- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

# Prove that the following are irrationals.

**(i)** $ \frac{1}{\sqrt{2}} $

**(ii)** $ 7 \sqrt{5} $

**(iii)** $ 6+\sqrt{2} $.

To prove:

Here we have to prove that, the given numbers are irrationals.

Solution:

(i) $\mathbf{\frac{1}{\sqrt{2}}}$

Let us assume that $\frac{1}{\sqrt{2}}$ is a rational number.

So, $\frac{1}{\sqrt{2}}$ can be written in the form $\frac{a}{b}$, where a and b are co prime and b is not equal to 0.

Hence,

$\frac{1}{\sqrt{2}}\ =\ \frac{a}{b}$

$\frac{b}{a}\ =\ \sqrt{2}$

Here, $\frac{b}{a}$ is rational but $\sqrt{2}$ is irrational.

Rational can't be equal to Irrational.

This contradicts our assumption, that $\frac{1}{\sqrt{2}}$ is a rational number.

Therefore, $\frac{1}{\sqrt{2}}$ is irrational number.

(ii) $\mathbf{7\sqrt{5}}$

Let us assume that $7\sqrt{5}$ is a rational number.

Hence, $7\sqrt{5}$ can be written in the form of $\frac{a}{b}$, where a and b are co prime and b is not equal to 0.

$7\sqrt{5}\ =\ \frac{a}{b}$

$\sqrt{5}\ =\ \frac{a}{7b}$

Here, $\sqrt{5}$ is irrational but, $\frac{a}{7b}$ is rational.

Rational can't be equal to Irrational.

This contradicts our assumption, that the number $7\sqrt{5}$ is a rational number.

Therefore, $7\sqrt{5}$ is irrational number.

(iii) $6\ +\ \sqrt{2}$

Let us assume, to the contrary, that $6\ +\ \sqrt{2}$ is rational.

So, we can find integers a and b ($≠$ 0) such that $6\ +\ \sqrt{2}\ =\ \frac{a}{b}$.

Where a and b are co-prime.

Now,

$6\ +\ \sqrt{2}\ =\ \frac{a}{b}$

$\sqrt{2}\ =\ \frac{a}{b}\ -\ 6$

$\sqrt{2}\ =\ \frac{a\ -\ 6b}{b}$

Here, $\frac{a\ -\ 6b}{b}$ 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 $6\ +\ \sqrt{2}$ is rational.

So, this proves that $6\ +\ \sqrt{2}$ is an irrational number.

- Related Questions & Answers
- PHP sqrt() Function
- sqrt() function in PHP
- Sqrt(x) in Python
- Java sqrt() method with Examples
- sqrt, sqrtl and sqrtf in C++
- Sqrt, sqrtl, and sqrtf in C++ programming
- sqrt ( ) function for complex number in C++
- Sum of series 1^2 + 3^2 + 5^2 + . . . + (2*n – 1)^2
- Sum of series 1^2 + 3^2 + 5^2 + . . . + (2*n - 1)^2 in C++
- Find sum of the series 1-2+3-4+5-6+7....in C++
- Sum of the Series 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ... in C++
- Use static Import for sqrt() and pow() methods in class Math in Java
- Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + + (1+3+5+7+....+(2n-1)) in C++
- Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + ...... + (1+3+5+7+...+(2n-1)) in C++
- C++ program to find the sum of the series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + … + (n*n)
- Sum of the series 1 / 1 + (1 + 2) / (1 * 2) + (1 + 2 + 3) / (1 * 2 * 3) + … + upto n terms in C++