# Trying to prove that for two numbers r,s in S, that r<q

Let $S=\{r \in \mathbb Q^+ : r^2<2\}$. Show that if $r \in S$, then there exists some $q \in S$ such that $r<q$.

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Let $r\in S$, that is $r$ is a positive rational with $r^2<2$. Also let $0<\epsilon<1$ with $$\epsilon<\frac{2-r^2}{1+2r}$$ (why does such an $\epsilon$ always exist?). Then, $$(r+\epsilon )^2=r^2+2r\epsilon+\epsilon^2<r^2+2r\epsilon+\epsilon<r^2+2-r^2=2$$ So if $q=r+\epsilon$ then obviously $r<q$ and by the above $q^2<2$

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Assuming $Q^+= \mathbb Q_{\geq 0}$:

Let $r \in S = \mathbb Q \cap [0,\sqrt{2})$. Then $r < \sqrt{2}$.

$\mathbb Q$ is dense in $\mathbb R$. Therefore there exists $q \in \mathbb Q$ with $r < q < \sqrt{2}$. It follows that $q \in S$ which proves the claim.

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First we prove that for every $r \in \mathbb{R}^+ - \{0\}$ there exists $q \in \mathbb{Q}$ such that $0 \lt q \lt r$. Suppose $a_0 = (1,+\infty)$ and for all $n \in \mathbb{N}: a_n = (\frac{1}{n+1},\frac{1}{n}]$. These sets are disjoint and $\cup a_i, 0 \le i = \mathbb{R}^+ - \{0\}$. So there exists $n$ such that $r \in a_n$. Define $q$ to be $\frac{1}{n+1}$. It's clear that $q \in \mathbb{Q}$ and $q \gt 0$ and $q \lt r$. $\square$

Now for every $s \in S$, define $r = \sqrt{2} - s$ and apply what we proved above to get a $q \in \mathbb{Q}, 0 \lt q \lt \sqrt{2} - s$. We have $(s+q)^2 < 2$ so $s+q \in S$ and $s \lt s+q$. $\square$

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Thanks that helps a lot – Math Student Feb 6 '13 at 23:01