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Math Help - inequality multiplication proof

  1. #1
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    inequality multiplication proof

    Please help with this proof...thanks

    use only addition and multiplication for proof

    x > y iff xz > yz
    Last edited by jay0148; December 4th 2007 at 04:27 PM.
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  2. #2
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    Quote Originally Posted by jay0148 View Post
    Please help with this proof...thanks

    x > y iff xz > yz
    What are x,y,z?
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  3. #3
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    Oops! Natural Numbers
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jay0148 View Post
    Please help with this proof...thanks

    x > y iff xz > yz
    I take \mathbb{N} to be \{  1,2,3,4,... \}. That seems to be how your other posts define the naturals.

    Proof: x > y \Longleftrightarrow xz > yz.

    The ( \implies) direction is obvious, just multiply both sides of the left hand side by z and you get the right.

    For the converse, suppose xz > yz. Since z \in \mathbb{N} we know that z > 0, thus we can divide by it. Dividing throughout the inequality by z yeilds the desired result.

    QED

    You may see my response to your other posts for a more rigorous write-up. The language I used here is not very appropriate. This is very similar to the x = y \Longleftrightarrow xz = yz proof I did.
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    I would like to see this proof using only multiplication and addition if anyone else can help...thanks.
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    x > y if and only if x = y + k has a solution in N for k (by definition). Then zx > zy if and only if zx = zy + j has a solution for some j in N. If you multiply x = y + k by z you get zx = zy + kz, thus choosing j = kz we have the solution to the equation.
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