# Deciding whether a number is rational (2 examples)

1) Prove that number irrational $\sqrt{7-\sqrt{2}}$

I created a polynomial $x=\sqrt{7-\sqrt{2}}$ so

$P(x)=x^4-14x^2+47$ and since $47$ is prime we check $P(x)$ for ${1,-1,47,-47}$ and since all of them are $P(x)\neq0$ it means our number is irrational.

Is my prof OK ?

2) Decide if the number $\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}$ is rational or irrational. I don't know how to tackle this one. I'd be grateful for hints

• The approach in your first proof is correct, though it could be a bit more polished: $P(x)$ certainly has this number as a root, and then you apply the Rational Root Theorem. As for the second, think about how you created the polynomial in 1) so that it'd include that root. – Semiclassical Jul 31 '14 at 14:20
• 1) That's the correct idea, but check your numbers. – Empy2 Jul 31 '14 at 14:20
• 2) Since there are four square-roots, I suspect the polynomial is going to be degree $2^4=16$! – Empy2 Jul 31 '14 at 14:23
• It should be $P(x) \equiv x^4 - 14 x^2 \color{red}{+ 47}$ and not $-47$ there. – hjpotter92 Jul 31 '14 at 16:23
• If you square your number, you get 7 - sqrt (2) which is irrational. Squaring a rational produces a rational, so you must have squared an irrational number. – gnasher729 Jul 31 '14 at 22:20

If $p=\sqrt{\sqrt 5+3}+\sqrt{\sqrt 5-2}$ is a rational number, then \begin{align}p-\sqrt{\sqrt5+3}=\sqrt{\sqrt5-2}&\Rightarrow p^2-2p\sqrt{\sqrt5+3}+\sqrt5+3=\sqrt5-2\\&\Rightarrow 2p\sqrt{\sqrt5+3}=p^2+5\\&\Rightarrow 4p^2(\sqrt 5+3)=(p^2+5)^2\\&\Rightarrow \sqrt5=\frac{(p^2+5)^2}{4p^2}-3\end{align} implies that $\sqrt 5$ is a rational number. This is a contradiction.
You don't need all that! If $\sqrt{7-\sqrt 2} = \dfrac{p}{q}$ is rational, then $\sqrt 2 = 7 - \dfrac{p^2}{q^2}$is rational. Which it isn't.