Multiplication cancellation proof

• 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 $x,y,z \in \mathbb{N}$

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