# 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.

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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.

neela | High School Teacher | (Level 3) Valedictorian

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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.