## Bee_see one year ago Prove that square root 2 + square root 2 is irrational. Your proof is allowed to use the fact that square root 2 is irrational.

1. anonymous

I think this might be similar to what you're looking for http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php It's not exact but you can follow the same process perhaps.

2. anonymous

Then again, if you're allowed to say that the sqrt of two is already irrational, you might not have to do that. Any number plus an irrational number is going to be irrational anyway

3. Bee_see

I think I understand it with one square root of 2, but I don't know what to do if I have two...

4. anonymous

Well it says that you're allowed to say that the sqrt of 2 is irrational, so I don't think you're suppose to prove it. https://www.khanacademy.org/math/algebra/rational-and-irrational-numbers/rational-and-irrational-expressions/v/proof-that-sum-of-rational-and-irrational-is-irrational This is what I think what you are supposed to prove ^^

5. jim_thompson5910

Hint: $\large \sqrt{2}+\sqrt{2} = \frac{a}{b}$ $\large 2\sqrt{2} = \frac{a}{b}$ $\large (2\sqrt{2})^2 = \left(\frac{a}{b}\right)^2$ $\large 8 = \frac{a^2}{b^2}$

6. anonymous

Oh I completely missed the sqrt in the first term. My apologies :(

7. Bee_see

Right so then 8b^2=a^2...so a is even and then change a to 2k then 8=(2k)^2/b^2?

8. Bee_see

then b^2=4k^2

9. freckles

$\sqrt{2}+\sqrt{2}=2 \sqrt{2} =\frac{a}{b} \\ \sqrt{2}=\frac{a}{2b} \text{ I think we could have stopped here }$

10. freckles

it says we can use that sqrt(2) is irrational

11. freckles

but here we have expressed sqrt(2) as an integer over an integer

12. freckles

13. Bee_see

8b^2=4k^2*

14. Bee_see

aren't I supposed to get rid of the 8 besides b^2? Because then I would get 1/2k^2 and that's not correct since the statement should say even number.....

15. freckles

you don't have to do all of that

16. jim_thompson5910

@freckles has a much better way to do it if you decide to go another route, you could say a = 4k (a is still even), so 8=a^2/b^2 8=(4k)^2/b^2 8b^2=16k^2 b^2 = 2k^2 which implies b is even. But if b is even, then a/b isn't fully reduced. So that's another contradiction

17. Loser66

We have a theorem : " If x is a solution of a polynomial, then either $$x\in\mathbb Z$$ or $$x \notin \mathbb Q$$." Are you allowed to use it?

18. Loser66

It allows us to prove this kind of problem nicely.

19. Bee_see

I'm not sure.

20. Bee_see

when are you supposed to use the proof by contradiction?

21. Loser66

Ok, whatever, I contribute what I know. Consider $$(x + 2\sqrt 2)(x-2\sqrt2)=0\\x^2 -8 =0$$ Hence either $$2\sqrt2 \in \mathbb Z~~or ~~2\sqrt2 \notin \mathbb Q$$ Suppose $$2\sqrt2\in \mathbb Z$$ we know that $$1< \sqrt 2<2\\2< 2\sqrt 2<4$$ Hence if $$2\sqrt2 \in\mathbb Z$$, then $$2\sqrt2 = 3$$ $$\implies 8 =9$$ contradiction. Hence $$2\sqrt2 \notin \mathbb Z$$ therefore $$2\sqrt2 \notin \mathbb Q$$

22. Bee_see

The class hasn't gone over what you wrote..The problems just mentions "prove," so how do I know for sure which proof to use between contradiction, indirect proof, and direct proof?