Thanks Walter! That was very helpful...


Corey...

Walter Vannini wrote:
>
> > 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...
>
>
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