Question about proof of the existence of square roots

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{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} 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 $n=1$?

• $n$ 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 $n > 1$. Jul 30, 2014 at 7:06
• @StevenVanGeluwe yeah, I see what you mean. Thanks! Jul 30, 2014 at 7:12

$n$ is any (arbitrarily large) number and thus $\frac 1 n$ is an arbitrarily small one.
• It's true that at $n=1$ we have an equality, but the author doesn't care, because $n$ is intended to be large. Jul 30, 2014 at 7:02
• Oh ok, thanks, so he basically choose $n$ such that the inequality $<$ is valid? Jul 30, 2014 at 7:03
• 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 $1$. Jul 30, 2014 at 7:09