Question

If p and q are two distinct irrational numbers, then which of the following is always an irrational number.
A. $ \dfrac{p}{q} $
B. $ pq $
C. $ {\left( {p + q} \right)^2} $
D. $ \dfrac{{{p^2}q + qp}}{{pq}} $

Answer 1

Hint: The variables p and q are irrational numbers which means they are not rational. Irrational numbers are repeating decimals or square roots or cube roots whose result cannot be written as rational numbers. Let p be $ a + \sqrt b $ and q be $ a - \sqrt b $ (as they are distinct). So perform operations on p and q as given in the options to find out which stays as an irrational number.

Complete step-by-step answer:
We are given that p and q are two distinct irrational numbers.
We have to find from among the options which is always an irrational number.
Let p be $ a + \sqrt b $ and q be $ a - \sqrt b $
First option is $ \dfrac{p}{q} $ , so divide p by q
 $ \dfrac{p}{q} = \dfrac{{a + \sqrt b }}{{a - \sqrt b }} $
Rationalizing the denominator, we get
 $ \Rightarrow \dfrac{p}{q} = \dfrac{{\left( {a + \sqrt b } \right) \times \left( {a + \sqrt b } \right)}}{{\left( {a - \sqrt b } \right) \times \left( {a + \sqrt b } \right)}} $
 $ \Rightarrow \dfrac{p}{q} = \dfrac{{{{\left( {a + \sqrt b } \right)}^2}}}{{{a^2} - {{\left( {\sqrt b } \right)}^2}}}\left( {\because \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right) $
 $ \Rightarrow \dfrac{p}{q} = \dfrac{{{a^2} + b + 2a\sqrt b }}{{{a^2} - b}} = \dfrac{1}{{{a^2} - b}}\left( {{a^2} + b + 2a\sqrt b } \right) $
As we can see the result also has a square root which we considered as irrational. So $ \dfrac{p}{q} $ is also irrational when p and q are irrational.
Second option is $ pq $ , this gives $ pq = \left( {a + \sqrt b } \right)\left( {a - \sqrt b } \right) $
 $ \Rightarrow pq = {a^2} - {\left( {\sqrt b } \right)^2}\left( {\because \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right) $
 $ \Rightarrow pq = {a^2} - b $
As we can see, the result of $ pq $ does not have any irrational roots, so $ pq $ is rational.
Third option is $ {\left( {p + q} \right)^2} $ , this means $ {\left( {a + \sqrt b + a - \sqrt b } \right)^2} $
On cancelling the similar terms with different signs, we get $ {\left( {p + q} \right)^2} = {\left( {a + a} \right)^2} = {\left( {2a} \right)^2} = 4{a^2} $ which is clearly not irrational.
Fourth option is $ \dfrac{{{p^2}q + qp}}{{pq}} $ , this gives
 $ \Rightarrow \dfrac{{{p^2}q + qp}}{{pq}} = \dfrac{{pq\left( {p + 1} \right)}}{{pq}} = p + 1 = \left( {a + \sqrt b } \right) + 1 = a + 1 + \sqrt b $
As we can see in $ p + 1 $ , p is irrational and 1 is rational. But we know that the sum of a rational and irrational number is always irrational. So $ \dfrac{{{p^2}q + qp}}{{pq}} $ is always irrational.
Hence, the correct option is Option D, because Option A can be rational sometimes when ‘a’ is also taken as a root but Option D is always irrational irrespective of a and b values.
So, the correct answer is “Option d”.

Note: Do not confuse irrational numbers with rational numbers. As irrational numbers can also be written as fractions like rational numbers for example $ \pi $ can also be written as a fraction $ \dfrac{{22}}{7} $ . But its decimal value is non-terminating. So all the non-terminating, non-repeating decimals are irrational