# What is the problem here (all integers are irrational proof...I think so)?

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 (I don't know 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.

• "a^2 is a factor b^2 and b" You are right $a^2$ divides $b^2$, but why would it divide $b$? Commented Jul 7, 2019 at 16:06
• First error I spotted is that $a^2$ need not be a factor of $b$ just because it is a factor of $b^2$. Indeed, if $a=2$ and $b=2$, then $a^2=4$ is a factor of $b^2=4$ but $a^2=4$ is not a factor of $b=2$.
– Dave
Commented Jul 7, 2019 at 16:06
• Also, it is clear that integers are rational because given any integer $a$ we can write $a=\frac{a}{1}$.
– Dave
Commented Jul 7, 2019 at 16:08
• If b and c are coprime, and a=b/c, then what does c have to be? Commented Jul 7, 2019 at 17:33
• @Toolazytothinkofaname It is rather simple to spot the error. Substitute the variables by actual integers and find where the proof is wrong. Commented Jul 8, 2019 at 14:18

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

• The symbol $|$ what means ?
– ESCM
Commented Jul 21, 2019 at 16:41
• @EduardoS. $a\mid b$ means that "$a$ divides $b$" i.e., there exists an integer $k$ such that $b=ak$ with $a\ne 0$. Commented Jul 21, 2019 at 18:43

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 $$a = 1$$).

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 $$b = d\prod p_i^{j_i}$$. 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 $$a$$ ($$2\mapsto 2; 3\mapsto 1$$) are less or equal to $$2$$ times the powers in $$b$$ ($$2\mapsto 1$$ and $$2 \le 2*1$$ and $$3\mapsto 2$$ and $$1 \le 2*2$$) but it doesnt mean the are less than or equal to the powers of $$b$$. (In $$a; 2\mapsto 2$$ but in $$b; 2\mapsto 1$$ 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.

• No, $\,p\mid b^k\,\Rightarrow\, p\mid b\,$ is true $\iff p\,$ is squarefree. Follow the link for a handful of characterizations of squarefree integers. Commented Jul 7, 2019 at 17:44
• Why do you say "no"? That is exactly what I said. Commented Jul 7, 2019 at 17:52
• Because the first paragraph was incorrect. Now it is correct after your edit. Commented Jul 7, 2019 at 17:54
• Okay. I used "only" colloquially. My bad. I'm pretty sure the OP was confusing the FTA with Euclid's lemma. So I said that only works if $p$ is prime. Colloquially that doesn't mean $p$ being prime is required and it is false otherwise. It means you can only cite that lemma if $p$ is prime. There are many other ways $a|b^k$ and $a|b$ can both be true but citing Euclid's lemma is usually reserved for $a$ prime. But in math I shouldn't have used the loaded word "only". Commented Jul 7, 2019 at 18:56

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