measure of irrational number I've once read a proof about this and I'm trying to remember how it went.
We want to show that if we randomly select a number $x$  from the set $[0,1],$ then
$P[  \text {x is irrational} ] = 1$
 A: This can be answered using measure theory. The idea is that, although dense, the measure of rationals in the interval is zero.
Informally, that is:
$$P(x\in[0,1]\backslash \mathbb{Q}) = \frac{\mu([0,1]\backslash \mathbb{Q})}{\mu ([0,1])} = \frac{\mu([0,1]) - \mu( [0,1]\cap\mathbb{Q})}{\mu ([0,1])}  = \frac{1-0}{1} = 1.$$
Even more formally, we use the Lebesgue integral, to integrate the characteristic function $$\chi_{\mathbb{Q}}(x) = \begin{cases}1, \, x\in \mathbb{Q}\\ 0 ,\, \text{else}\end{cases}$$
over the interval $[0,1]$. This is how you are to show that the measure of rationals is 0.
A: A real number can be defined as rolling a fair die with 10 faces. Write 0.209... if first time we get face 2, second time we get face 0, third time we get face 9 and so on. A rational number has periodic pattern, while irrational number has no pattern. Therefore, P[x is rational] = 0.
A: The crux is that $\mathbb Q \cap [0, 1]$ has measure 0. To show this, let $\epsilon > 0$ be arbitrary; we will show that $\mathbb Q \cap [0, 1]$ has measure no more than $\epsilon$. Since $\mathbb Q \cap [0, 1]$ is countable, suppose that $q_1, q_2, \dots$ is an enumeration of its elements. Put $q_1$ in an interval of size $\epsilon / 2$, which you can do because it's just a point; then put $q_2$ in an interval of size $\epsilon / 4$, put $q_3$ in an interval of size $\epsilon / 8$, and so forth. The union of these intervals will contain all of $\mathbb Q \cap [0, 1]$, and its measure will be at most $\sum \epsilon / 2^n = \epsilon$.
