Proof that odd + odd = even In math class today we started talking about proofs that odd + odd is even. We went over the basic proof (using 2k+1 and equations etc) and I realized that the only reason that this property exists is because the distance between two consecutive numbers divisible by 2 is one. So I wrote an alternate proof of why odd + odd = even and I wanted to share it here and ask if anyone has anything else to say. So my question basically is... did I get anywhere with this?

Proof that the sum of two odd numbers is an even number: 
      An even number is defined as a number that is divisible by 2. An odd number is a number that is not divisible by 2 but is divisible by 1. The reason that two odds are an even is that the difference between odd and even is only 1, and odd numbers are 1 more than even numbers. For example, we have the number 7. 7 is not divisible by 3. However, adding a random number that is not divisible by 3 either will only give you a number that is divisible by 3 half the time (because the gap between two numbers divisible by 3 is 2). Another example, the number 21. 21 is not divisible by 10, and the gap between consecutive numbers divisible by 10 is 9. That means that if a random number is added to 21 that is not divisible by 10 there is a 1/9 chance that the new number is divisible by 10. For odd numbers, the gap between two consecutive numbers divisible by 2 is only 1, which gives a 1/1 chance of creating an even number.

Thanks!
 A: It seems to me that you bring up examples within your proof in order to ensure that your point is getting across correctly.  However, while in spirit this is good, the actual proof steps should be unambiguous without the use of examples.  
Some of your terminology is also strange, such as "the difference between odd and even is only 1, and odd numbers are 1 more than even numbers".  If I understand your intentions, you're trying to say that in general an even number is of the form $2n$ for some natural number $n$, and that an odd number is of the form $2m+1$ for some natural number $m$.  While the form of an even number follows directly from the definition of being divisible by $2$, that every odd number of that form is not as immediate, and you should say something to prove this claim (Hint: Euclidean Division or Mathematical Induction!).  If I had not understood your intentions correctly, then the examples Thomas brought up would immediately show you that your proof was incorrect.
The probability arguments you bring in seem very hand-wavey, and seem to be trying to replace arguments concerning Euclidean Division, but the way in which they come off seems like you're going off on a tangent.  
In the end, it seems you are trying to get at the typical proof that goes along the lines of expressing two odd numbers as $2n+1$ and $2m+1$ (again, prove this!) and recognizing that
$$(2n+1)+(2m+1)=2(m+n)+2=2(m+n+1).$$
However, your proof does not make this immediately clear, and I would go so far as to say that in its current state that it is not a valid proof.  
A: Let b be odd and a is the even number before it c is odd and d is the even number before it so $$b+c=a+1+c+1=a+c+2$$ which is even since $even + even + even = even$
