# prove it!!!!!!!!!!!!!!!!!!!!(3)

Prove that $\sqrt{2}$ is irrational number

Note by Aman Sharma
3 years, 10 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Let us assume √2 is a rational number Then it can expressed in the form of p/q where p and q are integers and q is not equal to 0 √2=p/q Cancelling common factors √2=a/b where a and b are co primes Squaring on both sides 2=a square/ b square 2/b square= a square. .... (1) 2 divides a square Therefore 2 divides a a=2c for some integer c Putting a = 2c in (1) We get 2b square = 4c square 2 divides b square 2 divides b Therefore a and b have at least 2 as their common factors Therefore a and b are not co prime So, this contradiction has arisen due to our wrong assumption √2 is irrational

- 3 years, 10 months ago

Simply by long method of finding square root...we will find that the result is neither terminating nor repeating.

- 3 years, 9 months ago

Really how so?

- 3 years, 9 months ago

Firstly we have to know the definition of rational number. If we can write a number in the form p/ q, where p& q are integer and q>0. Then we can assure that is an rational number . Otherwise it is an iretional number.

- 3 years, 9 months ago

Let us assume √2 is a rational number Then it can expressed in the form of p/q where p and q are integers and q is not equal to 0 √2=p/q Cancelling common factors √2=a/b where a and b are co primes Squaring on both sides 2=a square/ b square 2/b square= a square. .... (1) 2 divides a square Therefore 2 divides a a=2c for some integer c Putting a = 2c in (1) We get 2b square = 4c square 2 divides b square 2 divides b Therefore a and b have at least 2 as their common factors Therefore a and b are not co prime So, this contradiction has arisen due to our wrong assumption √2 is irrational

- 3 years, 9 months ago

- 3 years, 9 months ago

what will be the value of 1

- 3 years, 10 months ago

What is 1 amit majumdar

- 3 years, 10 months ago