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Thread: Multiplication cancellation proof

  1. #1
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    Multiplication cancellation proof

    I need help with this proof also

    x = y <=> xz = yz

    use only addition and multiplication for proof

    x, y and z are natural numbers

    thanks.
    Last edited by jay0148; Dec 4th 2007 at 05:28 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jay0148 View Post
    I need help with this proof also

    x = y <=> xz = yz

    x, y and z are natural numbers

    thanks.
    I do not see a problem here, you just want to go from the right to the left and then from the left to the right. Only basic algebraic manipulation is required.

    Proof

    Let x,y,z \in \mathbb{N}

    Suppose x = y. Multiplying both sides by z we obtain: xz = yz

    Thus we have x = y \implies xz = yz

    For the converse, suppose xz = yz. Subtracting yz from both sides, we obtain: z(x - y) = 0. Now z \ne 0, since z \in \mathbb{N}, thus we must have x - y = 0. But this means x = y.

    Thus we have xz = yz \implies x = y.

    Therefore, if x,y,z \in \mathbb{N}, x = y \Longleftrightarrow xz = yz.

    QED
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  3. #3
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    If anyone else can show this proof using only multiplication and addition, that would be helpful...thanks.
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