Density of the Rationals

See that every interval, however narrow, hides a rational. Bigger denominators make the ticks denser.

Between any two numbers sits a fraction. Narrower gaps need bigger denominators. Chop the scale finely enough and some tick lands in every gap. The same trick works around irrationals.

For example, keep cutting a 🍰 cake in half, again and again. The pieces shrink without end. Pick any thin gap, and a thinner piece exists. That piece is the fraction landing inside.

Examples & Exercises

Prove completely, from the Archimedean property, that for every real $x$ there is an integer $n$ with $n \le x < n+1$ (Tao, Proposition 4.4.1).

\forall\, x \in \mathbb{R}\ \exists\, n \in \mathbb{Z}:\ n \le x < n+1

Interval

(1, 2)

n = \lfloor 1/(2-1) \rfloor + 1 = 2

m = \lfloor 2 \cdot 1 \rfloor + 1 = 3

, so

q = 3/2 = 1.5

lands inside. The proof below shows this recipe always works.

Show proof

Let $S = \{k \in \mathbb{Z} : k \le x\}$ . By the Archimedean property there is a natural number $K > -x$ ; then $-K < x$ , so $-K \in S$ — $S$ is nonempty and bounded above by $x$ .
A nonempty set of integers bounded above has a maximum: pick an integer $M > x$ (Archimedean); then $S'' = \{M - s : s \in S\}$ is a nonempty set of nonnegative integers, so by well-ordering it has a least element $M - n$ with $n \in S$ . Every $s \in S$ satisfies $M - s \ge M - n$ , i.e. $s \le n$ , so $n = \max S$ .
Since $n \in S$ , $n \le x$ .
Also $n + 1 > x$ : otherwise $n + 1 \le x$ would put $n + 1 \in S$ , contradicting the maximality of $n$ .
Hence $n \le x < n+1$ — every real sits exactly between two consecutive integer marks.

Prove completely that for reals $x < y$ there is a rational $q$ with $x < q < y$ — the rationals are dense in the reals (Tao, Proposition 4.4.3).

x < y \Rightarrow \exists\, q \in \mathbb{Q}:\ x < q < y

n = \left\lfloor\frac{1}{y-x}\right\rfloor + 1,\quad m = \lfloor nx \rfloor + 1,\quad q = \frac{m}{n}

Interval

(1, 2)

: taking

n = 2

with

1/n < y - x

, one of the

1/2

-spaced marks

\dots, 1, 3/2, 2, \dots

— namely

3/2

— falls inside the interval: the proof's recipe executed live.

Show proof

Since $y - x > 0$ , the Archimedean property gives a natural number $n$ larger than $\frac{1}{y-x}$ — i.e. $\frac{1}{n} < y - x$ . The model computes such an $n$ directly as $n = \left\lfloor\frac{1}{y-x}\right\rfloor + 1$ .
Applying problem 1 to $nx$ gives an integer $m$ with $m - 1 \le nx < m$ (the model computes $m = \lfloor nx \rfloor + 1$ ). Put $q = m/n$ ; from $m > nx$ we get $\frac{m}{n} > \frac{nx}{n} = x$ .
Also $m < nx + 1$ , and $\frac{1}{n} < y - x$ is equivalent to $1 < n(y - x)$ , so $nx + 1 < nx + n(y - x) = ny$ .
Hence $m < ny$ , i.e. $\frac{m}{n} < y$ .
Together: $x < \frac{m}{n} < y$ — the desired rational $q = m/n$ is found.
The key idea: once the marks are spaced finer than the interval ( $\frac{1}{n} < y - x$ ), no mark can jump clean over it. However narrow the interval (try the $(0, 0.1)$ and $(-3.5, -3.4)$ presets), finer marks always work — there are no exceptions.

Prove the irrationals are dense too: if $x < y$ there is an irrational $z$ with $x < z < y$ (hint: pick rationals $r < s$ inside $(x, y)$ and use $z = r + (s-r)\frac{\sqrt{2}}{2}$ ).

z = r + (s-r)\frac{\sqrt{2}}{2} \in (x, y),\qquad z \notin \mathbb{Q}

Interval

(1, 2)

: with rationals

r = 3/2

s = 7/4

z = \frac{3}{2} + \frac{1}{4}\cdot\frac{\sqrt{2}}{2} \approx 1.677

is an irrational inside

(1, 2)

— an irrational always squeezes between rational marks.

Show proof

Problem 2 alone gives only one rational, so apply it twice: with the midpoint $m = (x + y)/2$ , applying problem 2 in $(x, m)$ gives a rational $r$ with $x < r < m$ , and applying it in $(m, y)$ gives a rational $s$ with $m < s < y$ ; since $r < m < s$ , there are rationals $r < s$ inside $(x, y)$ .
Put $z = r + (s - r)\frac{\sqrt{2}}{2}$ .
Since $1 < 2 < 4$ and squaring is strictly increasing on $[0, \infty)$ (problem 1 of the sqrt2-irrational scene, plus the n-th root scene), $1 < \sqrt{2} < 2$ — hence $0 < \frac{\sqrt{2}}{2} < 1$ .
Since $s - r > 0$ , $0 < (s-r)\frac{\sqrt{2}}{2} < s - r$ ; adding $r$ gives $r < z < s$ . With $x < r$ and $s < y$ , $x < z < y$ — $z$ lies in the interval.
Now $z$ is irrational: if it were rational, $\sqrt{2} = \frac{2(z - r)}{s - r}$ would be rational too (the right side is arithmetic on rationals with nonzero denominator) — contradicting Proposition 4.4.4 that $\sqrt{2}$ is irrational (problem 1 of the sqrt2-irrational scene). Hence $z$ is irrational with $x < z < y$ .
Corollary: every open interval of the real line contains both rationals and irrationals. Density of the rationals does not mean "almost all" reals are rational — quite the opposite is true (the countability scene).