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

##### 2 Answers

See below.

#### Explanation:

The

Where

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

Examples:

and so on.

Because the

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

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.

Then there are positive integers

Without loss of generality, suppose

Given:

#p/q = sqrt(21)#

Square both sides to get:

#p^2/q^2 = 21#

Multiply both sides by

#p^2 = 21q^2#

Note that the right hand side is divisible by the prime numbers

So

Hence

So

Then:

#21q^2 = p^2 = (21k)^2 = 21(21k^2)#

Dividing both ends by

#q^2 = 21k^2#

and hence:

#q/k = sqrt(21)#

Note that

So

So there is no such pair of positive integers.