# Inequality Proof

1. Dec 22, 2014

### Bashyboy

Hello,

In Principles of Mathematical Analysis, the author is attempting to demonstrate that, if $x > 0$ and $y < z$, then $xy < xz$, which essentially states that multiplying by a positive number does not disturb the inequality.

I am hoping someone will quickly denounce this with an adequate explanation, but I feel as though the author is using the result to prove it.

He begins by noting that, if $z > y$, $z - y > 0$. He multiplies both sides by $x > 0$, and gets $x(z-y) > 0$.

This seems to be a special case of the theorem which we are trying to prove. Wouldn't this be an invalid step as we do not know what results from multiplying both sides of an inequality? Let $c = z - y$, and replace $y$ with zero in the theorem. This would give us

If $x > 0$ and $0 < c$, then $x \cdot 0 < xc$.

Am I mistaken?

2. Dec 22, 2014

### ShayanJ

He is assuming that if you multiply two positive numbers, you will get a positive number. I don't think for knowing this, you need to know the theorem mentioned, so I see no problem with the proof.

3. Dec 22, 2014

### Bashyboy

I am sorry, but I do not quite follow what you are saying here.

4. Dec 22, 2014

### ShayanJ

The product of two numbers is negative only when one of them is negative. You can't get a negative number by multiplying two positive numbers.