# are there more irrational numbers than rational numbers? or are they the same?

### 1 Answer | Add Yours

The answer here is that there are in fact far, far more irrational numbers than there are rational numbers. One way to think about this is that between any two rational numbers, there are an infinite number of irrational numbers.

There is quite a bit of background knowledge required to understand the answer to this question, and I will attempt to give an overview. We must first define a few terms. We call the number of elements in a set the *cardinality *of the set. For example, the set {1, 2, 3} has cardinality 3. Two sets are said to have the same cardinality if a bijection can be formed between the sets. We can extend the notion of cardinality to infinite sets, and we say that the set of natural numbers `NN` has cardinality `\aleph_0` (aleph null).

Any set that has cardinality `\aleph_0` is said to be *countably infinite. *The set of all real numbers `RR` was famously shown by Georg Cantor to have cardinality `\aleph_1 = 2^(\aleph_0)` . (See reference link "Cantor's diagonal argument"). This number is called the *cardinality of the continuum, *and a set with this cardinality is said to be *uncountably infinite.*

It has been shown that the irrational numbers are *uncountably infinite *(they have cardinality `\aleph_1` ).

However, the rational numbers are *countably infinite* (have cardinality `\aleph_0`). To show that a set has cardinality `\aleph_0`, you must construct a bijection (one-to-one and onto function) between the set in question and the natural numbers. An bijection between the rational numbers and the natural numbers is shown in reference [3]. As such, we know that the rational numbers have cardinality `\aleph_0`.

**It follows that there are more irrational numbers than rational numbers, since there are `\aleph_0` rational numbers and `\aleph_1` irrational numbers, and `\aleph_0 < \aleph_1 = 2^(\aleph_0)` .**