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.

If a number is rational, so is its square. And I think you know that $\sqrt 2$ is irrational. — MooS, Apr 10 '15 at 11:11
Consider asking two separate questions (or just remove the $\sqrt2+\sqrt3$ part because that has already been answered here). — Bart Michels, Apr 10 '15 at 11:13

score 3 · Answer 1 · answered Apr 10 '15 at 11:11

3

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.

answered Apr 10 '15 at 11:11

Jean-Claude Arbaut

23,277

There are nine answers already here. – Dietrich Burde Apr 10 '15 at 11:13
@DietrichBurde I know. There are dozens of questions of this kind on MSE. – Jean-Claude Arbaut Apr 10 '15 at 11:17
You are right, and sometimes one can answer it again, but here we could just copy and paste - this cannot be the idea of MSE. Why not mark it as a duplicate instead of repeating an answer ? – Dietrich Burde Apr 10 '15 at 11:18

score 2 · Answer 2 · answered Apr 10 '15 at 11:20

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.

answered Apr 10 '15 at 11:20

Bernard

175,478

Because I didn't know there was already a similar answer. Having checked, I give some more details (and also some less). Do you think I should withdraw my answer? – Bernard Apr 10 '15 at 11:28
@Jean-Claude Arbaut: I don't agree; it's only examining all possible cases. – Bernard Apr 10 '15 at 11:31
I don't think it's necessary. I know some would say so, however (see Dietrich' comment above). And some would even downvote (I was downvoted a couple of hours ago, on a correct answer). I don't like that kind of behaviour, even if I don't like either when students ask questions here without any work, and without even looking if the question was asked before. – Jean-Claude Arbaut Apr 10 '15 at 11:31
Btw, I was also downvoted here. How fun. – Jean-Claude Arbaut Apr 10 '15 at 16:08

score 1 · Answer 3 · answered Apr 10 '15 at 11:38

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.

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

3 Answers3