# I did a proof for 6th graders -- and they understood it

Let me preface this by saying that I completely understand that children don’t develop true abstraction skills until late teens/early 20s — and that some never develop a full ability for such.

That being said, guided prompts through a proof seem to work.

I have been substitute-teaching in a school district lately (I have loan payments coming up since I am post-graduation, and private tutoring, plus app-revenue doesn’t cover living expenses plus my loans), and I got a job for teaching 6th grade math for the day. There were a total of 5 classes of regular math and one advanced class (since I had to teach Reading and ELA in the morning, too). Let me say that I’m not clueless — I have (essentially) a psych degree completed (only missing the keystone thesis — I took a loooooooot of psych classes because I was always interested in computational brain simulations), as well as 5+ years as a TA and 7+ years as a private tutor. I’ve spent more than my fair share of time in front of a class and I’ve got an inkling of what works.

I know this anecdotal evidence, but the following worked for all 5 classes:

The students were working on ratios: the lesson for the non-advanced math students was talking about how equivalent ratios have the same value. That is to say, if A:B and C:D are equivalent ratios, then A/B = C/D. The lesson started with a reminder of how to find the values of a ratio: If A:B is a ratio, its value is A/B, and then you reduce to lowest terms.

Next, they were asked to find the equivalent ratios of the group: 1:2, 5:10, 6:16, 12:32. Then, I had them explain why each was equivalent, and how they knew; the key was to make sure they knew that there was a single factor you could pull out or multiply both of the numbers of one of the ratios to get the other. Then, they found the values of each of the 4, and were asked to tell me what they noticed (that the ultimate values of the fractions were the same).

Then, they were presented with a theorem: Equivalent ratios have equivalent values. I asked them to try and find a counter-example, specifically mentioning that a counter example would be equivalent ratios with nonequal values. Every time a student thought they had an example, I had them say it out loud. At first, I pointed out why they were wrong: they had the same values, or the ratios weren’t equal etc. After a few, I had some of their peers point out why they were wrong.

After 10 minutes, I decided to say that, there would never be such a thing. I explained what a theorem is (since it says “Theorem 2” on their sheets), in that it is a mathematical statement which we have proved to always be true (within ZFC, but you don’t mention that to a kid >.>) and will never be wrong. I said, as such, there will never be any counterexamples. A few kids looked straight-up confused at the statement, and some just looked surprised. Winging it, I decided to prove it.

The proof itself is simple:

Let us have 2 ratios, A:B and C:D, where B, D != 0. Then, suppose they are equal. Then we know A:B = C:D. Now, since they are equivalent, there is some factor such that A:B = xC:xD = x(C:D). Then, A:B = A/B and C:D = C/D, so xC:xD = xC/xD. Then:

A/B = xC/xD

A/B = (x/x)(C/D)

A/B = 1*(C/D) = C/D

This is what I ended up writing on the board. But, I didn’t write it out and present it like you see in an intro to analysis course or such. What happened was more like:

*To class*: Ok, let’s suppose we have 2 ratios: A:B and C:D. Ok? Everyone with me? (no one is lost yet) *writes the first line on the board*

*To class*: Ok, now, suppose they are equivalent. *Writes A:B = C:D* Who can tell me what that means?

*Goldfish around until they explain that there is a factor which makes it so that one is equal to another via multiplying or dividing both terms of the other. Makes sure that they understand that it means both terms.*

*To class*: Ok, so in that case, let’s let the factor be x. Don’t worry about the value of x, or solving for it. It is absolutely nothing more than the factor we pull out or multiply by. Everyone with me? Anyone confused, even the slightest?

*some kids may not get this entirely, but most seem to. If anyone doesn’t, just use an example: suppose you have 2:5 and 4:10, are they equal? Ok, how do you know? Well, you took 2 out of both terms of one right? Therefore, 1:5 = 2*2:2*5, right? so 1:5 = 2(1:5), so you know they’re the same. x in this case is just 2 — it’s whatever we pull out*

They usually get it then.

*To class*: Ok, so what’s the value of each of these ratios?

*Wait for A/B and C/D and that A/B = C/D*

Now, we know that x is multiplied into both terms right? *some confusion may apply, but reminding them of the example works* As such, we get that A/B = xC/xD.

So, what’s next? We pull out the x, since they’re in both C and D (trust me, they got it). Then, show that it’s A/B = (x/x)*(C/D).

Ask what happens when you divide any number by itself? *Wait for 1*. Then: A/B = 1*C/D. Ask what happens when you multiply by 1. *Wait for “It’s the same”* so, what’s A/B equal? *There’s usually a lot of “Ooooooooohs!” at this point.

And they got it. Every single class got it. And it blew my mind each time, so much, that I went back over it with them to reinforce it: pointing out a step and asking someone in the class to explain to me why that’s the case and what to do next (since kids listen to their peers better than the teachers), and then, by that point, no one was lost. Everyone got it. It worked so well, I did it in the next 4 classes. It blew my freaking mind.

Of course there was reinforcement afterwards: problems checking to make sure they understood the significance of the values and whatnot.

But, it worked. These 6th graders understood a proof.

\$40p/h

Tyler K.

Tutor for anything in the sciences!

if (isMyPost) { }