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

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To prove that if a ≤ b1 for every b1 > b, then a ≤ b, one can use proof by contradiction. Assume a > b, which implies there exists a value b1, specifically (a + b)/2, that is greater than b. This leads to the conclusion that a ≤ (a + b)/2, creating a contradiction since (a + b)/2 is less than a. Therefore, the initial assumption must be false, confirming that a ≤ b. The discussion emphasizes the validity of using averages to demonstrate the contradiction without needing infinitesimals.
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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