# measure of irrational number

We want to show that if we randomly select a number $$x$$ from the set $$[0,1],$$ then

$$P[ \text {x is irrational} ] = 1$$

• I know it's true because there are a countable infinity of rational numbers compared to the uncountable infinity of the irrational numbers, but I can't write it down neatly Commented Dec 2, 2015 at 16:09
• Im not sure but probably is ok cause cardinality of rationals and irrational sets.
– user173262
Commented Dec 2, 2015 at 16:09
• It's not that they are ALL irrational--there are infinitely many rationals there. But the "infinitely many" for irrationals is larger than the "infinitely many" for rationals. This is perhaps difficult to understand intuitively because infinite cardinals don't behave in exactly the same way as the finite cardinals (i.e., natural numbers) do.
– MPW
Commented Dec 2, 2015 at 16:11
• It does make sense to me intuitively. I am just looking for a well-written proof for it :) Commented Dec 2, 2015 at 16:12
• but if you pick $x$ from $[0,2]$ then the probability that it is irrational is $200\%$ :) Commented Dec 2, 2015 at 20:41

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.

• But how to show that $\mu(\Bbb Q)=0$? I think you've left the most important part out of your answer. Commented Dec 2, 2015 at 16:16
• There is a particular theorem involving the measure of a countable set inside of an uncountable set. Actually perhaps you would like to take a look at the answer to this post... math.stackexchange.com/questions/508217/… Commented Dec 2, 2015 at 16:20

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.

• Could you give more details in your answer? Commented Oct 23, 2017 at 13:39
• You should show why the set of eventually periodic sequences using digits $0..9$ has probability zero. Commented Oct 23, 2017 at 13:58
• Rewrite in more details. Question: If we randomly select a real number x from [0, 1], then P[ x is rational ] = 0. An intuitive explanation without measure theory. A real number in [0, 1] can be defined as rolling a fair die with 10 faces: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Write 0.209... if first rolling shows face 2, second rolling shows face 0, third rolling shows face 9 and so on. A rational number has periodic pattern, like 0.36207362073620736207.... While a real number has no pattern. Therefore, P[ x is rational ] = 0. Commented Oct 23, 2017 at 14:58

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 $\sum \epsilon / 2^n = \epsilon$.