1. ## Proof using contradiction!

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

2. Originally Posted by Vedicmaths
Let x, y be real numbers such that x<= y+c for every c>0. Then x<=y. Prove this by contradiction....
Assume to the contrary that x <= y + c but x > y

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).

Thus we have: x <= y + c < y + x - y = x. So that x < x. A contradiction!

3. 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!

4. Originally Posted by Vedicmaths
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!

Jhevon
is asking you to choose a positive c such that x - y > c, and moreover he has given you an example of (x-y)/2 in case you worry how to choose...

5. 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!

6. Originally Posted by Vedicmaths
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!
Yes thats right