Question:
Show that $(4+3 \sqrt{2})$ is irrational.
Solution:
Let (4 + 3√2) be a rational number.
Then both $(4+3 \sqrt{2})$ and 4 are rational.
$\Rightarrow(4+3 \sqrt{2}-4)=3 \sqrt{2}=$ rational $[\because$ Difference of two rational numbers is rational]
$\Rightarrow 3 \sqrt{2}$ is rational.
$\Rightarrow \frac{1}{3}(3 \sqrt{2})$ is rational. $\quad[\because$ Product of two rational numbers is rational]
$\Rightarrow \sqrt{2}$ is rational.
This contradicts the fact that $\sqrt{2}$ is irrational (when 2 is prime, $\sqrt{2}$ is irrational).
Hence, $(4+3 \sqrt{2})$ is irrational.