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From stephen abbott's book, there is a section in the text where he aims to show that There exists a real number $α ∈ \mathbb{R}$ satisfying $α^2 =2$

Consider the set $T=\{t∈R:t^2<2\}$

Set $\alpha = \sup T$

He approaches the proof by first showing that $\alpha < 2$ violates the condition that $\alpha$ is an upper bound. I need help in understanding the following:

" Assume $\alpha < 2$. In search of an element of $T$ that is larger than $\alpha$, write: $(\alpha + \frac{1}{n})^2 = \alpha ^2 + \frac{2 \alpha}{n} + \frac{1}{n^2} < \alpha^2 + \frac{2 \alpha + 1}{n}$ "

What justifies the choice of writing $(\alpha + \frac{1}{n})^2$? If I had to guess I'd say it's somehow related to the Archimedean property , but really I am unsure.

To continue on with my query, abbott says

"But now assuming $\alpha^2< 2$ gives us a little space in which to fit the $(2\alpha +1)/n$ term and keep the total less than 2. Specifically, choose $n_0 \in \mathbb{N}$ large enough so that $ \frac{1}{n_0} < \frac{2- \alpha^2}{2\alpha + 1}$

Why this final inequality? How is it even formulated. Again it looks like it may have to do with the Archimedean property, but I am unsure. If someone could shed light onto these two issues, I'd appreciate it.

Thanks.

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  • $\begingroup$ So, your question is how was the following inequality formed? $$\frac{1}{n_0} < \frac{2- \alpha^2}{2\alpha + 1}$$ $\endgroup$
    – Alice Ryhl
    Commented Feb 16, 2015 at 17:45
  • $\begingroup$ Hi @KristofferRyhl apologies for not being clear. Why did he choose $(\alpha + \frac{1}{n})^2 $ and also yes, what justifies the formulation of that inequality. $\endgroup$
    – elbarto
    Commented Feb 16, 2015 at 17:51

1 Answer 1

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In the authors proof of $\alpha^2 = 2$, he defines $\alpha$ to be the supremum of the set of numbers which have a square smaller than 2.
He now writes "In search of an element of $T$ that is larger than $\alpha$", since any number of the form $\alpha + 1/n$ is larger than $\alpha$, this is an appropiate choice.


The inequality

$$\frac{1}{n_0} < \frac{2- \alpha^2}{2\alpha + 1}$$

was formed when the author got the idea of keeping the total less than 2, here the total refers to $(\alpha + 1/n)^2$

$$\alpha^2+\frac{2\alpha+1}n<2$$

then solving for $1/n$ and renaming $n$ to $n_0$

$$\frac{2\alpha+1}{n_0}<2-\alpha^2$$ $$\frac{1}{n_0}<\frac{2-\alpha^2}{2\alpha+1}$$

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  • $\begingroup$ You are the best and I love you. $\endgroup$
    – elbarto
    Commented Feb 16, 2015 at 18:14

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