Question

Given that $\sqrt 2 $ is irrational, prove that $\left( {5 + 3\sqrt 2 } \right)$ is an irrational number.

Answer 1

Hint: This type of question can be solved by contradiction method .In contradiction method, firstly we assume that the given statement is wrong and solve it. After some steps a condition occurs where we get the result that our assumption is wrong and hence the given statement proved right.

Complete step-by-step answer:
$\left( {5 + 3\sqrt 2 } \right)$ is irrational , it can be proved by applying some basic theorems of irrationality.
Theorem 1: The product of any irrational number with a rational number is irrational.
Since we know that $3$ is a rational number and given that$\sqrt 2 $ is irrational, so $3\sqrt 2 $ is irrational.
Theorem 2: The sum of a rational number with an irrational number is irrational.
Since we know that 5 is a rational number and $3\sqrt 2 $ is irrational, so $\left( {5 + 3\sqrt 2 } \right)$ is an irrational number.
Hence Proved.

Note: An another method to solve this question is described below:
First assume that the negation of given statement is true, i.e., opposite of given statement is true. Therefore, we assume that $\left( {5 + 3\sqrt 2 } \right)$ is rational.
Hence, $\left( {5 + 3\sqrt 2 } \right) = \dfrac{a}{b}$, where $a$ and $b$ are co-prime positive integers.
$ \Rightarrow 3\sqrt 2 = \dfrac{a}{b} - 5$
$ \Rightarrow 3\sqrt 2 = \dfrac{{a - 5b}}{b}$
$ \Rightarrow \sqrt 2 = \dfrac{{a - 5b}}{{3b}}$
But we know that $\sqrt 2 $ is irrational.
So, our assumption is wrong.
Hence, $\left( {5 + 3\sqrt 2 } \right)$ is an irrational number.