# multiplication proof

• Dec 4th 2007, 06:13 AM
jay0148
multiplication proof
w > x and y > z then wy > xz

w,x, y and z are natural numbers

use only addition and multiplication for proof

I need help with this proof ASAP....please
• Dec 4th 2007, 10:36 AM
Jhevon
Quote:

Originally Posted by jay0148
w > x and y > z then wy > xz

w,x, y and z are natural numbers

I need help with this proof ASAP....please

The trouble here is that this is almost too obvious. The conclusion seems to follow immediately for natural numbers. So to make our proof seem more rigorous, let's take the long way around.

Proof

Let $\displaystyle w,x,y,z \in \mathbb{N}$.

Suppose $\displaystyle w \ge x$ and $\displaystyle y \ge z$. Then we have that $\displaystyle w - x \ge 0$ and $\displaystyle y - z \ge 0$. We now have two numbers that are non-negative, therefore, their product must be non-negative. Thus we have:

$\displaystyle 0 \le (w - x)(y - z) = wy - xz -xy - wz \Longleftrightarrow xz + (xy + wz) \le wy$.

Since $\displaystyle w,x,y,z \in \mathbb{N}$, $\displaystyle xy + wz \ge 0$. Thus we make the left side of the last inequality even smaller if we subtract it. Hence,

$\displaystyle wy \ge xz + (xy + wz) \ge xz \implies wy \ge xz$ as desired.

QED
• Dec 4th 2007, 10:47 AM
jay0148
Thanks!
:):):):) Thanks!!
• Dec 4th 2007, 06:52 PM
jay0148
I would like to do this proof using only multiplication and addition, if anyone else can help...thanks.