Answer to Question #340352 in Discrete Mathematics for bkay

Question

Prove the 3√7 is irrational 

Answer 1

Let $3\sqrt7$ be a rational Number in form of $\frac p q$ where $p$ , $q$ are coprimes:

$3\sqrt7=\frac p q$

Square on both sides:

$9*7=\frac {p^2} {q^2}$

$63q^2=p^2$

Since $p$ and $q$ are coprimes we can say that 63 is a factor of $p^2$, then $p$ is also a factor of 63 because it is a rational number.

Hence $p$ can be expressed as $k$ times 63 where k is some constant:

$p=63k$

Then $63q^2=(63k)^2$

$\implies q^2=63k^2$

From here we get

63 is a factor of $q^2$ and 63 is a factor of $q$.

We got that 63 is a factor of $p$ and $q$.

As we assumed $p$ and $q$ are coprimes and it proves our assumption was wrong and nereby we can say $3\sqrt7$ is an irrational number.