# Multiplication cancellation proof

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• Dec 4th 2007, 06:14 AM
jay0148
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.
• Dec 4th 2007, 10:24 AM
Jhevon
Quote:

Originally Posted by jay0148
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
• Dec 4th 2007, 06:53 PM
jay0148
If anyone else can show this proof using only multiplication and addition, that would be helpful...thanks.