Show that
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.

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