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# Prove that $5 + \sqrt 2$ is an irrational number.  Verified
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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 $\dfrac{a}{b}$

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.