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 a very special number, 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"
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.
(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