Question

Show that $\sqrt{p}+\sqrt{q}$ is an irrational number, where the numbers p and q are primes.

Answer 1

Hint: Use the fact that if p is a prime then $\sqrt{p}$ is irrational. Consider two cases. Case I when p = q and Case II when p and q are different primes. In case I use the fact that the product of a non-zero rational number and an irrational number is irrational. In case II prove by the method of contradiction. Assume $\sqrt{p}+\sqrt{q}$ is rational. Use the fact that the square of a rational number is rational. Square $\sqrt{p}+\sqrt{q}$ and use ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and arrive at a contradiction.

Complete step-by-step answer:
Consider the following cases:
Case I: When p=q.
When p = q, we have $\sqrt{p}+\sqrt{q}=2\sqrt{p}$. Since p is prime, we have $\sqrt{p}$ is irrational.
Now let us assume $2\sqrt{p}$ is rational.
Since the division of two rational numbers is rational if the divisor is non-zero, we have
$\dfrac{2\sqrt{p}}{2}=\sqrt{p}$ is rational.
But we know that $\sqrt{p}$ is irrational. Hence our assumption is incorrect. Hence $2\sqrt{p}$ is irrational.
Hence $\sqrt{p}+\sqrt{q}$ is irrational.
Case II: When $p\ne q$
Let $\sqrt{p}+\sqrt{q}$ is rational.
Since the square of a rational number is rational, we have
${{\left( \sqrt{p}+\sqrt{q} \right)}^{2}}$ is rational
Now, we have ${{\left( \sqrt{p}+\sqrt{q} \right)}^{2}}={{\left( \sqrt{p} \right)}^{2}}+{{\left( \sqrt{q} \right)}^{2}}+2\sqrt{p}\sqrt{q}=p+q+2\sqrt{pq}$
Hence, we have
$p+q+2\sqrt{pq}$ is rational
Since the difference of two rational numbers is rational, we have
$p+q+2\sqrt{pq}-\left( p+q \right)=2\sqrt{pq}$ is rational.
Since the division of two rational numbers is rational if the divisor is non-zero, we have
$\dfrac{2\sqrt{pq}}{2}=\sqrt{pq}$ is rational.
Now since $\sqrt{pq}$ is rational, we have
$\sqrt{pq}=\dfrac{m}{n},m,n\in \mathbb{Z},n\ne 0$ and m and n are coprime.
Squaring both sides, we get
$pq=\dfrac{{{m}^{2}}}{{{n}^{2}}}$
Multiplying both sides by ${{n}^{2}},$ we get
${{n}^{2}}pq={{m}^{2}}$
Hence, we have $p$ divides ${{m}^{2}}$. We know that if p is a prime and divides ${{a}^{2}}$ then p divides a.
Hence, we have p divides m.
So let $m=pk,k\in \mathbb{Z}$
Hence, we have
${{n}^{2}}pq={{p}^{2}}{{k}^{2}}\Rightarrow {{n}^{2}}q=pk$
Since p and q are different primes, we have
p does not divide q.
Hence p divides ${{n}^{2}}$
Hence, p divides n.
Hence there is a common factor p>1 between m and n.
But m and n are coprime.
Hence our assumption is incorrect.
Hence $\sqrt{p}+\sqrt{q}$ is irrational.
In both the cases, we have $\sqrt{p}+\sqrt{q}$ is irrational.
Hence $\forall \text{ prime }p,q,\sqrt{p}+\sqrt{q}$ is irrational.

Note: [1] The questions of the above type can be easily solved by the properties of rational numbers. We initially assume the incorrect statement to be true and then arrive at a contradiction. This is the method of proof of contradiction. Most of the questions of the above type are solved using the method of contradiction.