So far we have seen, among others, integers, *fractions* and *roots*. We have seen that these are all numbers. Numbers can be sorted into two categories, namely the rational numbers, such as #4#, #-1# and #\tfrac{1}{2}#, and irrational numbers, such as #\sqrt{2}# and #\pi#.

The **rational numbers** are all numbers that can be written as fractions of integers.

When we convert a rational number to a decimal number, we find either:

- a finite number of decimals (this number might be #0#)
- an infinite number of decimals, but we do find a pattern: some of the decimals repeat themselves.

We denote this with a line above the recurring numbers.

**Examples**

\[\begin{array}{rcl}\dfrac{2}{1}&=&2 \\ \\ \dfrac{1}{2}&=&0.5 \\ \\ \dfrac{1}{13}&=&0.\overline{076923} \\ \\ \dfrac{1}{3}&=&0.\overline{3}\end{array}\]

The **irrational numbers** are all numbers that have an infinite non-repeating decimal expansion.

This means that the number can be written as a decimal number with infinitely many decimals, that will not repeat themselves.

**Examples**

\[\begin{array}{rcl}\sqrt{2}&=&1.414213562373095... \\ \\ \pi&=& 3.1415926543589793... \end{array}\]

We prove that #\sqrt{2}# is irrational. We use a proof by contradiction. That means we assume #\sqrt{2}# is rational, then we will find a contradiction, and then we conclude that it cannot be true that #\sqrt{2}# is rational.

If we assume that #\sqrt{2}# is rational, we can write it as a fraction #\tfrac{\blue p}{\orange q}# for which #\blue p# and #\orange q# are integers. The integers #\blue p# and #\orange q# have no common divisors because if they did, we could further simplify the fraction.

Therefore:

\[\sqrt{2}=\frac{\blue p}{\orange q}\]

When squaring both sides, we find:

\[2=\frac{\blue p^2}{\orange q^2}\]

By multiplying both sides by #\orange q^2#, we find:

\[2\orange q^2=\blue p^2\]

The left-hand side is divisible by #2#, so the right-hand side should be too. Therefore, #\blue p^2# is divisible by #2# and therefore, so is #\blue p#.

Next, we note that if #\blue p# is divisible by #2#, we can write #\blue{p}# as #2 \cdot a#, for which #a# is an integer. This means:

\[2 \orange q^2=(2a)^2=4a^2\]

If we divide both sides by #2# we find:

\[\orange q^2=2a^2\]

This means that #\orange q^2# is divisible by #2# and therefore #\orange q# is also divisible by #2#.

So we see that both #\blue p# and #\orange q# are divisible by #2#. This contradicts the fact that #\blue{p}# and #\orange{q}# have no common divisors. Therefore, our assumption was wrong and we cannot write #\sqrt{2}# as a fraction. #\sqrt{2}# is therefore irrational.

Finally, we note that together, we call the rational and irrational numbers *real numbers*.

The **real** **numbers** are the rational and irrational numbers together.

These are all the numbers that are on the number line.

Is the number #\sqrt{6} # a rational or an irrational number?

Rational numbers can be written as a fraction. This means that the numbers have a finite number of decimals, or an infinite number of decimals but with a repeating pattern in these decimals. Irrational numbers cannot be written as a fraction and have an infinite number of decimals that do not repeat themselves.

In this case #\sqrt{6}# cannot be written as a fraction, so it is irrational .