Re: Proof by contradiction, is this right?

[David C. Ullrich]

On Fri, 29 Sep 2006 12:46:29 +0000 (UTC), magidin@xxxxxxxxxxxxxxxxx
(Arturo Magidin) wrote:

>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 
>
>As I suspected: You did NOT give the correct statement above. It
>should be:
>
>  If for every e>0,   a<=b + epsilon,   then a<b.

No, that's obviously false.

>Otherwise, you don't know anything about epsilon. You could have a=2,
>b=1, and epsilon = -1. 
>
>>choose epsilon = (a+b)/2 and then b>= a/3,
>
>Huh?
>
>Why?
>
>> which contradicts our
>>assumption that a>b.
>>
>>So, a<b.
>>
>>Is this right?
>
>I don't think so. I would try instead epsilon = (a-b)/2, but may
>that's just me. 


************************

David C. Ullrich