- Trending Categories
- 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.

- Related Articles
- Prove that $\sqrt{p} + \sqrt{q}$ is irrational, where $p$ and $q$ are primes.
- Prove that for any prime positive integer $p$, $\sqrt{p}$ is an irrational number.
- If p is a prime number and q is a positive integer such that $p + q = 1696$. If p and q are co prime numbers and their LCM is 21879, then find p and q.
- $0.5 \times 0.05 \times \sqrt{q}=\sqrt{0.5 \times 0.05 \times p} ; \frac{p}{q}=?$
- Given that \( \frac{4 p+9 q}{p}=\frac{5 q}{p-q} \) and \( p \) and \( q \) are both positive. The value of $\frac{p}{q}$ is
- Prove that $\sqrt{3}+\sqrt{5}$ is an irrational number.
- Prove that is $\sqrt{2}$ an irrational number.
- Prove that $2\ +\ 5\sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.
- Given that $\sqrt{2}$ is irrational, prove that$( 5+3\sqrt{2})$ is an irrational number.
- Given that $\sqrt{2}$ is irrational, prove that $(5\ +\ 3\sqrt{2})$ is an irrational number.
- Prove that $\frac{2\ +\ \sqrt{3}}{5}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.
- If p, q are real and p≠q, then show that the roots of the equation $(p-q)x^2+5(p+q)x-2(p-q)=0$ are real and unequal.
- If $p,\ q,\ r$ are in A.P., then show that $p^2( p+r),\ q^2( r+p),\ r^2( p+q)$ are also in A.P.
- Prove that $2-3\sqrt{5}$ is an irrational number.
- Prove that $4 − 5\sqrt{2}$ is an irrational number.