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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 04:28 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 $\displaystyle x,y,z \in \mathbb{N}$

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

    Thus we have $\displaystyle x = y \implies xz = yz$

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

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

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

    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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