2. A closer look at induction
2.10. Strong induction on a derived quantity: Bezout’s theorem
2.10.1. Strong induction on a derived quantity
In Section 2.5, we have seen how to use induction on a variable that does not explicitly appear in the claim. In the current section, we shall show the same for strong induction. This time, the fact that we shall be proving is the following:
Theorem 2.70. Let a ∈Nand b ∈ N. Then, there existg ∈ N, x ∈Z and y∈ Z such that g =ax+byand g | aand g |b.
Example 2.71. (a) If a = 3 and b = 5, then Theorem 2.70 says that there exist g ∈N, x ∈ Zand y ∈ Zsuch that g =3x+5y and g |3 and g | 5. And indeed, it is easy to find such g, x and y: for example, g =1, x = −3 and y =2 will do (since 1=3(−3) +5·2 and 1|3 and 1| 5).
(b) If a = 4 and b =6, then Theorem 2.70 says that there exist g ∈ N, x ∈ Z and y ∈ Z such that g = 4x+6y and g | 4 and g | 6. And indeed, it is easy to find such g, x and y: for example, g = 2, x = −1 and y = 1 will do (since 2=4(−1) +6·1 and 2|4 and 2| 6).
Theorem 2.70 is one form ofBezout’s theorem for integers, and its real significance might not be clear at this point; it becomes important when the greatest common divisor of two integers is studied. For now, we observe that the g in Theorem 2.70 is obviously a common divisor of a and b (that is, an integer that divides both a and b); but it is also divisible by every common divisor of a and b (because of Proposition 2.7).
Let us now focus on the proof of Theorem 2.70. It is natural to try proving it by induction (or perhaps strong induction) on a or on b, but neither option leads to success. It may feel like “induction on a and on b at the same time” could help, and this is indeed a viable approach69. But there is a simpler and shorter method available: strong induction on a+b. As in Section 2.5, the way to formalize such a
69Of course, it needs to be done right: An induction proof always requires choosingonevariable to do induction on; but it is possible to nest an induction proof inside the induction step (or inside the induction base) of a different induction proof. For example, imagine that we are trying to prove that
ab=ba for anya∈Nandb∈N.
We can prove this by induction on a. More precisely, for each a ∈ N, we let A(a) be the statement (ab=bafor allb∈N). We then prove A(a) by induction on a. In the induction step, we fix m ∈ N, and we assume that A(m) holds; we now need to prove that A(m+1) holds. In other words, we need to prove that (m+1)b = b(m+1) for all b ∈ N. We can now prove this statement by induction onb(although there are easier options, of course). Thus, the induction proof of this statement happens inside the induction step of another induction proof. This nesting of induction proofs is legitimate (and even has a name: it is called double induction), but tends to be rather confusing (just think about what the sentence “The induction base is complete” means: is it about the induction base of the first induction proof, or that of the second?), and is best avoided when possible.
strong induction is by introducing auxiliary statements A(n), which say as much as “Theorem 2.70 holds under the requirement that a+b =n”:
Proof of Theorem 2.70. First of all, let us forget that we fixed aand b. So we want to prove that if a ∈Nand b ∈ N, then there exist g ∈ N, x∈ Z and y∈ Z such that g =ax+by and g| aand g| b.
For eachn ∈N, we let A(n) be the statement
if a ∈Nand b ∈Nsatisfy a+b =n, then there exist g ∈N, x ∈Z and y∈ Zsuch that g=ax+byand g| a and g| b
. (108) We claim thatA(n) holds for alln ∈N.
Indeed, let us prove this by strong induction onnstarting at 0:
Induction step: Let m∈ N. Assume that
(A(n) holds for every n∈ Nsatisfyingn<m). (109) We must then show thatA(m) holds.
To do this, we shall prove the following claim:
Claim 1: Let a ∈ N and b ∈ N satisfy a+b = m. Then, there exist g ∈N, x ∈Zand y ∈Zsuch that g =ax+byand g| aand g |b.
Before we prove Claim 1, let us show a slightly weaker version of it, in which we rename aand b asuand v and add the assumption thatu≥v:
Claim 2: Letu ∈ Nand v ∈Nsatisfy u+v=m and u≥v. Then, there existg∈ N, x∈ Zandy ∈Zsuch that g=ux+vy and g| uand g |v.
[Proof of Claim 2: We are in one of the following two cases:
Case 1: We havev=0.
Case 2: We havev6=0.
Let us first consider Case 1. In this case, we have v =0. Hence, v = 0 = 0u, so that u| v. Also,u·1+v·0=u. Thus, u= u·1+v·0 and u| u and u | v. Hence, there exist g ∈ N, x ∈ Z and y ∈ Z such that g = ux+vy and g | u and g | v (namely, g =u, x =1 andy =0). Thus, Claim 2 is proven in Case 1.
Let us now consider Case 2. In this case, we have v 6= 0. Hence, v > 0 (since v ∈ N). Thus, u+v >u+0= u, so that u < u+v = m. Hence, (109) (applied to n=u) yields thatA(u)holds. In other words,
if a∈ Nand b∈ Nsatisfy a+b =u, then there exist g∈ N, x∈ Z andy ∈ Zsuch that g=ax+by and g| aand g| b
(110) (because this is what the statementA(u) says).
Also, u−v ∈ N (since u ≥ v) and (u−v) +v = u. Hence, (110) (applied to a = u−v and b = v) shows that there exist g ∈ N, x ∈ Z and y ∈ Z such that
g = (u−v)x+vy and g | u−v and g | v. Consider these g, x and y, and denote them by g0, x0 and y0. Thus, g0 is an element of N, and x0 and y0 are elements ofZ satisfyingg0 = (u−v)x0+vy0 and g0 | u−vand g0 | v.
Now, we have g0 | u−v; in other words, u ≡ vmodg0. Also, g0 | v; in other words, v ≡ 0 modg0. Hence, u ≡ v ≡ 0 modg0, so that u ≡ 0 modg0. In other words, g0 | u. Furthermore,
g0 = (u−v)x0+vy0=ux0−vx0+vy0 =ux0+v y0−x0 .
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: We are in one of the following two cases:
Case 1: We havea≥b.
Case 2: We havea<b.
Let us first consider Case 1. In this case, we havea ≥b. Hence, Claim 2 (applied to u = a and v = b) shows that there exist g ∈ N, x ∈ Z and y ∈ Z such that g =ax+by and g| aand g| b. Thus, Claim 1 is proven in Case 1.
Let us next consider Case 2. In this case, we have a < b. Hence, a ≤ b, so that b ≥a. Also,b+a=a+b =m. Hence, Claim 2 (applied tou=b andv =a) shows that there exist g ∈ N, x ∈ Zand y ∈ Z such that g =bx+ay and g | b and g | a.
Consider theseg, xand y, and denote them by g0, x0 and y0. Thus, g0 is an element ofN, and x0 and y0are elements of Zsatisfyingg0 =bx0+ay0 and g0 |b and g0 | a.
Now, g0 = bx0+ay0 = ay0+bx0. Hence, there exist g ∈ N, x ∈ Z and y ∈ Zsuch that g = ax+by and g | a and g | b (namely, g = g0, x = y0 and y = x0). Thus, Claim 1 is proven in Case 2.
We have now proven Claim 1 in each of the two Cases 1 and 2. Thus, Claim 1 always holds (since Cases 1 and 2 cover all possibilities).]
ButA(m) is defined as the statement
if a ∈Nand b∈ Nsatisfy a+b =m, then there exist g ∈N, x ∈Z and y∈ Zsuch that g=ax+byand g| a and g| b
. Thus, A(m) is precisely Claim 1. Hence, A(m) holds (since Claim 1 holds). This completes the induction step. Thus, we have proven by strong induction thatA(n) holds for all n ∈ N. In other words, the statement (108) holds for all n ∈ N(since this statement is preciselyA(n)).
Now, let a ∈ N and b ∈ N. Then, a+b ∈ N. Hence, we can apply (108) to n= a+b (sincea+b =a+b). We thus conclude that there existg ∈N, x∈ Zand y∈ Zsuch that g =ax+byand g | aand g |b. This proves Theorem 2.70.
2.10.2. Conventions for writing proofs by strong induction on derived quantities
Let us take a closer look at the proof we just gave. The statement A(n) that we defined was unsurprising: It simply says that Theorem 2.70 holds under the condi-tion thata+b =n. Thus, by introducingA(n), we have “sliced” Theorem 2.70 into a sequence of statements A(0),A(1),A(2), . . ., which then allowed us to prove these statements by strong induction on n even though no “n” appeared in Theo-rem 2.70 itself. This strong induction can be simply called a “strong induction on a+b”. More generally:
Convention 2.72. Let B be a logical statement that involves some variables v1,v2,v3, . . .. (For example, B can be the statement of Theorem 2.70; then, these variables are aand b.)
Let g ∈Z. (This g has nothing to do with the gfrom Theorem 2.70.)
Letqbe some expression (involving the variablesv1,v2,v3, . . . or some of them) that has the property that whenever the variablesv1,v2,v3, . . . satisfy the assump-tions of B, the expression q evaluates to some element of Z≥g. (For example, if Bis the statement of Theorem 2.70 andg =0, thenqcan be the expressiona+b, because a+b ∈N=Z≥0whenever aand b are as in Theorem 2.70.)
Assume that you want to prove the statement B. Then, you can proceed as follows: For each n∈ Z≥g, define A(n) to be the statement saying that70
(the statementB holds under the condition that q =n). Then, prove A(n) by strong induction on nstarting at g. Thus:
• Theinduction step consists in fixingm∈ Z≥g, and showing that if A(n) holds for everyn∈ Z≥g satisfyingn<m
, (111)
then
(A(m) holds). (112)
In other words, it consists in fixing m∈ Z≥g, and showing that if
(the statementB holds under the condition that q <m), (113) then
(the statementB holds under the condition that q =m). (114) (Indeed, the previous two sentences are equivalent, because of the logical
equivalences
A(n) holds for every n∈ Z≥g satisfyingn <m
⇐⇒
(the statementB holds under the condition that q =n) holds for every n∈ Z≥g satisfyingn <m
since the statement A(n) is defined as
(the statementB holds under the condition that q =n)
⇐⇒ (the statementB holds under the condition that q <m) and
(A(m) holds)
⇐⇒ (the statementB holds under the condition that q =m) since the statementA(m) is defined as
(the statementB holds under the condition that q=m)
. )
In practice, this induction step will usually be organized as follows: We fix m ∈ Z≥g, then we assume that the statement B holds under the condition that q < m (this is the induction hypothesis), and then we prove that the statementB holds under the condition that q =m.
Once this induction proof is finished, it immediately follows that the statement B always holds (because the induction proof has shown that, whatevern ∈ Z≥g
is, the statement B holds under the condition that q =n).
This strategy of proof is called “strong induction on q” (or “strong induction over q”). Once you have specified what q is, you don’t need to explicitly define A(n), nor do you ever need to mention n.
Using this convention, we can rewrite our above proof of Theorem 2.70 as follows (remembering once again thatZ≥0 =N):
Proof of Theorem 2.70 (second version). Let us prove Theorem 2.70 by strong induc-tion ona+bstarting at 0:
Induction step: Let m∈ N. Assume that Theorem 2.70 holds under the condition that a+b < m. We must then show that Theorem 2.70 holds under the condition that a+b =m. This is tantamount to proving the following claim:
Claim 1: Let a ∈ N and b ∈ N satisfy a+b = m. Then, there exist g ∈N, x ∈Zand y ∈Zsuch that g =ax+byand g| aand g |b.
70We assume that no variable named “n” appears in the statementB; otherwise, we need a different letter for our new variable in order to avoid confusion.
Before we prove Claim 1, let us show a slightly weaker version of it, in which we rename aand b asuand v and add the assumption thatu≥v:
Claim 2: Letu ∈ Nand v ∈Nsatisfy u+v=m and u≥v. Then, there existg∈ N, x∈ Zandy ∈Zsuch that g=ux+vy and g| uand g |v.
[Proof of Claim 2: We are in one of the following two cases:
Case 1: We havev=0.
Case 2: We havev6=0.
Let us first consider Case 1. In this case, we have v =0. Hence, v = 0 = 0u, so that u| v. Also,u·1+v·0=u. Thus, u= u·1+v·0 and u| u and u | v. Hence, there exist g ∈ N, x ∈ Z and y ∈ Z such that g = ux+vy and g | u and g | v (namely, g =u, x =1 andy =0). Thus, Claim 2 is proven in Case 1.
Let us now consider Case 2. In this case, we have v 6= 0. Hence, v > 0 (since v ∈ N). Thus, u+v > u+0 = u, so that u < u+v = m. Also, u−v ∈ N(since u≥v) and(u−v) +v=u.
But we assumed that Theorem 2.70 holds under the condition that a+b < m.
Thus, we can apply Theorem 2.70 to a = u−v and b = v (since u−v ∈ N and (u−v) +v =u < m). We thus conclude that there exist g ∈ N, x ∈ Zand y ∈ Z such that g = (u−v)x+vy and g | u−v and g | v. Consider these g, x and y, and denote them by g0, x0 and y0. Thus, g0 is an element of N, and x0 and y0 are elements ofZsatisfyingg0 = (u−v)x0+vy0 and g0 | u−vand g0 | v.
Now, we have g0 | u−v; in other words, u ≡ vmodg0. Also, g0 | v; in other words, v ≡ 0 modg0. Hence, u ≡ v ≡ 0 modg0, so that u ≡ 0 modg0. In other words, g0 | u. Furthermore,
g0 = (u−v)x0+vy0=ux0−vx0+vy0 =ux0+v y0−x0 .
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.