# Thread: inequality multiplication proof

1. ## inequality multiplication proof

use only addition and multiplication for proof

x > y iff xz > yz

2. Originally Posted by jay0148

x > y iff xz > yz
What are x,y,z?

3. Oops! Natural Numbers

4. Originally Posted by jay0148

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

Proof: $\displaystyle x > y \Longleftrightarrow xz > yz$.

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

For the converse, suppose $\displaystyle xz > yz$. Since $\displaystyle z \in \mathbb{N}$ we know that $\displaystyle z > 0$, thus we can divide by it. Dividing throughout the inequality by $\displaystyle z$ 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 $\displaystyle x = y \Longleftrightarrow xz = yz$ proof I did.

5. I would like to see this proof using only multiplication and addition if anyone else can help...thanks.

6. 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.