# 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.
– user296602
Apr 16, 2018 at 21:38
• Well, I think it fits both, but math educator site is much slower to respond... Apr 16, 2018 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. Apr 16, 2018 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. Apr 16, 2018 at 22:23
• @RobArthan That is exactly how* students think! Apr 16, 2018 at 22:24

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 $\square\square...\square\square\quad\square = \blacksquare\blacksquare...\blacksquare\blacksquare...\blacksquare\blacksquare\quad\square\square...\square\square\quad\blacksquare \square$ Apr 16, 2018 at 22:27
• Yes you are right, I've improved this part!
– user
Apr 16, 2018 at 22:37

Why is an empirical proof not acceptable?

While I agree an infinite number of examples does not constitute proof, are an infinite number of examples necessary since the one's digit can only be zero thru nine?

In other words:

• there are only ten (base-10) digits, 0-9
• 0,2,4,6,8 are even.
• 1,3,5,7,9 are odd.
• There are fifty possible additions of two of these digits – one half of the total possible sums of all digits. (11, 13, 15, 17, 19, 21, 23 ... 95,97,99.)
• Each sum of these fifty additions results in an even number.
• Therefore the sum of any two numbers, regardless of magnitude, both having an odd number as the one's digit, will result in an even number.

Is this not an adequate proof?