Prove that

Question:

Prove that $(3+5 \sqrt{2})$ is an irrational number, given that $\sqrt{2}$ is an irrational number.

 

Solution:

Let us assume that $(3+5 \sqrt{2})$ is a rational number.

Thus, $(3+5 \sqrt{2})$ 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.

$3+5 \sqrt{2}=\frac{p}{q}$

$\Rightarrow 5 \sqrt{2}=\frac{p}{q}-3$

$\Rightarrow 5 \sqrt{2}=\frac{p-3 q}{q}$

$\Rightarrow \sqrt{2}=\frac{p-3 q}{5 q}$

Since, $\frac{p-3 q}{5 q}$ is rational $\Rightarrow \sqrt{2}$ is rational

But, it is given that $\sqrt{2}$ is an irrational number.

Therefore, our assumption is wrong.

Hence, $3+5 \sqrt{2}$ is an irrational number.

 

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