Use contraposition to prove that if n^2 is a multiple of 3, then n is a multiple of 3.Please show the solution in detail.

Let p and q  be
two statements.

Then  the contrapositive p--> q is 
notq-->notp.

Thus  the if  n^2 is a mutiple of 3, 
then the contrapositive statement is:

If a number  is  not
a multiple of 3, then n^2 is (also) not a multiple of
3.

Proof

We take a number  n
such that n is not divissible by 3 and n is a number which gives remainder if divided by
3.

We know that any whole number can be of the for 3x,3x+1
and 3x+2, where x is 0,1,2,3,4,.....

So let n = 3x+1 or n =
3x+2, where x=0,1,2...

Then n^2 =  (3x+1)^2 = 9n^2+2*3x+1
.

Therefore  n^2 divided by 3 = [(3x+1)^2]/3 = (9^2+6x+1)/3
= (3x^2+2x)+ 1/3 Or 3x+2x is quotient and 1 is remainder. So if n is not a multiple of
3, then n^2 is not a multiple of 3.

Now let us take a
number n  of the type = 3x+2 , x  = 0,1,2,3..., the set of numbers which give a
remainder 2 when divided by 3.

Then n^2=(3x+2)^2 =
9x^2+2*3*2x+2^2 = 9x^2+12x+4

Therefore n^2 divided by 3 =
[(3x+2)^2]/3 = (9x^2+12x+4) = 3x^2+4x+4/3 = 3x^2+4x+1+1/3 = (3x^2+4x+1) quotient  and 1
remainder.

So if n is not a  multiple of 3, then n ^2 is
not a multiple of 3.

Which is of the form  not p-->
not q.