3 Answer each of the following a Disprove If n2 is a multipl
(3) Answer each of the following:
(a) Disprove: If n2 is a multiple of 9, then n is a multiple of 9. [5 marks] (b) Prove: If n2 is a multiple of 3, then n is a multiple of 3. [5 marks]
(Hint: RecallthatifaZisnotamultipleof3,thena=3b+1ora=3b+2for some b Z.)
Solution
(a) Lets take one example, take n=6, 36 is multiple of 9, but 6 is not a multiple of 9. hence proved.
(b)
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.
OR
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.
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