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Consider the set $A=\{\,x\in\mathbb{Q} : x^2<2\,\}.$

I want to prove that $\sqrt2$ is the least upper bound for $A$.

To do this, I think I need to prove that if I take any $\alpha\in\mathbb{R}$ with $\alpha<\sqrt2$, then there exists $y\in A$ with $y>\alpha$.

I'm not sure how to choose $y\in A$ to satisfy this property.

MHW
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    You can use archimedian property: between any two different numbers there is always a rational (observe you can assume $\alpha>0$) – Tito Eliatron Feb 13 '21 at 11:37
  • Minor point: You also do need to show that $\sqrt 2$ is an upper bound to begin with. – Aryaman Maithani Feb 13 '21 at 11:40
  • @AryamanMaithani Does this not just follow directly from the fact that $x^2<2$? If $x^2<2$ then $-\sqrt2<x<\sqrt2$. – MHW Feb 13 '21 at 11:42
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    Density of the rationals has been discussed here. – J.G. Feb 13 '21 at 11:51
  • @MHW: Yes, it is easy indeed. I just was pointing out that the solution for something is a least upper bound should also show that it is indeed an upper bound. (Even if you just state it as "That it is an upper bound is clear".) – Aryaman Maithani Feb 13 '21 at 14:15
  • @AryamanMaithani Thanks—that's useful to know to make sure I construct the proof properly. – MHW Feb 13 '21 at 14:16

2 Answers2

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I'll expand my comment.

WLOG one may assume that $\alpha>0$. If $\alpha\le 0$ it is clear that $y:=\frac{1}{2}\in A$ and $\alpha<\frac12<\sqrt2$.

Assume $0<\alpha<\sqrt2$ is an upper bound of $A$. Archimedean property (or the density of rationals) guarantees the existence of $y\in\Bbb Q$ such that $\alpha<y<\sqrt2$, so (since $0<y<\sqrt2$) $y^2<2$ and $y\in A$.

Tito Eliatron
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Let $0\lt d\lt\sqrt2$, where $d\in\mathbb{Q}$.

Then we can prove the existence of a rational between $d$ and $\sqrt2$ by noting that such a rational, $\frac{a}{b}$, satisfies:

$$d\lt \frac{a}{b}\lt \sqrt2$$ $$bd\lt a\lt b\sqrt2$$

and we can make $b$ as large as required in order to find such an $a$.

JMP
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