Assume $\exists x\in\mbb Q$ such that $x^2=3$, we can express as a simplified fraction $\dfrac{p}{q}$
now $x^2=\dfrac{p^2}{q^2}=3\riff p^2=3q^2\riff p^2$ is divisible by 3
Also, we've proved (其实这一问是在第二小题里的,第一小题就证明了以下真命题,在这里就不证了...) that "if $q$ is not divisible by 3, then neither is $q^2$", which is equivalent to "if $q^2$ is divisible by 3, then also is $q$", since the latter is the contrapositive of the former statement.
Thus, $p$ is divisible by 3 as well.
So, let $p=3k$ for some $k\in\mbb Z$, now $p^2=(3k)^2=9k^2=3q^2\riff q^2=3k^2\riff q^2$ is divisible by 3.
Again, similarly, we can know that $q$ is divisible by 3 as well.
Therefore, $p$ and $q$ are both divisible by 3, so $\dfrac{p}{q}$ is not a simplified fraction.
CONTRADICTION$\riff$ initial assumption is false$\riff \nexists x\in\mbb Q$ such that $x^2=3$