I realize that this is a quite simple proof. I can reason the answer but cannot prove it.

problem Statement:

Say c is a real number and has the property Q if x <= y whenever x, y in real numbers with cx <= cy. Which real numbers have property Q?

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- Jun 7th 2009, 06:22 AMpberardihow to prove?
I realize that this is a quite simple proof. I can reason the answer but cannot prove it.

problem Statement:

Say c is a real number and has the property Q if x <= y whenever x, y in real numbers with cx <= cy. Which real numbers have property Q? - Jun 7th 2009, 06:41 AMPlato
- Jun 7th 2009, 06:49 AMpberardi
The problem statement is exactly as I have written it. Perhaps it is not a proof and that simply the answer would suffice. My other rationale is that we need to use a certain property that we know is true i.e. (x-y)^2 > 0 for x != y. If you Plato cannot formalize a proof for this then I am going to quit feeling bad about it. My only problem is that I would expect the prof to say "prove it".

- Jun 7th 2009, 07:04 AMPlato
- Jun 7th 2009, 09:41 AMGamma
Suppose $\displaystyle x \leq y$, $\displaystyle c\in \mathbb{R}$.

$\displaystyle x \leq y$

$\displaystyle 0\leq (y-x)$

Case 1: $\displaystyle c\in \mathbb{R}^+$

$\displaystyle c0\leq c(y-x)\Rightarrow 0\leq cy-cx \Rightarrow cx \leq cy$

Which is property Q.

Case 2: c=0

$\displaystyle 0\cdot 0\leq 0(y-x)\Rightarrow 0\leq 0$

Which is trivially property Q.

Case 3: $\displaystyle c\in \mathbb{R}^-$

$\displaystyle c0\geq c(y-x)\Rightarrow 0\geq cy-cx \Rightarrow cx \geq cy$

Which is only property Q if x=y, but Q is for all $\displaystyle x,y \in \mathbb{R}$, so case 3 does not satisfy property Q.

Thus property Q is being non-negative.