Discrete Mathematics An Open Introduction, 3rd edition

Perhaps in investigating the problem above you picked some amounts of postage, and then figured out whether you could make that amount using just 8-cent and 5-cent stamps. Perhaps you did this in order: can you make 1 cent of postage? Can you make 2 cents? 3 cents? And so on. If this is what you did, you were actually answering a sequence of questions. We have methods for dealing with sequences. Let’s see if that helps.

Let’s think a bit about how we could find the value of $P(n)$ for some specific $n$ (the “value” will be either $T$ or $F$). How did we find the value of the $n$th term of a sequence of numbers? How did we find $a_n\text{?}$ There were two ways we could do this: either there was a closed formula for $a_n\text{,}$ so we could plug in $n$ into the formula and get our output value, or we had a recursive definition for the sequence, so we could use the previous terms of the sequence to compute the $n$th term. When dealing with sequences of statements, we could use either of these techniques as well. Maybe there is a way to use $n$ itself to determine whether we can make $n$ cents of postage. That would be something like a closed formula. Or instead we could use the previous terms in the sequence (of statements) to determine whether we can make $n$ cents of postage. That is, if we know the value of $P(n-1)\text{,}$ can we get from that to the value of $P(n)\text{?}$ That would be something like a recursive definition for the sequence. Remember, finding recursive definitions for sequences was often easier than finding closed formulas. The same is true here.

Suppose I told you that $P(43)$ was true (it is). Can you determine from this fact the value of $P(44)$ (whether it true or false)? Yes you can. Even if we don’t know how exactly we made 43 cents out of the 5-cent and 8-cent stamps, we do know that there was some way to do it. What if that way used at least three 5-cent stamps (making 15 cents)? We could replace those three 5-cent stamps with two 8-cent stamps (making 16 cents). The total postage has gone up by 1, so we have a way to make 44 cents, so $P(44)$ is true. Of course, we assumed that we had at least three 5-cent stamps. What if we didn’t? Then we must have at least three 8-cent stamps (making 24 cents). If we replace those three 8-cent stamps with five 5-cent stamps (making 25 cents) then again we have bumped up our total by 1 cent so we can make 44 cents, so $P(44)$ is true.

Notice that we have not said how to make 44 cents, just that we can, on the basis that we can make 43 cents. How do we know we can make 43 cents? Perhaps because we know we can make $42$ cents, which we know we can do because we know we can make 41 cents, and so on. It’s a recursion! As with a recursive definition of a numerical sequence, we must specify our initial value. In this case, the initial value is “$P(1)$ is false.” That’s not good, since our recurrence relation just says that $P(k+1)$ is true if $P(k)$ is also true. We need to start the process with a true $P(k)\text{.}$ So instead, we might want to use “$P(28)$ is true” as the initial condition.

Suppose we have done this. Then we know that the 28th term of the sequence above is a $T$ (using step 1, the initial condition or base case), and that every term after the 28th is $T$ also (using step 2, the recursive part or inductive case). Here is what the proof would actually look like.

What we did in the stamp example above works for many types of problems. Proof by induction is useful when trying to prove statements about all natural numbers, or all natural numbers greater than some fixed first case (like 28 in the example above), and in some other situations too. In particular, induction should be used when there is some way to go from one case to the next – when you can see how to always “do one more.”

This is a big idea. Thinking about a problem inductively can give new insight into the problem. For example, to really understand the stamp problem, you should think about how any amount of postage (greater than 28 cents) can be made (this is non-inductive reasoning) and also how the ways in which postage can be made changes as the amount increases (inductive reasoning). When you are asked to provide a proof by induction, you are being asked to think about the problem dynamically; how does increasing $n$ change the problem?

But there is another side to proofs by induction as well. In mathematics, it is not enough to understand a problem, you must also be able to communicate the problem to others. Like any discipline, mathematics has standard language and style, allowing mathematicians to share their ideas efficiently. Proofs by induction have a certain formal style, and being able to write in this style is important. It allows us to keep our ideas organized and might even help us with formulating a proof.

Here is the general structure of a proof by mathematical induction:

Sometimes the statement $P(n)$ will only be true for values of $n\ge 4\text{,}$ for example, or some other value. In such cases, replace all the 0’s above with 4’s (or the other value).

The other advantage of formalizing inductive proofs is it allows us to verify that the logic behind this style of argument is valid. Why does induction work? Think of a row of dominoes set up standing on their edges. We want to argue that in a minute, all the dominoes will have fallen down. For this to happen, you will need to push the first domino. That is the base case. It will also have to be that the dominoes are close enough together that when any particular domino falls, it will cause the next domino to fall. That is the inductive case. If both of these conditions are met, you push the first domino over and each domino will cause the next to fall, then all the dominoes will fall.

Induction is powerful! Think how much easier it is to knock over dominoes when you don’t have to push over each domino yourself. You just start the chain reaction, and the rely on the relative nearness of the dominoes to take care of the rest.

Think about our study of sequences. It is easier to find recursive definitions for sequences than closed formulas. Going from one case to the next is easier than going directly to a particular case. That is what is so great about induction. Instead of going directly to the (arbitrary) case for $n\text{,}$ we just need to say how to get from one case to the next.

When you are asked to prove a statement by mathematical induction, you should first think about why the statement is true, using inductive reasoning. Explain why induction is the right thing to do, and roughly why the inductive case will work. Then, sit down and write out a careful, formal proof using the structure above.

Here are some examples of proof by mathematical induction.

Note that in the part of the proof in which we proved $P(k+1)$ from $P(k)\text{,}$ we used the equation $P(k)\text{.}$ This was the inductive hypothesis. Seeing how to use the inductive hypotheses is usually straight forward when proving a fact about a sum like this. In other proofs, it can be less obvious where it fits in.

We had to be a little bit clever (i.e., use some algebra) to locate the $6^k-1$ inside of $6^{k+1}-1$ before we could apply the inductive hypothesis. This is what can make inductive proofs challenging.

In the two examples above, we started with $n=1$ or $n=0\text{.}$ We can start later if we need to.

The previous example might remind you of the racetrack principle from calculus, which says that if $f(a)<g(a)\text{,}$ and $f^{\prime }(x)<g^{\prime }(x)$ for $x>a\text{,}$ then $f(x)<g(x)$ for $x>a\text{.}$ Same idea: the larger function is increasing at a faster rate than the smaller function, so the larger function will stay larger. In discrete math, we don’t have derivatives, so we look at differences. Thus induction is the way to go.

Sometimes, to prove that $P(k+1)$ is true, it would be helpful to know that $P(k)$ and $P(k-1)$ and $P(k-2)$ are all true. Consider the following puzzle:

At first, this question might seem impossible. Perhaps I meant to ask for the smallest number of breaks needed? Let’s investigate.

Start with some small cases. If $n=2\text{,}$ you must have a $1\times 2$ rectangle, which can be reduced to single pieces in one break. With $n=3\text{,}$ we must have a $1\times 3$ bar, which requires two breaks: the first break creates a single square and a $1\times 2$ bar, which we know takes one (more) break.

What about $n=4\text{?}$ Now we could have a $2\times 2$ bar, or a $1\times 4$ bar. In the first case, break the bar into two $2\times 2$ bars, each which require one more break (that’s a total of three breaks required). If we started with a $1\times 4$ bar, we have choices for our first break. We could break the bar in half, creating two $1\times 2$ bars, or we could break off a single square, leaving a $1\times 3$ bar. But either way, we still need two more breaks, giving a total of three.

It is starting to look like no matter how we break the bar (and no matter how the $n$ squares are arranged into a rectangle), we will always have the same number of breaks required. It also looks like that number is one less than $n\text{:}$

It makes sense to prove this by induction because after breaking the bar once, you are left with smaller chocolate bars. Reducing to smaller cases is what induction is all about. We can inductively assume we already know how to deal with these smaller bars. The problem is, if we are trying to prove the inductive case about a $(k+1)$-square bar, we don’t know that after the first break the remaining bar will have $k$ squares. So we really need to assume that our conjecture is true for all cases less than $k+1\text{.}$

Is it valid to make this stronger assumption? Remember, in induction we are attempting to prove that $P(n)$ is true for all $n\text{.}$ What if that were not the case? Then there would be some first $n_0$ for which $P(n_0)$ was false. Since $n_0$ is the first counterexample, we know that $P(n)$ is true for all $n<n_0\text{.}$ Now we proceed to prove that $P(n_0)$ is actually true, based on the assumption that $P(n)$ is true for all smaller $n\text{.}$

This is quite an advantage: we now have a stronger inductive hypothesis. We can assume that $P(1)\text{,}$ $P(2)\text{,}$ $P(3)\text{,}$ … $P(k)$ is true, just to show that $P(k+1)$ is true. Previously, we just assumed $P(k)$ for this purpose.

It is slightly easier if we change our variables for strong induction. Here is what the formal proof would look like:

Of course, it is acceptable to replace 0 with a larger base case if needed.

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Technically, strong induction does not require you to prove a separate base case. This is because when proving the inductive case, you must show that $P(0)$ is true, assuming $P(k)$ is true for all $k<0\text{.}$ But this is not any help so you end up proving $P(0)$ anyway. To be on the safe side, we will always include the base case separately.

Let’s prove our conjecture about the chocolate bar puzzle:

Here is a more mathematically relevant example:

Whether you use regular induction or strong induction depends on the statement you want to prove. If you wanted to be safe, you could always use strong induction. It really is stronger, so can accomplish everything “weak” induction can. That said, using regular induction is often easier since there is only one place you can use the induction hypothesis. There is also something to be said for elegance in proofs. If you can prove a statement using simpler tools, it is nice to do so.

As a final contrast between the two forms of induction, consider once more the stamp problem. Regular induction worked by showing how to increase postage by one cent (either replacing three 5-cent stamps with two 8-cent stamps, or three 8-cent stamps with five 5-cent stamps). We could give a slightly different proof using strong induction. First, we could show five base cases: it is possible to make 28, 29, 30, 31, and 32 cents (we would actually say how each of these is made). Now assume that it is possible to make $k$ cents of postage for all $k<n$ as long as $k\ge 28\text{.}$ As long as $n>32\text{,}$ this means in particular we can make $k=n-5$ cents. Now add a 5-cent stamp to get make $n$ cents.