Mdthbs

Let us assume $a$ is an integer which is rational which implies $a=p/q$ (where $p$ and $q$ are integers and $q$ not equal to $0$). If $p$ and $q$ are not coprime let us simplify the fraction so this it is.(idk how to talk like mathematicians)
Which implies, $$a=b/c$$ (where $b$ and $c$ are coprime integers). Squaring on both sides,
begin{align}
a^2&=b^2/c^2\
a^2c^2&=b^2
end{align}
So $a^2$ is a factor of $b^2$, and also of $b$, due to the uniqueness of the fundamental theorem of arithmetic.
So,
begin{align}
b &=a^{2}d tag{where $d$ is an integer}\
b^2 &= a^{4}d^{2}
end{align}
But $b^2=a^2c^2$ So,
begin{align}
a^2c^2 &= a^4d^2\
c^2 &= a^2d^2
end{align}
So, $a^2$ is a factor of $c^2$ and $c$ due to the fundamental theorem of arithmetic. So $b$ and $c$ have $a^2$ as a common factor. But this contradicts the fact that $b$ and $c$ are coprime. This is because we have taken $a$ as a rational integer so $a$ cannot be a rational integer.

Whats wrong here?(genuinely asking)

The problem in the proof is that $a^2|b^2nRightarrow a^2|b$. For instance, take $a=2$ and $b=6$. Clearly, $4|36$ but $4nmid 6$.

I think you are confusing that if $p$ is prime and $p$ divides $b^k$ then $p|b$. That is true if $p$ is prime.

Actually it's also true for a composite $a|b^k$ then $a|b$ if $a$ has no square factors. But if $a$ as any prime factors to a power greater than $1$ it need not be true.

And in fact its obviously not true as $a^2$ divides $a^2$ but $a^2$ doesn't divide $a$ (unless $a = 1$).

Read on....

It most certainly is not true if $a|b^k$ that $a|b$ It means that the prime factors of $a$ are prime factors of $b$. And it means that the powers of those prime factors of $a$ are at most equal to $k$ times the powers of the same prime factors of $b$ but because $k$ is larger than .....

Oh let me put it this way.

Suppose $a = prod p_i^{m_i}$ be the prime factorization of $a$. Suppose $a|b^k$. Then that means that $p_i$ are prime factors of $b$ and that $b = dprod p_i^{j_i}$. And it means that $b^k = d^k prod p_i^{k*j_i}$.

And as $a|b^k$ that means each $m_i le k*j_i$. But that does not mean $m_i le j_i$ which would mean $a|b$.

You statement $a|b^k$ means $a|b$ if $a$ has square free and all the prime factor powers were $1$ but not other wise.

Simple example if $a = 12 = 2^2*3$ and $b= 90 = 2*3^2*5$. Now $a|b^2 = 8100 = 2^2*3^4*5^2$.

This means the prime factors of $a$ ($2,3$) are also prime factors of $b$. And it means that the powers of the prime factors of $a$ ($2mapsto 2; 3mapsto 1$) are less or equal to $2$ times the powers in $b$ ($2mapsto 1$ and $2 le 2*1$ and $3mapsto 2$ and $1 le 2*2$) but it doesnt mean the are less than or equal to the powers of $b$. (In $a; 2mapsto 2$ but in $b; 2mapsto 1$ and $2 not le 1$).

So $12 not mid 90$.

It's certainly can't be the case that $a|b implies a^2| b^2 implies a^2|b$! That would mean every time you have $a|b$ you can just keep squaring and reducing to get $a^{m}|b$ for any power of $m$.

That would mean if $3|6$ then $3^2|6$ and $3^4|6$ and $3^{2048}|6$ and so on.

Or in this case as $a = b$ (and $c=1$.... because $a$ is an integer) you would have $a|a$ so $a^2|a$? And $a^4|a$. That's .... simply not true.

Basic facts missing $ac=b$ is a lot easier to use. $a^2$ does not need to divide $b$. A fraction sharing no common factor other than 1, between the number on top ( numerator), and the number on the bottom ( denominator), is said to be in lowest terms .

Anyways starting from $a={bover c}$ we get $ac=b$ showing c divides b, sharing no factor other than 1, and therefore, $c=1$, implying $a=b$ so $a={aover 1}$ it Also can be used to show :$a={-aover -1}$