Exercise 0.2.7 ($\sqrt{p}$ is not a rational number)

Proof. Assume for the sake of contradiction that $a^2=p{b}^2$, where $b\ne 0$.

Let $d=g.c.d(a,b)$. Then $d>0$, since $b\ne 0$. Put $A=a∕d$ and $B=b∕d$. Then $g.c.d(A,B)=1$. Moreover ${(a∕d)}^2=p{(b∕d)}^2$, thus

Then $p\mid A^2$, where $p$ is a prime, therefore $p\mid A$. So there is some integer $k$ such that $A=\mathit{𝑝𝑘}$. Then $p^2{k}^2=p{B}^2$. Simplifying by $p\ne 0$, we obtain $p{k}^2=B^2$, so $p\mid B^2$. Since $p$ is a prime, this shows that $p\mid B$. Hence $p\mid g.c.d(A,B)=1$. But $p>1$, thus $p\nmid 1$. This is a contradiction, which proves that the only solution of $a^2=p{b}^2$ is $(0,0)$ (i.e., $\sqrt{p}$ is not a rational number). □