Question

Show that the reciprocal of an irrational number is irrational

Answer 1

Hint: Here we will assume any number as the irrational number then will find its reciprocal and its equivalent form and then will check as per our requirement whether it is a rational or irrational number.

Complete step-by-step answer:
Let us assume that the given expression $ 2\sqrt 5 + \sqrt 7 $ is irrational.
This means that- the given number can be expressed in the form of $ \dfrac{p}{q} $ where p and q belong to the integers and it can be composite or prime numbers.
So, $ 2\sqrt 5 + \sqrt 7 = \dfrac{p}{q} $
Now, the reciprocal of an above expression is –
 $ \dfrac{1}{{2\sqrt 5 + \sqrt 7 }} = \dfrac{q}{p} $
The above equation can be re-written as –
 $ \dfrac{q}{p} = \dfrac{1}{{2\sqrt 5 + \sqrt 7 }} $
Take conjugate on the right hand side of the equation –
 $ \dfrac{q}{p} = \dfrac{1}{{2\sqrt 5 + \sqrt 7 }} \times \dfrac{{2\sqrt 5 - \sqrt 7 }}{{2\sqrt 5 - \sqrt 7 }} $
Simplify the above equation using the identity $ {a^2} - {b^2} = (a - b)(a + b) $
 $ \dfrac{q}{p} = \dfrac{{2\sqrt 5 - \sqrt 7 }}{{{{\left( {2\sqrt 5 } \right)}^2} - {{\left( {\sqrt 7 } \right)}^2}}} $
Simplify the above equation –
 $ \dfrac{q}{p} = \dfrac{{2\sqrt 5 - \sqrt 7 }}{{\left( {4(5)} \right) - \left( 7 \right)}} $
 $ \Rightarrow \dfrac{q}{p} = \dfrac{{2\sqrt 5 - \sqrt 7 }}{{20 - 7}} $
Simplify the above equation –
 $ \Rightarrow \dfrac{q}{p} = \dfrac{{2\sqrt 5 - \sqrt 7 }}{{13}} $
Split the numerator –
 $ \Rightarrow \dfrac{q}{p} = \dfrac{{2\sqrt 5 }}{{13}} - \dfrac{{\sqrt 7 }}{{13}} $
Therefore, if $ 2\sqrt 5 + \sqrt 7 $ is irrational then its reciprocal is also irrational. (Proved).
So, the correct answer is “ $ \dfrac{{2\sqrt 5 }}{{13}} - \dfrac{{\sqrt 7 }}{{13}} $ ”.

Note: : Always remember that between any two given numbers there are infinite rational and irrational numbers irrespective of how small or large the difference between the two may be. In irrational numbers the decimal form is in the non-repeating and non-terminating numbers.