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 .
Irrational numbers
