Show that there is no rational number $x$ such that $x^2=12$.
This is an exercise from Rudin, yes. I am posting this to ask you to verify my proof please.
Let $x=m/n$ and assume it is in its lowest form. Then $(m/n)^2=12 \,\,\,(*)\,$ and $m^2=12n^2$.
Now observe that $12n^2$ must be even, since any even number times another number is even again. Since the square of an uneven number is always uneven and the square of an even number is always even, we conclude that $m$ is even.
Now let us rewrite $m^2$ as another even number $12 q$ and substitute it into $(*)$: $12q=12n^2$. We can cancel the $12$ to get $q=n^2$. Thus $n^2$ is even and as such $n$ is even. Now we have that both $m$ and $n$ are even, which is in contradiction with $x$ being in its lowest form and as such the desired result is proven.
Best Answer
Your proof doesn't work. Yes, you can say that $m^2=12q$ (of course! Since $m^2=12n^2$ ...), but how do you know $q$ is even just because $m$ is even? All you know is that $12q$ is even, and that is also true if $q$ is odd.
In fact, by your logic, there is also no rational number such that $x^2=4$. Say it is $x=\frac{m}{n}$ in its lowest form. Then $m^2=4n^2$, so $m^2$ is even and thus $m$ is even. We can write $m=4q$ for some $q$ ... and if $q$ is even (again, here is your mistake!) then $q = n^2$ is even, and thus $n^2$ is even and thus $n$ is even, and thus both $m$ and $n$ are even, and thus $\frac{m}{n}$ is not in its lowest form, thus a contradiction is reached, and thus there is also no rational number such that $x^2=4$.