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Is multiplication of real numbers uniquely defined as being distributive over addition?

Are the addition and multiplication of real numbers, as we know them, unique?Real Numbers as Well Defined SetsHow is addition different than multiplication?Binary operation commutative, associative, and distributive over multiplicationA foundation where objects are unlabelledCommutative binary operations on $Bbb C$ that distribute over both multiplication and additionSo there's addition, multiplication, exponentiation and tetration, but is there a level before addition?Are the addition and multiplication of real numbers, as we know them, unique?Binary multiplication as combination of addition and left shiftChecking a set with the Axiom of IntegersIs addition on $mathbb{R}$ unique up to automorphism?

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5

$begingroup$

In the set of real numbers, I wonder whether the distributive law uniquely determines multiplication.
Suppose that for a function $f$: $Bbb{R}timesBbb{R}$ $to$ $Bbb{R}$ the following hold for every $x,y,z$, where $+$ is the usual addition
(as defined via Cauchy sequences of rationals), and $1$ is the known natural number:

1. $f(x+y,z) = f(x,z) + f(y,z)$


2. $f(x, y+z) = f(x,y) + f(x,z)$


3. $f(1,x) = f(x,1) = x $

From the above does it follow that $f(x,y) = xy$, the usual multiplication?

In this post: Are the addition and multiplication of real numbers, as we know them, unique?
a somewhat related "dual" question is asked concerning addition, and a simple solution is given in the form of $(x^3+y^3)^{1/3}$. So am I missing something obvious here?

foundations binary-operations

share|cite|improve this question

edited 8 hours ago

José Carlos Santos

206k2626 gold badges159159 silver badges283283 bronze badges

asked 8 hours ago

exp8jexp8j

12499 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We can do that for natural numbers then for rationals hence for real numbers
  $endgroup$
  – Ameryr
  8 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

In the set of real numbers, I wonder whether the distributive law uniquely determines multiplication.
Suppose that for a function $f$: $Bbb{R}timesBbb{R}$ $to$ $Bbb{R}$ the following hold for every $x,y,z$, where $+$ is the usual addition
(as defined via Cauchy sequences of rationals), and $1$ is the known natural number:

1. $f(x+y,z) = f(x,z) + f(y,z)$


2. $f(x, y+z) = f(x,y) + f(x,z)$


3. $f(1,x) = f(x,1) = x $

From the above does it follow that $f(x,y) = xy$, the usual multiplication?

In this post: Are the addition and multiplication of real numbers, as we know them, unique?
a somewhat related "dual" question is asked concerning addition, and a simple solution is given in the form of $(x^3+y^3)^{1/3}$. So am I missing something obvious here?

foundations binary-operations

share|cite|improve this question

edited 8 hours ago

José Carlos Santos

206k2626 gold badges159159 silver badges283283 bronze badges

asked 8 hours ago

exp8jexp8j

12499 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We can do that for natural numbers then for rationals hence for real numbers
  $endgroup$
  – Ameryr
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

In the set of real numbers, I wonder whether the distributive law uniquely determines multiplication.
Suppose that for a function $f$: $Bbb{R}timesBbb{R}$ $to$ $Bbb{R}$ the following hold for every $x,y,z$, where $+$ is the usual addition
(as defined via Cauchy sequences of rationals), and $1$ is the known natural number:

1. $f(x+y,z) = f(x,z) + f(y,z)$


2. $f(x, y+z) = f(x,y) + f(x,z)$


3. $f(1,x) = f(x,1) = x $

From the above does it follow that $f(x,y) = xy$, the usual multiplication?

In this post: Are the addition and multiplication of real numbers, as we know them, unique?
a somewhat related "dual" question is asked concerning addition, and a simple solution is given in the form of $(x^3+y^3)^{1/3}$. So am I missing something obvious here?

foundations binary-operations

share|cite|improve this question

edited 8 hours ago

José Carlos Santos

206k2626 gold badges159159 silver badges283283 bronze badges

asked 8 hours ago

exp8jexp8j

12499 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

José Carlos Santos

206k2626 gold badges159159 silver badges283283 bronze badges

asked 8 hours ago

exp8jexp8j

12499 bronze badges

share|cite|improve this question

edited 8 hours ago

José Carlos Santos

206k2626 gold badges159159 silver badges283283 bronze badges

asked 8 hours ago

exp8jexp8j

12499 bronze badges

edited 8 hours ago

José Carlos Santos

206k2626 gold badges159159 silver badges283283 bronze badges

edited 8 hours ago

José Carlos Santos

206k2626 gold badges159159 silver badges283283 bronze badges

asked 8 hours ago

exp8jexp8j

12499 bronze badges

asked 8 hours ago

exp8jexp8j

12499 bronze badges

2 Answers
2

active

oldest

votes

6

$begingroup$

For one thing $Bbb R$ is a $Bbb Q$-vector space, therefore by choosing a Hamel basis it is possible to define uncountably many symmetric $Bbb Q$-bilinear maps $Bbb RtimesBbb RtoBbb R$, even with the restriction $phi(1,bullet)=phi(bullet,1)=id$. The only continuous one among these is the usual product, though.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Gae. S.Gae. S.

1,27455 silver badges1414 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good answer! Would be good to detail the way to define a $mathbb Q$-bilinear map from a Hamel basis and what a Hamel basis is, as the OP is probably not familiar with it.
  $endgroup$
  – mathcounterexamples.net
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Gae. S. Could you suggest a good reference where I can study the concepts that you mention? In fact, what motivated me to ask this question is that I've just started studying calculus in normed spaces on my own, and I bumped into bilinear maps...
  $endgroup$
  – exp8j
  8 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

First for a natural $n$
$f(x,n)= underbrace{f(x,1)+...+f(x,1)}_{n text{. times}}=nx$

Then for a rational $1/n$
$x=f(x,1)=f(x,n/n) = n f(x,1/n) $ so $f(x,1/n) =x/n$

Now for a $y$ which is a Cauchy sequence of rational numbers $y=lim r_n$
$f(x,y) = f(x,lim r_n) = lim f(x,r_n)= lim x r_n = xy$ pulling limit outside require the continuity of $f$

share|cite|improve this answer

answered 8 hours ago

AmeryrAmeryr

84233 silver badges1212 bronze badges

$endgroup$

add a comment | 

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6

$begingroup$

For one thing $Bbb R$ is a $Bbb Q$-vector space, therefore by choosing a Hamel basis it is possible to define uncountably many symmetric $Bbb Q$-bilinear maps $Bbb RtimesBbb RtoBbb R$, even with the restriction $phi(1,bullet)=phi(bullet,1)=id$. The only continuous one among these is the usual product, though.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Gae. S.Gae. S.

1,27455 silver badges1414 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good answer! Would be good to detail the way to define a $mathbb Q$-bilinear map from a Hamel basis and what a Hamel basis is, as the OP is probably not familiar with it.
  $endgroup$
  – mathcounterexamples.net
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Gae. S. Could you suggest a good reference where I can study the concepts that you mention? In fact, what motivated me to ask this question is that I've just started studying calculus in normed spaces on my own, and I bumped into bilinear maps...
  $endgroup$
  – exp8j
  8 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

For one thing $Bbb R$ is a $Bbb Q$-vector space, therefore by choosing a Hamel basis it is possible to define uncountably many symmetric $Bbb Q$-bilinear maps $Bbb RtimesBbb RtoBbb R$, even with the restriction $phi(1,bullet)=phi(bullet,1)=id$. The only continuous one among these is the usual product, though.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Gae. S.Gae. S.

1,27455 silver badges1414 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good answer! Would be good to detail the way to define a $mathbb Q$-bilinear map from a Hamel basis and what a Hamel basis is, as the OP is probably not familiar with it.
  $endgroup$
  – mathcounterexamples.net
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Gae. S. Could you suggest a good reference where I can study the concepts that you mention? In fact, what motivated me to ask this question is that I've just started studying calculus in normed spaces on my own, and I bumped into bilinear maps...
  $endgroup$
  – exp8j
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

For one thing $Bbb R$ is a $Bbb Q$-vector space, therefore by choosing a Hamel basis it is possible to define uncountably many symmetric $Bbb Q$-bilinear maps $Bbb RtimesBbb RtoBbb R$, even with the restriction $phi(1,bullet)=phi(bullet,1)=id$. The only continuous one among these is the usual product, though.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Gae. S.Gae. S.

1,27455 silver badges1414 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Gae. S.Gae. S.

1,27455 silver badges1414 bronze badges

answered 8 hours ago

Gae. S.Gae. S.

1,27455 silver badges1414 bronze badges

answered 8 hours ago

Gae. S.Gae. S.

1,27455 silver badges1414 bronze badges

1

$begingroup$

First for a natural $n$
$f(x,n)= underbrace{f(x,1)+...+f(x,1)}_{n text{. times}}=nx$

Then for a rational $1/n$
$x=f(x,1)=f(x,n/n) = n f(x,1/n) $ so $f(x,1/n) =x/n$

Now for a $y$ which is a Cauchy sequence of rational numbers $y=lim r_n$
$f(x,y) = f(x,lim r_n) = lim f(x,r_n)= lim x r_n = xy$ pulling limit outside require the continuity of $f$

share|cite|improve this answer

answered 8 hours ago

AmeryrAmeryr

84233 silver badges1212 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

First for a natural $n$
$f(x,n)= underbrace{f(x,1)+...+f(x,1)}_{n text{. times}}=nx$

Then for a rational $1/n$
$x=f(x,1)=f(x,n/n) = n f(x,1/n) $ so $f(x,1/n) =x/n$

Now for a $y$ which is a Cauchy sequence of rational numbers $y=lim r_n$
$f(x,y) = f(x,lim r_n) = lim f(x,r_n)= lim x r_n = xy$ pulling limit outside require the continuity of $f$

share|cite|improve this answer

answered 8 hours ago

AmeryrAmeryr

84233 silver badges1212 bronze badges

$endgroup$

add a comment | 

$begingroup$

First for a natural $n$
$f(x,n)= underbrace{f(x,1)+...+f(x,1)}_{n text{. times}}=nx$

Then for a rational $1/n$
$x=f(x,1)=f(x,n/n) = n f(x,1/n) $ so $f(x,1/n) =x/n$

Now for a $y$ which is a Cauchy sequence of rational numbers $y=lim r_n$
$f(x,y) = f(x,lim r_n) = lim f(x,r_n)= lim x r_n = xy$ pulling limit outside require the continuity of $f$

share|cite|improve this answer

answered 8 hours ago

AmeryrAmeryr

84233 silver badges1212 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

AmeryrAmeryr

84233 silver badges1212 bronze badges

answered 8 hours ago

AmeryrAmeryr

84233 silver badges1212 bronze badges

answered 8 hours ago

AmeryrAmeryr

84233 silver badges1212 bronze badges