Exercise 1.17

Proof. Suppose that there exists some $r\in \mathbb{Q},r>0$, such that $r^2=2$. Then $r=a∕b,a\in {\mathbb{N}}^{\text{∗}},b\in {\mathbb{N}}^{\text{∗}}$. With $d=a\wedge b$, $a=d{a}^{\prime },b=d{b}^{\prime },a^{\prime }\wedge b^{\prime }=1$, so $r=a^{\prime }∕b^{\prime },a^{\prime }\wedge b^{\prime }=1$, so we may suppose $r=a∕b,a>0,b>0,a\wedge b=1$ and $a^2=2b^2$.

$a^2$ is even, then $a$ is even (indeed, if $a$ is odd, $a=2k+1,k\in \mathbb{Z}$, $a^2=4k^2+4k+1=2(2k^2+2k)+1$ is odd).

So $a=2A,A\in \mathbb{N}$, then $4A^2=2b^2$, $2A^2=b^2$.

With the same reasoning, $b^2$ is even, then $b$ is even, so $b=2B,B\in \mathbb{N}$. Thus $2\mid a,2\mid b$, $2\mid a\wedge b$, in contradiction with $a\wedge b=1$.

Conclusion : $\sqrt{2}$ is irrational. □