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Math Help - Can you help me find my error?

  1. #1
    Junior Member
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    Can you help me find my error?

    Question Details:
    , is given, along with the fact that m and n are natural numbers.
    I have to prove:
    Here's what I have:

    so
    what!? I proved the opposite of what I needed to.
    I see that I can do:

    so
    which is the desired result, but what am I missing in the first part? why can I prove it is both greater than and less than 2?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Both your methods are incorrect!

    We need to prove:

    \frac{(m+2n)^2}{(m+n)^2}>2

    (m+2n)^2>2(m+n)^2

    m^2+4n^2+4mn>2(m^2+n^2+2mn)

    m^2+4n^2+4mn>2m^2+2n^2+4mn

    m^2<2n^2

    \frac{m^2}{n^2}<2, which is true (given).
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  3. #3
    Junior Member
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    could you please explain why they are incorrect? Thank you!

    (I realize your method is correct, but why are mine incorrect?)
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  4. #4
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    m<\sqrt{2}n\rightarrow m+2n<\sqrt{2}n+2n

    (m+2n)^2<(\sqrt{2}n+2n)^2\rightarrow \frac{(m+2n)^2}{(m+n)^2}<\frac{(\sqrt{2}n+2n)^2}{(  m+n)^2}>\frac{(\sqrt{2}n+2n)^2}{(\sqrt{2}n+n)^2}=2 (when we put \sqrt{2}n back in the denominator gets bigger, so the fraction gets smaller)

    as you can see, the < > combo means this analysis is useless
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