Epsilon Argument: Does "a<r+eps" Imply "a<=r"?

[magnumartus]

Hi,

Want to show: a<=r

However I ended up with such an argument: For each eps>0, a<r+eps. Does this statement imply a<=r?

 

[tanujkush]

Hi,

Want to show: a<=r

However I ended up with such an argument: For each eps>0, a<r+eps. Does this statement imply a<=r?

No, this implies that you can find any epsilon that when added to r makes it larger than a. This would work even when a is not equal to r, and is lesser. for eg a = 1, r = 2.

 

[Office_Shredder]

Hi,

Want to show: a<=r

However I ended up with such an argument: For each eps>0, a<r+eps. Does this statement imply a<=r?

This does show a <= r. If a>r, then a-r is a positive number. In particular, you then showed that

a<r+(a-r) (picking eps=a-r)

Hence a<a. That doesn't make any sense, so it must have been a<=r the whole time

 

[magnumartus]

hey, it helped a lot. thank you all