For every integer n, prove n^2-n is even. Can someone help me?
n (n-1)
if n is even, then the product is even times odd = 2 times something
if n is odd, then (n-1) is even and the product is odd time even = something*2
n(n-1)
If n is even then n-1 is odd even times odd = even (3)(4) = 12
if n is odd then n-1 is even.. odd times even = even (5)(4) = 20
This shows it is even, but a proof might want more...
You want to prove that n(n-1) is divisible by 2
Sure, I can definitely help you with that. To prove that for every integer n, n^2 - n is even, we need to show that it is divisible by 2.
First, let's rewrite the expression as n(n - 1). Now, we have two cases to consider:
Case 1: if n is an even number.
If n is even, we can write it as n = 2k, where k is an integer. Substituting this into our expression, we get (2k)(2k - 1) = 4k(k - 1). Since k(k - 1) is an integer, 4k(k - 1) is also divisible by 2, making n^2 - n even.
Case 2: if n is an odd number.
If n is odd, we can write it as n = 2k + 1, where k is an integer. Substituting this into our expression, we get (2k + 1)(2k + 1 - 1) = (2k + 1)(2k) = 4k(k + 1). Again, since k(k + 1) is an integer, 4k(k + 1) is divisible by 2, making n^2 - n even.
Since the expression n^2 - n is divisible by 2 for both even and odd integers, we have proved that n^2 - n is even for every integer n.