Prove that : n! (n+2) = n! + (n+1)!

Proof:
n! + (n+1)!=
n! + (n+1)*(n)(n-1)(n-2)...=
n! + (n+1)n!=
n![1+(n+1)]=
n!(n+2)


or...

n!(n+2)=
n![(n+1)+1]=
n!*(n+1)+1*n!=
(n+1)!+n!=
n!+(n+1)!

Prove:

Prove that a^n/b^n=(a/b)^n

How to prove that 1/4 < log2 < 1/3 ?

m12: 1/4 < log2 < 1/3 

3 < 12log2 < 4 

3 < log2^12 < 4 

3 < log 4096 < 4 

10^3 < 10^log 4096 < 10^4 

1,000 < 4096 < 10,000 

:)

Why are the infinite sets: set of natural numbers, set of integers, set of odd numbers and set of even numbers equal? Prove it.

These sets are not equal. We call them equivalent (isomorphic) because they have the same number of elements. The proof is that we can find a bijection between the natural numbers and each other set. Please search for examples on your own - it's not too tough 

:)

Given: N is the midpoint of line MP, Q is the midpoint of line RP, and line PQ is approximately or equal to line NM. Prove line PN is approximately of equal to line NM

There seems to be extraneous information included (and we look at lengths of segments, not lines - be careful geometry must be precisely written). By definition of midpoint the segment MN and PN are congruent. It seems that this question is either mis-stated, or a "trick question". Please repost if you need to :)

Show that there are infinitely many positive primes.

Suppose that there are only n-prime numbers and denote the product of all of the primes as A. Next let B be the product of all of the primes plus 1. Now 1 is not a prime number by definition so for any particular prime p>1. By the fundamental theorem of algebra, our p had to be a divisor of A as well as a divisor of B. Furthermore it also must divide their difference, B-A - which is 1. Since the difference is 1, p must be a divisor of 1 and this is impossible.

Our contradiction tells us that our first statement must be false. There are infinitely many primes. 

Derive the Sum of Cubes Identity

Square Root of a product