Prove irrationality of $\sqrt{2+\sqrt{2}}$ and $\sqrt{2}+\sqrt{3}$

Question

I am trying to prove the irrationality of the above two numbers, but I don't know how. What would be a general strategy for problems like these?

My current strategy is trying to reach a contradiction by starting with $\sqrt{2+\sqrt{2}}=\frac{a}{b}$ with integers $a$ and $b$ and somehow reaching that $a$ needs to be both even and odd for example. But I keep getting stuck.

Answer 1

Let $x=\sqrt{2}+\sqrt{3}$. Then

$$(x-\sqrt{2})^2=3=x^2-2\sqrt{2}x+2$$

$$\sqrt{2}=\frac{x^2-1}{2x}$$

Then if $x$ is rational, so is $\sqrt{2}$, contradiction.

If $x=\sqrt{2+\sqrt{2}}$, then $x^2-2=\sqrt{2}$, same idea then.

Answer 2

No proof by contradiction is required: the minimal polynomials of $x$ are, respectively: $x^4-10x^2+1\,$ and $\,x^4-4x^2+2$.

By the Rational Root theorem, if $x$ were rational, it would be an integer that is a divisor of the constant term. Test the possibilities, and check none is a root.

Answer 3

To use the even/odd idea, note that $$2b^2+b^2\sqrt 2=a^2$$

Now $$(a^2-2b^2)^2=2b^4=a^4-4a^2b^2+4b^4$$ whence $$a^4-4a^2b^2+2b^2=0$$ so that $$a^4=4a^2b^2-2b^4$$From this $a$ must be even ($2c$, say) $$16c^4=16c^2b^2-2b^4$$ or $$8c^4=8c^2b^2-b^4$$ and $b$ must also be even.

The rational root theorem amounts to a systematic generalisation of this idea.