![Trending Articles on Technical and Non Technical topics](/images/trending_categories.jpeg)
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
If $p$, $q$ are prime positive integers, prove that $\sqrt{p} + \sqrt{q}$ is an irrational number.
Given: $p$, $q$ are prime positive integers.
To prove: Here we have to provethat $\sqrt{p}\ +\ \sqrt{q}$ is an irrational number.
Solution:
Let us assume, to the contrary, that $\sqrt{p}\ +\ \sqrt{q}$ is rational.
So, we can find integers $a$ and $b$ ($≠$ 0) such that $\sqrt{p}\ +\ \sqrt{q}\ =\ \frac{a}{b}$.
Where $a$ and $b$ are co-prime.
Now,
$\sqrt{p}\ +\ \sqrt{q}\ =\ \frac{a}{b}$
$\sqrt{p}\ =\ \frac{a}{b}\ -\ \sqrt{q}$
Squaring both sides:
$(\sqrt{p})^2\ =\ (\frac{a}{b}\ -\ \sqrt{q})^2$
$p\ =\ (\frac{a}{b})^2\ +\ (\sqrt{q})^2\ -\ 2(\frac{a}{b})(\sqrt{q})$
$p\ =\ \frac{a^2}{b^2}\ +\ q\ -\ 2(\frac{a}{b})(\sqrt{q})$
$2(\frac{a}{b})(\sqrt{q})\ =\ \frac{a^2}{b^2}\ +\ q\ -\ p$
$\sqrt{q}\ =\ (\frac{b}{2a})(\frac{a^2}{b^2}\ +\ q\ -\ p)$
Here, $(\frac{b}{2a})(\frac{a^2}{b^2}\ +\ q\ -\ p)$ is a rational number but $\sqrt{2}$ is irrational number.
But, Irrational number $≠$ Rational number.
This contradiction has arisen because of our incorrect assumption that $\sqrt{p}\ +\ \sqrt{q}$ is rational.