Let x, y be real numbers such that x<= y+c for every c>0. Then x<=y. Prove this by contradiction....

Printable View

- May 20th 2008, 06:16 PMVedicmathsProof using contradiction!
Let x, y be real numbers such that x<= y+c for every c>0. Then x<=y. Prove this by contradiction....

- May 20th 2008, 06:34 PMJhevon
- May 20th 2008, 06:46 PMVedicmaths
Thank you very much Sir..

but I could not understand one point here:

"since x > y we have x - y > 0. Choose c > 0 such that c < x - y (this is of course possible, c = (x - y)/2 is an example)."..

How could you say that c < x - y....example is not very clear!

Thanks! - May 20th 2008, 06:50 PMIsomorphism
- May 20th 2008, 06:58 PMVedicmaths
yeah that makes sense perfectly now...that is what I was wondering for..how do you choose that c = (x - y)/2 and hence c < x - y...but now I understand that c is any number greater than 0 and that is we choosing it to be equal to (x-y)/2...

Thank you! - May 20th 2008, 06:59 PMIsomorphism