Question

Prove that square root 2 + square root 2 is irrational. Your proof is allowed to use the fact that square root 2 is irrational.

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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 ^^

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

then b^2=4k^2

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

contradiction

Answer 13

8b^2=4k^2*

Answer 14

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.....

Answer 15

you don't have to do all of that

Answer 16

@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

Answer 17

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?

Answer 18

It allows us to prove this kind of problem nicely.

Answer 19

I'm not sure.

Answer 20

when are you supposed to use the proof by contradiction?

Answer 21

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 \(2\sqrt2 \notin \mathbb Q\)

Answer 22

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?