Answer to Question #337133 in Discrete Mathematics for bkay

Question

Prove the 3√7 is irrational

Answer 1

Let us first prove that √7 is an irrational number:

let √7 be rational, then it can be represented as

√7 = p/q - irreducible fraction, where p,q are natural numbers

then

$7=\frac{p^{2}}{q^{2}} , 7q^2=p^2$

Because $7q^2$ is divisible by 7, then $p^2$ is also divisible by 7,

then p=7k, where k is a natural number.

We get

$7q^2=(7k)^2, 7q^2=49k^2, q^2=7k^2$ - so q is divisible by 7.

It turns out that p - is divisible by 7 and q - is divisible by 7, i.e. contradiction,

because p/q is an irreducible fraction. So there is no rational number that is equal to √7.

And accordingly there is no rational number that is equal to 3√7.