# Lecture 6

# Induction

## The basics of induction

The most obvious interesting thing about the natural numbers is that it’s natural to start listing them, one after the other: $\mathbb{N}= \left\{ 0, 1, 2, 3, 4, \ldots \right\}.$ This, of course, is how *counting* works.

It turns out that this way of thinking about the integers gives us a very powerful tool for proving things one integer at a time: the *principle of mathematical induction*, usually known to mathematicians simply as *induction*.

Informally, I like to think of the following example:

*If* I can reach the bottom (rung number zero?) of a ladder,

*and* if I’m on any rung I can reach the next rung up,

*then* I can reach any rung on the ladder.

Why, for example, can I reach the fourth rung? One can imagine a detailed proof of this, as follows:

I can reach rung zero;

Because I can reach rung zero, I can reach rung one;

Because I can reach rung one, I can reach rung two;

Because I can reach rung two, I can reach rung three;

Because I can reach rung three, I can reach rung four.

The connection with counting is obvious: our proof visibly counts up to four.

Writing that out was okay, but you are probably glad I didn’t write out a proof that we could reach the hundred and seventy-eighth rung. I suppose that we could do so, writing “and so on” at some point: but that’s a little vague (what about situations where it isn’t obvious what “and so on” means)?

It’s helpful to have a way which isn’t vague.

So here’s a formal version:

**Definition:** [Induction] Let $P(n)$ be a statement that depends on a natural number $n$. Then, if

the statement $P(0)$ is true, and

for all $k\in\mathbb{N}$, if $P(k)$ is true, then $P(k+1)$ is true,

then the statement $P(n)$ is true for all $n\in\mathbb{N}$.

Here are some useful words:

We call part () the *base case*, and part () the *induction step*. These words agree quite well with our mental picture of a ladder!

When we are trying to prove the induction step $P(k)\Rightarrow
P(k+1)$ we refer to $P(k)$ as the *induction hypothesis*.

## Example of induction

We’ll prove many things by induction in this course, but this is one:

#### Proposition

For any natural number $n$, we have the following formula for the sum of the first $n$ positive integers: $1+2+\cdots+(n-1)+n = \frac{n(n+1)}{2}.$

Let $P(n)$ be the statement above for some particular $n$.

So $P(3)$ is the statement that says $1+2+3 = 3\times 4/2$, and $P(10)$ is the statement that $1+2+3+4+5+6+7+8+9+10 = (10\times 11)/2.$ Notice that $P(n)$ is *not a number*, it’s a *statement*.

#### Proof

We will prove $P(n)$, which says that $1+2+\cdots+(n-1)+n = \frac{n(n+1)}{2}$ for all $n$ by induction.

For our base case, $P(0)$ says that the sum of *no integers at all* is $0\times 1/2$, which is true, as the sum of no integers is zero.

Now we will do our induction step, proving $P(k)\Rightarrow P(k+1)$ for all $k$. Suppose $P(k)$ is true: we need to show that $P(k+1)$ is true.

The statement $P(k)$ tells us that $1+2+\cdots+(k-1)+k = \frac{k(k+1)}{2}.$ We need to prove $P(k+1)$, which would say that $1+2+\cdots+(k-1)+k+(k+1) = \frac{(k+1)(k+2)}{2}.$ Now note that $\begin{aligned} & 1+2+\cdots+(k-1)+k+(k+1)\\ =& \left(1+2+\cdots+(k-1)+k\right)+(k+1)\\ =& \frac{k(k+1)}{2}+(k+1)\quad\text{(by the induction hypothesis)}\\ =& \frac{k(k+1)+2(k+1)}{2}\\ =& \frac{(k+1)(k+2)}{2}.\end{aligned}$ This is exactly the statement $P(k+1)$, which is what we needed for the induction step, and that completes the proof.

You may know other ways of proving that. (I can think of a few.) But I hope you’re impressed with this as a strong potential method for proving identities.

## Nonexamples of induction

Let’s now try proving some *completely false* statements using induction. The plan is (of course) not to succeed, but to understand where we need to be careful.

Clearly this statement is complete and utter rubbish.

If you believe that induction is a reliable method of proof (and I do, and I hope you do too), then it had better be the case that we’re not using induction correctly.

Anyway, here’s an induction “proof”. Suppose that $k = k+83$ for some $k$. We’ll prove that $(k+1) = (k+1)+83$. But we have $\begin{aligned} {}k+1 &= (k+83)+1\qquad\text{(by assumption)}\\ {}&= (k+1)+83\qquad\text{(by rearrangement)}.\end{aligned}$ This completes the proof.

What’s the problem with the argument above?

There’s no base case.

If you don’t have a base case, such as $P(0)$, then it’s of no use to prove that $P(k)\Rightarrow P(k+1)$ for all $k$. It’s no use to be able to climb a ladder if the bottom of the ladder is unreachable.

Here’s another, more subtle example:

Again, we find ourselves hoping very strongly that there’s a mistake in the use of induction in what follows. I’ll write it out and we can see if we can spot it.

In order to do this, we’ll let $P(n)$ be the statement “Given any $n$ horses, all of them have the same colour”. We’ll prove $P(n)$ for all $n$ by induction: that will give us what we want, because we can take $n$ to be the number of horses in the world.

We’ll take $P(1)$ as the base case of the induction. This is the statement “Given any one horse, all of them have the same colour”: this is obviously true.

Now we’ll prove the induction step. We will assume that $P(k)$ is true (“given any $k$ horses, all of them have the same colour”): our job is to prove that $P(k+1)$ is true (“given any $(k+1)$ horses, all of them have the same colour”).

So suppose we have $(k+1)$ horses. Name two of them Alice and Zebedee.

Excluding Alice, there are $k$ horses, which all have the same colour, by the induction hypothesis. So all the horses except Alice have the same colour as Zebedee.

Also, excluding Zebedee, there are $k$ horses, which all have the same colour, again by the induction hypothesis. So all the horses except Zebedee have the same colour as Alice.

Hence all the horses except Alice and Zebedee have the same colour as both Alice and Zebedee, which says that all the horses have the same colour. That ends the proof.

What’s wrong with this?

The particular case $P(1)\Rightarrow P(2)$ doesn’t work.

I find this surprisingly subtle.

In fact, it’s a parody of a *valid* style of argument. If it is the case that *any two things are the same*, then we could prove using exactly this method that they’re *all the same*. In fact, this is something you already know, since “all are alike” and “no two differ” are synonymous phrases.

## Variants

There are other techniques which use the same *idea* of induction, but not quite the same formal principle as I’ve written out above.

We can start with a base case which isn’t $P(0)$. For example, if

$P(15)$ is true, and

$P(k)$ implies $P(k+1)$ for all $k\geq 15$,

then $P(n)$ is true for all $n\geq 15$.

Perhaps you want to think of that as saying “if have a door which leads to the fifteenth rung of a ladder, and you know how to climb ladders, then you can get to every rung above the fifteenth”.

Actually, this is really just ordinary induction in disguise.

Indeed, if we define $Q(n)$ to be $P(n+15)$, then proving $Q(n)$ for all $n\in\mathbb{N}$ by induction is the same as proving $P(n)$ for all integers $n\geq 15$.

Perhaps you want to think of that as “ignoring all the bits of the ladder below the fifteenth rung, imagining the ladder starts outside your door, and starting counting rungs from there”. Or perhaps you’re bored of the ladder analogy now.

Actually, you should have been prepared for this variant: my induction proof that “all horses have the same colour” started with $1$, not $0$. (Okay, that proof was wrong. But there was nothing wrong with *that bit* of the proof: there’s nothing wrong with induction starting from $1$. It was something else that was wrong).