Question

Prove that $6\sqrt{3}$ is irrational.
 

Hint:

The product of a nonzero rational number and an irrational number is always irrational.

Answer 1

Step 1: Assume the contrary.
Suppose $6\sqrt{3}$ is rational. Then it can be expressed as \[ 6\sqrt{3} = \frac{m}{n}, \] where $m,n \in \mathbb{Z},\ n \neq 0,\ \gcd(m,n)=1$.

Step 2: Divide by the rational number 6.
Since 6 is a nonzero rational, we get \[ \sqrt{3} = \frac{m}{6n}. \] This implies that $\sqrt{3}$ is rational.

Step 3: Contradiction.
It is already known that $\sqrt{3}$ is irrational. Hence, our assumption is false.
Conclusion:
Therefore, $6\sqrt{3}$ is irrational.