2. A closer look at induction
2.12. Strong induction in an interval
2.12.1. The strong induction principle for intervals
We shall next state yet another induction principle – one that combines the idea of strong induction (as in Theorem 2.60) with the idea of working inside an interval {g,g+1, . . . ,h} (as in Theorem 2.74):
Theorem 2.77. Let g ∈ Z and h∈ Z. For each n ∈ {g,g+1, . . . ,h}, letA(n) be a logical statement.
Assume the following:
Assumption 1: Ifm ∈ {g,g+1, . . . ,h} is such that
(A(n) holds for everyn ∈ {g,g+1, . . . ,h} satisfying n<m), then A(m) holds.
Then, A(n) holds for eachn ∈ {g,g+1, . . . ,h}.
Our proof of Theorem 2.77 will be similar to the proof of Theorem 2.74, except that we shall be using Theorem 2.60 instead of Corollary 2.61. Or, to be more precise, we shall be using the following restatement of Theorem 2.60:
Corollary 2.78. Let g ∈Z. For each n∈ Z≥g, let B(n) be a logical statement.
Assume the following:
Assumption A:If p∈ Z≥gis such that
B(n) holds for every n∈ Z≥g satisfyingn < p , then B(p) holds.
Then, B(n) holds for eachn ∈Z≥g.
Proof of Corollary 2.78. Corollary 2.78 is exactly Theorem 2.60, except that some names have been changed:
• The statements A(n) have been renamed asB(n).
• Assumption 1 has been renamed as Assumption A.
• The variable min Assumption A has been renamed as p.
Thus, Corollary 2.78 holds (since Theorem 2.60 holds).
We can now prove Theorem 2.77:
Proof of Theorem 2.77. 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 Assumption A from Corollary 2.78. We claim that this assumption is satisfied.
Indeed, let p ∈Z≥g be such that
B(n) holds for everyn ∈Z≥g satisfyingn< p
. (120)
We shall now show that B(p) holds.
Indeed, assume that p ∈ {g,g+1, . . . ,h}. Thus, p≤h.
Now, let n ∈ {g,g+1, . . . ,h} be such that n < p. Then, n ∈ {g,g+1, . . . ,h} ⊆ {g,g+1,g+2, . . .} = Z≥g and n < p. Hence, (120) shows that B(n) holds.
In other words, (ifn ∈ {g,g+1, . . . ,h}, thenA(n) holds) (because the statement B(n) is defined as (if n∈ {g,g+1, . . . ,h}, thenA(n) holds)). Therefore, A(n) holds (since we know that n∈ {g,g+1, . . . ,h}).
Now, forget that we fixed n. We thus have proven that
(A(n) holds for every n∈ {g,g+1, . . . ,h} satisfyingn< p). Hence, Assumption 1 (applied tom= p) yields thatA(p)holds.
Now, forget that we assumed that p ∈ {g,g+1, . . . ,h}. We thus have proven that
(if p∈ {g,g+1, . . . ,h}, thenA(p) holds). In other words,B(p) holds (since the statementB(p)was defined as (if p ∈ {g,g+1, . . . ,h}, thenA(p) holds)).
Now, forget that we fixed p. We thus have shown that if p∈ Z≥g is such that B(n) holds for every n∈ Z≥g satisfying n< p
, then B(p) holds. In other words, Assumption A is satisfied.
Hence, Corollary 2.78 shows that
B(n) holds for each n∈ Z≥g. (121) Now, letn ∈ {g,g+1, . . . ,h}. Thus,n≥ g, so thatn ∈ Z≥g. Hence, (121) 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 2.77.
Theorem 2.77 is called theprinciple of strong induction starting at g and ending at h, and proofs that use it are usually called proofs by strong induction. Once again, we usually don’t explicitly cite Theorem 2.77 in such proofs, and we usually don’t say explicitly what gandhare and what the statementsA(n) are when it is clear from the context. But (as with all the other induction principles considered so far) we shall be explicit about all these details in our first example:
Proposition 2.79. 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 that for each p ∈ {g+1,g+2, . . . ,h},
there exists somej ∈ {g,g+1, . . . ,p−1} such that bp ≥bj−1. (122) (Of course, the jcan depend on p.) Then, bn >0 for each n ∈ {g,g+1, . . . ,h}. Proposition 2.79 is a more general (although less intuitive) version of Proposition 2.75; indeed, it is easy to see that the condition (116) is stronger than the condition (122) (when required for all p∈ {g+1,g+2, . . . ,h}).
Example 2.80. For this example, set g = 3 and h = 7. Then, if we set (b3,b4,b5,b6,b7) = (4, 5, 3, 4, 2), then the condition (122) holds for all p ∈ {g+1,g+2, . . . ,h}. (For example, it holds for p = 5, since b5 = 3 ≥ 4−1 = b1 −1 and 1 ∈ {g,g+1, . . . , 5−1}.) On the other hand, if we set (b3,b4,b5,b6,b7) = (4, 5, 2, 4, 3), then this condition does not hold (indeed, it fails for p=5, since b5 =2 is neither ≥4−1 nor ≥5−1).
Let us now prove Proposition 2.79 using Theorem 2.77:
Proof of Proposition 2.79. 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
).
Now, we make the following claim:
Claim 1: Ifm∈ {g,g+1, . . . ,h} is such that
(A(n) holds for every n∈ {g,g+1, . . . ,h} satisfyingn<m), then A(m) holds.
[Proof of Claim 1: Let m∈ {g,g+1, . . . ,h} be such that
(A(n) holds for everyn ∈ {g,g+1, . . . ,h} satisfyingn <m). (123) We must show thatA(m)holds.
If m = g, then this follows from the fact that A(g) holds. Thus, for the rest of the proof of Claim 1, we WLOG assume that we don’t have m = g. Hence, m6=g. Combining this withm∈ {g,g+1, . . . ,h}, we obtainm ∈ {g,g+1, . . . ,h} \ {g} ⊆ {g+1,g+2, . . . ,h}. Hence, (122) (applied to p = m) shows that there exists some j ∈ {g,g+1, . . . ,m−1} such that bm ≥ bj−1. Consider this j. From m∈ {g+1,g+2, . . . ,h}, we obtainm ≤h.
Fromj ∈ {g,g+1, . . . ,m−1}, we obtain j≤m−1 <m. Also,
j ∈ {g,g+1, . . . ,m−1} ⊆ {g,g+1, . . . ,h} (since m−1 ≤ m ≤ h). Thus, (123) (applied ton = j) yields that A(j) holds. In other words, bj >0 holds (sinceA(j) is defined to be the statement bj >0). Thus, bj ≥1 (sincebjis an integer), so that bj−1≥0. But recall that bm ≥bj−1 ≥0.
But all the h−g+1 integersbg,bg+1, . . . ,bh are nonzero (by assumption). Thus, in particular, bm is nonzero. In other words, bm 6= 0. Combining this withbm ≥0, we obtain bm > 0. But this is precisely the statement A(m) (because A(m) is defined to be the statement (bm >0)). Thus, the statement A(m) holds. This completes the proof of Claim 1.]
Claim 1 says that Assumption 1 of Theorem 2.77 is satisfied. Thus, Theorem 2.77 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 2.79.
2.12.2. Conventions for writing strong induction proofs in intervals
Next, we shall introduce some standard language that is commonly used in proofs by strong induction starting at g and ending at h. This language closely imitates the one we use for proofs by “usual” strong induction:
Convention 2.81. 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 2.77 offers the following strategy for proving this: Show that As-sumption 1 of Theorem 2.77 is satisfied; then, Theorem 2.77 automatically com-pletes your proof.
A proof that follows this strategy is called a proof by strong induction on n starting at g and ending at h. Most of the time, the words “starting at g and ending at h” are omitted. The proof that Assumption 1 is satisfied is called the induction stepof the proof. This kind of proof has no “induction base”.
In order to prove that Assumption 1 is satisfied, you will usually want to fix an m ∈ {g,g+1, . . . ,h} such that
(A(n) holds for everyn ∈ {g,g+1, . . . ,h} satisfying n<m), (124) and then prove that A(m) holds. In other words, you will usually want to fix m∈ {g,g+1, . . . ,h}, assume that (124) holds, and then prove that A(m) holds.
When doing so, it is common to refer to the assumption that (124) holds as the induction hypothesis(orinduction assumption).
Unsurprisingly, this language parallels the language introduced in Convention 2.63.
As before, proofs using strong induction can be shortened by leaving out some uninformative prose. In particular, the explicit definition of the statement A(n) can often be omitted when this statement is precisely the claim that we are proving (without the “for each n ∈ {g,g+1, . . . ,h}” part). The values of g and h can also be inferred from the statement of the claim, so they don’t need to be specified explicitly. And once again, we don’t need to write “Induction step:”, since our strong induction has no induction base.
This leads to the following abridged version of our above proof of Proposition 2.79:
Proof of Proposition 2.79 (second version). We claim that
bn >0 (125)
for eachn ∈ {g,g+1, . . . ,h}.
Indeed, we shall prove (125) by strong induction onn:
Let m∈ {g,g+1, . . . ,h}. Assume that (125) holds for everyn∈ {g,g+1, . . . ,h} satisfying n < m. We must show that (125) also holds for n = m. In other words, we must show thatbm >0.
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.
We have assumed that (125) holds for every n ∈ {g,g+1, . . . ,h} satisfying n <
m. In other words, we have
bn >0 for everyn ∈ {g,g+1, . . . ,h} satisfying n<m. (126) Recall that we must prove that bm >0. If m = g, then this follows frombg >0.
Thus, for the rest of this induction step, we WLOG assume that we don’t have m = g. Hence, m 6= g. Combining this with m ∈ {g,g+1, . . . ,h}, we obtain m ∈ {g,g+1, . . . ,h} \ {g} ⊆ {g+1,g+2, . . . ,h}. Hence, (122) (applied to p = m) shows that there exists some j ∈ {g,g+1, . . . ,m−1} such that bm ≥ bj−1.
Consider this j. From m∈ {g+1,g+2, . . . ,h}, we obtainm≤h.
Fromj ∈ {g,g+1, . . . ,m−1}, we obtain j≤m−1 <m. Also,
j ∈ {g,g+1, . . . ,m−1} ⊆ {g,g+1, . . . ,h} (since m−1 ≤ m ≤ h). Thus, (126) (applied to n = j) yields that bj > 0. Thus, bj ≥ 1 (since bj is an integer), so that bj−1≥0. But recall that bm ≥bj−1 ≥0.
But all the h−g+1 integersbg,bg+1, . . . ,bh are nonzero (by assumption). Thus, in particular, bm is nonzero. In other words, bm 6= 0. Combining this withbm ≥0, we obtainbm >0.
Thus, we have proven that bm > 0. In other words, (125) holds for n = m.
This completes the induction step. Thus, (125) is proven by strong induction. This proves Proposition 2.79.