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Math Help - Prooving inequalities

  1. #1
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    Prooving inequalities

    I'm supposed to proove the proposition that for all integers k>=2, that k^2<k^3. The book says you can use induction or another method. I can't seem to get either fully proved. Any help would be great. Thanks!!
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  2. #2
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    Quote Originally Posted by nataliemarie View Post
    I'm supposed to proove the proposition that for all integers k>=2, that k^2<k^3. The book says you can use induction or another method. I can't seem to get either fully proved. Any help would be great. Thanks!!
    k^2<k^3 is clearly equivalent to k^3-k^2=k^2(k-1)>0.
    Can you prove that if k\ge 2?
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  3. #3
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    Ok... well I had started doing it, but I'm not sure.

    k>=2 so (k-2) > 0

    I can multiply it by k^2 because k^2 >0

    k^3 - 2k^2 > 0
    so
    k^3 > 2k^2
    but I'm missing something
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  4. #4
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    Quote Originally Posted by nataliemarie View Post
    Ok... well I had started doing it, but I'm not sure.

    k>=2 so (k-2) > 0

    I can multiply it by k^2 because k^2 >0

    k^3 - 2k^2 > 0
    so
    k^3 > 2k^2
    but I'm missing something
    If I were you, I would say if k=2 then 2^2<2^3.

    If k>2 then k-1>0 so k^2(k-1)>0
    So k^3>k^2.
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  5. #5
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    This is actually easier without induction

    Basically, 1<n since n>2 and we "know" that for p>0, if m<n then pm<pn

    Also, n^2>0 since n>2

    So letting n^2=p and 1<n, then n^2\cdot 1<n\cdot n^2

    Now of course, you might not acutally "know" the proposition that I referenced if you haven't proven it yet
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  6. #6
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    Thank you both very much. I can't believe I missed the fact that k >1.
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  7. #7
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    You may also do it this way:

    k^2<k^3 is equivalent to \frac{k^2}{k^3} < 1 \Rightarrow \frac{1}{k}<1 (we can do that since k\neq 0)

    Now, \frac{1}{k}<1 is obviously true for any k>1...
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