# Set of Rational numbers a countable set?

How can we say that rational numbers is a countable set? I can divide a rational number by infinite different number of natural numbers so shouldn't there be infinite rational numbers.

http://www.homeschoolmath.net/teaching/irrational_numbers.php

They say here in this site that the number of irrational numbers is greater than rational numbers. But aren't they both infinite in number?

• Do you know what "countable" means? It's not the same as "finite." (And, yes, the rationals and irrationals are both infinite. And there are more irrationals. There are many different types of infinity, some bigger than others!) – Akiva Weinberger Feb 19 '16 at 14:17
• @akiva Weinberg actually I don't can you explain? – N.S.JOHN Feb 19 '16 at 14:18
• Look at these three links: One, two, three. They explain it. – Akiva Weinberger Feb 19 '16 at 14:23

A set is countable if there exists an injective function, or injection, from that set, the domain, into the natural numbers, the codomain. An injection preserves uniqueness, that is, no number from the codomain is used more than once.

If no injection exists then the set is called uncountable.

For example, are the natural numbers countable? We know they are, but let's apply the definition. We have to find an injection from the natural numbers to themselves. One such injection is the identity function mapping each element of the natural numbers to itself. Since this injection exists, the natural numbers are countable.

How about the positive rational numbers? Does there exist an injection from the positive rational numbers into the natural numbers. Georg Cantor, who provided the first proofs of this, used a function called the "Cantor pairing function" that is best viewed geometrically.

Let the positive rational numbers be represented as the lattice points on the $xy$ plane with positive coordinates where the $y$ coordinate represents the numerator and the $x$ coordinate represents the denominator. Any rational number would be represented by at least one lattice point. Choose the lattice point that represents the rational number in reduced form. That is, for the rational number $\frac{2}{3}$ let the lattice point $(3,2)$ represent the rational number and skip lattice points such as $(6,4)$. This represents an injection of the rational numbers into the lattice points. Then start at the bottom left corner and associate a natural number with each lattice point proceeding in a zig-zag manner across the lattice points.

We would intuitively suspect that the positive rational numbers are bigger than the natural numbers, but since an injection exists, the rational numbers are countable just like the natural numbers. They both meet the conditions required by the definition.

Now consider the positive irrational numbers. The claim is that these are not countable. If that is true, then there does not exist any injection from the irrationals into the natural numbers. Cantor showed that by assuming such an injection existed and then he derived a contraction. His method is called "Cantor's diagonal argument".

Assume the injection exists. Each natural number is associated with a positive irrational number that can be represented as an infinite decimal expansion. Construct an irrational number by adding $1$ to the $n$th position of the decimal expansion modulo $10$ for the irrational number associated with the $n$th natural number. That irrational number is different from each irrational number in the list since it differs in the $n$th position. That means, the injection did not use all the irrational numbers as we assumed. That means it is not an injection from the positive irrational numbers to the natural numbers as assumed. That means there does not exist an injection from the irrational numbers to the natural numbers. By definition, the positive irrational numbers are uncountable.

Although I used the sets of positive rational and irrational numbers, the argument could be extended to all rational and irrational numbers.