Mathematical Proof help!

1)
Let n = p*q*..*r where p,q,..,r are prime numbers with p<=q<=..<=r, and n is larger than 1. Then,

n^2 = (p*p)*(q*q)*..*(r*r)

and as n^2 is a multiple of 2 by supposition, p=2 and hence

n=2*q*..*r,

i.e., n is a multiple of 2.
Conversely, if n is a multiple of 2, it is clear that n^2 is a multiple of 2.


2) If n=6, then n cannot be divided by 12.
However if n is divided by 12, then n is the form of

n = 2*2*3*p*..*r (p,..,r are prime numbers)

and hence n is divided by 6.

1) Given: n^2 is a multiple of 2
Assume n is not a multiple of 2.

n = 2x+1
n^2 = (2x+1)^2 = 4x^2 + 4x + 1 = 4(x^2 + x) + 1
4(x^2 + x) + 1 is not a multiple of 2.
The assumption was false.

oh thanks guys! I get the how to prove number 1) but I'm still wary over the 2nd question. I just cant seem to get my head round it

This is the example we were given:

Decide whether this condition is necessary, sufficient or both.

If n is a positive integer and n is divisible by 6, n is divisible by 3.

p => q but q does not imply p as n = 3 is a counter example.

So p => q therefore:

=> n = 6a
=> n = 3b where b is 2a
=> n is divisible by 3

therefore q is necessary but not sufficient for p.

For the 2nd question I posted up, how do I prove it in that format?

Substitute 12 for 6 and 6 for 3.

ohh that simple?

Would the same apply to da format if:

statement a = n is divisible by 6 and

statement b = n is divisible by 9?

a => b but b does not => a since n = 9 is a counter example

Looks like you'll have to rely on yourself cookie.

Can you think of any multiples of 6 that aren't multiples of 9?

Multiples of 6 are of the form

n = 2*3*p*..*q (p,..,q are primes)

and multiples of 9 are of the form

m = 3*3*r*..*s (r,..,s are primes)

and you can find multiples of 6 that are not multiples of 9 easily, e.g., 6, 12, 24, ..