Exercise 6.3.3 ( $\sqrt{2}+ \sqrt{3}$ is irrational)

Proof. Put $\alpha =\sqrt{2}+\sqrt{3}$. Then

thus ${\alpha }^2-2\sqrt{2}\alpha +2=3$, so

This implies

therefore

so $\alpha$ is a root of $x^4-10x^2+1$.

Assume, for the sake of contradiction, that $\alpha$ is rational. Then $\alpha =p∕q$, where $p,q$ are integers, $q>0$ and $p\wedge q=1$. Since $\alpha =p∕q$ is a root of $x^4-10x^2+1$, multiplying by $q^4$, we obtain

Then $q\mid p^4$, and $q\wedge p=1$, thus $q\mid 1$, where $q>0$, so $q=1$, and $\alpha =p$ is an integer satisfying $p^4-10p^2+1=0$. Hence $p\mid 1$, and since $p=\alpha >0$, $p=1$. But $1$ is not a root of $x^4-10x^2+1$. This is a contradiction, so $\alpha =\sqrt{2}+\sqrt{3}$ is irrational. (Alternatively, we may use directly Corollary 6.14.) □