Bee_see
  • Bee_see
Prove that square root 2 + square root 2 is irrational. Your proof is allowed to use the fact that square root 2 is irrational.
Discrete Math
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • 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.
anonymous
  • 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
Bee_see
  • 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...

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anonymous
  • 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 ^^
jim_thompson5910
  • jim_thompson5910
Hint: \[\large \sqrt{2}+\sqrt{2} = \frac{a}{b}\] \[\large 2\sqrt{2} = \frac{a}{b}\] \[\large (2\sqrt{2})^2 = \left(\frac{a}{b}\right)^2\] \[\large 8 = \frac{a^2}{b^2}\]
anonymous
  • anonymous
Oh I completely missed the sqrt in the first term. My apologies :(
Bee_see
  • 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?
Bee_see
  • Bee_see
then b^2=4k^2
freckles
  • freckles
\[\sqrt{2}+\sqrt{2}=2 \sqrt{2} =\frac{a}{b} \\ \sqrt{2}=\frac{a}{2b} \text{ I think we could have stopped here }\]
freckles
  • freckles
it says we can use that sqrt(2) is irrational
freckles
  • freckles
but here we have expressed sqrt(2) as an integer over an integer
freckles
  • freckles
contradiction
Bee_see
  • Bee_see
8b^2=4k^2*
Bee_see
  • 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.....
freckles
  • freckles
you don't have to do all of that
jim_thompson5910
  • 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
Loser66
  • 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?
Loser66
  • Loser66
It allows us to prove this kind of problem nicely.
Bee_see
  • Bee_see
I'm not sure.
Bee_see
  • Bee_see
when are you supposed to use the proof by contradiction?
Loser66
  • 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 \(2\sqrt2 \notin \mathbb Q\)
Bee_see
  • 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?

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