sum of two even numbers is an even number

First of all, we have to know what an even number is.

DEFINITION

X is an even number if and only if there exists an INTEGER m such that X = 2m. (NOTE : it is easy to prove that the choice of m is unique. In other words, if there exists two integers m and n such that X = 2m and X = 2n, then m =n.)

PROOF

LET X and Y be any two even numbers. By definition, there exists integers k and l such that X = 2k and Y = 2l.

Then, as all of us know, we have X + Y = 2k + 2l = 2 (k+l). Since k and l are both integers, (k+l) is an integer too. Let P = (k+l), then (X+Y) = 2P where P is an integer. By defintion, (X+Y) is an integer.

As such, sum of any two even numbers is an even number.

<<<<< HOPE THIS IS OF ASSISTANCE TO YOU>>>>>>>

2007-07-29 12:24:45 補充：
ANSWER to QUESTION No. 2 : FALSECounter Examples : 2 (prime no.) + 3 (prime no.) = 5 (prime no.);2 (prime no.) + 5 (prime no.) = 7 (prime no.);2 (prime no.) + 11 (prime no.) = 13 (prime no.);2 (prime no.) + 17 (prime no.) = 19 (prime no.).

Q1. Sum of two even numbers is an even number?
Suppose we have two even numbers, say n1 and n2. We can express n1 = 2k1, n2 = 2k2 for some integer k1, k2. Then,
n1 + n2 = 2k1 + 2k2 = 2 (k1 + k2)
As (k1 + k2) is an integer, 2 (k1 + k2) is an even number, implying that n1 + n2 is an even number.

Q2. Sum of two prime numbers must be a composite number?
False, we can prove it by providing a counter example.
2, 3 are prime number, 2 + 3 = 5 is a prime number also.

sum of two even numbers is an even number
ture/false?

True. Reason same as above.

sum of two prime numbers must be a composite number
ture/false?

False. e.g. 2+5=7

only sum of two odd primes is composite...

True.

Examples:

2 + 2 = 4
6 + 14 = 20

True sum of two even numbers is an even number