Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 2 Exercises 2.3

Our main tool is Corollary 2.3.

Solution: First, we need to find an equation with integer coefficients such that $\sqrt{2-\sqrt{2}}$ is a solution to it.

Let $x=\sqrt{2-\sqrt{2}}$. Taking square for both sides, we obtain that $x^2=2-\sqrt{2}$. Hence $x^2-2=-\sqrt{2}$. Squaring both sides, we get

$$

(x^2-2)^2=2,

$$ which is

$$

x^4-4x^2+2=0.

$$ Consider the rational solution of $x^4-4x^2+2=0$. It follows from Corollary 2.3 that those rationa solutions can only be $\pm 1$ and $\pm 2$.

If we plug in $\pm 1$, we get $$1-4+2=-1\ne 0.$$ If we plug in $\pm 2$, we get $$16-16+2=2\ne 0.$$ Hence $x^4-4x^2+2=0$ has no rational solutions. Therefore, as a solution to $x^4-4x^2+2=0$, $\sqrt{2-\sqrt{2}}$ is irrational.