2. A closer look at induction
2.11. Induction in an interval
Hence, there exist g ∈ N, x ∈ Z and y ∈ Zsuch that g = ux+vy and g | u and g |v (namely, g =g0, x =x0 andy =y0−x0). Thus, Claim 2 is proven in Case 2.
We have now proven Claim 2 in each of the two Cases 1 and 2. Thus, Claim 2 always holds (since Cases 1 and 2 cover all possibilities).]
Now, we can prove Claim 1 as well:
[Proof of Claim 1: Claim 1 can be derived from Claim 2 in the same way as we derived it in the first version of the proof above. We shall not repeat this argument, since it just applies verbatim.]
But Claim 1 is simply saying that Theorem 2.70 holds under the condition that a+b = m. Thus, by proving Claim 1, we have shown that Theorem 2.70 holds under the condition that a+b = m. This completes the induction step. Thus, Theorem 2.70 is proven by strong induction.
2.11. Induction in an interval
2.11.1. The induction principle for intervals
The induction principles we have seen so far were tailored towards proving state-ments whose variables range over infinite sets such asNandZ≥g. Sometimes, one
instead wants to do an induction on a variable that ranges over a finite interval, such as{g,g+1, . . . ,h}for some integersgand h. We shall next state an induction principle tailored to such situations. First, we make an important convention:
Convention 2.73. If g and h are two integers such that g > h, then the set {g,g+1, . . . ,h}is understood to be the empty set.
Thus, for example,{2, 3, . . . , 1} =∅and{2, 3, . . . , 0} =∅and {5, 6, . . . ,−100} =
∅. (But{5, 6, . . . , 5}={5} and {5, 6, . . . , 6}={5, 6}.) We now state our induction principle for intervals:
Theorem 2.74. Let g ∈ Z and h∈ Z. For each n ∈ {g,g+1, . . . ,h}, letA(n) be a logical statement.
Assume the following:
Assumption 1: If g≤h, then the statement A(g) holds.
Assumption 2: If m ∈ {g,g+1, . . . ,h−1} is such that A(m) holds, then A(m+1) also holds.
Then, A(n) holds for eachn ∈ {g,g+1, . . . ,h}.
Theorem 2.74 is, in a sense, the closest one can get to Theorem 2.53 when having only finitely many statements A(g),A(g+1), . . . ,A(h) instead of an infinite se-quence of statements A(g),A(g+1),A(g+2), . . .. It is easy to derive Theorem 2.74 from Corollary 2.61:
Proof of Theorem 2.74. For eachn∈ Z≥g, we defineB(n)to be the logical statement (if n∈ {g,g+1, . . . ,h}, thenA(n) holds).
Now, let us consider the Assumptions A and B from Corollary 2.61. We claim that both of these assumptions are satisfied.
Assumption 1 says that if g ≤ h, then the statement A(g) holds. Thus, B(g) holds71. In other words, Assumption A is satisfied.
Next, we shall prove that Assumption B is satisfied. Indeed, let p ∈Z≥gbe such thatB(p)holds. We shall now show that B(p+1) also holds.
Indeed, assume that p+1 ∈ {g,g+1, . . . ,h}. Thus, p+1 ≤ h, so that p ≤ p+1 ≤ h. Combining this with p ≥ g (since p ∈ Z≥g), we conclude that p ∈ {g,g+1, . . . ,h} (since p is an integer). But we have assumed thatB(p) holds. In other words,
if p ∈ {g,g+1, . . . ,h}, then A(p) holds
71Proof. Assume that g∈ {g,g+1, . . . ,h}. Thus, g≤ h. But Assumption 1 says that ifg ≤h, then the statementA(g)holds. Hence, the statementA(g)holds (sinceg≤h).
Now, forget that we assumed that g ∈ {g,g+1, . . . ,h}. We thus have proven that if g ∈ {g,g+1, . . . ,h}, then A(g)holds. In other words,B(g)holds (because the statement B(g)is defined as(if g∈ {g,g+1, . . . ,h}, thenA(g) holds)). Qed.
(because the statementB(p)is defined as(if p∈ {g,g+1, . . . ,h}, then A(p) holds)).
Thus, A(p) holds (since we have p ∈ {g,g+1, . . . ,h}). Also, from p+1 ≤ h, we obtain p ≤ h−1. Combining this with p ≥ g, we find p ∈ {g,g+1, . . . ,h−1}. Thus, we know that p ∈ {g,g+1, . . . ,h−1} is such that A(p) holds. Hence, Assumption 2 (applied tom= p) shows that A(p+1)also holds.
Now, forget that we assumed that p+1∈ {g,g+1, . . . ,h}. We thus have proven that if p+1 ∈ {g,g+1, . . . ,h}, then A(p+1) holds. In other words, B(p+1) holds (since the statementB(p+1) is defined as
(if p+1∈ {g,g+1, . . . ,h}, thenA(p+1) holds)).
Now, forget that we fixed p. We thus have proven that if p ∈ Z≥g is such that B(p) holds, then B(p+1) also holds. In other words, Assumption B is satisfied.
We now know that both Assumption A and Assumption B are satisfied. Hence, Corollary 2.61 shows that
B(n) holds for each n∈ Z≥g. (115) Now, letn ∈ {g,g+1, . . . ,h}. Thus,n≥ g, so thatn ∈ Z≥g. Hence, (115) shows thatB(n)holds. In other words,
ifn ∈ {g,g+1, . . . ,h}, then A(n) holds
(since the statementB(n)was defined as(if n∈ {g,g+1, . . . ,h}, thenA(n) holds)).
Thus, A(n) holds (since we haven ∈ {g,g+1, . . . ,h}).
Now, forget that we fixed n. We thus have shown that A(n) holds for each n∈ {g,g+1, . . . ,h}. This proves Theorem 2.74.
Theorem 2.74 is called the principle of induction starting at g and ending at h, and proofs that use it are usually calledproofs by inductionorinduction proofs. As with all the other induction principles seen so far, we don’t usually explicitly cite Theorem 2.74, but instead say certain words that signal that it is being applied and that (ideally) also indicate what integers g and h and what statementsA(n) it is being applied to72. However, we shall reference it explicitly in our very first example of the use of Theorem 2.74:
Proposition 2.75. Let g and h be integers such that g ≤h. Let bg,bg+1, . . . ,bh be any h−g+1 nonzero integers. Assume that bg≥0. Assume further that
|bi+1−bi| ≤ 1 for everyi∈ {g,g+1, . . . ,h−1}. (116) Then, bn >0 for each n∈ {g,g+1, . . . ,h}.
Proposition 2.75 is often called the “discrete intermediate value theorem” or the
“discrete continuity principle”. Its intuitive meaning is that if a finite list of nonzero integers starts with a nonnegative integer, and every further entry of this list differs
72We will explain this in Convention 2.76 below.
from its preceding entry by at most 1, then all entries of this list must be positive.
An example of such a list is(2, 3, 3, 2, 3, 4, 4, 3, 2, 3, 2, 3, 2, 1). Notice that Proposition 2.75 is, again, rather obvious from an intuitive perspective: It just says that it isn’t possible to go from a nonnegative integer to a negative integer by steps of 1 without ever stepping at 0. The rigorous proof of Proposition 2.75 is not much harder – but because it is a statement about elements of {g,g+1, . . . ,h}, it naturally relies on Theorem 2.74:
Proof of Proposition 2.75. For each n ∈ {g,g+1, . . . ,h}, we let A(n) be the state-ment(bn >0).
Our next goal is to prove the statement A(n) for eachn ∈ {g,g+1, . . . ,h}. All the h−g+1 integers bg,bg+1, . . . ,bh are nonzero (by assumption). Thus, in particular, bg is nonzero. In other words, bg 6= 0. Combining this with bg ≥ 0, we obtainbg >0. In other words, the statementA(g)holds (since this statementA(g) is defined to be bg >0
). Hence,
if g ≤h, then the statementA(g) holds. (117) Now, we claim that
if m∈ {g,g+1, . . . ,h−1} is such that A(m) holds, thenA(m+1) also holds.
(118) [Proof of (118): Let m ∈ {g,g+1, . . . ,h−1} be such that A(m) holds. We must show thatA(m+1) also holds.
We have assumed that A(m) holds. In other words, bm > 0 holds (since A(m) is defined to be the statement (bm >0)). Now, (116) (applied to i = m) yields
|bm+1−bm| ≤ 1. But it is well-known (and easy to see) that every integer x sat-isfies −x ≤ |x|. Applying this to x = bm+1−bm, we obtain −(bm+1−bm) ≤
|bm+1−bm| ≤1. In other words, 1 ≥ −(bm+1−bm) = bm−bm+1. In other words, 1+bm+1 ≥ bm. Hence, 1+bm+1 ≥bm >0, so that 1+bm+1 ≥1 (since 1+bm+1 is an integer). In other words,bm+1 ≥0.
But all the h−g+1 integersbg,bg+1, . . . ,bh are nonzero (by assumption). Thus, in particular, bm+1 is nonzero. In other words, bm+1 6= 0. Combining this with bm+1 ≥ 0, we obtain bm+1 > 0. But this is precisely the statement A(m+1) (because A(m+1)is defined to be the statement (bm+1 >0)). Thus, the statement A(m+1) holds.
Now, forget that we fixedm. We thus have shown that ifm ∈ {g,g+1, . . . ,h−1} is such thatA(m) holds, thenA(m+1) also holds. This proves (118).]
Now, both assumptions of Theorem 2.74 are satisfied (indeed, Assumption 1 holds because of (117), whereas Assumption 2 holds because of (118)). Thus, The-orem 2.74 shows that A(n) holds for each n ∈ {g,g+1, . . . ,h}. In other words, bn > 0 holds for each n ∈ {g,g+1, . . . ,h} (since A(n) is the statement (bn >0)).
This proves Proposition 2.75.
2.11.2. Conventions for writing induction proofs in intervals
Next, we shall introduce some standard language that is commonly used in proofs by induction starting at g and ending at h. This language closely imitates the one we use for proofs by standard induction:
Convention 2.76. Let g ∈ Zand h ∈ Z. For each n∈ {g,g+1, . . . ,h}, let A(n) be a logical statement. Assume that you want to prove that A(n) holds for each n∈ {g,g+1, . . . ,h}.
Theorem 2.74 offers the following strategy for proving this: First show that Assumption 1 of Theorem 2.74 is satisfied; then, show that Assumption 2 of The-orem 2.74 is satisfied; then, TheThe-orem 2.74 automatically completes your proof.
A proof that follows this strategy is called a proof by induction on n(orproof by induction over n)starting at g and ending at h or (less precisely) an inductive proof.
Most of the time, the words “starting at g and ending at h” are omitted, since they merely repeat what is clear from the context anyway: For example, if you make a claim about all integersn ∈ {3, 4, 5, 6}, and you say that you are proving it by induction onn, it is clear that you are using induction onnstarting at 3 and ending at 6.
The proof that Assumption 1 is satisfied is called the induction base (or base 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 ∈ {g,g+1, . . . ,h−1} such that A(m) holds, and then prove that A(m+1) holds. In other words, you will usually want to fix m ∈ {g,g+1, . . . ,h−1}, assume that A(m) holds, and then prove that A(m+1) holds. When doing so, it is common to refer to the assumption that A(m) holds as the induction hypothesis(orinduction assumption).
Unsurprisingly, this language parallels the language introduced in Convention 2.3 and in Convention 2.56.
Again, we can shorten our inductive proofs by omitting some sentences that convey no information. In particular, we can leave out the explicit definition of the statementA(n) when this statement is precisely the claim that we are proving (without the “for each n ∈ {g,g+1, . . . ,h}” part). Furthermore, it is common to leave the “Ifg≤h” part of Assumption 1 unsaid (i.e., to pretend that Assumption 1 simply says that A(g) holds). Strictly speaking, this is somewhat imprecise, since A(g) is not defined when g > h; but of course, the whole claim that is being proven is moot anyway when g>h (because there exist no n∈ {g,g+1, . . . ,h} in this case), so this imprecision doesn’t matter.
Thus, we can rewrite our above proof of Proposition 2.75 as follows:
Proof of Proposition 2.75 (second version). We claim that
bn >0 (119)
for eachn ∈ {g,g+1, . . . ,h}.
Indeed, we shall prove (119) by induction onn:
Induction base: All the h−g+1 integersbg,bg+1, . . . ,bh are nonzero (by assump-tion). Thus, in particular, bg is nonzero. In other words, bg 6= 0. Combining this withbg ≥0, we obtainbg >0. In other words, (119) holds forn= g. This completes the induction base.
Induction step: Let m ∈ {g,g+1, . . . ,h−1}. Assume that (119) holds for n =m.
We must show that (119) also holds forn =m+1.
We have assumed that (119) holds forn=m. In other words,bm >0. Now, (116) (applied to i = m) yields |bm+1−bm| ≤ 1. But it is well-known (and easy to see) that every integer x satisfies −x ≤ |x|. Applying this to x = bm+1−bm, we obtain
−(bm+1−bm)≤ |bm+1−bm| ≤1. In other words, 1≥ −(bm+1−bm) = bm−bm+1. In other words, 1+bm+1 ≥ bm. Hence, 1+bm+1 ≥ bm > 0, so that 1+bm+1 ≥ 1 (since 1+bm+1 is an integer). In other words,bm+1≥0.
But all the h−g+1 integersbg,bg+1, . . . ,bh are nonzero (by assumption). Thus, in particular, bm+1 is nonzero. In other words, bm+1 6= 0. Combining this with bm+1 ≥ 0, we obtain bm+1 > 0. In other words, (119) holds for n = m+1. This completes the induction step. Thus, (119) is proven by induction. This proves Proposition 2.75.