Answer

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**Hint:**Here we will first take an assumption that \[5 + \sqrt 2 \] is an rational number and then write it in the form of \[\dfrac{a}{b}\] and then solve both the sides to get the desired answer.

An irrational number is a number which non terminating as well as non recurring and cannot be expressed in the form of \[\dfrac{a}{b}\]

**Complete step-by-step answer:**

Let us assume that \[5 + \sqrt 2 \] is rational. Then, there exist coprime positive integers a and b such that

\[

5 + \sqrt 2 = \dfrac{a}{b},b \ne 0 \\

\Rightarrow a = \left( {5 + \sqrt 2 } \right)b \\

\]

Solving it further we get:-

\[

a = 5b + \sqrt 2 b \\

\Rightarrow \sqrt 2 b = a - 5b \\

\Rightarrow \sqrt 2 = \dfrac{{a - 5b}}{b} \\

\]

Now since we know that a and b are coprime and \[b \ne 0\]

Therefore the quantity \[\dfrac{{a - 5b}}{b}\] is rational.

Also it is a known fact that \[\sqrt 2 \] is an irrational number as it cannot be expressed in the form of \[\dfrac{a}{b}\] and is non-terminating as well as non-recurring number.

So, as we know that a rational number cannot be equal to an irrational number i.e,

\[{\text{irrational number}} \ne {\text{rational number}}\]

Hence, this implies

\[\sqrt 2 \ne \dfrac{{a - 5b}}{b}\]

But this is a contradiction to our assumption

Therefore, our assumption is wrong

Hence \[5 + \sqrt 2 \] is not a rational number

Therefore, \[5 + \sqrt 2 \] is an irrational number.

Hence proved.

**Note:**Students should keep in mind that the only numbers which can be expressed in the form of \[\dfrac{a}{b}\] where a and b are co-prime numbers which means they have only 1 as their common factor and \[b \ne 0\] rest all are irrationals.

Also, a rational number and an irrational number can never be equal and when a rational number is added or subtracted from an irrational number then it always gives an irrational number.

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