Question

Prove that $\sqrt p + \sqrt q $ is irrational, where p, q are primes.

Answer 1

Hint: In this question, we have been asked to prove that $\sqrt p + \sqrt q $ is irrational. In this question, we will use the method of contradiction to prove what is required. We will assume that $\sqrt p + \sqrt q $ is a rational number and we will write it equal to $\dfrac{a}{b}$. Then we will shift one term of the expression $\sqrt p + \sqrt q $ to the right. And then, square both the sides and next, shift all the terms to the other side, leaving only the root term on one side. You will notice that one side is irrational, whereas the other is rational. This is not possible. Hence, our assumption will be proved wrong.

Complete step-by-step solution:
We have to prove that $\sqrt p + \sqrt q $ is irrational. We will start this question by assuming that $\sqrt p + \sqrt q $ is rational and then we will use some properties of rational numbers to prove our assumption wrong.
Let $\sqrt p + \sqrt q $ be a rational number. If it is a rational number, it can be expressed in the form of $\dfrac{a}{b},b \ne 0$, where a and b are co-primes (every rational number can be expressed in this form).
Therefore, $\sqrt p + \sqrt q $$ = \dfrac{a}{b}$.
We will now shift $\sqrt q $ to the other side.
$ \Rightarrow \sqrt p = \dfrac{a}{b} - \sqrt q $
Making denominator same on RHS,
$ \Rightarrow \sqrt p = \dfrac{{a - b\sqrt q }}{b}$
Now, we will square both the sides.
$ \Rightarrow {\left( {\sqrt p } \right)^2} = {\left( {\dfrac{{a - b\sqrt q }}{b}} \right)^2}$
Using ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ on RHS,
$ \Rightarrow p = \dfrac{{{a^2} - 2ab\sqrt q + {b^2}q}}{{{b^2}}}$
Now, we will shift everything to LHS in such a way that only $\sqrt q $ remains on the RHS.
$ \Rightarrow - \left( {\dfrac{{p{b^2} - {b^2}q - {a^2}}}{{2ab}}} \right) = \sqrt q $
As we can observe that RHS is irrational whereas LHS is rational. This is not possible. Hence, this is a contradiction. Our assumption was wrong.

$\therefore $ $\sqrt p + \sqrt q $ is an irrational number.

Note: We have to mind that the method of contradiction is used to prove a statement right as we cannot prove it right with taking an example. Taking an example method can only be used to prove a statement wrong. This is because if one example is taken to prove something right, it might be wrong in other examples. Hence, that is why we use contradiction methods.