How Does the Epsilon Argument Prove \( a \leq b \)?

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Homework Help Overview

The discussion revolves around the epsilon argument in the context of proving the inequality \( a \leq b \). Participants are examining the implications of the statement that for any \( \epsilon > 0 \), \( a < b + \epsilon \) leads to the conclusion \( a \leq b \).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the meaning of "any \( \epsilon > 0 \)" and questioning the validity of specific choices for \( \epsilon \). There is a discussion about the implications of choosing different values for \( \epsilon \), particularly in relation to the original inequality.

Discussion Status

Some participants have offered clarifications regarding the interpretation of the epsilon argument, noting that \( \epsilon \) can be chosen to be arbitrarily small. Others are reflecting on their understanding of the proof and its implications, indicating a productive exchange of ideas without reaching a definitive consensus.

Contextual Notes

There is a mention of the original poster's confusion regarding the proof and the choice of \( \epsilon \), as well as the context of the discussion occurring late at night in the UK.

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Homework Statement


Suppose that , for any $$ \epsilon > 0, a < b + \epsilon $$ . Then $$ a\le b $$

The Attempt at a Solution


I have the proof, its not a question that was assigned to me, it was an example used.
According to the proof i can choose ANY epsilon greater than 0, so let's choose 10. then 2 < 1 + 10 = 11. Right?
Then $$ 2 \le 1 $$ . NOT right.

it's late here in the UK, is it me?
 
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It isn't saying for anyone epsilon, but for any epsilon you could possibly choose. Therefore, you can take epsilon as small (positive) as you like and the relation will still be true.
 
##2 < 1 + \epsilon ## for ##1 < \epsilon ## but, you can choose any epsilon, so choose ##\epsilon = 1/2##, then ##2 < 1+ \epsilon ## is false.
 
Thanks for clearing that one up! The penny/cent/nickel/kuna has dropped!
 

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