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Proof underroot N is not a …

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Proof underroot N is not a rational number if 'n' the s not a perfect square
  • 1 answers

Gaurav Seth 6 years, 7 months ago

Let , us assume that √n is rational .

so, √n = a/b where a and b are integers and b is not equation to zero .

Let, a/b are co- prime 
taking square both side ,we get 

=> n = a^2/b^2
=> nb^2 = a^2 ......(1)
so, n divide a^2 
it means n also divide a 
for some integer c
a = nc 

now squaring both side 
a^2 = n^2c^2
=> nb^2 = n^2c^2 [ from (1) ]
=> b^2 = nc^2

so , n divide b^2 

it means b also divide b

so, a and b have n as a prime factor 

but this contradict the fact that a and b are co- prime .

therefore , our assumption is wrong .
hence, √n is irrational 
 

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