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Math Help - Induction Inequality

  1. #1
    Member jacs's Avatar
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    Induction Inequality

    Prove by Mathematical Induction that n^2-11n+30 \geq 0 for n\geq1.

    Step 1: Prove true for n = 1
    1^2-11(1)+30=20
    hence true for n = 1

    Step 2: Assume true for n = k
    k^2-11k+30 \geq 0

    Step 3: Prove true for n = k + 1
    (k+1)^2-11(k+1)+30 \geq 0

    LHS = k^2+2k+1-11k-11+30

    not really sure what to do from here, tried this approach

    k^2-11k+30+2k-10
    since from step 2 we know that k^2-11k+30 \geq 0, but not sure if this is right as I do not know what to do with the 2k - 10

    any insight appreciated
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  2. #2
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    Quote Originally Posted by jacs View Post
    Prove by Mathematical Induction that n^2-11n+30 \geq 0 for n\geq1.

    Step 1: Prove true for n = 1
    1^2-11(1)+30=20
    hence true for n = 1

    Step 2: Assume true for n = k
    k^2-11k+30 \geq 0

    Step 3: Prove true for n = k + 1
    (k+1)^2-11(k+1)+30 \geq 0

    LHS = k^2+2k+1-11k-11+30

    not really sure what to do from here, tried this approach

    k^2-11k+30+2k-10
    since from step 2 we know that k^2-11k+30 \geq 0, but not sure if this is right as I do not know what to do with the 2k - 10

    any insight appreciated
    Note that 2k-10\ge 0 for all k\ge 5. You can check n= 1, 2, 3, and 4 "by hand", then start your induction at 5.
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  3. #3
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    This is clearly not true. This is a quadratic that has a minimum turning point.

    Note that \displaystyle n^2 - 11n + 30 = (n - 5)(n - 6).

    So the roots are n = 5 and n = 6.

    That means that the function is < 0 for 5 < n < 6 and \geq 0 everywhere else.
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  4. #4
    Member jacs's Avatar
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    thanks for that,
    Wasnt sure if you could use that type of approach, or if it was thought of as a sort of cheat way around.

    I did try k^2-9k+20

    using the discriminant : \Delta=(-9)^2-4(1)(20)
    = 1

    but that proves that it is not positive definite (which is what i would have expected), it has two distinct rational roots, so it would NOT be larger than or equal to 0 for all k since it is indefinite, unless something is dodgy with my logic here
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  5. #5
    Member jacs's Avatar
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    Quote Originally Posted by Prove It View Post
    This is clearly not true. This is a quadratic that has a minimum turning point.

    Note that \displaystyle n^2 - 11n + 30 = (n - 5)(n - 6).

    So the roots are n = 5 and n = 6.

    That means that the function is < 0 for 5 < n < 6 and \geq 0 everywhere else.
    your reply not posted yet when i did my last post, never even thought of checking that, duh, the question is set up dodgy then?
    My discriminant argument confirms what you are saying?
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