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**Proof that Q is "not complete"**

Hello.

I have recently started taking a course in the foundations of analysis. We started off with the completeness "axiom": Every upper bounded non-empty subset of R has a supremum (least upper bound). The lecturer then went on demonstrating that Q does not satisfy this. He also gave a proof for it, and that is where he lost me a bit.

Let [tex]E= \{x \in Q: x^2 < 2 \}[/tex]. E is upper bounded, for example by 2. However it has no supremum. Assume that [tex]s \in Q[/tex] is a supremum for E. Clearly [tex]s>0[/tex]. Also [tex]s \neq 2[/tex] since 2 has no square root in Q.

If [tex]s^2 < 2[/tex], notice that [tex](s+t)^2 = s^2 + 2st + t^2 = s^2 + (2s+t)t[/tex], where [tex]t \in Q[/tex].

If t>0 is chosen to be small, [tex]s^2 +(2s+t)t<2[/tex]. This contradicts our hypothesis that s is a supremum for E.

I am well aware that the proof is not complete (s^2 >2), but I would like to do and comprehend that by myself.

My only problem with this is where he chooses t to be small so that the expression is less than 2. How can one guarantee that it will be? Would anybody care to elaborate?