Download The natural numbers

The natural numbers
For a subset A of R, A is called inductive (or inductively closed) if for all
x, x ∈ A implies x + 1 ∈ A. One inductive set is R itself.
The set of natural numbers is defined to be the set
N=
\
{A : A is inductive with 1 ∈ A}.
Theorem. N is an inductive subset of R containing 1. If A is any inductive set
that contains 1, then A ⊃ N.
Proof. First, there is an inductive subset of R containing 1, namely R itself, so N
exists and is a subset of R.
T
From the definition of intersection, 1 ∈ {A : A is inductive and 1 ∈ A} = N.
To see that N is inductive, let x ∈ N. Then,
x ∈ A, for all inductive sets A with 1 ∈ A.
Hence
x + 1 ∈ A, for all inductive sets A with 1 ∈ A.
Therefore,
x+1∈
\
{A : A is inductive with 1 ∈ A} = N,
by definition. Thus x ∈ N =⇒ x + 1 ∈ N, so N is inductive.
For the second statement, let A be inductive with 1 belonging to it. If x ∈ N,
then x ∈ A, by definition of intersection. So N ⊂ A. The end of the proof is a special case of the general fact that the intersection
of a family of sets is always contained in each set in the family.
The above immediately yields the Principle of Mathematical Induction:
Theorem. (PMI)
(1) If A is an inductive set of natural numbers including the element 1, then A = N.
(2) If P (n) is a statement about natural numbers n, such that
(i) P (1) is true, and
(ii) whenever P (n) is true, P (n + 1) is true,
then P (n) is true for all n ∈ N.
Proof.
(1) By hypothesis, A ⊂ N; Also A is inductive with 1 ∈ A, so N ⊂ A by the
previous result. Thus A = N.
(2) Let A = {n ∈ N : P (n) is true }. By hypothesis, 1 ∈ A and x ∈ A =⇒
x + 1 ∈ A. Thus A is inductive, so A = N, by part (1). That is, for all n ∈ N,
P (n) is true. 13/9/2008 843 mam
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THE NATURAL NUMBERS
Theorem. For all n ∈ N, n ≥ 1.
Proof. Here the statement to be proved by induction on n is:
n ≥ 1.
Now,
(i) 1 ≥ 1 is true of course and,
(ii) if n ≥ 1, then n + 1 ≥ 1 + 0 ≥ 1.
Thus, by PMI n ≥ 1, for all n ∈ N. Corollary. 0 ∈
/ N.
Proof. If 0 ∈ N we would have 0 ≥ 1, by the previous theorem. But 0 < 1, so this
would be a contradiction. So, if n = 1, its predecessor n − 1 is not a natural number, but this is the
only such case.
Theorem. If n ∈ N and n > 1, then n − 1 ∈ N.
Proof. Since every natural number is greater than or equal to 1, saying n > 1 is
the same as saying n 6= 1. Let A be {n ∈ N : n = 1 or n − 1 ∈ N}.
Then A contains 1, by definition.
Suppose now n ∈ A. Then n ∈ N; hence, (n + 1) − 1 = n ∈ N.
Thus n ∈ A =⇒ n + 1 ∈ A, so A is inductive.
Hence N = A, by PMI. Thus, for all n ∈ N, n = 1 or n − 1 ∈ N, In other words,
n 6= 1 implies n − 1 ∈ N. The last step of the proof used the fact, from elementary logic, that the statement p ∨ q is equivalent to the statement (not p) =⇒ q. (∨ means “or” and =⇒
means “implies”.)
Theorem. If n ∈ N and n < m ∈ N, then n + 1 ≤ m.
Proof. We use induction on n: So let
A = {n ∈ N : for all m ∈ N, if n < m then
n + 1 ≤ m}.
Let m ∈ N. If 1 < m, then m − 1 ∈ N, so 1 ≤ m − 1, hence 1 + 1 ≤ m. Since m
was arbitrary, 1 ∈ A.
Suppose n ∈ A, and let m ∈ N again be arbitrary. If n + 1 < m, then 1 < m,
so m − 1 ∈ N and n < m − 1; hence, n + 1 ≤ m − 1, by the inductive hypothesis
(that is, since n ∈ A). Thus, (n + 1) + 1 ≤ m. And this was true for arbitrary m,
so A is inductive, and A = N, as required. In the above proof we used a general principle. If we have a statement to prove
about two natural numbers n and m, we try to do induction on just one of them (in
the above case, n). We get more power by making the inductive hypothesis strong,
saying “for all m ∈ N”. The idea is that if you assume more, you can prove more.
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THE NATURAL NUMBERS
19
Corollary. If n ∈ N and n < a < n + 1, then a ∈
/ N.
In other words, there is no natural number between successive ones.
Proof. Let n ∈ N and n < a < n + 1. If a ∈ N, then by the theorem n + 1 ≤ a,
which contradicts a < n + 1. Well-ordering property. Each non-∅ subset of N has a least element.
If A is a set of real numbers, to say that a0 is a least element of A means
a0 ∈ A and for each a ∈ A, a0 ≤ a. Clearly, a set A can have no more than one
least element (why?).
Proof. Let M ⊂ N and let M be non-empty.
Suppose M has no least element. We will show that this leads to a contradiction.
Let B be the set of n ∈ N for which n ≤ m, for all m ∈ M . We claim that
1 ∈ B and B is inductive.
First, 1 ∈ B. Indeed, 1 ≤ m, for all m ∈ M , since M is a set of natural
numbers.
Now let n ∈ B. Then
n ≤ m, for all m ∈ M ,
But n ∈
/ M , for otherwise n would be the least element of M which we are assuming
doesn’t exist. Thus,
n < m, for all m ∈ M ,
so
n + 1 ≤ m, for all m ∈ M ,
by the previous result, that is n + 1 ∈ B. Thus, n ∈ B implies n + 1 ∈ B.
This shows B is inductive, so contains all the elements of N. In other words,
for all natural numbers n,
n ≤ m, for all m ∈ M .
Since M 6= ∅ it contains some element, say m0 . But m0 is a natural number so
m0 ≤ m, for all m ∈ M .
So m0 is the least member of M after all, a contradiction. 13/9/2008 843 mam
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THE NATURAL NUMBERS
There is a strengthening of the Principle of Mathematical induction which is
often more convenient, known as the 2nd Principle of Mathematical Induction
or the Principle of Complete Induction. As usual {1, . . . , n} means the set of
all natural numbers k with 1 ≤ k ≤ n.
Theorem. (2nd PMI)
(1) Let B be a set of natural numbers such that 1 ∈ B and that for all n, {1, . . . , n} ⊂
B implies n + 1 ∈ B, then B = N.
(2) If P (n) is a statement about natural numbers n such that
(i) P (1) is true, and
(ii) whenever P (k) is true for all natural numbers k ≤ n, P (n + 1) is also true,
then P (n) is true for all n ∈ N.
Proof. Let A be the set of all n ∈ N such that {1, . . . , n} ⊂ B. We will show that
A is inductive.
Since 1 ∈ B, {1} ⊂ B, so 1 ∈ A.
Let n ∈ A. Then {1, . . . , n} ⊂ B, so n + 1 ∈ B. But then, {1, . . . , n + 1} ⊂ B
so that n + 1 ∈ A.
Thus, by the Principle of Mathematical Induction, A = N, so that for all n ∈ N,
{1, . . . , n} ⊂ B, in particular, N ⊂ B. Since B ⊂ N , B = N. Theorem. N is closed under addition.
The statement means that for all m, n ∈ N, m + n ∈ N. In the proof, we will
use the general principle mentioned earlier: We will choose one of m and n and
prove the statement with the other universally quantified.
Proof. Let
A = {n ∈ N :
m + n ∈ N,
for all m ∈ N}.
Then 1 ∈ A, because this just says m + 1 ∈ N for all m ∈ N, which we already
know.
Now let n ∈ A. Then, for each m ∈ N, m + n ∈ N, so
m + (n + 1) = (m + n) + 1 ∈ N,
so that n + 1 ∈ A.
Thus, A is inductive, so each n ∈ N belongs to A, which says that for all n ∈ N
and for all m ∈ N, m + n ∈ N, as required. In the above proof, how did we know to add n + 1 to the m? Because, we
assumed n ∈ A and we were trying to show n + 1 in A. For that, we needed to
show n + 1 satisfied the definition of the members of the set A. Thus, we fixed one
m and added n + 1 to it to see if m + (n + 1) still belonged to N. It did.
We also followed a common custom of not bothering to write out the line:
n ∈ A implies n + 1 ∈ A.
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THE NATURAL NUMBERS
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Theorem. N is closed under multiplication.
Proof. Let
A = {n ∈ N : mn ∈ N,
for all m ∈ N}
Then, 1 ∈ A, since 1 is a multiplicative identity, so suppose n ∈ A and fix m ∈ N.
Then, m(n + 1) = mn + m, by the distributive law. Since n ∈ A, mn ∈ N and thus,
by closure under addition, mn + m ∈ N. This proves n + 1 ∈ A, so A is inductive.
Therefore by PMI, for all n ∈ N and for all m ∈ N mn ∈ N. Definition. For n ∈ N and x ∈ R we define xn using the idea of induction: x1 = x,
and assuming xn defined we let xn+1 = xn x. A definition of this type is called a
recursive definition or a definition by recursion.
One also lets x0 = 1. (In some books, 00 is left undefined, but these still tend
to use it as though its definition were 00 = 1, for example in an expression such as
P
n
k
0
k=0 x , when x is allowed to be 0. We take 0 = 1 always. We have to be very
careful, though, not to give this expression properties that it doesn’t have.)
1. Sometimes the 2nd principle of induction is stated in only one step: If B is a set of natural
numbers such that n ∈ N and k ∈ B, for all k ∈ N with k < n implies n ∈ B, then B = N.
Show that in using this version one still has to prove 1 ∈ N.
2. If m, n ∈ N, with m > n, then m − n ∈ N.
3. The Binomial Theorem: Let a, b ∈ R (or in any field), for any n ∈ N, (a + b)n =
` ´
Pn `n´ k n−k
n!
. Here n is the binomial coefficient k!(n−k)!
. It satisfies Pascal’s
k=0 k a b
`n+1´ `n´ ` n ´k
Law: k+1 = k + k+1 .
4. The Bernoulli inequalities: For n ∈ N, if b > −1, (1+b)n ≥ 1+nb; if b < 1, (1−b)n ≥ 1−nb.
P
n−i bi−1 .
5. For each n ∈ N, an − bn = (a − b) n
i=1 a
6. If S is a non-empty finite subset of R, then S has a maximum (that is, a largest element) and
a minimum ( smallest element). (This can be proved by induction on the number of elements
of S.)
7. Prove that if S is a non-empty finite subset of R, then S has a maximum, by induction on
the number of elements of S.
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