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I'm self-studying from the book Understanding Analysis by Stephen Abbott and I'm stuck on Theorem 1.4.5 on page 21. The aim of this theorem is to prove that $\sqrt{2}$ exists.

He starts by considering the set: $$ T = \{ t \in \mathbb{R} \mid t^2 < 2 \} $$ and set $\alpha = \mathrm{sup} \; T$. He then assumes that $\alpha^2 <2$ and writes: \begin{equation} \begin{aligned} \left( \alpha + \frac{1}{n} \right)^2 & = \alpha^2 + \frac{2 \alpha}{n} + \frac{1}{n^2} \\& < \alpha^2 + \frac{2 \alpha}{n} + \frac{1}{n} \end{aligned} \end{equation} He hasn't defined what $n$ is so I'm quite confused what the inequality exactly means. But normally, using his notation, $n \in \mathbb{N}$; if this is correct, then shouldn't the inequality sign actually be $\leq$ for the situation that $n=1$?

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    $\begingroup$ $n$ is indeed a natural number. You're right about the inequality sign. I think his next step will be to increase $n$ to infinity, so then you can safely assume that $n > 1$. $\endgroup$
    – Steven Van Geluwe
    Jul 30, 2014 at 7:06
  • $\begingroup$ @StevenVanGeluwe yeah, I see what you mean. Thanks! $\endgroup$
    – Hunter
    Jul 30, 2014 at 7:12

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$n$ is any (arbitrarily large) number and thus $\frac 1 n$ is an arbitrarily small one.

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    $\begingroup$ It's true that at $n=1$ we have an equality, but the author doesn't care, because $n$ is intended to be large. $\endgroup$
    – Karolis Juodelė
    Jul 30, 2014 at 7:02
  • $\begingroup$ Oh ok, thanks, so he basically choose $n$ such that the inequality $<$ is valid? $\endgroup$
    – Hunter
    Jul 30, 2014 at 7:03
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    $\begingroup$ Yes. Most authors don't do the difference between $<$ and $\leq$ in real analysis because we always pick $n$ to be as large as we want, and thus always greater than $1$. $\endgroup$
    – Hubble
    Jul 30, 2014 at 7:09
  • $\begingroup$ @iHubble thanks for letting me know, that is useful information. $\endgroup$
    – Hunter
    Jul 30, 2014 at 7:11

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