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anonymous

  • 4 years ago

How does \( x^2 = 2y^2\) imply that x is an even number?

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  1. anonymous
    • 4 years ago
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    I think this is under the assumption that y is an integer, right?

  2. anonymous
    • 4 years ago
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    Assuming these are integers, whether or not y^2 is even or odd, 2*y^2 will be even. An even times an odd is even.

  3. anonymous
    • 4 years ago
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    If this is the case, then it's obvious that \(y^2\) would have to be an integer too, and then 2 times any integer is an even number (that would be your x).

  4. anonymous
    • 4 years ago
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    Yes, I agree if it was \(x^2\) but only the \(x\).

  5. anonymous
    • 4 years ago
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    Oh right, didn't look carefully. If the assumption was that BOTH \(x\) and \(y\) are integers, then both \(x^2\) and \(y^2\) have to be perfect squares. Add to that \(x^2\) is even then we can easily say that it follows that \(x\) is also even because the square root of a "perfect even square" is an even number.

  6. anonymous
    • 4 years ago
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    Ahhhhh, yes your absolutely right :) Thanks a lot. Since \(x^2 = 2k\) that also means that you have to be able to do a square root of the even number, to be able to do that, the factor has to be 4 not 2. The k just has to have a whole square root.

  7. anonymous
    • 4 years ago
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    Exactly!

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