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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!

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    $\begingroup$ Is it really true that "the distance between two consecutive numbers divisible by 2 is one?" For example, $2$ and $4$ are two cnsecutive numbers divisible by $2$.... $\endgroup$
    – Thomas Andrews
    Commented Aug 29, 2014 at 0:45
  • $\begingroup$ Is the difference between and odd $7$ and even $4$ "only one?" You need to strive for more precision of language - statements you have made above, such as this one, are simply wrong. $\endgroup$
    – Thomas Andrews
    Commented Aug 29, 2014 at 0:47
  • $\begingroup$ Ah, you are using "distance" to mean the number of integers strictly between them? That's a very strange usage of the term. $\endgroup$
    – Thomas Andrews
    Commented Aug 29, 2014 at 0:48

2 Answers 2

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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.

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  • $\begingroup$ You can also prove odds are of the form $2n+1$ by induction - showing that every integer is of the form $2n$ or $2n+1$. (Induction only covers the non-negative cases, but the negative cases follow directly.) $\endgroup$
    – Thomas Andrews
    Commented Aug 29, 2014 at 0:52
  • $\begingroup$ @ThomasAndrews That is true. Personally I like the Euclidean Division argument simply because it matches my own intuition of why the claim is true. I'll add that as an alternative hint though, thanks! $\endgroup$
    – Hayden
    Commented Aug 29, 2014 at 0:53
  • $\begingroup$ Yeah, but this exercise often comes up before the students have learned Euclidean division, and applying division requires you to prove the only integers $n$ with $0\leq n<2$ are $0$ and $1$, which is "obvious," but actually requires proof. $\endgroup$
    – Thomas Andrews
    Commented Aug 29, 2014 at 0:55
  • $\begingroup$ Thanks for all the comments and suggestions. By the way this is the first paragraph proof i've written so the feedback is really helpful. I'm in high school right now so my math understanding goes up through Algebra II. Euclidean Division has not been covered yet but from what i've seen it seems similar to what I was trying to say. $\endgroup$
    – thequantumguy01
    Commented Aug 29, 2014 at 0:57
  • $\begingroup$ @user2520995 Glad to help, and it's great you're trying to improve. Best of luck $\endgroup$
    – Hayden
    Commented Aug 29, 2014 at 1:03
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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 $even + even + even = even$

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