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

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.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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