e.g., $mathbb{Z}[x].$ What makes integers numbers, while polynomials (or power series!) aren't? I know of no satisfactory answer to this. My point being, user21820's comment seems misguided, in my opinion. I see no simple explanation as to why we consider complex numbers to be "numbers"; "they form a field" certainly doesn't cut it.
$endgroup$
– Will R
2 days ago
$begingroup$
@XanderHenderson: Bearing in mind the obvious point about "field of numbers" being undefined, I presume you mean that $mathbb{Z}$ can be seen as a subset of a field, and that's what makes it a system of numbers. But the same logic applies to any integral domain, e.g., $mathbb{Z}[x].$ What makes integers numbers, while polynomials (or power series!) aren't? I know of no satisfactory answer to this. My point being, user21820's comment seems misguided, in my opinion. I see no simple explanation as to why we consider complex numbers to be "numbers"; "they form a field" certainly doesn't cut it.
$endgroup$
– Will R
2 days ago
|
show 6 more comments
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
It is set theory that allows us to give a rigorous foundation for complex numbers. In particular, as explained here the axiom of pairing plays a crucial role, It allows us to construct the product set $,Bbb R^2,$ and then reduce complex arithmetic to arithmetic on pairs of reals - as Hamilton did when he gave the first rigorous construction of $,Bbb C,,$ representing $,a + b,i $ by the pair $,(a,b),$ with operations
$$begin{align}
(a!+!bi) + (c!+!di) &= , a!+!c!+! (b!+!d)i\[.2em]
rightsquigarrow, (a, b) + (c, d) &= (a!+!c, b+d)\[.4em]
(a!+!bi)times (c!+!di) &= ac!-!bd!+!(ad!+!bc)i\[.2em]
rightsquigarrow, (a, b) times (c, d), &= (ac!-!bd, ad!+!bc) end{align}qquadqquad$$
This reduces the consistency of $,Bbb C,$ to the consistency of $,Bbb R,$ i.e. any contradiction derived in $,Bbb C,$ would yield a contradiction on such pairs of reals, so a contradiction in $,Bbb R.$
Further, a major accomplishment of the set-theoretical construction of $,Bbb C,$ (and
algebraic structures) is that it eliminates imprecise syntax and semantics in informal approaches. The imprecise term $, a + b, i, $
is replaced it by its rigorous set-theoretic representation
$,(a,b),$ - which eliminates many ambiguities, e.g. doubts about the meaning of symbols $,i,$ and $,+,$ and $,=,$ in complex arithmetic. Such questions were rampant in the early development of complex numbers, and without set theory or any other rigorous foundation it was difficult to provide convincing precise answers. For example below is how Cauchy tried to explain them.
In analysis, we call a symbolic expression any combination of symbols or algebraic signs which means nothing by itself but which one attributes a value different from the one it should naturally be [...] Similarly, we call symbolic equations those that, taken literally and interpreted according to conventions generally established, are inaccurate or have no meaning, but from which can be deduced accurate results, by changing and altering, according to fixed rules, the equations or symbols within [...] Among the symbolic expressions and equations whose theory is of considerable importance in analysis, one distinguishes especially those that have been called imaginary. -- Cauchy, Cours d'analyse,1821, S.7.1
It's no surprise that Cauchy's peers were not persuaded by such handwaving, e.g. Hankel replied
If one were to give a critique of this reasoning, we can not
actually see where to start. There must be something "which
means nothing," or "which is assigned a different value than
it should naturally be" something that has "no sense" or is
"incorrect", coupled with another similar kind, producing
something real. There must be "algebraic signs" - are these
signs for quantities or what? as a sign must designate something
- combined with each other in a way that has "a meaning." I do
not think I'm exaggerating in calling this an unintelligible
play on words, ill-becoming of mathematics, which is proud
and rightly proud of the clarity and evidence of its concepts. $quad$-- Hankel
Hamilton's elimination of such "meaningless" symbols - in favor of pairs of reals - served as a major step forward in placing complex numbers on a foundation more amenable to his contemporaries. Although there was not yet any theory of sets in which to rigorously axiomatize the notion of pairs, they were far easier to accept naively - esp. given the already known closely associated geometric interpretation of complex numbers.
See said answer for further discussion of this and related topics (above is excerpted from there).
$endgroup$
$begingroup$
The axiom of pairing is not really important. For one thing, it can be proven from the other axioms of set theory. For another, even in the first-order theory of the real numbers where you have no sets at all, you can still reason about pairs of real numbers without needing a single object in the syntax to represent them (you just always represent them with two separate variables).
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
You could make a reasonable case that being able to talk about $mathbb{R}^2$ as a single object is crucial, but again this doesn't actually really have to do with the axiom of pairing. The concept of pairs is obviously very important, but that does not translate into importance for the axiom of pairing in the technical development of set theory.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
@Eric Of course it goes without saying that judgments about such foundational matters and the role they played in historical development of algebra will be subjective. Please read the full linked article for more on my viewpoint (above is only an excerpt).
$endgroup$
– Bill Dubuque
yesterday
add a comment |
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
It is set theory that allows us to give a rigorous foundation for complex numbers. In particular, as explained here the axiom of pairing plays a crucial role, It allows us to construct the product set $,Bbb R^2,$ and then reduce complex arithmetic to arithmetic on pairs of reals - as Hamilton did when he gave the first rigorous construction of $,Bbb C,,$ representing $,a + b,i $ by the pair $,(a,b),$ with operations
$$begin{align}
(a!+!bi) + (c!+!di) &= , a!+!c!+! (b!+!d)i\[.2em]
rightsquigarrow, (a, b) + (c, d) &= (a!+!c, b+d)\[.4em]
(a!+!bi)times (c!+!di) &= ac!-!bd!+!(ad!+!bc)i\[.2em]
rightsquigarrow, (a, b) times (c, d), &= (ac!-!bd, ad!+!bc) end{align}qquadqquad$$
This reduces the consistency of $,Bbb C,$ to the consistency of $,Bbb R,$ i.e. any contradiction derived in $,Bbb C,$ would yield a contradiction on such pairs of reals, so a contradiction in $,Bbb R.$
Further, a major accomplishment of the set-theoretical construction of $,Bbb C,$ (and
algebraic structures) is that it eliminates imprecise syntax and semantics in informal approaches. The imprecise term $, a + b, i, $
is replaced it by its rigorous set-theoretic representation
$,(a,b),$ - which eliminates many ambiguities, e.g. doubts about the meaning of symbols $,i,$ and $,+,$ and $,=,$ in complex arithmetic. Such questions were rampant in the early development of complex numbers, and without set theory or any other rigorous foundation it was difficult to provide convincing precise answers. For example below is how Cauchy tried to explain them.
In analysis, we call a symbolic expression any combination of symbols or algebraic signs which means nothing by itself but which one attributes a value different from the one it should naturally be [...] Similarly, we call symbolic equations those that, taken literally and interpreted according to conventions generally established, are inaccurate or have no meaning, but from which can be deduced accurate results, by changing and altering, according to fixed rules, the equations or symbols within [...] Among the symbolic expressions and equations whose theory is of considerable importance in analysis, one distinguishes especially those that have been called imaginary. -- Cauchy, Cours d'analyse,1821, S.7.1
It's no surprise that Cauchy's peers were not persuaded by such handwaving, e.g. Hankel replied
If one were to give a critique of this reasoning, we can not
actually see where to start. There must be something "which
means nothing," or "which is assigned a different value than
it should naturally be" something that has "no sense" or is
"incorrect", coupled with another similar kind, producing
something real. There must be "algebraic signs" - are these
signs for quantities or what? as a sign must designate something
- combined with each other in a way that has "a meaning." I do
not think I'm exaggerating in calling this an unintelligible
play on words, ill-becoming of mathematics, which is proud
and rightly proud of the clarity and evidence of its concepts. $quad$-- Hankel
Hamilton's elimination of such "meaningless" symbols - in favor of pairs of reals - served as a major step forward in placing complex numbers on a foundation more amenable to his contemporaries. Although there was not yet any theory of sets in which to rigorously axiomatize the notion of pairs, they were far easier to accept naively - esp. given the already known closely associated geometric interpretation of complex numbers.
See said answer for further discussion of this and related topics (above is excerpted from there).
$endgroup$
$begingroup$
The axiom of pairing is not really important. For one thing, it can be proven from the other axioms of set theory. For another, even in the first-order theory of the real numbers where you have no sets at all, you can still reason about pairs of real numbers without needing a single object in the syntax to represent them (you just always represent them with two separate variables).
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
You could make a reasonable case that being able to talk about $mathbb{R}^2$ as a single object is crucial, but again this doesn't actually really have to do with the axiom of pairing. The concept of pairs is obviously very important, but that does not translate into importance for the axiom of pairing in the technical development of set theory.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
@Eric Of course it goes without saying that judgments about such foundational matters and the role they played in historical development of algebra will be subjective. Please read the full linked article for more on my viewpoint (above is only an excerpt).
$endgroup$
– Bill Dubuque
yesterday
add a comment |
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
It is set theory that allows us to give a rigorous foundation for complex numbers. In particular, as explained here the axiom of pairing plays a crucial role, It allows us to construct the product set $,Bbb R^2,$ and then reduce complex arithmetic to arithmetic on pairs of reals - as Hamilton did when he gave the first rigorous construction of $,Bbb C,,$ representing $,a + b,i $ by the pair $,(a,b),$ with operations
$$begin{align}
(a!+!bi) + (c!+!di) &= , a!+!c!+! (b!+!d)i\[.2em]
rightsquigarrow, (a, b) + (c, d) &= (a!+!c, b+d)\[.4em]
(a!+!bi)times (c!+!di) &= ac!-!bd!+!(ad!+!bc)i\[.2em]
rightsquigarrow, (a, b) times (c, d), &= (ac!-!bd, ad!+!bc) end{align}qquadqquad$$
This reduces the consistency of $,Bbb C,$ to the consistency of $,Bbb R,$ i.e. any contradiction derived in $,Bbb C,$ would yield a contradiction on such pairs of reals, so a contradiction in $,Bbb R.$
Further, a major accomplishment of the set-theoretical construction of $,Bbb C,$ (and
algebraic structures) is that it eliminates imprecise syntax and semantics in informal approaches. The imprecise term $, a + b, i, $
is replaced it by its rigorous set-theoretic representation
$,(a,b),$ - which eliminates many ambiguities, e.g. doubts about the meaning of symbols $,i,$ and $,+,$ and $,=,$ in complex arithmetic. Such questions were rampant in the early development of complex numbers, and without set theory or any other rigorous foundation it was difficult to provide convincing precise answers. For example below is how Cauchy tried to explain them.
In analysis, we call a symbolic expression any combination of symbols or algebraic signs which means nothing by itself but which one attributes a value different from the one it should naturally be [...] Similarly, we call symbolic equations those that, taken literally and interpreted according to conventions generally established, are inaccurate or have no meaning, but from which can be deduced accurate results, by changing and altering, according to fixed rules, the equations or symbols within [...] Among the symbolic expressions and equations whose theory is of considerable importance in analysis, one distinguishes especially those that have been called imaginary. -- Cauchy, Cours d'analyse,1821, S.7.1
It's no surprise that Cauchy's peers were not persuaded by such handwaving, e.g. Hankel replied
If one were to give a critique of this reasoning, we can not
actually see where to start. There must be something "which
means nothing," or "which is assigned a different value than
it should naturally be" something that has "no sense" or is
"incorrect", coupled with another similar kind, producing
something real. There must be "algebraic signs" - are these
signs for quantities or what? as a sign must designate something
- combined with each other in a way that has "a meaning." I do
not think I'm exaggerating in calling this an unintelligible
play on words, ill-becoming of mathematics, which is proud
and rightly proud of the clarity and evidence of its concepts. $quad$-- Hankel
Hamilton's elimination of such "meaningless" symbols - in favor of pairs of reals - served as a major step forward in placing complex numbers on a foundation more amenable to his contemporaries. Although there was not yet any theory of sets in which to rigorously axiomatize the notion of pairs, they were far easier to accept naively - esp. given the already known closely associated geometric interpretation of complex numbers.
See said answer for further discussion of this and related topics (above is excerpted from there).
$endgroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
It is set theory that allows us to give a rigorous foundation for complex numbers. In particular, as explained here the axiom of pairing plays a crucial role, It allows us to construct the product set $,Bbb R^2,$ and then reduce complex arithmetic to arithmetic on pairs of reals - as Hamilton did when he gave the first rigorous construction of $,Bbb C,,$ representing $,a + b,i $ by the pair $,(a,b),$ with operations
$$begin{align}
(a!+!bi) + (c!+!di) &= , a!+!c!+! (b!+!d)i\[.2em]
rightsquigarrow, (a, b) + (c, d) &= (a!+!c, b+d)\[.4em]
(a!+!bi)times (c!+!di) &= ac!-!bd!+!(ad!+!bc)i\[.2em]
rightsquigarrow, (a, b) times (c, d), &= (ac!-!bd, ad!+!bc) end{align}qquadqquad$$
This reduces the consistency of $,Bbb C,$ to the consistency of $,Bbb R,$ i.e. any contradiction derived in $,Bbb C,$ would yield a contradiction on such pairs of reals, so a contradiction in $,Bbb R.$
Further, a major accomplishment of the set-theoretical construction of $,Bbb C,$ (and
algebraic structures) is that it eliminates imprecise syntax and semantics in informal approaches. The imprecise term $, a + b, i, $
is replaced it by its rigorous set-theoretic representation
$,(a,b),$ - which eliminates many ambiguities, e.g. doubts about the meaning of symbols $,i,$ and $,+,$ and $,=,$ in complex arithmetic. Such questions were rampant in the early development of complex numbers, and without set theory or any other rigorous foundation it was difficult to provide convincing precise answers. For example below is how Cauchy tried to explain them.
In analysis, we call a symbolic expression any combination of symbols or algebraic signs which means nothing by itself but which one attributes a value different from the one it should naturally be [...] Similarly, we call symbolic equations those that, taken literally and interpreted according to conventions generally established, are inaccurate or have no meaning, but from which can be deduced accurate results, by changing and altering, according to fixed rules, the equations or symbols within [...] Among the symbolic expressions and equations whose theory is of considerable importance in analysis, one distinguishes especially those that have been called imaginary. -- Cauchy, Cours d'analyse,1821, S.7.1
It's no surprise that Cauchy's peers were not persuaded by such handwaving, e.g. Hankel replied
If one were to give a critique of this reasoning, we can not
actually see where to start. There must be something "which
means nothing," or "which is assigned a different value than
it should naturally be" something that has "no sense" or is
"incorrect", coupled with another similar kind, producing
something real. There must be "algebraic signs" - are these
signs for quantities or what? as a sign must designate something
- combined with each other in a way that has "a meaning." I do
not think I'm exaggerating in calling this an unintelligible
play on words, ill-becoming of mathematics, which is proud
and rightly proud of the clarity and evidence of its concepts. $quad$-- Hankel
Hamilton's elimination of such "meaningless" symbols - in favor of pairs of reals - served as a major step forward in placing complex numbers on a foundation more amenable to his contemporaries. Although there was not yet any theory of sets in which to rigorously axiomatize the notion of pairs, they were far easier to accept naively - esp. given the already known closely associated geometric interpretation of complex numbers.
See said answer for further discussion of this and related topics (above is excerpted from there).
edited 2 days ago
answered 2 days ago
Bill DubuqueBill Dubuque
221k30 gold badges210 silver badges682 bronze badges
221k30 gold badges210 silver badges682 bronze badges
$begingroup$
The axiom of pairing is not really important. For one thing, it can be proven from the other axioms of set theory. For another, even in the first-order theory of the real numbers where you have no sets at all, you can still reason about pairs of real numbers without needing a single object in the syntax to represent them (you just always represent them with two separate variables).
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
You could make a reasonable case that being able to talk about $mathbb{R}^2$ as a single object is crucial, but again this doesn't actually really have to do with the axiom of pairing. The concept of pairs is obviously very important, but that does not translate into importance for the axiom of pairing in the technical development of set theory.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
@Eric Of course it goes without saying that judgments about such foundational matters and the role they played in historical development of algebra will be subjective. Please read the full linked article for more on my viewpoint (above is only an excerpt).
$endgroup$
– Bill Dubuque
yesterday
add a comment |
$begingroup$
The axiom of pairing is not really important. For one thing, it can be proven from the other axioms of set theory. For another, even in the first-order theory of the real numbers where you have no sets at all, you can still reason about pairs of real numbers without needing a single object in the syntax to represent them (you just always represent them with two separate variables).
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
You could make a reasonable case that being able to talk about $mathbb{R}^2$ as a single object is crucial, but again this doesn't actually really have to do with the axiom of pairing. The concept of pairs is obviously very important, but that does not translate into importance for the axiom of pairing in the technical development of set theory.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
@Eric Of course it goes without saying that judgments about such foundational matters and the role they played in historical development of algebra will be subjective. Please read the full linked article for more on my viewpoint (above is only an excerpt).
$endgroup$
– Bill Dubuque
yesterday
$begingroup$
The axiom of pairing is not really important. For one thing, it can be proven from the other axioms of set theory. For another, even in the first-order theory of the real numbers where you have no sets at all, you can still reason about pairs of real numbers without needing a single object in the syntax to represent them (you just always represent them with two separate variables).
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
The axiom of pairing is not really important. For one thing, it can be proven from the other axioms of set theory. For another, even in the first-order theory of the real numbers where you have no sets at all, you can still reason about pairs of real numbers without needing a single object in the syntax to represent them (you just always represent them with two separate variables).
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
You could make a reasonable case that being able to talk about $mathbb{R}^2$ as a single object is crucial, but again this doesn't actually really have to do with the axiom of pairing. The concept of pairs is obviously very important, but that does not translate into importance for the axiom of pairing in the technical development of set theory.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
You could make a reasonable case that being able to talk about $mathbb{R}^2$ as a single object is crucial, but again this doesn't actually really have to do with the axiom of pairing. The concept of pairs is obviously very important, but that does not translate into importance for the axiom of pairing in the technical development of set theory.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
@Eric Of course it goes without saying that judgments about such foundational matters and the role they played in historical development of algebra will be subjective. Please read the full linked article for more on my viewpoint (above is only an excerpt).
$endgroup$
– Bill Dubuque
yesterday
$begingroup$
@Eric Of course it goes without saying that judgments about such foundational matters and the role they played in historical development of algebra will be subjective. Please read the full linked article for more on my viewpoint (above is only an excerpt).
$endgroup$
– Bill Dubuque
yesterday
add a comment |
$begingroup$
The "necessary and sufficient" axioms to define the complex numbers are
$$(a,b)+(a',b')=(a+a',b+b')$$
$$(a,b)cdot(a',b')=(aa'-bb',ab'+a'b).$$
(Subtraction and division can be defined as the inverses of addition and multiplication, as usual.)
In particular,
$$(a,b)+(0,0)=(a,b)$$ so that $(0,0)$ is the zero and
$$(a,b)cdot(1,0)=(a,b)$$ so that $(1,0)$ is the unity.
As you can check, $(a,b)$ can also be represented as the expression $a+ib$, where $i$ is a reserved symbol, with the usual computation rules on polynomials (with $i$ seen as the variable).
Using this notation,
$$(0,1)cdot(0,1)=(-1,0)$$
translates to the famous
$$i^2=-1.$$
As you can check, the "pair" representation and the "$i$" representation are completely interchangeable. $i$ has a simple geometric interpretation: in a 2D plane, multiplication by $i$ corresponds to a rotation around the origin by a quarter turn.
Note that there are absolutely no undefined operations here.
$endgroup$
4
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
@BillDubuque: take it informally.
$endgroup$
– Yves Daoust
2 days ago
add a comment |
$begingroup$
The "necessary and sufficient" axioms to define the complex numbers are
$$(a,b)+(a',b')=(a+a',b+b')$$
$$(a,b)cdot(a',b')=(aa'-bb',ab'+a'b).$$
(Subtraction and division can be defined as the inverses of addition and multiplication, as usual.)
In particular,
$$(a,b)+(0,0)=(a,b)$$ so that $(0,0)$ is the zero and
$$(a,b)cdot(1,0)=(a,b)$$ so that $(1,0)$ is the unity.
As you can check, $(a,b)$ can also be represented as the expression $a+ib$, where $i$ is a reserved symbol, with the usual computation rules on polynomials (with $i$ seen as the variable).
Using this notation,
$$(0,1)cdot(0,1)=(-1,0)$$
translates to the famous
$$i^2=-1.$$
As you can check, the "pair" representation and the "$i$" representation are completely interchangeable. $i$ has a simple geometric interpretation: in a 2D plane, multiplication by $i$ corresponds to a rotation around the origin by a quarter turn.
Note that there are absolutely no undefined operations here.
$endgroup$
4
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
@BillDubuque: take it informally.
$endgroup$
– Yves Daoust
2 days ago
add a comment |
$begingroup$
The "necessary and sufficient" axioms to define the complex numbers are
$$(a,b)+(a',b')=(a+a',b+b')$$
$$(a,b)cdot(a',b')=(aa'-bb',ab'+a'b).$$
(Subtraction and division can be defined as the inverses of addition and multiplication, as usual.)
In particular,
$$(a,b)+(0,0)=(a,b)$$ so that $(0,0)$ is the zero and
$$(a,b)cdot(1,0)=(a,b)$$ so that $(1,0)$ is the unity.
As you can check, $(a,b)$ can also be represented as the expression $a+ib$, where $i$ is a reserved symbol, with the usual computation rules on polynomials (with $i$ seen as the variable).
Using this notation,
$$(0,1)cdot(0,1)=(-1,0)$$
translates to the famous
$$i^2=-1.$$
As you can check, the "pair" representation and the "$i$" representation are completely interchangeable. $i$ has a simple geometric interpretation: in a 2D plane, multiplication by $i$ corresponds to a rotation around the origin by a quarter turn.
Note that there are absolutely no undefined operations here.
$endgroup$
The "necessary and sufficient" axioms to define the complex numbers are
$$(a,b)+(a',b')=(a+a',b+b')$$
$$(a,b)cdot(a',b')=(aa'-bb',ab'+a'b).$$
(Subtraction and division can be defined as the inverses of addition and multiplication, as usual.)
In particular,
$$(a,b)+(0,0)=(a,b)$$ so that $(0,0)$ is the zero and
$$(a,b)cdot(1,0)=(a,b)$$ so that $(1,0)$ is the unity.
As you can check, $(a,b)$ can also be represented as the expression $a+ib$, where $i$ is a reserved symbol, with the usual computation rules on polynomials (with $i$ seen as the variable).
Using this notation,
$$(0,1)cdot(0,1)=(-1,0)$$
translates to the famous
$$i^2=-1.$$
As you can check, the "pair" representation and the "$i$" representation are completely interchangeable. $i$ has a simple geometric interpretation: in a 2D plane, multiplication by $i$ corresponds to a rotation around the origin by a quarter turn.
Note that there are absolutely no undefined operations here.
edited 2 days ago
answered 2 days ago
Yves DaoustYves Daoust
144k10 gold badges89 silver badges245 bronze badges
144k10 gold badges89 silver badges245 bronze badges
4
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
@BillDubuque: take it informally.
$endgroup$
– Yves Daoust
2 days ago
add a comment |
4
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
@BillDubuque: take it informally.
$endgroup$
– Yves Daoust
2 days ago
4
4
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
@BillDubuque: take it informally.
$endgroup$
– Yves Daoust
2 days ago
$begingroup$
@BillDubuque: take it informally.
$endgroup$
– Yves Daoust
2 days ago
add a comment |
$begingroup$
At the risk of sounding like a postmodernist: all numbers are imaginary.
Long ago, someone abstracted: what is the thing that this collection of sheep has in common with the number of fingers on my left hand, and called that thing "five." No inconsistencies were introduced and there were great simplifications to be made.
Someone asked how to divide two pies among three people and the abstraction of fractions was born. Someone thought about debt and the abstraction of negative numbers was born. Someone realized that positive and negative fractions didn't describe the intuitive nature of a continuum and the abstraction of the reals was born.
And eventually someone abstracted solutions to $x^2 + 1 = 0$; they're not any more imaginary than any of the other abstractions, they're all products of human imagination. The name "imaginary numbers" is unfortunate.
You say: why can't I abstract a solution to $0*z = 1$, i.e. $1/0$? The problem is that your abstraction will be incompatible with your other abstractions, i.e. you will break arithmetic. But there are areas of geometry (e.g. Mobius transformations of the plane) where there is a consistent way to do a little arithmetic with an idea of $1/0 = infty$ (although one has to be careful to remain consistent).
$endgroup$
add a comment |
$begingroup$
At the risk of sounding like a postmodernist: all numbers are imaginary.
Long ago, someone abstracted: what is the thing that this collection of sheep has in common with the number of fingers on my left hand, and called that thing "five." No inconsistencies were introduced and there were great simplifications to be made.
Someone asked how to divide two pies among three people and the abstraction of fractions was born. Someone thought about debt and the abstraction of negative numbers was born. Someone realized that positive and negative fractions didn't describe the intuitive nature of a continuum and the abstraction of the reals was born.
And eventually someone abstracted solutions to $x^2 + 1 = 0$; they're not any more imaginary than any of the other abstractions, they're all products of human imagination. The name "imaginary numbers" is unfortunate.
You say: why can't I abstract a solution to $0*z = 1$, i.e. $1/0$? The problem is that your abstraction will be incompatible with your other abstractions, i.e. you will break arithmetic. But there are areas of geometry (e.g. Mobius transformations of the plane) where there is a consistent way to do a little arithmetic with an idea of $1/0 = infty$ (although one has to be careful to remain consistent).
$endgroup$
add a comment |
$begingroup$
At the risk of sounding like a postmodernist: all numbers are imaginary.
Long ago, someone abstracted: what is the thing that this collection of sheep has in common with the number of fingers on my left hand, and called that thing "five." No inconsistencies were introduced and there were great simplifications to be made.
Someone asked how to divide two pies among three people and the abstraction of fractions was born. Someone thought about debt and the abstraction of negative numbers was born. Someone realized that positive and negative fractions didn't describe the intuitive nature of a continuum and the abstraction of the reals was born.
And eventually someone abstracted solutions to $x^2 + 1 = 0$; they're not any more imaginary than any of the other abstractions, they're all products of human imagination. The name "imaginary numbers" is unfortunate.
You say: why can't I abstract a solution to $0*z = 1$, i.e. $1/0$? The problem is that your abstraction will be incompatible with your other abstractions, i.e. you will break arithmetic. But there are areas of geometry (e.g. Mobius transformations of the plane) where there is a consistent way to do a little arithmetic with an idea of $1/0 = infty$ (although one has to be careful to remain consistent).
$endgroup$
At the risk of sounding like a postmodernist: all numbers are imaginary.
Long ago, someone abstracted: what is the thing that this collection of sheep has in common with the number of fingers on my left hand, and called that thing "five." No inconsistencies were introduced and there were great simplifications to be made.
Someone asked how to divide two pies among three people and the abstraction of fractions was born. Someone thought about debt and the abstraction of negative numbers was born. Someone realized that positive and negative fractions didn't describe the intuitive nature of a continuum and the abstraction of the reals was born.
And eventually someone abstracted solutions to $x^2 + 1 = 0$; they're not any more imaginary than any of the other abstractions, they're all products of human imagination. The name "imaginary numbers" is unfortunate.
You say: why can't I abstract a solution to $0*z = 1$, i.e. $1/0$? The problem is that your abstraction will be incompatible with your other abstractions, i.e. you will break arithmetic. But there are areas of geometry (e.g. Mobius transformations of the plane) where there is a consistent way to do a little arithmetic with an idea of $1/0 = infty$ (although one has to be careful to remain consistent).
answered yesterday
hunterhunter
16.8k3 gold badges27 silver badges44 bronze badges
16.8k3 gold badges27 silver badges44 bronze badges
add a comment |
add a comment |
$begingroup$
We can always use the usual rules for doing arithmetic with real numbers on complex numbers too, provided we always substitute $-1$ for $i^2$ whenever we encounter it. It follows that once we allow the imaginary unit, we have a consistent algebra which obeys the usual laws.
Thus, we discover that if we extend all the usual laws of addition and multiplication on binomials of the form $a+ib,$ with $i^2=-1,$ everything goes on smoothly. One way to formally do this is by using ordered pairs, as William Hamilton first did, but the idea I think you need is that if we allow ourselves to calculate with complex numbers as we did with real numbers, only remembering to replace $i^2$ with $-1,$ then we have a consistent algebra. Mathematicians usually call the system so defined a field. There are many other fields apart from the ones formed by the real or complex numbers with addition and multiplication as usually defined, but that's another story. The gist is just that we can define two operations over the complex numbers similar in behavior to the usual addition and multiplication over the real numbers -- and with a certain lightness we may think of these as extensions of the usual addition and multiplication. Thus, the $+$ and $×$ of the complex system is technically different from that of the reals, but of course for symbolic economy (and also because they behave very similarly), we retain the same symbols. In general we use these symbols for the operations in any field, too, whether the members of the field are numbers or people.
$endgroup$
add a comment |
$begingroup$
We can always use the usual rules for doing arithmetic with real numbers on complex numbers too, provided we always substitute $-1$ for $i^2$ whenever we encounter it. It follows that once we allow the imaginary unit, we have a consistent algebra which obeys the usual laws.
Thus, we discover that if we extend all the usual laws of addition and multiplication on binomials of the form $a+ib,$ with $i^2=-1,$ everything goes on smoothly. One way to formally do this is by using ordered pairs, as William Hamilton first did, but the idea I think you need is that if we allow ourselves to calculate with complex numbers as we did with real numbers, only remembering to replace $i^2$ with $-1,$ then we have a consistent algebra. Mathematicians usually call the system so defined a field. There are many other fields apart from the ones formed by the real or complex numbers with addition and multiplication as usually defined, but that's another story. The gist is just that we can define two operations over the complex numbers similar in behavior to the usual addition and multiplication over the real numbers -- and with a certain lightness we may think of these as extensions of the usual addition and multiplication. Thus, the $+$ and $×$ of the complex system is technically different from that of the reals, but of course for symbolic economy (and also because they behave very similarly), we retain the same symbols. In general we use these symbols for the operations in any field, too, whether the members of the field are numbers or people.
$endgroup$
add a comment |
$begingroup$
We can always use the usual rules for doing arithmetic with real numbers on complex numbers too, provided we always substitute $-1$ for $i^2$ whenever we encounter it. It follows that once we allow the imaginary unit, we have a consistent algebra which obeys the usual laws.
Thus, we discover that if we extend all the usual laws of addition and multiplication on binomials of the form $a+ib,$ with $i^2=-1,$ everything goes on smoothly. One way to formally do this is by using ordered pairs, as William Hamilton first did, but the idea I think you need is that if we allow ourselves to calculate with complex numbers as we did with real numbers, only remembering to replace $i^2$ with $-1,$ then we have a consistent algebra. Mathematicians usually call the system so defined a field. There are many other fields apart from the ones formed by the real or complex numbers with addition and multiplication as usually defined, but that's another story. The gist is just that we can define two operations over the complex numbers similar in behavior to the usual addition and multiplication over the real numbers -- and with a certain lightness we may think of these as extensions of the usual addition and multiplication. Thus, the $+$ and $×$ of the complex system is technically different from that of the reals, but of course for symbolic economy (and also because they behave very similarly), we retain the same symbols. In general we use these symbols for the operations in any field, too, whether the members of the field are numbers or people.
$endgroup$
We can always use the usual rules for doing arithmetic with real numbers on complex numbers too, provided we always substitute $-1$ for $i^2$ whenever we encounter it. It follows that once we allow the imaginary unit, we have a consistent algebra which obeys the usual laws.
Thus, we discover that if we extend all the usual laws of addition and multiplication on binomials of the form $a+ib,$ with $i^2=-1,$ everything goes on smoothly. One way to formally do this is by using ordered pairs, as William Hamilton first did, but the idea I think you need is that if we allow ourselves to calculate with complex numbers as we did with real numbers, only remembering to replace $i^2$ with $-1,$ then we have a consistent algebra. Mathematicians usually call the system so defined a field. There are many other fields apart from the ones formed by the real or complex numbers with addition and multiplication as usually defined, but that's another story. The gist is just that we can define two operations over the complex numbers similar in behavior to the usual addition and multiplication over the real numbers -- and with a certain lightness we may think of these as extensions of the usual addition and multiplication. Thus, the $+$ and $×$ of the complex system is technically different from that of the reals, but of course for symbolic economy (and also because they behave very similarly), we retain the same symbols. In general we use these symbols for the operations in any field, too, whether the members of the field are numbers or people.
answered 2 days ago
AllawonderAllawonder
3,9498 silver badges18 bronze badges
3,9498 silver badges18 bronze badges
add a comment |
add a comment |
$begingroup$
Why "adding a new element" $i$ with $i^2 = -1$ to $Bbb R$ "works", the technical cause.
What means "works": we want a field extension of $Bbb R$. Intuitively, a set greater than $Bbb R$ (this is the easy part) that is also a field: we have two operations $+$ and $cdot$ extending the operations sum and product of $Bbb R$ and verifying the same properties (see https://en.wikipedia.org/wiki/Field_(mathematics)). This is the hard part.
The "addition of a new element" $i$ with $i^2 = -1$ makes perfect sense in algebra: we are taking the quotient
$$Bbb R[x]/(x^2 + 1)$$
where $Bbb R[x]$ is the polynomial ring in one indeterminate and $(x^2 + 1)$ is the ideal generated by $x^2 + 1$.
The essential fact: a quotient like this will be a field iff the ideal is maximal. You can check easily that $(x^2 + 1)$ is maximal because $x^2 + 1$ is a polynomial of degree 2 without roots in $Bbb R$. Also important: the quotient contains (an isomorphic copy of) $Bbb R$ strictly.
Two examples where "adding a new element" does not "work":
$Bbb R[x]/(x)$ is a field because $(x)$ is a maximal ideal, but the quotient is isomorphic to $Bbb R$.
$Bbb R[x]/(x^2)$ isn't a field because $(x^2)$ isn't a maximal ideal.
$endgroup$
add a comment |
$begingroup$
Why "adding a new element" $i$ with $i^2 = -1$ to $Bbb R$ "works", the technical cause.
What means "works": we want a field extension of $Bbb R$. Intuitively, a set greater than $Bbb R$ (this is the easy part) that is also a field: we have two operations $+$ and $cdot$ extending the operations sum and product of $Bbb R$ and verifying the same properties (see https://en.wikipedia.org/wiki/Field_(mathematics)). This is the hard part.
The "addition of a new element" $i$ with $i^2 = -1$ makes perfect sense in algebra: we are taking the quotient
$$Bbb R[x]/(x^2 + 1)$$
where $Bbb R[x]$ is the polynomial ring in one indeterminate and $(x^2 + 1)$ is the ideal generated by $x^2 + 1$.
The essential fact: a quotient like this will be a field iff the ideal is maximal. You can check easily that $(x^2 + 1)$ is maximal because $x^2 + 1$ is a polynomial of degree 2 without roots in $Bbb R$. Also important: the quotient contains (an isomorphic copy of) $Bbb R$ strictly.
Two examples where "adding a new element" does not "work":
$Bbb R[x]/(x)$ is a field because $(x)$ is a maximal ideal, but the quotient is isomorphic to $Bbb R$.
$Bbb R[x]/(x^2)$ isn't a field because $(x^2)$ isn't a maximal ideal.
$endgroup$
add a comment |
$begingroup$
Why "adding a new element" $i$ with $i^2 = -1$ to $Bbb R$ "works", the technical cause.
What means "works": we want a field extension of $Bbb R$. Intuitively, a set greater than $Bbb R$ (this is the easy part) that is also a field: we have two operations $+$ and $cdot$ extending the operations sum and product of $Bbb R$ and verifying the same properties (see https://en.wikipedia.org/wiki/Field_(mathematics)). This is the hard part.
The "addition of a new element" $i$ with $i^2 = -1$ makes perfect sense in algebra: we are taking the quotient
$$Bbb R[x]/(x^2 + 1)$$
where $Bbb R[x]$ is the polynomial ring in one indeterminate and $(x^2 + 1)$ is the ideal generated by $x^2 + 1$.
The essential fact: a quotient like this will be a field iff the ideal is maximal. You can check easily that $(x^2 + 1)$ is maximal because $x^2 + 1$ is a polynomial of degree 2 without roots in $Bbb R$. Also important: the quotient contains (an isomorphic copy of) $Bbb R$ strictly.
Two examples where "adding a new element" does not "work":
$Bbb R[x]/(x)$ is a field because $(x)$ is a maximal ideal, but the quotient is isomorphic to $Bbb R$.
$Bbb R[x]/(x^2)$ isn't a field because $(x^2)$ isn't a maximal ideal.
$endgroup$
Why "adding a new element" $i$ with $i^2 = -1$ to $Bbb R$ "works", the technical cause.
What means "works": we want a field extension of $Bbb R$. Intuitively, a set greater than $Bbb R$ (this is the easy part) that is also a field: we have two operations $+$ and $cdot$ extending the operations sum and product of $Bbb R$ and verifying the same properties (see https://en.wikipedia.org/wiki/Field_(mathematics)). This is the hard part.
The "addition of a new element" $i$ with $i^2 = -1$ makes perfect sense in algebra: we are taking the quotient
$$Bbb R[x]/(x^2 + 1)$$
where $Bbb R[x]$ is the polynomial ring in one indeterminate and $(x^2 + 1)$ is the ideal generated by $x^2 + 1$.
The essential fact: a quotient like this will be a field iff the ideal is maximal. You can check easily that $(x^2 + 1)$ is maximal because $x^2 + 1$ is a polynomial of degree 2 without roots in $Bbb R$. Also important: the quotient contains (an isomorphic copy of) $Bbb R$ strictly.
Two examples where "adding a new element" does not "work":
$Bbb R[x]/(x)$ is a field because $(x)$ is a maximal ideal, but the quotient is isomorphic to $Bbb R$.
$Bbb R[x]/(x^2)$ isn't a field because $(x^2)$ isn't a maximal ideal.
answered yesterday
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
36.3k4 gold badges31 silver badges73 bronze badges
36.3k4 gold badges31 silver badges73 bronze badges
add a comment |
add a comment |
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
The definition of the complex numbers? There are really no new axioms involved since complex numbers are defined in terms of real ones and their behavior is completely derivable from definitions.
In the beginning, when there were just reals, these operations were defined for them.
See, there is your mistake. "there were just reals". Reals are imaginary constructs with no inherent existence. They are the consequence of choosing convenient definitions (and some axioms).
Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
That is a common misconception and actually untenable. You cannot "define" i as "the square root of -1" since -1 has two square roots in the complex numbers. Instead, complex numbers are defined via their arithmetic properties and then the real numbers are embedded into the complex numbers. Since that embedding retains basically all operations on the real numbers (logarithms and exponentiation get a lot more iffy than they are in the reals, though, even though exp(x) remains perfectly well defined), one can continue using the same symbols and operators without introducing trouble, just like embedding whole numbers into rational numbers and rational numbers into reals worked without necessitating different operators for writing down relations.
$endgroup$
add a comment |
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
The definition of the complex numbers? There are really no new axioms involved since complex numbers are defined in terms of real ones and their behavior is completely derivable from definitions.
In the beginning, when there were just reals, these operations were defined for them.
See, there is your mistake. "there were just reals". Reals are imaginary constructs with no inherent existence. They are the consequence of choosing convenient definitions (and some axioms).
Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
That is a common misconception and actually untenable. You cannot "define" i as "the square root of -1" since -1 has two square roots in the complex numbers. Instead, complex numbers are defined via their arithmetic properties and then the real numbers are embedded into the complex numbers. Since that embedding retains basically all operations on the real numbers (logarithms and exponentiation get a lot more iffy than they are in the reals, though, even though exp(x) remains perfectly well defined), one can continue using the same symbols and operators without introducing trouble, just like embedding whole numbers into rational numbers and rational numbers into reals worked without necessitating different operators for writing down relations.
$endgroup$
add a comment |
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
The definition of the complex numbers? There are really no new axioms involved since complex numbers are defined in terms of real ones and their behavior is completely derivable from definitions.
In the beginning, when there were just reals, these operations were defined for them.
See, there is your mistake. "there were just reals". Reals are imaginary constructs with no inherent existence. They are the consequence of choosing convenient definitions (and some axioms).
Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
That is a common misconception and actually untenable. You cannot "define" i as "the square root of -1" since -1 has two square roots in the complex numbers. Instead, complex numbers are defined via their arithmetic properties and then the real numbers are embedded into the complex numbers. Since that embedding retains basically all operations on the real numbers (logarithms and exponentiation get a lot more iffy than they are in the reals, though, even though exp(x) remains perfectly well defined), one can continue using the same symbols and operators without introducing trouble, just like embedding whole numbers into rational numbers and rational numbers into reals worked without necessitating different operators for writing down relations.
$endgroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
The definition of the complex numbers? There are really no new axioms involved since complex numbers are defined in terms of real ones and their behavior is completely derivable from definitions.
In the beginning, when there were just reals, these operations were defined for them.
See, there is your mistake. "there were just reals". Reals are imaginary constructs with no inherent existence. They are the consequence of choosing convenient definitions (and some axioms).
Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
That is a common misconception and actually untenable. You cannot "define" i as "the square root of -1" since -1 has two square roots in the complex numbers. Instead, complex numbers are defined via their arithmetic properties and then the real numbers are embedded into the complex numbers. Since that embedding retains basically all operations on the real numbers (logarithms and exponentiation get a lot more iffy than they are in the reals, though, even though exp(x) remains perfectly well defined), one can continue using the same symbols and operators without introducing trouble, just like embedding whole numbers into rational numbers and rational numbers into reals worked without necessitating different operators for writing down relations.
answered 2 days ago
user696180
add a comment |
add a comment |
$begingroup$
In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Not quite: in the beginning were the natural numbers, and addition over the natural numbers, and everything was good.
But then people wanted to "undo" addition, and subtraction turned out to be really useful, and related to many different real-world examples (taking things away from a collection). And suddenly a problem arose: you could add any two natural numbers and get a third natural number, but when you subtracted, the result might be undefined. What natural number is five minus seven?
Some people were okay with this being undefined, but others weren't, and the ones who weren't started wondering: what if we define a new quantity that's equal to five minus seven? And so the negative numbers were born! (And for quite a long time, mainstream mathematicians derided them and didn't consider them real numbers, because can you show me negative two apples, or people, or coins? No you can't.)
Similarly, division led to the introduction of rational numbers ("one divided by two is defined to be a new type of number, which I'll just write as 1/2"), and square roots led to the introduction of irrational numbers, and eventually we ended up with the real numbers that we all know and love today. "Imaginary" numbers are just another extension of the concept of "number", giving it new capabilities.
So then, would you like to define 1/0? Go for it! Look at the projectively extended real numbers, which do exactly that: augmenting the real numbers with a single "point at infinity". In the projectively-extended reals, 1/0=∞, and there's no problem there. As Tanner Swett put it, you can do anything you want! The real question in mathematics is: does this lead to anything interesting? And the only way to answer that is, try it and see!
P.S. The summary in this answer isn't meant to be chronologically accurate—the Greeks were dealing with "real numbers" in the form of lengths long before anyone came up with the Peano axioms. Instead, take this as a summary of how axiomatic definitions of "numbers" have evolved over time.
P.P.S. There's a legend that one ancient mathematician was either murdered or struck down by the gods for the hubris of constructing an irrational number. I haven't heard of this happening with the projectively-extended reals, but watch your back, just in case.
P.P.P.S. If you're curious about the complex numbers in particular, this website has a better explanation than I could ever give. Geometrically, $i$ can be considered as a 90-degree rotation in the plane; in geometric algebra, it would be called a unit bivector. But that's a can of worms too big for an answer here to encompass.
$endgroup$
add a comment |
$begingroup$
In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Not quite: in the beginning were the natural numbers, and addition over the natural numbers, and everything was good.
But then people wanted to "undo" addition, and subtraction turned out to be really useful, and related to many different real-world examples (taking things away from a collection). And suddenly a problem arose: you could add any two natural numbers and get a third natural number, but when you subtracted, the result might be undefined. What natural number is five minus seven?
Some people were okay with this being undefined, but others weren't, and the ones who weren't started wondering: what if we define a new quantity that's equal to five minus seven? And so the negative numbers were born! (And for quite a long time, mainstream mathematicians derided them and didn't consider them real numbers, because can you show me negative two apples, or people, or coins? No you can't.)
Similarly, division led to the introduction of rational numbers ("one divided by two is defined to be a new type of number, which I'll just write as 1/2"), and square roots led to the introduction of irrational numbers, and eventually we ended up with the real numbers that we all know and love today. "Imaginary" numbers are just another extension of the concept of "number", giving it new capabilities.
So then, would you like to define 1/0? Go for it! Look at the projectively extended real numbers, which do exactly that: augmenting the real numbers with a single "point at infinity". In the projectively-extended reals, 1/0=∞, and there's no problem there. As Tanner Swett put it, you can do anything you want! The real question in mathematics is: does this lead to anything interesting? And the only way to answer that is, try it and see!
P.S. The summary in this answer isn't meant to be chronologically accurate—the Greeks were dealing with "real numbers" in the form of lengths long before anyone came up with the Peano axioms. Instead, take this as a summary of how axiomatic definitions of "numbers" have evolved over time.
P.P.S. There's a legend that one ancient mathematician was either murdered or struck down by the gods for the hubris of constructing an irrational number. I haven't heard of this happening with the projectively-extended reals, but watch your back, just in case.
P.P.P.S. If you're curious about the complex numbers in particular, this website has a better explanation than I could ever give. Geometrically, $i$ can be considered as a 90-degree rotation in the plane; in geometric algebra, it would be called a unit bivector. But that's a can of worms too big for an answer here to encompass.
$endgroup$
add a comment |
$begingroup$
In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Not quite: in the beginning were the natural numbers, and addition over the natural numbers, and everything was good.
But then people wanted to "undo" addition, and subtraction turned out to be really useful, and related to many different real-world examples (taking things away from a collection). And suddenly a problem arose: you could add any two natural numbers and get a third natural number, but when you subtracted, the result might be undefined. What natural number is five minus seven?
Some people were okay with this being undefined, but others weren't, and the ones who weren't started wondering: what if we define a new quantity that's equal to five minus seven? And so the negative numbers were born! (And for quite a long time, mainstream mathematicians derided them and didn't consider them real numbers, because can you show me negative two apples, or people, or coins? No you can't.)
Similarly, division led to the introduction of rational numbers ("one divided by two is defined to be a new type of number, which I'll just write as 1/2"), and square roots led to the introduction of irrational numbers, and eventually we ended up with the real numbers that we all know and love today. "Imaginary" numbers are just another extension of the concept of "number", giving it new capabilities.
So then, would you like to define 1/0? Go for it! Look at the projectively extended real numbers, which do exactly that: augmenting the real numbers with a single "point at infinity". In the projectively-extended reals, 1/0=∞, and there's no problem there. As Tanner Swett put it, you can do anything you want! The real question in mathematics is: does this lead to anything interesting? And the only way to answer that is, try it and see!
P.S. The summary in this answer isn't meant to be chronologically accurate—the Greeks were dealing with "real numbers" in the form of lengths long before anyone came up with the Peano axioms. Instead, take this as a summary of how axiomatic definitions of "numbers" have evolved over time.
P.P.S. There's a legend that one ancient mathematician was either murdered or struck down by the gods for the hubris of constructing an irrational number. I haven't heard of this happening with the projectively-extended reals, but watch your back, just in case.
P.P.P.S. If you're curious about the complex numbers in particular, this website has a better explanation than I could ever give. Geometrically, $i$ can be considered as a 90-degree rotation in the plane; in geometric algebra, it would be called a unit bivector. But that's a can of worms too big for an answer here to encompass.
$endgroup$
In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Not quite: in the beginning were the natural numbers, and addition over the natural numbers, and everything was good.
But then people wanted to "undo" addition, and subtraction turned out to be really useful, and related to many different real-world examples (taking things away from a collection). And suddenly a problem arose: you could add any two natural numbers and get a third natural number, but when you subtracted, the result might be undefined. What natural number is five minus seven?
Some people were okay with this being undefined, but others weren't, and the ones who weren't started wondering: what if we define a new quantity that's equal to five minus seven? And so the negative numbers were born! (And for quite a long time, mainstream mathematicians derided them and didn't consider them real numbers, because can you show me negative two apples, or people, or coins? No you can't.)
Similarly, division led to the introduction of rational numbers ("one divided by two is defined to be a new type of number, which I'll just write as 1/2"), and square roots led to the introduction of irrational numbers, and eventually we ended up with the real numbers that we all know and love today. "Imaginary" numbers are just another extension of the concept of "number", giving it new capabilities.
So then, would you like to define 1/0? Go for it! Look at the projectively extended real numbers, which do exactly that: augmenting the real numbers with a single "point at infinity". In the projectively-extended reals, 1/0=∞, and there's no problem there. As Tanner Swett put it, you can do anything you want! The real question in mathematics is: does this lead to anything interesting? And the only way to answer that is, try it and see!
P.S. The summary in this answer isn't meant to be chronologically accurate—the Greeks were dealing with "real numbers" in the form of lengths long before anyone came up with the Peano axioms. Instead, take this as a summary of how axiomatic definitions of "numbers" have evolved over time.
P.P.S. There's a legend that one ancient mathematician was either murdered or struck down by the gods for the hubris of constructing an irrational number. I haven't heard of this happening with the projectively-extended reals, but watch your back, just in case.
P.P.P.S. If you're curious about the complex numbers in particular, this website has a better explanation than I could ever give. Geometrically, $i$ can be considered as a 90-degree rotation in the plane; in geometric algebra, it would be called a unit bivector. But that's a can of worms too big for an answer here to encompass.
answered 22 hours ago
DraconisDraconis
6704 silver badges9 bronze badges
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$begingroup$
What allow us to use anything in mathematics are the axioms. Real numbers form a complete ordered field, if you take the set of real numbers and "add" $i$ to it (considering its usual behavior), you lose this order (that is: They won't be a complete ordered field anymore). You can still define some order but it will no longer be compatible with the ordering of the reals. (It is a good exercise to verify what axioms this "addition" of $i$ violates. See this or this).
In the previous case, the natural ordering was broken but mathematicians found a way to cope with it by using a new system in which the broken axioms were removed. These axioms give us a plethora of stuff we can do in the real numbers, it was natural to think: "In this new system, how can we do something that is at least remotely similar to what we could do with the ordering in the real numbers?"
There is no "moral ground" to define and use things in mathematics, you can define anything you want. There is - for example - a number system in which division by zero is meaningful. In the same way, you may run into trouble (such as the aforementioned loss of ordering) when you compare some two systems.
$endgroup$
add a comment |
$begingroup$
What allow us to use anything in mathematics are the axioms. Real numbers form a complete ordered field, if you take the set of real numbers and "add" $i$ to it (considering its usual behavior), you lose this order (that is: They won't be a complete ordered field anymore). You can still define some order but it will no longer be compatible with the ordering of the reals. (It is a good exercise to verify what axioms this "addition" of $i$ violates. See this or this).
In the previous case, the natural ordering was broken but mathematicians found a way to cope with it by using a new system in which the broken axioms were removed. These axioms give us a plethora of stuff we can do in the real numbers, it was natural to think: "In this new system, how can we do something that is at least remotely similar to what we could do with the ordering in the real numbers?"
There is no "moral ground" to define and use things in mathematics, you can define anything you want. There is - for example - a number system in which division by zero is meaningful. In the same way, you may run into trouble (such as the aforementioned loss of ordering) when you compare some two systems.
$endgroup$
add a comment |
$begingroup$
What allow us to use anything in mathematics are the axioms. Real numbers form a complete ordered field, if you take the set of real numbers and "add" $i$ to it (considering its usual behavior), you lose this order (that is: They won't be a complete ordered field anymore). You can still define some order but it will no longer be compatible with the ordering of the reals. (It is a good exercise to verify what axioms this "addition" of $i$ violates. See this or this).
In the previous case, the natural ordering was broken but mathematicians found a way to cope with it by using a new system in which the broken axioms were removed. These axioms give us a plethora of stuff we can do in the real numbers, it was natural to think: "In this new system, how can we do something that is at least remotely similar to what we could do with the ordering in the real numbers?"
There is no "moral ground" to define and use things in mathematics, you can define anything you want. There is - for example - a number system in which division by zero is meaningful. In the same way, you may run into trouble (such as the aforementioned loss of ordering) when you compare some two systems.
$endgroup$
What allow us to use anything in mathematics are the axioms. Real numbers form a complete ordered field, if you take the set of real numbers and "add" $i$ to it (considering its usual behavior), you lose this order (that is: They won't be a complete ordered field anymore). You can still define some order but it will no longer be compatible with the ordering of the reals. (It is a good exercise to verify what axioms this "addition" of $i$ violates. See this or this).
In the previous case, the natural ordering was broken but mathematicians found a way to cope with it by using a new system in which the broken axioms were removed. These axioms give us a plethora of stuff we can do in the real numbers, it was natural to think: "In this new system, how can we do something that is at least remotely similar to what we could do with the ordering in the real numbers?"
There is no "moral ground" to define and use things in mathematics, you can define anything you want. There is - for example - a number system in which division by zero is meaningful. In the same way, you may run into trouble (such as the aforementioned loss of ordering) when you compare some two systems.
edited yesterday
answered yesterday
Billy RubinaBilly Rubina
10.8k14 gold badges62 silver badges139 bronze badges
10.8k14 gold badges62 silver badges139 bronze badges
add a comment |
add a comment |
$begingroup$
I think you already have a good interpretation of complex numbers algebraic view. So I propose here the geometric approach of this numbers, following Hamilton, Clifford and Grassmann geometric view of algebra.
First, we need to separate the meaning about the number (1) and the number (-1). You can do this for trying to count negative numbers as one does with positive numbers, e.g. one cup, two notes, three boxes, minus four houses???
The later seems meaningless because we have not added a good definition about (-). For this, we need to think in the number not only as a quantification, but as a quantification AND a direction. So (-1) is not just a number as (4), but a number that one can interpret as a quantity in a certain direction.
So the geometric view of (+) and (-) is that of symbols whose when fixed with number defines a direction in a certain quantity, e.g. If (+3) means 3 steps of unity in the certain direction, say right, then (-3) means 3 steps in the opposite direction. This could be demonstrated with the axiomatic approaches that my colleagues presented above, but I want to propose here just a geometric view of all this.
In the Euclidean Geometry, we can assert more than the opposite notion. If we suppose a segment $AB$, with a certain magnitude, when applying (-1) in the segment, we rotate it $pi$ rad. So $AB=-BA$.
This notion can be extended to "when applying positive number to a segment, it contracts (when between 0 and 1), maintain or stretch this segment. But when negative number are applied, besides the homothety, it rotates through $pi$ rad."
It's just a geometric view, I'm not proposing here any rigorous justifications.
But this is just in the $mathbb{R}$ line. Let's extend this thought to $mathbb{R}^2$ plane.
We already know that $(-1)$ means rotates in $pi$ rad, but as we now have this two lines orthogonally positioned, we also need to define an application (a number) which could rotate the segment $pi/2$ rad. Let's suppose a new type of number, whose faces i don't see, but algebraically i can give him a name (a letter), say $i$. So as we can see (and you could prove it using the axioms of above answers), when applying twice $i$ to this segment, it rotates to $pi$ rad and we can assert that $i^2=-1$. In this view, the meaning $i=sqrt{-1}$ is just a consequence of $i^2=-1$.
Now, as Kline say here, Hamilton pointed out that
A complex number $a+bi$ is not a genuine sum in the sense that $2 + 3$ is. The
use of the plus sign is a historical accident and $bi$ cannot be added to $a$. The
complex number $a + bi$ is no more than an ordered couple $(a, b)$ of real
numbers.
i suppose this answer your saying "In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined"
This allows us to use the knowledge of vectors. Indeed, Hamilton knew that a complex number is nothing more that a ratio between two segments. If $AB$ and $AC$ are two segments conveniently positioned at the origin $A$, a complex number $z=a+bi=(a,b)=e^{itheta}$, with $theta$ the angle between the complex number $z$ and the origin, is the operation necessary to rotate and stretch the segment $AB$ to give $AC$, or $ABcdot z=ACto z=frac{AC}{AB}$, which allows us to use complex numbers as a type of rotational numbers, as the negative numbers are a directional system of numbers.
And Hamilton gave latter a full usefulness of the imaginary numbers to also rotate in $mathbb{R}^3$, where the Quaternions, a four dimensional algebra, was born. And you could ask "why a four dimensional system of numbers is required to rotate three dimensional directed segments (vectors)?"
I'm not allowed to answer this question here, but i can tell you to read about the Clifford Algebras, the Geometric Algebra, which are an excellent extension of Hamilton's Quaternions with the concept of Grassmann's exterior algebra. In the geometric algebra, real numbers, complex numbers, quaternions, biquaternions, and other n-dimensional systems of numbers are synthesized in a rigorous way giving to all this abstract algebraic views a good geometrical interpretation.
$endgroup$
add a comment |
$begingroup$
I think you already have a good interpretation of complex numbers algebraic view. So I propose here the geometric approach of this numbers, following Hamilton, Clifford and Grassmann geometric view of algebra.
First, we need to separate the meaning about the number (1) and the number (-1). You can do this for trying to count negative numbers as one does with positive numbers, e.g. one cup, two notes, three boxes, minus four houses???
The later seems meaningless because we have not added a good definition about (-). For this, we need to think in the number not only as a quantification, but as a quantification AND a direction. So (-1) is not just a number as (4), but a number that one can interpret as a quantity in a certain direction.
So the geometric view of (+) and (-) is that of symbols whose when fixed with number defines a direction in a certain quantity, e.g. If (+3) means 3 steps of unity in the certain direction, say right, then (-3) means 3 steps in the opposite direction. This could be demonstrated with the axiomatic approaches that my colleagues presented above, but I want to propose here just a geometric view of all this.
In the Euclidean Geometry, we can assert more than the opposite notion. If we suppose a segment $AB$, with a certain magnitude, when applying (-1) in the segment, we rotate it $pi$ rad. So $AB=-BA$.
This notion can be extended to "when applying positive number to a segment, it contracts (when between 0 and 1), maintain or stretch this segment. But when negative number are applied, besides the homothety, it rotates through $pi$ rad."
It's just a geometric view, I'm not proposing here any rigorous justifications.
But this is just in the $mathbb{R}$ line. Let's extend this thought to $mathbb{R}^2$ plane.
We already know that $(-1)$ means rotates in $pi$ rad, but as we now have this two lines orthogonally positioned, we also need to define an application (a number) which could rotate the segment $pi/2$ rad. Let's suppose a new type of number, whose faces i don't see, but algebraically i can give him a name (a letter), say $i$. So as we can see (and you could prove it using the axioms of above answers), when applying twice $i$ to this segment, it rotates to $pi$ rad and we can assert that $i^2=-1$. In this view, the meaning $i=sqrt{-1}$ is just a consequence of $i^2=-1$.
Now, as Kline say here, Hamilton pointed out that
A complex number $a+bi$ is not a genuine sum in the sense that $2 + 3$ is. The
use of the plus sign is a historical accident and $bi$ cannot be added to $a$. The
complex number $a + bi$ is no more than an ordered couple $(a, b)$ of real
numbers.
i suppose this answer your saying "In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined"
This allows us to use the knowledge of vectors. Indeed, Hamilton knew that a complex number is nothing more that a ratio between two segments. If $AB$ and $AC$ are two segments conveniently positioned at the origin $A$, a complex number $z=a+bi=(a,b)=e^{itheta}$, with $theta$ the angle between the complex number $z$ and the origin, is the operation necessary to rotate and stretch the segment $AB$ to give $AC$, or $ABcdot z=ACto z=frac{AC}{AB}$, which allows us to use complex numbers as a type of rotational numbers, as the negative numbers are a directional system of numbers.
And Hamilton gave latter a full usefulness of the imaginary numbers to also rotate in $mathbb{R}^3$, where the Quaternions, a four dimensional algebra, was born. And you could ask "why a four dimensional system of numbers is required to rotate three dimensional directed segments (vectors)?"
I'm not allowed to answer this question here, but i can tell you to read about the Clifford Algebras, the Geometric Algebra, which are an excellent extension of Hamilton's Quaternions with the concept of Grassmann's exterior algebra. In the geometric algebra, real numbers, complex numbers, quaternions, biquaternions, and other n-dimensional systems of numbers are synthesized in a rigorous way giving to all this abstract algebraic views a good geometrical interpretation.
$endgroup$
add a comment |
$begingroup$
I think you already have a good interpretation of complex numbers algebraic view. So I propose here the geometric approach of this numbers, following Hamilton, Clifford and Grassmann geometric view of algebra.
First, we need to separate the meaning about the number (1) and the number (-1). You can do this for trying to count negative numbers as one does with positive numbers, e.g. one cup, two notes, three boxes, minus four houses???
The later seems meaningless because we have not added a good definition about (-). For this, we need to think in the number not only as a quantification, but as a quantification AND a direction. So (-1) is not just a number as (4), but a number that one can interpret as a quantity in a certain direction.
So the geometric view of (+) and (-) is that of symbols whose when fixed with number defines a direction in a certain quantity, e.g. If (+3) means 3 steps of unity in the certain direction, say right, then (-3) means 3 steps in the opposite direction. This could be demonstrated with the axiomatic approaches that my colleagues presented above, but I want to propose here just a geometric view of all this.
In the Euclidean Geometry, we can assert more than the opposite notion. If we suppose a segment $AB$, with a certain magnitude, when applying (-1) in the segment, we rotate it $pi$ rad. So $AB=-BA$.
This notion can be extended to "when applying positive number to a segment, it contracts (when between 0 and 1), maintain or stretch this segment. But when negative number are applied, besides the homothety, it rotates through $pi$ rad."
It's just a geometric view, I'm not proposing here any rigorous justifications.
But this is just in the $mathbb{R}$ line. Let's extend this thought to $mathbb{R}^2$ plane.
We already know that $(-1)$ means rotates in $pi$ rad, but as we now have this two lines orthogonally positioned, we also need to define an application (a number) which could rotate the segment $pi/2$ rad. Let's suppose a new type of number, whose faces i don't see, but algebraically i can give him a name (a letter), say $i$. So as we can see (and you could prove it using the axioms of above answers), when applying twice $i$ to this segment, it rotates to $pi$ rad and we can assert that $i^2=-1$. In this view, the meaning $i=sqrt{-1}$ is just a consequence of $i^2=-1$.
Now, as Kline say here, Hamilton pointed out that
A complex number $a+bi$ is not a genuine sum in the sense that $2 + 3$ is. The
use of the plus sign is a historical accident and $bi$ cannot be added to $a$. The
complex number $a + bi$ is no more than an ordered couple $(a, b)$ of real
numbers.
i suppose this answer your saying "In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined"
This allows us to use the knowledge of vectors. Indeed, Hamilton knew that a complex number is nothing more that a ratio between two segments. If $AB$ and $AC$ are two segments conveniently positioned at the origin $A$, a complex number $z=a+bi=(a,b)=e^{itheta}$, with $theta$ the angle between the complex number $z$ and the origin, is the operation necessary to rotate and stretch the segment $AB$ to give $AC$, or $ABcdot z=ACto z=frac{AC}{AB}$, which allows us to use complex numbers as a type of rotational numbers, as the negative numbers are a directional system of numbers.
And Hamilton gave latter a full usefulness of the imaginary numbers to also rotate in $mathbb{R}^3$, where the Quaternions, a four dimensional algebra, was born. And you could ask "why a four dimensional system of numbers is required to rotate three dimensional directed segments (vectors)?"
I'm not allowed to answer this question here, but i can tell you to read about the Clifford Algebras, the Geometric Algebra, which are an excellent extension of Hamilton's Quaternions with the concept of Grassmann's exterior algebra. In the geometric algebra, real numbers, complex numbers, quaternions, biquaternions, and other n-dimensional systems of numbers are synthesized in a rigorous way giving to all this abstract algebraic views a good geometrical interpretation.
$endgroup$
I think you already have a good interpretation of complex numbers algebraic view. So I propose here the geometric approach of this numbers, following Hamilton, Clifford and Grassmann geometric view of algebra.
First, we need to separate the meaning about the number (1) and the number (-1). You can do this for trying to count negative numbers as one does with positive numbers, e.g. one cup, two notes, three boxes, minus four houses???
The later seems meaningless because we have not added a good definition about (-). For this, we need to think in the number not only as a quantification, but as a quantification AND a direction. So (-1) is not just a number as (4), but a number that one can interpret as a quantity in a certain direction.
So the geometric view of (+) and (-) is that of symbols whose when fixed with number defines a direction in a certain quantity, e.g. If (+3) means 3 steps of unity in the certain direction, say right, then (-3) means 3 steps in the opposite direction. This could be demonstrated with the axiomatic approaches that my colleagues presented above, but I want to propose here just a geometric view of all this.
In the Euclidean Geometry, we can assert more than the opposite notion. If we suppose a segment $AB$, with a certain magnitude, when applying (-1) in the segment, we rotate it $pi$ rad. So $AB=-BA$.
This notion can be extended to "when applying positive number to a segment, it contracts (when between 0 and 1), maintain or stretch this segment. But when negative number are applied, besides the homothety, it rotates through $pi$ rad."
It's just a geometric view, I'm not proposing here any rigorous justifications.
But this is just in the $mathbb{R}$ line. Let's extend this thought to $mathbb{R}^2$ plane.
We already know that $(-1)$ means rotates in $pi$ rad, but as we now have this two lines orthogonally positioned, we also need to define an application (a number) which could rotate the segment $pi/2$ rad. Let's suppose a new type of number, whose faces i don't see, but algebraically i can give him a name (a letter), say $i$. So as we can see (and you could prove it using the axioms of above answers), when applying twice $i$ to this segment, it rotates to $pi$ rad and we can assert that $i^2=-1$. In this view, the meaning $i=sqrt{-1}$ is just a consequence of $i^2=-1$.
Now, as Kline say here, Hamilton pointed out that
A complex number $a+bi$ is not a genuine sum in the sense that $2 + 3$ is. The
use of the plus sign is a historical accident and $bi$ cannot be added to $a$. The
complex number $a + bi$ is no more than an ordered couple $(a, b)$ of real
numbers.
i suppose this answer your saying "In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined"
This allows us to use the knowledge of vectors. Indeed, Hamilton knew that a complex number is nothing more that a ratio between two segments. If $AB$ and $AC$ are two segments conveniently positioned at the origin $A$, a complex number $z=a+bi=(a,b)=e^{itheta}$, with $theta$ the angle between the complex number $z$ and the origin, is the operation necessary to rotate and stretch the segment $AB$ to give $AC$, or $ABcdot z=ACto z=frac{AC}{AB}$, which allows us to use complex numbers as a type of rotational numbers, as the negative numbers are a directional system of numbers.
And Hamilton gave latter a full usefulness of the imaginary numbers to also rotate in $mathbb{R}^3$, where the Quaternions, a four dimensional algebra, was born. And you could ask "why a four dimensional system of numbers is required to rotate three dimensional directed segments (vectors)?"
I'm not allowed to answer this question here, but i can tell you to read about the Clifford Algebras, the Geometric Algebra, which are an excellent extension of Hamilton's Quaternions with the concept of Grassmann's exterior algebra. In the geometric algebra, real numbers, complex numbers, quaternions, biquaternions, and other n-dimensional systems of numbers are synthesized in a rigorous way giving to all this abstract algebraic views a good geometrical interpretation.
answered yesterday
Lucas BarbiereLucas Barbiere
817 bronze badges
817 bronze badges
add a comment |
add a comment |
$begingroup$
Well, coming from the algebraic side, consider field extensions of the form
$${Bbb Q}(sqrt n)={a+bsqrt nmid a,bin{Bbb Q}},$$
where $nne0,1$ is a square-free integer.
The addition is ''componentwise'',
$$(a+bsqrt n) + (c+dsqrt n) = (a+c) + (b+d)sqrt n$$
and the multiplication is given by noting that $sqrt n^2 = n$,
$$(a+bsqrt n) cdot (c+dsqrt n) = (ac + bdn) + (ad + bc)sqrt n.$$
Here the case $n=-1$ comes up naturally and writing $i=sqrt{-1}$, we obtain
${Bbb Q}(i)={a+bimid a,bin{Bbb Q}}$ and also ${Bbb C} = {Bbb R}(i)={a+bimid a,bin{Bbb R}}$, the field of complex numbers.
$endgroup$
add a comment |
$begingroup$
Well, coming from the algebraic side, consider field extensions of the form
$${Bbb Q}(sqrt n)={a+bsqrt nmid a,bin{Bbb Q}},$$
where $nne0,1$ is a square-free integer.
The addition is ''componentwise'',
$$(a+bsqrt n) + (c+dsqrt n) = (a+c) + (b+d)sqrt n$$
and the multiplication is given by noting that $sqrt n^2 = n$,
$$(a+bsqrt n) cdot (c+dsqrt n) = (ac + bdn) + (ad + bc)sqrt n.$$
Here the case $n=-1$ comes up naturally and writing $i=sqrt{-1}$, we obtain
${Bbb Q}(i)={a+bimid a,bin{Bbb Q}}$ and also ${Bbb C} = {Bbb R}(i)={a+bimid a,bin{Bbb R}}$, the field of complex numbers.
$endgroup$
add a comment |
$begingroup$
Well, coming from the algebraic side, consider field extensions of the form
$${Bbb Q}(sqrt n)={a+bsqrt nmid a,bin{Bbb Q}},$$
where $nne0,1$ is a square-free integer.
The addition is ''componentwise'',
$$(a+bsqrt n) + (c+dsqrt n) = (a+c) + (b+d)sqrt n$$
and the multiplication is given by noting that $sqrt n^2 = n$,
$$(a+bsqrt n) cdot (c+dsqrt n) = (ac + bdn) + (ad + bc)sqrt n.$$
Here the case $n=-1$ comes up naturally and writing $i=sqrt{-1}$, we obtain
${Bbb Q}(i)={a+bimid a,bin{Bbb Q}}$ and also ${Bbb C} = {Bbb R}(i)={a+bimid a,bin{Bbb R}}$, the field of complex numbers.
$endgroup$
Well, coming from the algebraic side, consider field extensions of the form
$${Bbb Q}(sqrt n)={a+bsqrt nmid a,bin{Bbb Q}},$$
where $nne0,1$ is a square-free integer.
The addition is ''componentwise'',
$$(a+bsqrt n) + (c+dsqrt n) = (a+c) + (b+d)sqrt n$$
and the multiplication is given by noting that $sqrt n^2 = n$,
$$(a+bsqrt n) cdot (c+dsqrt n) = (ac + bdn) + (ad + bc)sqrt n.$$
Here the case $n=-1$ comes up naturally and writing $i=sqrt{-1}$, we obtain
${Bbb Q}(i)={a+bimid a,bin{Bbb Q}}$ and also ${Bbb C} = {Bbb R}(i)={a+bimid a,bin{Bbb R}}$, the field of complex numbers.
answered yesterday
WuestenfuxWuestenfux
9,2682 gold badges5 silver badges16 bronze badges
9,2682 gold badges5 silver badges16 bronze badges
add a comment |
add a comment |
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$begingroup$
Would it bother you less simply to think of $a + bi$ as shorthand for the matrix $$begin{pmatrix}a & -b \ b & aend{pmatrix}?$$
$endgroup$
– Bungo
2 days ago
40
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You may be surprised to learn that the real numbers are not really older: when $i$ was first considered, negative numbers were not generally accepted as numbers either.
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– Robert Israel
2 days ago
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Interestingly, one can introduce $sqrt{-1}$ so that all operations 'behaves the usual way', while introducing $1/0$ will definitely break some familiar laws. Nevertheless, it can be done..
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– Berci
2 days ago
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One particular turning point was the search for a cubic formula similar to the familiar quadratic. (See Wikipedia's "Cubic Function" entry.) The mathematician Tartaglia had a method for solving certain cubics; the method required taking square roots of negatives; this didn't really bother him, since the those values would cancel in the end. Eventually, it was found that, for cubics in general, those values didn't always cancel, and so mathematics needed to come to grips with them. Then Gauss came along and put everything together.
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– Blue
2 days ago
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Also, in the beginning there were only rational numbers. The discovery that $sqrt{2}$ was irrational apparently cost the discoverer his life.
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– Lee Mosher
2 days ago