# how to prove?

• Jun 7th 2009, 07:22 AM
pberardi
how 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, 07:41 AM
Plato
Quote:

Originally Posted by pberardi
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?

I frankly don’t see what there is to prove here.
The number property Q is clearly membership in the positive real numbers.
Are you to prove this: $x \leqslant y\;\& \,c > 0\, \Rightarrow \,cx \leqslant cy$ ?
• Jun 7th 2009, 07:49 AM
pberardi
Quote:

Originally Posted by Plato
I frankly don’t see what there is to prove here.
The number property Q is clearly membership in the positive real numbers.
Are you to prove this: $x \leqslant y\;\& \,c > 0\, \Rightarrow \,cx \leqslant cy$ ?

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, 08:04 AM
Plato
Quote:

Originally Posted by pberardi
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".

Once again, as I said above, I do not see what there is to formalize other than what I posted.

BTW: I think the statement is very poorly written.
• Jun 7th 2009, 10:41 AM
Gamma
Suppose $x \leq y$, $c\in \mathbb{R}$.

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

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

$c0\leq c(y-x)\Rightarrow 0\leq cy-cx \Rightarrow cx \leq cy$
Which is property Q.
Case 2: c=0

$0\cdot 0\leq 0(y-x)\Rightarrow 0\leq 0$
Which is trivially property Q.

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

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

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

Thus property Q is being non-negative.