Prove that $\frac{(3-4 \sqrt{2})}{7}$ is an irrational number, given that $\sqrt{2}$ is an irrational number.
Let us assume that $\frac{(3-4 \sqrt{2})}{7}$ is a rational number.
Thus, $\frac{(3-4 \sqrt{2})}{7}$ can be represented in the form of $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0, p$ and $q$ are co-prime numbers.
$\frac{3-4 \sqrt{2}}{7}=\frac{p}{q}$
$\Rightarrow 3-4 \sqrt{2}=\frac{7 p}{q}$
$\Rightarrow 4 \sqrt{2}=3-\frac{7 p}{q}$
$\Rightarrow 4 \sqrt{2}=\frac{3 q-7 p}{q}$
$\Rightarrow \sqrt{2}=\frac{3 q-7 p}{4 q}$
Since, $\frac{3 q-7 p}{4 q}$ is rational $\Rightarrow \sqrt{2}$ is rational
But, it is given that $\sqrt{2}$ is an irrational number.
Therefore, our assumption is wrong.
Hence, $\frac{3-4 \sqrt{2}}{7}$ is an irrational number.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.