Teaching Fermat to year 11 students

Most people involved in education have probably noticed the (now traditional and annual, it seems) furore over exams being too hard. This is usually punctuated with comments like “kids can’t possibly be expected to know/be able to/understand/etc.”.

 

It is in this context that I wrote a tweet which read: “Yesterday I taught Fermat’s little theorem to Y11s in half an hour at lunchtime. Don’t tell me what kids can/can’t do.” Whilst the tweet itself may appear somewhat flippant, it is nevertheless making a valid point: in my admittedly very limited experience, we are often far too culpable of assuming what kids can and can’t do, and lowering expectations accordingly.

 

My tweet was picked up on by a number of people, and one suggested I write a longer post expanding on what it was I did. So don’t blame me for this post, blame that person.

 

To give a bit of context then, I teach a top set Y11 class at a very good school and also run lunchtime booster sessions. These sessions aren’t aimed specifically at the GCSE curriculum, in fact very much the opposite; their aim is to enthuse the students about more challenging maths, and introduce them to some of the ideas they will encounter at A-levels and university.

 

The sessions are just over half an hour long and vary immensely: some sessions we will be looking at applying knowledge they have to harder problems, for example looking at some Olympiad problems. Other sessions I will introduce them to a new area of maths, like combinatorics (which we started with by wondering what our chances of winning at black-jack were). Other times one of the students will arrive with something they want to go over and we will go from there. On one memorable occasion they wanted to learn about calculus.

 

I plan the sessions thoroughly. I have a clear idea of what I want to achieve and how we are going to get there. As with lessons they very rarely go according to plan. Sometimes (as with the time they asked about calculus) I have to improvise completely. On rare occasions they face me with a problem which I can’t solve instantly, so we look at it together. The students love it: they appreciate that I am giving up my lunchtime to introduce them to extra maths; they appreciate that we are looking at hard and challenging problems which we have to think about to solve; they love the fact that they are pushing and stretching themselves and expanding their knowledge and skill-set.

 

This Tuesday I set them a challenge: we were going to attempt to understand and prove Fermat’s little theorem (the less famous one…) in half an hour. For those who don’t know what the theorem says, a statement of it goes:

If p is a prime number and $latex a$ is any number not divisible by p, then $latex a^{p-1}-1$ is divisible by p.

 

Fermat’s little theorem relies on the concept of modular arithmetic (something my Y11s had never heard of before Tuesday). The easiest proof is an induction argument. My Y11s had never come across mathematical induction before. It uses the binomial theorem, in a particular case that if p is prime then the middle terms are all divisible by p. My Y11s hadn’t seen the binomial theorem before.

 

I won’t bore you with the precise details of how we did all of this in just over half an hour (I’ve written it up as a lengthy post-script for the geeky mathmos out there). The point is though that we did it. By the end of the session, all the students there understood binomial expansions and how to implement them. They all understood Fermat’s little theorem and the concept of mathematical induction, and how to use it in this case. Their proofs if written up would have been rudimentary and lacking the polished language and notation of university students, but essentially correct.

 

The point is that our students are capable sometimes of the most remarkable feats of learning. I know the easy counter-argument: these are the top students in the top set at a very good school, however in my admitedly extremely limited experience this is true not just of the top sets, but all the way through the school. Children are hungry for knowledge, and if you can hook them in, show them you care, and teach them well they will by and large respond to this.

 

So let’s have a little bit less rhetoric about “hard” or “unfair” tests. Instead how about we raise our own expectations and aspirations about what we can teach students? How about we focus on teaching what we can and well, and let the results take care of themselves?

 

As a brief footnote, of the 6 students who have attended my lunchtime boosters most regularly, I am fairly certain we will get at least 4 A*’s come August (we may get more, certainly everyone will be close). Of those only one was on A*’s before they started attending the sessions. Maybe this is not coincidence.

 

Don’t underestimate the capacity of your students. And teach well.

 

Postscript: for those interested in the technical side of that particular half hour. I am not suggesting this is the only or even the best way to teach Fermat’s little theorem from scratch in half an hour. But it is a way, and it worked, so here goes…

 

We started with the binomial theorem. We investigated what happened when you looked at $latex (x+y)^1, (x+y)^2, (x+y)^3$ and so on. We had to do the first few bracket expansions together because they were a bit lazy to start, but by the time we reached $latex (x+y)^4$ they were able to tell me that we expected $latex x^4, x^3y, x^2y^2, xy^3, y^4$ in the expansion.

 

We then looked at the coefficients and they very quickly spotted Pascal’s triangle (they didn’t know what it was called, so I told them). I explained why this idea of adding the two numbers above it was what you needed for the expansion and they were comfortable with this. They could then expand $latex {\left(x+y\right)^{10}}$ although as it turns out we struggled to add numbers correctly enough times and made some mistakes.

 

Using this idea that repeated addition was really rather tedious, I moved onto the binomial coefficients as combinatorial quantities . It is true that we had already seen the idea of choosing 4 elements from a list of 8 (say) where the order doesn’t matter, so they were fairly confident with what that meant and how you worked it out; I explained why this was the coefficient we needed and they were happy, and could then work out the coefficient of $latex x^{11}y^4$ in the expansion of $latex (\x+y)^{15}$.

 

I then returned to Pascal’s triangle and pointed out the prime rows. I asked what was special about these as opposed to the others. Everyone immediately spotted that all the middle numbers were divisible by the prime whose row they were in. I asked whether someone could explain why and amazingly more than one came up with the link with the binomial coefficient. The argument is that the numerator is  and the denominator is , so assuming all the factors of the denominator are less than p ; it follows that none of the factors in the denominator have any common factors with p because p is prime, so the whole thing is a multiple of p. Whilst the students who saw the link weren’t quite capable of articulating their argument like this, they had entirely the right idea. Amazing!

 

Having therefore got what we needed from the binomial theorem one of the students asked “what is this useful for?” which was exactly what was needed to move the session on. So I asked them to take a prime p (we chose 7; it is probably easier with 5) and to look at $latex a^{p-1}-1$ for various $latex a$. We did this (I allowed them, on this rare occasion, to use calculators) and pretty much immediately they told me they were all multiples of 7 (this in itself is fairly remarkable, given that the 3rd smallest number involved was 4,095…).

 

Good, said I, can we prove this?

 

This then required a bit of a diversion as I talked them through the principle of mathematical induction, but only for a minute or two as they got their heads around it fairly quickly. I then introduced them to the concept of modular arithmetic (I introduced this as “well, essentially all we care about numbers for this purpose is their remainder when divided by 7, so let’s make some maths around this”) and even got them used to using formal notation. We then restated Fermat’s as:

$latex a^p \equiv a$  (mod p)

Proving this by induction was then the challenge for the students. They understood that they were trying to show that $\latex (a+1)^p\equiv a+1$; I needed to prod them a bit to remember the binomial expansion we had done at the start, and “what do we know about the middle numbers?” but by the end they all understood that yes it did work!

 

The question was then put to me (there were about 2 minutes left at this point) “but what’s this useful for?” so I told them it was basically the foundation of all modern cryptography (a bit of a stretch, but not that much). “How?” they asked; “well, that will have to wait for another time…” I’ll let you know how that goes.