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Hint- Here, we will assume the opposite and will prove that the assumption contradicts.

Let us suppose that $3\sqrt 2 $ is a rational number.

As we know that any rational number can be represented in the form of $\dfrac{a}{b}$ where $a$ and $b$ are two co-prime positive integers.

$\therefore 3\sqrt 2 = \dfrac{a}{b} \Rightarrow \sqrt 2 = \dfrac{a}{{3b}}{\text{ }} \to {\text{(1)}}$

Now if we observe the RHS of equation (1) carefully, we can say it is always a rational number because $a$ and $b$ are two co-prime positive integers and 3 is also an integer.

Also, the LHS of the equation (1) i.e., $\sqrt 2 $ is an irrational number.

Therefore, equation (1) is contradicting since LHS is irrational and RHS is rational.

So, our assumption is not correct.

Hence, $3\sqrt 2 $ is irrational.

Note- In these types of problems, the number which needs to be proved irrational is assumed as rational and then equated to the general form of any rational number and if this equation holds true then the given number is rational else irrational.

Let us suppose that $3\sqrt 2 $ is a rational number.

As we know that any rational number can be represented in the form of $\dfrac{a}{b}$ where $a$ and $b$ are two co-prime positive integers.

$\therefore 3\sqrt 2 = \dfrac{a}{b} \Rightarrow \sqrt 2 = \dfrac{a}{{3b}}{\text{ }} \to {\text{(1)}}$

Now if we observe the RHS of equation (1) carefully, we can say it is always a rational number because $a$ and $b$ are two co-prime positive integers and 3 is also an integer.

Also, the LHS of the equation (1) i.e., $\sqrt 2 $ is an irrational number.

Therefore, equation (1) is contradicting since LHS is irrational and RHS is rational.

So, our assumption is not correct.

Hence, $3\sqrt 2 $ is irrational.

Note- In these types of problems, the number which needs to be proved irrational is assumed as rational and then equated to the general form of any rational number and if this equation holds true then the given number is rational else irrational.

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