Why is the cardinality of irrational numbers greater than rational numbers?

Question

This was asked by blogegog on a YouTube comment (gasp!):

[Regarding Cantor's diagonal argument:]

Couldn't I just make the same statement about rational numbers and say, 'take the largest number [sic he probably meant the one with the most digits] on the chart and append to the end of it a 1 if I'm hungry, or a 2 if I'm not' to show that his list of rational numbers doesn't contain them all?

Answer 1

Well. The real numbers are uncountable. But, the real numbers are the union of the rationals and the irrationals. The rational numbers are countable. If the irrationals were also countable, then so would be the real numbers (why?). Hence, the set of irrational numbers is uncountable.

Note that countable means that there exists a bijection between the natural numbers and the set in question.

Answer 2

Another way of looking at this is via continued fractions. Note the following facts:

1. Every rational number finite continued fraction representation of the form $[a;b_1,b_2,\ldots,b_n]$ -- although every rational number has two such representations. Also, every such finite continued fraction represents a rational number.
2. Every irrational number has a unique infinite continued fraction representation of the form $[a;b_1,b_2,\ldots]$, and every such infinite continued fraction represents an irrational number in this interval.

If we restrict our attention to the interval $[0,1)$, we may assume that $a=0$ in all of these representations.

It follows easily that the family of rational numbers in $[0,1)$ has cardinality not greater than the family of all finite sequences of positive integers. Similarly, the family of all irrational numbers in $[0,1)$ has cardinality exactly equal to the family of all infinite sequences of positive integers. We have thus reduced our problem to comparing the families of finite and infinite sequences of positive integers.

It is not too difficult to show that the family of all finite sequences of positive integers is equinumerous to the family of all positive integers itself. From this and the above observation we have that $\mathbb{Q} \cap [0,1)$ is countable.

Exactly using Cantor's diagonalisation technique one can show that the family of infinite sequences of positive integers is uncountable, and therefore $[0,1) \setminus \mathbb{Q}$ is also uncountable.

Answer 3

Rationals are Countably Infinite (Proof : http://theoremoftheweek.wordpress.com/2010/02/24/theorem-18-the-rational-numbers-are-countable/) , while Irrationals ($\Bbb R-\Bbb Q$) are uncountably Infinite as Real Numbers are uncountably infinite (Proof: http://germain.its.maine.edu/~farlow/sec25.pdf)

Answer 4

A sequence of all possible rational numbers though infinite (regardless of their magnitude) has a definite unbreakable order. Between two rational numbers in such sequence, you can't squeeze in more rational numbers. But that is not the case with irrational numbers. In any given sequence of irrational numbers, you can squeeze in infinite irrational numbers between any two irrational numbers. So irrational numbers are uncountable and thus have greater cardinality beyond the countable rational numbers.