> Can someone here demonstrate this?

Sure, I'll give it a go. There's lots of ways to do it.
Here's the most direct one I can think of:

Assume the conclusion is false.
That means, assume it is false that a > 0.
Well, that means that a <= 0.
If that's so, and we're given that -a < 0,
then a+(-a) < 0+0, (since w<=x and y<z implies w+y<x+z).
Simplifying the two sides, this says that "0<0"
is true. But, this isn't true, and so the initial
assumption that the conclusion is false was wrong.
End of Proof by Contradiction.

Of course, if you already know that w<=x and y<z implies w+y<x+z,
why not substitute w=a, x=a, y=-a, z=0 and immediately get
that a<=a (always true) and -a<0 (given) implies that a+(-a)<a+0,
i.e. 0<a, which is a>0 (the desired conclusion).
This bypasses proof by contradiction altogether.
Then again, who says you can't use a knife as a screwdriver.
It's clumsy, but it works, as I've proved many times.

Hope that helps,

Walter

--- Corey Bray wrote:
> Hello Everyone,
>
> Given that -a < 0, prove that a > 0 by assuming the conclusion
> is false and prove that it must be the case by contradiction.
> I was not exactly certain how to show this to be the case using
> a proof by contradiction method.
> Can someone here demonstrate this?
>
> Thanks in advance...
> Corey...