Question

Prove that $5-3\sqrt{2}$ is an irrational number.

Answer 1

We know that, $\sqrt{2}$ has non-terminating, non-repeating decimal expansion.

Hence it is an irrational number.

Using the properties of irrational numbers, we can say that the product of a non-zero rational number and an irrational number is always an irrational number.

Hence, $3\sqrt{2}$ is an irrational number.

Further, we know that the difference of a rational number and an irrational number is always an irrational number.

Therefore, $5-3\sqrt{2}$ is also an irrational number. $\left[\text{Hence proved}\right]$