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

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

Note by Aman Sharma
2 years, 1 month 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 · 2 years, 1 month ago

Simply by long method of finding square root...we will find that the result is neither terminating nor repeating. · 2 years, 1 month ago

Really how so? · 2 years 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. · 2 years 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 · 2 years ago

copied!!! at your own risk · 2 years ago

what will be the value of 1 · 2 years, 1 month ago