Courses for Kids
Free study material
Offline Centres
Store Icon

Prove that \[5 + \sqrt 2 \] is an irrational number.

Last updated date: 20th Jun 2024
Total views: 415.2k
Views today: 10.15k
415.2k+ views
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.