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    cowboys190's Avatar
    cowboys190 Posts: 1, Reputation: 1
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    #1

    Oct 2, 2012, 02:51 PM
    Prove by Contradiction
    If A is the average of three positive real numbers then one of the three numbers is less than or equal to A.
    ArcSine's Avatar
    ArcSine Posts: 969, Reputation: 106
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    #2

    Oct 3, 2012, 04:16 AM
    This'll get it teed up for you: To prove by contradiction, first assume the premise is true; i.e...

    • A is the average of three positive reals s, t, and u.

    But then also assume the negation of the consequent is true...

    • s > A, t > A, u > A

    Then show how the second assumption renders the first assumption impossible. Hint: add the three inequalities together.
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