# Question about the proof in "Understanding Analysis" by Stephen Abbot

My question is rather simple and involves a step that is taken when proving that there is a number squared that equals two. He uses the least upper bound theorem. The first few steps make sense, but how he gets between these two highlighted steps confuses me. I do not know how he derives that second inequality that I have highlighted.

Thank you in advance for your help.

• He is just saying that if $\alpha^2 < 2$ then there is $n$ large enough so that $(\alpha^ + \frac{1}{n})^2 < 2$ Dec 21, 2022 at 20:54
• It should be noted that the author (Abbot) is trying to prove that $\alpha = \sup_{t\in\mathbb{R}}\{t^2 < 2\} = \sqrt{2}$ by contradiction. The post above is concerned with trying to prove that $\alpha^2$ cannot be strictly less than $2$. Dec 21, 2022 at 21:16
• The author is trying to argue that if $T=\left \lbrace x \in \mathbb R: x^2<2 \right \rbrace$ and if $\alpha=\sup(T),$ then $\alpha=\sqrt{2}.$ Suppose $\alpha<\sqrt 2$ i.e. $\alpha^2<2.$ The author is asserting if we pick any $n_0 \in \mathbb N$ such that $\frac{1}{n_0} < \frac{2-\alpha^2}{2 \alpha +1},$ then we will get that $\alpha + \frac{1}{n_0} \in T.$ But can that be possible knowing $\alpha=\sup(T)$ ? By the way, why is $\alpha > \sqrt{2}$ impossible ? Dec 22, 2022 at 7:02

He's trying to prove that if $$\alpha^2\lt 2$$ then he can find a number that is larger than $$\alpha$$ but still less than $$2$$.

He is trying to find a positive number $$n$$ such that $$\alpha$$ so that the $$\left(\alpha+\frac{1}{n}\right)^2\lt 2$$

He then expands that to get $$\left(\alpha+\frac{1}{n}\right)^2=\alpha^2+\frac{2\alpha}{n}+\frac{1}{n^2}$$ and notices that if $$\frac{1}{n}\lt 1$$ then $$\frac{1}{n^2}\lt \frac{1}{n}$$ which gives us that $$\left(\alpha+\frac{1}{n}\right)^2=\alpha^2+\frac{2\alpha}{n}+\frac{1}{n^2}\lt\alpha^2+\frac{2\alpha}{n}+\frac{1}{n}=\alpha^2+\frac{2\alpha + 1}{n}$$

We want $$\alpha^2+\frac{2\alpha+1}{n}\lt 2$$ so we solve for $$n$$ to get $$\frac{2\alpha+1}{n}\lt 2-\alpha^2\implies\frac{2\alpha+1}{2-\alpha^2}\lt n\implies\frac{1}{n}\lt\frac{2-\alpha^2}{2\alpha+1}$$

He uses $$n_0$$ instead of $$n$$ when he solves for the specific value of $$n$$ to emphasize that $$n$$ is a variable and $$n_0$$ is the specific value that keeps $$\left(\alpha+\frac{1}{n_0}\right)^2\lt 2$$

I do not know how he derives that second inequality that I have highlighted.

You question contains a false premise. He does not derive the second inequality. The second inequality is not a statement about an already-known $$n_0$$, and it is not arrived at from the previous inequality (at least not directly, see below), it is a constraint describing what $$n_0$$. He says to pick $$n_0$$, and then gives the inequality as a description of what sort of $$n_0$$ to choose. If I say "Find a turkey that's smaller than my oven", then turkey<oven isn't an inequality that I've "derived", it's a description telling you what kind of turkey I want. Now, he obviously has a motivation for why he wants to choose an $$n_0$$ that satisfies that inequality, and that motivation comes from the previous inequality, but that's a bit different from the second inequality being "derived" from the first.