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

Note by Aman Sharma
3 years, 2 months ago

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  Easy Math Editor

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**bold** or __bold__ bold

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1. numbered
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[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} \( \boxed{123} \)

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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

Sanchet Katakol - 3 years, 2 months ago

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

Gunit Varshney - 3 years, 2 months ago

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Really how so?

Ali Caglayan - 3 years, 2 months ago

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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.

Tarek Aziz - 3 years, 2 months ago

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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

Vaibhav Sharma - 3 years, 2 months ago

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copied!!! at your own risk

Sanchet Katakol - 3 years, 2 months ago

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what will be the value of 1

Amit Majumdar - 3 years, 2 months ago

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What is 1 amit majumdar

Sanchet Katakol - 3 years, 2 months ago

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