# Rudin's Proof that $\nexists x \in \mathbb{Q}, x^2 = 12$

An exercise in Rudin is:

There is no rational number whose square is $12$.

He proceeds in two alternate ways. In one solution, he posits that such an $x$ exists with the property that $x^2 = 12$, letting $x = \frac{m}{n}$ for $m, n \in \mathbb{Z}$, and finding a contradiction via algebraic manipulation and the assumption, without sacrificing generality, that $m$ and $n$ are coprime. This ends up generating the absurd conclusion that $4$ divides $2$. This strategy makes sense to me, even though it seems somewhat "out of a hat," to quote a phrase from another commenter. The other proof I'm struggling with.

In that proof, he begins by noting that $\sqrt{12} = 2 \sqrt{3}$, and a previous theorem states the product of a non-zero rational and an irrational number is irrational, in which case it suffices to show that the $\sqrt{3}$ is irrational. The proof of this is very standard, and proceeds similarly to the above proof, finding in the process that $m$ and $n$ have a common factor of $3$, a contradiction.

Where I am confused, however, is the logic of why this proves our result. It seems to me that we should be assuming, if I am not mistaken, that $\exists x, x^2 = 12$. So, we have $x = \pm \sqrt{12}$. Rudin has ruled out that $\sqrt{12}$ is rational, surely, but not that $- \sqrt{12}$ is rational. This sounds trivial, and perhaps we could even argue that $- \sqrt{12} = -2 \sqrt{3}$, so because $-2 \in \mathbb{Q}$ by closure, we end up with precisely the same result by proving that $\sqrt{3}$ is irrational.

The more I think about it, the more I think that I'm making a rather meaningless distinction. But, I have seen proofs worded in this exact same way about, say, the square root of $2$, and it almost seems that wording to the effect of

There is no rational number whose square is 2

is taken to be equivalent to

The square root of $2$ is irrational

but this doesn't quite make sense to me.

Any helpful insights would be greatly appreciated.

• $x\in\Bbb Q\iff -x\in\Bbb Q$ – Lord Shark the Unknown Aug 27 '18 at 8:29
• What is your question? – W. mu Aug 27 '18 at 8:29
• My question, essentially, concerns why the two formulations of the theorem are equivalent. Specifically, I'm wondering whether the logic I laid out about closure, which I believe is what Lord Shark might be saying, is the reason that establishing the theorem for the principal root is sufficient. – Matt.P Aug 27 '18 at 8:31
• More generally you have that the two roots of a quadratic polynomial with coefficients in $\Bbb Q$ are either both rational or both irrational. Even more generally, if you have a degree $n$ polynomial $P(x)$ with rational coefficients with $n-1$ roots in $\Bbb Q$ also the last root is in $\Bbb Q$. – Andrea Mori Aug 27 '18 at 8:35
• This is a nice question to occur to a student. The answer may be trivial, but that this question occurred at all is appreciable. :) – Brahadeesh Aug 27 '18 at 9:19

A number $x$ is rational if and only if $x=\frac mn$, with $m\in\mathbb Z$ and $n\in\mathbb N$. Therefore, $x$ is rational if and only if $-x$ is rational, since $-\frac mn=\frac{-m}n$. So, $\sqrt{12}\in\mathbb{Q}\iff-\sqrt{12}\in\mathbb{Q}$.

• Thank you for the answer. This logic does make sense, though it seems to me that, while the rationals are closed under multiplication, the irrational numbers are not. So, we might establish that $\sqrt{12} \in \mathbb{I} = \mathbb{R} \setminus \mathbb{Q}$, but this wouldn't necessarily necessitate that $- \sqrt{12} \in \mathbb{I}$, unless we reasoned that $-2 \in \mathbb{Q}$ and reapplied the prior theorem in Rudin. – Matt.P Aug 27 '18 at 8:33
• @Matt.P Indeed, the irrational numbers are not closed under multiplication. I never claimed they were. All that I claimed (and proved) was that if $x$ is rational, then $-x$ is rational too. It follows from this that if $x$ is irrational, then $-x$ is irrational too. Did you detect some flaw in my argument? If you did, please tell me where it is. – José Carlos Santos Aug 27 '18 at 8:36
• My apologies. I didn't mean to imply that your argument was wrong: it surely isn't, and you didn't even reference the irrational numbers. Rather, I still find myself struggling to apply this logic the problem, where Rudin's proof seems to argue that $\sqrt{12} \in \mathbb{I} \implies \sqrt{12} \in \mathbb{I}$. If you could comment on this, I would greatly appreciate it. – Matt.P Aug 27 '18 at 8:42
• @Matt.P I am sorry, but I don't see in Rudin's proof any attemp of asserting that $\sqrt{12}\in\mathbb{I}\implies\sqrt{12}\in\mathbb{I}$. Besides, Rudin never mentions $-\sqrt{12}$ and he doesn't have to, since what he wants to prove is that $\sqrt{12}\notin\mathbb Q$. Why should he mention $-\sqrt{12}$ at all? – José Carlos Santos Aug 27 '18 at 8:45
• Well, I think that gets back to the distinction between the two formulations of the proof. I seem to be reading "there is no rational whose square is $12$" as saying that there does not exist an $x$ such that $x^2 = 12$, so $x = \pm \sqrt{12}$. So, that would require proving that $\sqrt{12} \not \in \mathbb{Q}$ and $- \sqrt{12} \not \in \mathbb{Q}$. Is this interpretation of the theorem incorrect? – Matt.P Aug 27 '18 at 8:50

If $-x$ is rational and $x^2=12$, then $x$ is rational and $x^2=12$, so the sign does not matter.

Also, if $x$ is rational and $x^2=12$, then $\dfrac x2$ is rational and $\left(\dfrac x2\right)^2=3$.

And if $x$ is irrational, there is no rational equal to $x$.

• This is the correct explanation of Rudin's approach without using the $\sqrt{}$ symbol or any facts about irrational / real numbers. +1 – Paramanand Singh Aug 27 '18 at 15:41