Irrationality proof

Question

Prove that $\sqrt{7-\sqrt{2}} $ is irrational My idea is the following

$\sqrt{7-\sqrt{2}} \in\mathbb{Q}$

$w^2=7+\sqrt{2}$

$w^2-7=\sqrt{2}$ thus w cannot be rational is this correct?

edit: or maybe i can assume $\sqrt{7-\sqrt{2}} =\frac{m}{n}$ and then square both sides but the i am not sure what to do next

Yes. If $\sqrt{7-\sqrt 2}$ was rational then so would be $\sqrt 2$, your proof is correct. (Only the square is $7 - \sqrt 2$.) — k.stm, Nov 08 '14 at 18:48

score 1 · Accepted Answer · answered Nov 08 '14 at 18:48

1

We can even show $7-\sqrt2$ is irrational

as if $7-\sqrt2=a$ is rational

$$7^2+2-14\sqrt2=a^2\iff \sqrt2=(51-a^2)/(14a)$$ which is rational unlike $\sqrt2$

answered Nov 08 '14 at 18:48

lab bhattacharjee

2

Wow, that's a roundabout way to prove that if $a$ is rational, then $7-a$ is rational... – Thomas Andrews Nov 08 '14 at 18:49

1 Answers1