Question

prove that √n is not a rational number, if n is not a perfect square

Answer 1

Is this useful my dear friend

Answer 2

Howdy , your answer is here--

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


hope it help you