Hello,

Huh

But here you proved that if n is even, then n^3 is even... while you're asked to prove the converse.

Actually, it's more complicated, you proved that if is in a form 2k, with k avery specialnumber, then n is even. Hehe that was unfair

So...what you can do is to prove the contrapositive :

"if n is odd, then n^3 is odd"

Let .

So

It's your method, you keep it because it's okay !1b) is an irrational number.

My proof(by contradiction):

is a positive number such that its cube is 2.

Assume is rational.

integers P, Q such that = , fully simplified.

(Corollary: If is even, then n is even. (proven in 1a))

An integer n is even if there exists an integer k such that n = 2k.

is an integer, therefore Q is even, along with P.

Since P and Q are even, they share a common factor. Since we defined that was fully simplified, we have a contradiction.

Difference of two squares ^^Now the one I am having a hard time with is

2) Prove or disprove the following proposition: If x and y are positive integers such that x > y + 1, then is not prime.

x²-y²=(x-y)(x+y)

(note that x-y < x+y since x and y are positive integers)

The only way for it to be a prime is that x-y=1 and x+y is a prime number.

But... It is said that x > y+1; that is to say x-y>1

Got it ?

Blop.Thanks in advance

James