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Math Help - This calculus lark has gone right over my head!

  1. #1
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    This calculus lark has gone right over my head!

    This is the most recent problem I can't do!

    State the contrapositive and give a proof of the contropositive:

    If 1/2(a^2 + b^2) >ab then a/=/b (a not equal to b)

    I guess the contrapositve is If Not a/=/b then Not 1/2(a^2 + b^2) > ab


    But now what?!
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  2. #2
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    Can someone please go over what they'd do for this answer, possibly even the full answer so I can go through it, learn how it's to be done then move on with the next 5 questions I have!

    Thanks in advance

    Chris
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  3. #3
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    Quote Originally Posted by chr91 View Post
    This is the most recent problem I can't do!

    State the contrapositive and give a proof of the contropositive:

    If 1/2(a^2 + b^2) >ab then a/=/b (a not equal to b)

    I guess the contrapositve is If Not a/=/b then Not 1/2(a^2 + b^2) > ab


    But now what?!
    \frac12(a^2+b^2)>ab~\implies~a^2+b^2>2ab~\implies~  a^2-2ab+b^2>0

    (a-b)^2>0 is true if a-b\neq 0

    And now state the contrapositive more precisely.
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  4. #4
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    Quote Originally Posted by earboth View Post
    \frac12(a^2+b^2)>ab~\implies~a^2+b^2>2ab~\implies~  a^2-2ab+b^2>0

    (a-b)^2>0 is true if a-b\neq 0

    And now state the contrapositive more precisely.
    Thanks.

    Well now we have (a-b)^2 > 0 implies a/=/b

    Taking the contropositive we get Not a/=/b implies not (a-b)^2 > 0

    a=b implies (a-b)^2 < or the same as 0

    If a=b, (a-b)^2 = 0 therefore the contropositive has been proved.


    Have I done that correctly?
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