Chapter 2. The Real Number System
5. Natural Numbers Induction
At the end of §2, we showed how to select from E1 the natural numbers 1, 2, 3, . . ., starting with 1 and then adding 1 to each preceding number to get the following one. This process also applies to any other field F; the ele- ments so selected are called the natural elements of F, and the set of all such elements (obtained by continuing the process indefinitely) is denoted by N. Note that, by this construction, we always have n+ 1∈N if n∈N.
∗A more precise approach to natural elements is as follows.1 A subsetS of a field F is called inductive iff
(i) 1∈S (S contains the unity element of F) and
(ii) (∀x∈S) x+ 1∈S (S contains x+ 1 wheneverx is in S).2
DefineN to be the intersection ofall such subsets. We then obtain the follow- ing.
∗Theorem 1. The set N so defined is inductive itself. In fact, it is the
“smallest” inductive set in F (i.e., contained in any other such set). Proof. We have to show that, with our new definition,
(i) 1∈N and
1The beginner may omit all “starred” passages and simply assume Theorems 1′
and 2′
below as additional axioms without proof.
2Such subsets do exist; e.g., the entire fieldFis inductive since 1∈Fand (∀x∈F)x+1∈
(ii) (∀x∈N)x+ 1∈N.
Now, by definition, the unity 1 is ineach inductive set; hence it also belongs to the intersection of such sets, i.e., to N. Thus 1∈N, as claimed.
Next, take any x ∈ N. Then, by our new definition of N, x is in every inductive set S. Hence, by property (ii) of such sets, alsox+ 1 is in every such S; thus x+ 1 is in the intersection of all inductive sets, i.e., x+ 1∈N, and so N is inductive, indeed.
Finally, by definition,N is thecommon part of all such sets, hence contained in each.
For applications, Theorem 1 is usually expressed as follows.
Theorem 1′ (First induction law). A proposition P(n) involving a natural n
holds for all n∈N in a field F if
(i) it holds for n= 1 [P(1) is true]; and
(ii) whenever P(n) holds for n = m, it holds for n = m+ 1; [P(m) =⇒
P(m+ 1)].
∗Proof. Let S be the set of all those n ∈ N for which P(n) is true; that is, S = {n ∈ N | P(n)}. We must show that actually each n ∈ N is in S, i.e., N ⊆S.
First, we show that S is inductive. By our assumption (i), P(1) is true, so 1∈S.
Next, supposex∈S. This means thatP(x) is true. But by assumption (ii), this implies P(x+ 1), i.e., x+ 1∈ S. Thus (∀x ∈ S) x+ 1 ∈S and 1∈S; so S is inductive. But then, by Theorem 1 (second clause), N ⊆S.
This theorem is widely used to prove general propositions on natural ele- ments, as follows. In order to show that some formula or proposition P(n) is true for every natural n, we first verify P(1), i.e., show that P(n) holds for n= 1. We then show that
(∀m∈N) P(m) =⇒ P(m+ 1);
that is, if P(n) holds for some value n=m, then it also holds forn=m+ 1. Once these two facts are established, Theorem 1′ ensures that P(n) holds for
all natural n.
Proofs of this kind are called inductive, or proofs by induction. Note that every such proof consists of two steps:
(i) P(1) and (ii) P(m) =⇒ P(m+ 1).
Special caution must be applied in step (ii). Here we temporarily assume thatP(n) has already been verified for someparticular (but unspecified) value
§5. Natural Numbers. Induction 65
n = m.3 From this assumption, we then try to deduce that P(n) holds for
n=m+ 1 as well. This fact must be proved; it would be a bad error to simply substitute m+ 1 for min the assumed formula P(m) since it was assumed for a particular value m, not for m+ 1. The following examples illustrate this procedure.4
Examples.
(A) If m and n are natural elements, so are m+n and mn. To prove it, fix any m ∈ N. Let P(n) mean that m+n ∈ N. We now verify the following:
(i) P(1) is true; for m ∈ N is given. Hence, by the very definition of N, m+ 1∈ N. But this means exactly that P(n) holds for n = 1, i.e., P(1) is true.
(ii) P(k) =⇒ P(k+1) (here we use a different letter,k, sincemis fixed already). Suppose that P(n) holds for some particular n=k. This means thatm+k ∈N. Hence, by the definition ofN, (m+k) + 1∈
N; or, by associativity, m+ (k + 1) ∈ N. But this means exactly thatP(k+ 1) is true (ifP(k) is). Thus, indeed,P(k) =⇒ P(k+ 1). Since (i) and (ii) have been established, induction is complete; that is, Theorem 1′ shows thatP(n) holds foreach n∈N, and this
means thatm+n∈N. Asmandnarearbitrary naturals, our first assertion is proved.5
To show that mn∈N also, we now let P(n) mean that mn∈N (for a fixed m ∈ N) and proceed similarly. We leave this to the reader.
(B) If n ∈ N, then n−1 = 0 or n−1 ∈ N. Indeed, let P(n) mean that n−1 = 0or n−1∈N (one of the two is required). We again verify the two steps:
(i) P(1) is true; for if n= 1, then n−1 = 1−1 = 0. Thus one of the two desired alternatives, namely n−1 = 0, holds if n = 1. Hence P(1) is true.
(ii) P(m) =⇒ P(m+1). SupposeP(n) holds for someparticular value n=m(inductive hypothesis). This means that either m−1 = 0 or m−1∈N.
In the first case, we have (m−1) + 1 = 0 + 1 = 1 ∈ N. But (m−1)+1 = (m+1)−1 by associativity and commutativity (verify!). Thus (m+ 1)−1∈N.
3This temporary assumption is called theinductive hypothesis.
4Actually, these examples are basic theorems on naturals, to be well noted.
5Note the technique we applied here. Faced withtwo variablesmandn, we fixed mand carried out the induction onn. This is a common procedure.
In the second case, m−1 ∈ N implies (m−1) + 1∈ N by the very definition of N. Thus, in both cases, (m+ 1)−1∈N, and this shows that P(m+ 1) is true if P(m) is.
As (i) and (ii) have been established, induction is complete. (C) In an orderedfield, all naturals are ≥1. Indeed, letP(n) now mean that
n≥1. As before, we again carry out the two inductive steps.
(i) P(1) holds; for if n = 1, then certainly n ≥ 1; so P(n) holds for n= 1.
(ii) P(m) =⇒ P(m+ 1). We make the inductive hypothesis thatP(m) holds for some particular m. This means that m ≥ 1. Hence, by monotonicity of addition and transitivity (Axioms II and VIII), we have m+ 1 ≥ 1 + 1 > 1 (the latter follows by adding 1 on both sides of 1 > 0). Thus m+ 1 > 1 and certainly m+ 1 ≥ 1, that is, P(m+ 1) holds (if P(m) does). Induction is complete.
(D) In an ordered field, m, n ∈ N and m > n implies m−n ∈ N. Indeed, fixing an arbitrary m ∈ N, let P(n) mean “m−n ≤ 0 or m−n∈ N.” Then we have the following:
(i) P(1)is true; for ifn= 1, thenm−n=m−1. But, by Example (B), m−1 = 0 or m−1 ∈ N. This shows that P(n) holds for n = 1; P(1) is true.
(ii) P(k) =⇒ P(k+1). SupposeP(k) holds for someparticular k ∈N. This means that
m−k ≤0 or m−k ∈N. By Example (B), it easily follows that either
(m−k)−1≤0 or (m−k)−1∈N; that is,
either m−(k+ 1)≤0 or m−(k+ 1)∈N.
But this shows that P(k + 1) holds (if P(k) does). By induction, then, P(n) holds for everyn∈N; that is,
either m−n≤0 or m−n∈N for every n∈N.
Lemma. For no naturals m, n in an ordered field is m < n < m+ 1.
For, by Example (D), n > m would imply n−m∈N, hencen−m≥1 (by Example (C)). Butn−m≥1, or n≥m+ 1, excludesn < m+ 1 (trichotomy). Thus m < n < m+ 1 is impossible for naturals.
§5. Natural Numbers. Induction 67 Theorem 2. In an ordered field, every nonempty subset of N (the naturals) has a least element, i.e., one not exceeding any other of its members.6
∗Proof. Given ∅ 6= A ⊆ N, we want to show that A has a least element. To do this, let
An ={x∈A |x≤n} n= 1, 2 . . . .
That is, An consists of those elements of A that are ≤n (An may be empty).
Now let P(n) mean
“either An =∅ or An has a least element.” (1)
We show by induction that P(n) holds for each n ∈ N. Indeed, we have the following:
(i)P(1)is true; for, by construction, A1 consists of all naturals fromA that
are ≤1 (if any). But, by Example (C), the only such natural is 1. ThusA1, if
not empty, consists of 1 alone, and so 1 is also its least member. We see that either A1 =∅ or A1 has a least element; i.e., P(1) is true.
(ii)P(m) =⇒ P(m+ 1). Suppose P(m) holds for someparticular m. This means that Am =∅ or Am has a least element (call it m0). In the latter case,
m0 is also the least member of Am+1; for, by the lemma, Am+1 differs from
Am by the element m+ 1 at most, which is greater than all members of Am.
If, however, Am = ∅, then for the same reason, Am+1 (if 6= ∅) consists of
m+ 1 alone; so m+ 1 is also its least element.
This shows that P(m+ 1) is true (if P(m) is). Thus the inductive proof is complete, and (1) holds for every An.
Now, by assumption,A 6=∅; so we fix some n∈A. Then the set An ={x∈A|x ≤n}
contains n, and henceAn 6=∅. Thus by (1),An must have a least elementm0,
m0 ≤ n. But A differs from An only by elements > n (if any), which are all
> m0. Thus m0 is the desired least element ofA as well.
Theorem 2 yields a new form of the induction law for ordered fields.
Theorem 2′ (Second induction law). A proposition P(n) holds for each nat-
ural n in an ordered field if (i′) P(1) holds, and
(ii′) whenever P(n) holds for all naturals n less than some m ∈ N, it also
holds for n=m.
6This is the so-calledwell-ordering property of N. A simpler proof for E1 will be given in§10. Thus the present proof may be omitted.
∗Proof. We use a so-called indirect proof or proof by contradiction. That is, instead of proving our assertion directly, we shall show that the opposite is false, and so our theorem must be true.
Thus assume (i′) and (ii′) and, seeking a contradiction, supposeP(n)failsfor
some n∈N (call suchn “bad”). Then these “bad” naturals form a nonempty subset ofN, call itA. By Theorem 2,A has a least memberm. Thusm is the least natural for which P(n) fails. It follows that all n less than m do satisfy P(n) (among them is 1 by (i′)). But then, by our assumption (ii′), P(n) also
holds for n=m, which is impossible sincem is “bad” by construction.
This contradiction shows that there cannot be any “bad” naturals, and the theorem is proved.
Note. In inductive proofs, Theorem 2′ is used in much the same manner
as Theorem 1′, but it leaves us more freedom in step (ii): instead of assuming
that justP(m) is true, we may assume thatP(1), P(2), . . . , P(m−1) are true.
Problem. Verify Example (A) for mn. (See other problems in §6.)