# 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.

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Let n = 1, 2, 3, 4, 5,6, ......

==> n^2 = 1, 4, **9**, 16, 25, **36**, 49, 64, **81**, 100, 121, **144**........

But we know that n^2 is a multiply of 3:

==> n^2 = 9, 36, 81, 144, ........

let n^2 forms a series ak such that:

a1= 1*9 = 9

a2= 4*9 = 2^2 *9 = 36

a3= 9*9 = 3^2 *9= 81

a4= 16*9 = 4^2 *9 = 144

a5 = 25*9 = 5^2 *9 = 225

==> aK = (k^2)*9 k = (1,2,3,....)

Then n^2 could be written as the formula :

n^2 = (k^2)*9 (k= 1,2,....)

Now let us take square root for both sides:

==> n = k*3

Then n is a multiple of 3.

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.