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.
Best Answer
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 $x$.