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
·
2 years, 4 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
·
2 years, 3 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.
–
Tarek Aziz
·
2 years, 3 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
·
2 years, 3 months ago

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TopNewestLet 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 · 2 years, 4 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 · 2 years, 3 months ago

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– Ali Caglayan · 2 years, 3 months ago

Really how so?Log in to reply

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 · 2 years, 3 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 · 2 years, 3 months ago

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– Sanchet Katakol · 2 years, 3 months ago

copied!!! at your own riskLog in to reply

what will be the value of 1 – Amit Majumdar · 2 years, 3 months ago

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– Sanchet Katakol · 2 years, 3 months ago

What is 1 amit majumdarLog in to reply