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  • Prove that suareroot p + squareroot q is irrational, where p, q are primes.

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

Answers (1)

We will do it by method of contradiction: We will assume \sqrt{p}+\sqrt{q} is a rational number. If it leads to some absurd outcome then it is wrong assumption.
Let \sqrt{p}+\sqrt{q} is a rational number
\sqrt{p}+\sqrt{q}= \frac{a}{b},b\neq a,a,b\epsilon z
\sqrt{q}= \frac{a}{b}-\sqrt{p}
Squaring both sides
\left ( \sqrt{q} \right )^{2}=\left ( \frac{a}{b}-\sqrt{p} \right )^{2}
q= \left ( \frac{a}{b} \right )^{2}+\left ( \sqrt{p} \right )^{2}-2\left ( \frac{a}{b} \right )\left ( \sqrt{p} \right )
(Using (a – b)2 = a2 + b2 – 2ab)

\frac{a^{2}}{b^{2}}+p-q= 2\sqrt{p}\frac{a}{b}

2\sqrt{p}\frac{a}{b}= \frac{a^{2}}{b^{2}}+p-q

\sqrt{p}= \frac{b\left ( a^{2}+b^{2}\left ( p-q \right ) \right )}{2ab^{2}}

\sqrt{p}= \frac{a^{2}+b^{2}\left ( p-q \right )}{2ab}
RHS is a rational number but LHS is an irrational number.
This is not possible, hence our assumption was wrong.

Here \sqrt{p}+\sqrt{q} has to be irrational.

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