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.