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Thread: Order of Numbers Proofs

  1. April 28th 2009, 08:29 PM #1
    noles2188
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    Order of Numbers Proofs

    Prove that if u < v and x < y, then u+x < v+y.

    Also if x < y and z < 0, then y*z < x*z.
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  2. April 28th 2009, 08:36 PM #2
    Gamma
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    Quote Originally Posted by noles2188 View Post
    Prove that if u < v and x < y, then u+x < v+y.

    Also if x < y and z < 0, then y*z < x*z.
    I dunno what axioms you can use.
    0 < v-u
    0 < y-x
    0< v-u+y-x = (v+y) - (u+x)
    u+x < v+y




    0<y-x
    0(z)>(y-x)z
    0>yz-xz
    xz>yz
    Last edited by Gamma; April 28th 2009 at 08:37 PM. Reason: Quote
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