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

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Homework Help Overview

The discussion revolves around the implications of the statement "for each eps>0, a

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore whether the statement about epsilon implies the inequality a<=r, with some questioning the validity of the implication and others providing reasoning based on the properties of inequalities.

Discussion Status

The discussion includes various interpretations of the epsilon argument, with some participants suggesting that the implication does not hold, while others argue that it does. There is an acknowledgment of differing perspectives on the reasoning involved.

Contextual Notes

Participants are working under the assumption that they need to prove the inequality a<=r, and there is a focus on the implications of the epsilon argument without reaching a definitive conclusion.

magnumartus
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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?
 
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magnumartus said:
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.
 
magnumartus said:
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
 
hey, it helped a lot. thank you all
 

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