Prove that the sum of two odd numbers is an even number. How can I prove this? Should I take $x$ and $x+2$ or not ? I am confused. 
 A: Any even number has the form $2n$. (Why? No matter what you make $n$ to be, $2n$ will, be divisible by $2$.).  
Any odd number has the form $2n+1$. (Why? Play with this by plugging numbers into $n$.). 
So, add two odd numbers:  
$$(2n+1)+(2n+1)=4n+2=2(2n+1)$$  
Is your result always divisible by $2$? Why or why not?  
Would you be able to reproduce the above with understanding?
A: Other option: modular arithmetic,

even number $\pmod 2 \equiv 0$ and
odd number $\pmod 2 \equiv 1$, then

(odd+odd) $\pmod 2 \equiv \ ?$
Basically you can continue from there:
((odd $\pmod 2$) + (odd $\pmod 2$)) $\pmod 2$ $\equiv \ ?$
A: Suppose there is greatest even integer N
Then
For every even integer n, N ≥ n.
Now suppose M = N + 2. Then, M is an even integer. [Because it is a sum of even integers.] Also, M > N [since M = N + 2]. Therefore, M is an integer that is greater than the greatest integer. This contradicts the supposition that N ≥ n for every even integer n. [Hence, the supposition is false and the statement is true.]
A: Hint : With your definition of odd numbers : "All numbers that ends with 1, 3, 5, 7, or 9 are odd numbers." (consequently, even numbers are the numbers that end with 0,2,4,6 or 8). Take two odd numbers, what are the possible ends for this sum?
A: Using your approach, let $x$ be odd, and consider the other odd number as $x+2k$. Then the sum is $x+(x+2k)=2x+2k=2(x+k)$, which is even.
A: Let m and n be odd integers. Then, m and n can be expressed as 2r + 1 and 2s + 1 respectively, where r and s are integers. This only means that any odd number can be written as the sum of some even integer and one.
when substituting lets have m + n = (2r + 1) + 2s + 1 = 2r + 2s + 2.
