# The sum of two odd numbers is even

I know formal ways of proving this, but I am doing research on how middle grade students create arguments to support conjectures. I would like to anticipate many more solutions before I enact the task. Can anyone help me find ways that middle grade students might approach proving this?

I have come to two approaches thus far from how they might approach it:

1. Formal: Let k, m be any integer, (2k+1)+(2m+1) and so on ....

2. An odd number is equal to even+1. So, we have (even+1)+(even+1). We know two even sum to an even number and 1+1 sums to two which is even. Again, the sum of two even numbers is even. Therefore, two odds add to an even.

The last approach assumes that even+even=even is accepted by the mathematical community which is a conversation we will have. Of course, students will come up with empirical arguments, but I'm looking for other ways students might try to justify this.

• This seems like it would be a far better fit on the mathematics educators stack exchange site. – T. Bongers Apr 16 '18 at 21:38
• Well, I think it fits both, but math educator site is much slower to respond... – MathGuy Apr 16 '18 at 21:40
• Even numbers have decimal expansions that end in 0, 2, 4, 6 or 8 and odd numbers have decimal expansions that end in 1, 3, 5, 7 or 9. So as $1 + 1 = 2$, $1 + 3 = 4$, etc. the sum of two odd numbers will have a decimal expansion that ends in 0, 2, 4, 6, or 8. – Rob Arthan Apr 16 '18 at 21:44
• I personally find $even + even = even$ less of an assumption then $odd = 2m + 1$... Actually these are the same thing. $2k + 1 = (2k) + 1=even + 1$ and $2m + 1 = (2m) + 1 = even +1$ and $even + even = 2k + 2m = 2(k+m)$ is justification. – fleablood Apr 16 '18 at 22:23
• @RobArthan That is exactly how* students think! – fleablood Apr 16 '18 at 22:24

## 1 Answer

As simple way to explain could be made by the naive interpretation of integers as collections of countable objects.

Indeed we said that an amount of object is odd if we can arrange them by pairs with the exclusion of one object, thus if we consider two distinct set of objects (black and withe box)

$\blacksquare\blacksquare...\blacksquare\blacksquare...\blacksquare\blacksquare\quad\blacksquare$

$\square\square...\square\square\quad\square$

we can rearrange all the objects in pairs and thus they are an even number

$\blacksquare\blacksquare...\blacksquare\blacksquare...\blacksquare\blacksquare\quad\square\square...\square\square\quad\blacksquare \square$

• Unfortunately that image looks like the two odd numbers are equal to each other. Maybe you should have done some thing like:$\blacksquare\blacksquare...\blacksquare\blacksquare...\blacksquare\blacksquare\quad\blacksquare$ and $\square\square...\square\square\quad\square = \blacksquare\blacksquare...\blacksquare\blacksquare...\blacksquare\blacksquare\quad\square\square...\square\square\quad\blacksquare \square$ – fleablood Apr 16 '18 at 22:27
• Yes you are right, I've improved this part! – gimusi Apr 16 '18 at 22:37