2
$\begingroup$

Let $S:=(-\sqrt{2},\sqrt{2})\cap \mathbb{Q}$. It is clear that $\sup(S) = \sqrt{2}$, but I don't know how to prove this.

Suppose that $\sup(S)<\sqrt{2}$, then $\forall a\in S$ and $\forall\varepsilon>0$, $\sup(S)-\varepsilon

  • 1
    Have you shown or seen something like the following? "For any two real numbers $a < b$, there is a rational number between them." – 2017-01-08

2 Answers 2

6

It is easy to see that $\sqrt{2}$ is an upper bound of $S$.

Now use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$, one can show that $\sqrt{2}$ must be the least upper bound of $S$ also:

To show that $\sqrt{2}$ is the least upper bound of $S$, suppose otherwise there exists $s<\sqrt{2}$ as an upper bound of $S$. By the density of $\mathbb{Q}$, there exists $r\in S$ such that $s

  • 0
    Thanks, but still not sure. – 2017-01-08
  • 3
    @sequence Say that $s = \sqrt{2}$ is an upper bound. Now, suppose that $s'. – 2017-01-08
  • 0
    Let $s=\sqrt{2}$ be an upper bound for $s$. Let $s' := \sup(S)$, then $s'$\mathbb{Q}$ is dense in $\mathbb{R}$, $\exists q\in \mathbb{Q}$ such that $s'$q\in S$, a contradiction. Thus $\sup(S)=\sqrt{2}$. Do you think this one is correct? – 2017-01-08
  • 0
    The first sentence is not needed and the second one is wrong. See edited answer. – 2017-01-08
2

Since it is obvious that $x<\sqrt 2$ for all $x$ in the interval, and it is easy to construct a sequence in the interval which converges to $\sqrt 2$ (namely, its truncated decimal expansión), we have the result.