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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.

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  1. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Oh I completely missed the sqrt in the first term. My apologies :(

  7. Bee_see
    • one year ago
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    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
    • one year ago
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    then b^2=4k^2

  9. freckles
    • one year ago
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    \[\sqrt{2}+\sqrt{2}=2 \sqrt{2} =\frac{a}{b} \\ \sqrt{2}=\frac{a}{2b} \text{ I think we could have stopped here }\]

  10. freckles
    • one year ago
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    it says we can use that sqrt(2) is irrational

  11. freckles
    • one year ago
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    but here we have expressed sqrt(2) as an integer over an integer

  12. freckles
    • one year ago
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    contradiction

  13. Bee_see
    • one year ago
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    8b^2=4k^2*

  14. Bee_see
    • one year ago
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    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
    • one year ago
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    you don't have to do all of that

  16. jim_thompson5910
    • one year ago
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    @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
    • one year ago
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    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
    • one year ago
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    It allows us to prove this kind of problem nicely.

  19. Bee_see
    • one year ago
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    I'm not sure.

  20. Bee_see
    • one year ago
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    when are you supposed to use the proof by contradiction?

  21. Loser66
    • one year ago
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    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
    • one year ago
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    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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