The Density of Q in R

The set $Q$ is an extension of $N$ , and $R$ is in turn an extension of $Q$ . The next few results indicate how $N, Q$ sit within $R$ .

Archimedean Property

i. Given any $x \in R$ $\exists n \in N$ where $n > x$ .
ii. Given any real number $y > 0$ , $\exists n \in N$ where $\frac{1}{n} < y$ .

Proof
(i) Says that $N$ is not bounded above. Assume for contradiction that $N$ is instead bounded above. By the axiom of completeness, then $N$ should have a least upper bound, denoted $α = sup (N)$ . If we consider $α - 1$ then we no longer have an upper bound so then $\exists n \in N$ where $α - 1 < n$ . But then $α < n + 1 \in N$ so clearly we have a contradiction since $α$ needs to be an upper bound.

(ii): Apply (i) except have $x = \frac{1}{y}$ .
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This leads to an important property about $R$ :

Density of

Q

R

$\forall a, b \in R$ where $a < b$ , then $\exists r \in Q$ where $a < r < b$ .

Proof
Here we need $r = \frac{m}{n}$ for some $m, n \in Z$ . We want to show:

a < \frac{m}{n} < b \Rightarrow n a < m < n b

The idea is that we want to choose $n$ to be large enough such that an increment of $\frac{1}{n}$ is small enough to not 'step over' $b$ from $a$ or around it.

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Now we know by the archimedean property that since $b - a > 0$ then $\exists n$ such that $\frac{1}{n} < b - a$ . We want our inequality above, so then just pick $m \in Z$ such that:

m - 1 \leq n a < m

Where the second inequality comes from using the archimedean property (i). Notice then that:

\begin{aligned} m & \leq n a + 1 \\ < n (b + \frac{1}{n}) + 1 \\ = n b \end{aligned}

Because $m < n b ⟺ \frac{m}{n} < b$ , so then $a < \frac{m}{n} < b$ as needed.
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We say that $Q$ is dense in $R$ . We can show that the irrationals are also dense in $R$ as well:

Irrationals dense in

R

Given any two $a, b \in R$ with $a < b$ , $\exists t$ where $t$ is irrational and satisfies $a < t < b$ .

Proof
See Suprema + Density Practice#1.4.5.
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