Proof underroot N is not a …

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Posted by Samarth Jain 6 years, 7 months ago
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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
2Thank You