2. A closer look at induction
2.1. Standard induction
2.1.1. The Principle of Mathematical Induction
We first recall the classical principle of mathematical induction30: Theorem 2.1. For eachn∈ N, letA(n) be a logical statement.
Assume the following:
Assumption 1: The statementA(0) holds.
Assumption 2: If m∈ Nis such thatA(m) holds, thenA(m+1)also holds.
Then, A(n) holds for eachn ∈N.
Theorem 2.1 is commonly taken to be one of the axioms of mathematics (the
“axiom of induction”), or (in type theory) as part of the definition ofN. Intuitively, Theorem 2.1 should be obvious: For example, if you want to prove (under the assumptions of Theorem 2.1) thatA(4) holds, you can argue as follows:
• By Assumption 1, the statement A(0)holds.
• Thus, by Assumption 2 (applied to m=0), the statement A(1)holds.
• Thus, by Assumption 2 (applied to m=1), the statement A(2)holds.
• Thus, by Assumption 2 (applied to m=2), the statement A(3)holds.
• Thus, by Assumption 2 (applied to m=3), the statement A(4)holds.
A similar (but longer) argument shows that the statementA(5) holds. Likewise, you can show that the statement A(15) holds, if you have the patience to apply Assumption 2 a total of 15 times. It is thus not surprising thatA(n)holds for each n∈ N; but if you don’t assume Theorem 2.1 as an axiom, you would need to write down a different proof for each value ofn (which becomes the longer the larger n is), and thus would never reach the general result (i.e., that A(n) holds for each n∈ N), because you cannot write down infinitely many proofs. What Theorem 2.1
29If not, introductions can be found in [LeLeMe16, Chapter 5], [Day16], [Vellem06, Chapter 6], [Hammac15, Chapter 10], [Vorobi02] and various other sources.
30Keep in mind thatNmeans the set{0, 1, 2, . . .}for us.
does is, roughly speaking, to apply Assumption 2 for you as many times as it is needed for each n∈ N.
(Authors of textbooks like to visualize Theorem 2.1 by envisioning an infinite sequence of dominos (numbered 0, 1, 2, . . .) placed in row, sufficiently close to each other that if domino m falls, then domino m+1 will also fall. Now, assume that you kick domino 0 over. What Theorem 2.1 then says is that each domino will fall. See, e.g., [Hammac15, Chapter 10] for a detailed explanation of this metaphor.
Here is another metaphor for Theorem 2.1: Assume that there is a virus that infects nonnegative integers. Once it has infected some m ∈ N, it will soon spread to m+1 as well. Now, assume that 0 gets infected. Then, Theorem 2.1 says that each n∈ Nwill eventually be infected.)
Theorem 2.1 is called the principle of induction or principle of complete induction or principle of mathematical induction, and we shall also call it principle of standard induction in order to distinguish it from several variant “principles of induction”
that we will see later. Proofs that use this principle are calledproofs by induction or induction proofs. Usually, in such proofs, we don’t explicitly cite Theorem 2.1, but instead say certain words that signal that Theorem 2.1 is being applied and that (ideally) also indicate what statements A(n) it is being applied to31. However, for our very first example of a proof by induction, we are going to use Theorem 2.1 explicitly. We shall show the following fact:
Proposition 2.2. Let q and d be two real numbers such that q 6= 1. Let (a0,a1,a2, . . .) be a sequence of real numbers. Assume that
an+1=qan+d for eachn ∈N. (42) Then,
an =qna0+q
n−1
q−1 d for each n∈ N. (43)
Proof of Proposition 2.2. For each n∈ N, we let A(n)be the statement
an =qna0+q
n−1 q−1 d
. Thus, our goal is to prove the statement A(n) for each n∈ N.
We first notice that the statementA(0) holds32.
31We will explain this in Convention 2.3 below.
32Proof. This is easy to verify: We haveq0=1, thusq0−1 =0, and therefore q0−1 q−1 = 0
q−1 =0.
Now,
q0
|{z}
=1
a0+q
0−1 q−1
| {z }
=0
d=1a0+0d=a0,
so that a0 = q0a0+q
0−1
q−1 d. But this is precisely the statement A(0)(sinceA(0)is defined to
Now, we claim that
Now, forget that we fixed m. We thus have shown that if m ∈ N is such that A(m)holds, then A(m+1) also holds. This proves (44).]
Now, both assumptions of Theorem 2.1 are satisfied (indeed, Assumption 1 holds because the statement A(0) holds, whereas Assumption 2 holds because of (44)). Thus, Theorem 2.1 shows that A(n) holds for each n ∈ N. In other
33becauseA(m)is defined to be the statement
2.1.2. Conventions for writing induction proofs
Now, let us introduce some standard language that is commonly used in proofs by induction:
Convention 2.3. For each n ∈ N, let A(n) be a logical statement. Assume that you want to prove that A(n) holds for each n∈ N.
Theorem 2.1 offers the following strategy for proving this: First show that As-sumption 1 of Theorem 2.1 is satisfied; then, show that AsAs-sumption 2 of Theorem 2.1 is satisfied; then, Theorem 2.1 automatically completes your proof.
A proof that follows this strategy is called a proof by induction on n(orproof by induction over n) or (less precisely) an inductive proof. When you follow this strat-egy, you say that you are inducting on n (or over n). The proof that Assumption 1 is satisfied is called theinduction base (orbase case) of the proof. The proof that Assumption 2 is satisfied is called the induction stepof the proof.
In order to prove that Assumption 2 is satisfied, you will usually want to fix an m ∈ N such that A(m) holds, and then prove that A(m+1) holds. In other words, you will usually want to fix m ∈ N, assume that A(m)holds, and then prove that A(m+1) holds. When doing so, it is common to refer to the assumption thatA(m)holds as the induction hypothesis(orinduction assumption).
Using this language, we can rewrite our above proof of Proposition 2.2 as follows:
Proof of Proposition 2.2 (second version). For each n ∈ N, we let A(n) be the state-ment
an =qna0+ q
n−1 q−1 d
. Thus, our goal is to prove the statement A(n) for eachn ∈N.
We shall prove this by induction onn:
Induction base: We haveq0=1, thusq0−1=0, and therefore q0−1
q−1 = 0
q−1 =0.
Now,
q0
|{z}
=1
a0+q
0−1 q−1
| {z }
=0
d =1a0+0d=a0,
so that a0 = q0a0+q
0−1
q−1d. But this is precisely the statement A(0) (since A(0) is defined to be the statement
a0 =q0a0+q
0−1 q−1 d
). Hence, the statementA(0) holds. This completes the induction base.
Induction step: Let m ∈ N. Assume that A(m) holds. We must show that A(m+1) also holds.
We have assumed that A(m) holds (this is our induction hypothesis). In other
words,am =qma0+ q
Now, forget that we fixed m. We thus have shown that if m ∈ N is such that A(m)holds, then A(m+1) also holds. This completes the induction step.
Thus, we have completed both the induction base and the induction step. Hence, by induction, we conclude thatA(n)holds for eachn ∈N. This proves Proposition 2.2.
The proof we just gave still has a lot of “boilerplate” text. For example, we have explicitly defined the statement A(n), but it is not really necessary, since it is clear what this statement should be (viz., it should be the claim we are proving, without the “for eachn ∈N” part). Allowing ourselves some imprecision, we could say this statement is simply (43). (This is a bit imprecise, because (43) contains the words
“for eachn∈ N”, but it should be clear that we don’t mean to include these words, since there can be no “for each n ∈ N” in the statement A(n).) Furthermore, we don’t need to write the sentence
“Thus, we have completed both the induction base and the induction step”
35becauseA(m)is defined to be the statement
before we declare our inductive proof to be finished; it is clear enough that we have completed them. We also can remove the following two sentences:
“Now, forget that we fixedm. We thus have shown that ifm∈ Nis such thatA(m)holds, then A(m+1) also holds.”.
In fact, these sentences merely say that we have completed the induction step;
they carry no other information (since the induction step always consists in fixing m ∈Nsuch that A(m)holds, and proving that A(m+1) also holds). So once we say that the induction step is completed, we don’t need these sentences anymore.
So we can shorten our proof above a bit further:
Proof of Proposition 2.2 (third version). We shall prove (43) by induction onn:
Induction base: We haveq0=1, thusq0−1=0, and therefore q0−1
q−1 = 0
q−1 =0.
Now,
q0
|{z}
=1
a0+q
0−1 q−1
| {z }
=0
d =1a0+0d=a0,
so that a0 = q0a0+ q
0−1
q−1 d. In other words, (43) holds for n = 0. 37 This completes the induction base.
Induction step: Let m∈ N. Assume that (43) holds forn=m. 38 We must show that (43) holds for n=m+1. 39
We have assumed that (43) holds forn =m. In other words,am =qma0+q
m−1 q−1 d
37Note that the statement “(43) holds forn =0” (which we just proved) is precisely the statement A(0)in the previous two versions of our proof.
38Note that the statement “(43) holds forn=m” (which we just assumed) is precisely the statement A(m)in the previous two versions of our proof.
39Note that this statement “(43) holds forn = m+1” is precisely the statementA(m+1) in the previous two versions of our proof.
holds. Now, (42) (applied ton=m) yields n = m+1. This completes the induction step. Hence, (43) is proven by induction.
This proves Proposition 2.2.