The Student Room Group

Lowest rational upper bound help

"Consider the set S=xR:x2<2S = {{x \in \mathbb{R} : x^2 < 2}}. Show that if H is any upper bound for S that is a rational number, then there exists another rational number hQh \in \mathbb{Q} which is also an upper bound for S but satisfi es h < H."

I'm guessing proof by contradiction is the way to go with this, but don't have a clue how to start it off. Assume h doesn't exist or something similar?

Any hints?
Reply 1
What (real) number is it "necessary and sufficient" for H to be bigger than if H is to be an upper bound?

So, ...
Reply 2
2\sqrt{2}... So, proceed with 2+ϵ=H\sqrt{2} + \epsilon = H?
Reply 3
Somewhat depends what you can or can't assume. If you can quote the result that between any distinct real numbers there's at least one rational, then that directly tells you you can find h between 2\sqrt{2} and H.

If you don't have it as a known result, that's probably a little like "cheating" though. But you almost certainly have been given an axiom along the lines of ϵ>0,nNs.t.1n<ϵ\forall \epsilon > 0, \exists n \in {\mathbb N} s.t. \dfrac{1}{n} < \epsilon. At which point it should be obvious how to finish.
Reply 4
Yep. Got it thanks.

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