Given a prime number $p$, show that there is no rational number a such that $a^2 = p$.
I tried assuming the equality is true then I took $a= \frac{m}{n}$ such that $m$ and $n$ are relatively prime then $$a^2= \frac{m^2}{n^2}= p\implies m^2=p \cdot n^2$$ then I got stuck.
From $m^2=pn^2$ we obtain a contradiction since
$$p|m^2 \implies p|m$$
that is
$$p^2l^2=pn^2 \implies n^2=pl^2\implies p|n$$
Then $p\mid m^2$ but since $p$ prime then $p\mid m$
You can write $m=pm'$
So $p^2m'^2=pn^2\iff n^2=pm'^2$ and similarly $p$ divides also $n$ contradiction since $\gcd(m,n)=1$.
You know that $m,n$ are relatively prime. So there is no prime that divides them both. Notice that $p$ divides the right side, $pn^2$, so it must divide the left side, $m^2$.
But because $p$ divides $m^2= m\cdot m$, and $p$ is prime, $p$ must divide either $m$ or $m$....so it divides $m$! Then we can write $m= kp$ for some $k$, i.e. $m$ is a multiple of $p$. But then $$ \begin{split} m^2&= pn^2 \\ (kp)^2&= pn^2 \\ k^2p^2&= pn^2 \\ k^2p&= n^2 \end{split} $$
But we can apply the same logic as above. Notice that $p$ divides the left side, $k^2p$, so it divides the right. Then $p$ divides $n^2= n \cdot n$ so that $p$ must divide $n$. But then $p$ divides both $m,n$, so that they share a common prime factor, which is not possible because you assumed that you reduced properly so that $m$ and $n$ share no common factor! A contradiction. Therefore, the square root of any prime cannot be rational.
