# How do you prove that sqrt(21) is irrational?

Sep 27, 2017

See below.

#### Explanation:

The $\sqrt{21}$. when evaluated produces a number that has a non terminating and non repeating decimal part. Numbers that are rational can all be written in the form:

$\frac{a}{b}$

$b \ne 0$

Where $a$ and $b$ are integers.

Non terminating decimal number that have repeating digits can be written in this form so are called rational numbers.

Examples:

$0.3 \overline{3}$ can be written $\frac{1}{3}$

$0.78 \overline{78}$ can be written $\frac{26}{33}$

$5$ can be written $\frac{5}{1}$

and so on.

Because the $\sqrt{21}$ and many many other square roots produce non repeating digits we can't represent them in this way.

This is the square root of 21 to 49 .d.p. as you can see there is no repeating pattern of digits. 4.5825756949558400065880471937280084889844565767680

The proof that values like this can't be written as $\frac{a}{b}$ was found by the Greek mathematician Pythagoras, and can be found in text books or online.

Sep 27, 2017

Here's a sketch of a proof...

#### Explanation:

For a formal proof you really need the unique factorisation theorem, but here's a sketch of a proof by contradiction.

$\textcolor{g r e e n}{\text{Suppose " sqrt(21) " is rational.}}$

Then there are positive integers $p , q$ such that $\frac{p}{q} = \sqrt{21}$

Without loss of generality, suppose $p > q > 0$ are the smallest such pair of integers.

Given:

$\frac{p}{q} = \sqrt{21}$

Square both sides to get:

${p}^{2} / {q}^{2} = 21$

Multiply both sides by ${q}^{2}$ to get:

${p}^{2} = 21 {q}^{2}$

Note that the right hand side is divisible by the prime numbers $3$ and $7$.

So ${p}^{2}$ is divisible by both $3$ and $7$.

Hence $p$ is divisible by both $3$ and $7$ and hence by $3 \cdot 7 = 21$.

So $p = 21 k$ for some positive integer $k$.

Then:

$21 {q}^{2} = {p}^{2} = {\left(21 k\right)}^{2} = 21 \left(21 {k}^{2}\right)$

Dividing both ends by $21$ we get:

${q}^{2} = 21 {k}^{2}$

and hence:

$\frac{q}{k} = \sqrt{21}$

Note that $p > q > k > 0$.

So $q > k > 0$ is a smaller pair of positive integers with quotient $\sqrt{21}$ contradicting our assertion that $p > q > 0$ was the smallest such pair.

So there is no such pair of positive integers.