Proving a<=b when a<=b1 for all b1>b in Real Analysis

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Homework Help Overview

The problem involves proving a relationship between two real numbers, a and b, specifically showing that if a is less than or equal to every b1 that is greater than b, then a must also be less than or equal to b. This falls under the subject area of real analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use infinitesimals in their reasoning but expresses uncertainty about how to apply axioms. Another participant suggests a proof by contradiction, proposing to consider the midpoint of a and b as a candidate for b1. This leads to a contradiction if a is assumed to be greater than b.

Discussion Status

The discussion has progressed with one participant confirming the correctness of the reasoning presented by another. There is an exploration of different approaches, including the use of contradiction and the consideration of specific values for b1.

Contextual Notes

There is a mention of the original poster's struggle with axioms and the use of infinitesimals, which may indicate constraints in their understanding or the rules of the homework assignment.

phygiks
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Hey guys, got stuck on this question while doing homework. I would appreciate any help.
Let a,b exist in reals. Show that if a<=b1 for every b1 > b. then a <= b.

I really got nowhere. I tried letting b1(n)=b+nE where E is a infinitesimal. Then a <= b+nE for all n. Don't really know how to use any axioms here either.
 
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Do a proof by contradiction. Suppose it's NOT true that a<=b. Then a>b. Where is (a+b)/2? Use that for b1. You certainly don't need infinitesimals.
 
I think I got it, so a>b, let b1 = (a+b)/2, then b1 > b, but a<= (b1=(a+b)/2). Contradiction
 
You've got it.
 

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