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In article
<1159490080.238757.87950@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"jennifer" <scrilla_12_1999@xxxxxxxxx> wrote:
> if a<= b + epsilon then a<b
>
> Proof by contradiction
>
> Assume a<= b + epsilon and a>b
>
> a>b implies that (a-b)>0
>
> choose epsilon = (a+b)/2 and then b>= a/3, which contradicts our
> assumption that a>b.
>
> So, a<b.
>
> Is this right?
Now that this is answered consider a direct proof.
If for each eps > 0 implies a <= b + eps, then a <= b.
There is eps' such that 0 < eps' < eps, hence
0 < eps-eps'. Then a <= b + (eps-eps'). Adding 0<eps' to
this we have a < b + eps for all eps>0. This means that
-eps < b-a, which is to say that for all x<0, x < b-a. Now
sup {x st x < 0} = 0, therefore 0 <= b-a or a <= b.
--
Michael Press
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