Download Section 1.2-1.3

36
Chapter 1. The Integers
Exercises 1.1
1. Prove the following theorem:
Theorem. Let m and a be integers. If m | a and a | m, then m = ±a.
2. Prove the following theorem:
Theorem. For all integers a, b and k, with a 6= 0 and k 6= 0, we have that a | b if and only if
ak | bk.
1.2
The Well-Ordering Principle
An important property of the set of natural numbers N is that any nonempty2 subset of N has a
least element. This property follows from a principle that is often applied in mathematical proofs.
Well-Ordering Principle. Let S be a non-empty set of integers all of which are larger than some
fixed integer. Then S has a least element.
Proof Strategy 1.2.1 (Well-ordering proof strategy). Let b be a fixed integer and let n be an
integer variable. Given a diagram of the form
Prove (8n
b) P (n)
use the diagram
To prove (8n b) P (n):
Assume that ¬P (n) holds for some integer n b.
Let N b be the least such integer satisfying ¬P (N ).
Derive “a contradiction.”
Definition 1.2.2. A natural number p > 1 is a prime number if and only if for all natural
numbers a and b, if p = ab then either a = 1 or b = 1.
When a natural number n > 1 is not a prime, then we shall say that n is a composite number.
Definition 1.2.3. A natural number n is called a composite number if and only if there are
natural numbers a and b such that n = ab where a 6= 1 and b 6= 1.
Euclid used proof strategy 1.2.1 to prove the following lemma.3
Lemma 1.2.4. Every natural number n > 1 is divisible by a prime number.
Proof. Suppose, for a contradiction, that there is a natural number n > 1 that is not divisible by a
prime. By the well-ordering principle, there is a least natural number N > 1 that is not divisible
by a prime. Thus we have that
(i) N is not divisible by a prime,
(ii) for any natural number k, if 1 < k < N then k is divisible by a prime.
2
3
A set is nonempty if it has at least one element in it.
In mathematics, a lemma is a ‘little theorem’ that is usually used in the proof of a more important theorem.
1.3 Proof by Mathematical Induction
37
We now consider two cases: either N is a prime or it is not a prime.
Case 1: N is a prime. Since N is divisible by itself, this contradicts (i).
Case 2: N is not a prime. Thus N is a composite number. So, N = ab for some natural numbers
a and b with 1 < a < N and 1 < b < N . Since 1 < a < N , assertion (ii) implies that p | a for some
prime p. Because a | N , Theorem 1.1.2(c) implies that p | N . Hence, N is divisible by a prime. This
contradicts (i) and completes the proof.
Exercises 1.2
1. Prove the statement: For all natural numbers n
k and m where k 0 and m is odd.
2. Prove that for every natural number n
1, we can write n = 2k · m for some integers
0, either n is even or n is odd.
3. Let P (n) be a statement where n stands for a natural number. Suppose that the following two
conditions hold: (a) P (1) is true, and (b) for all n 1, if P (n) holds then P (n + 1) also holds.
Use the well-ordering principle to prove that P (n) must be true for all natural numbers n 1.
Exercise Notes: For Exercise 1, use the well-ordering proof strategy 1.2.1 where P (n) is the statement “n = 2k · m for some integers k and m where k 0 and m is odd.” There are two cases to
consider about N . Case 1: N is even; and Case 2: N is odd. If N is even, then N = 2i for some
natural number i where 1  i < N . If N is odd, then N = 20 N . For Exercise 2, use the proof
strategy 1.2.1. Show that N > 0. Note that either N 1 is even or N 1 is odd. For Exercise 3, if
N is the least natural number satisfying ¬P (N ), then explain why N = n + 1 for some n 1 and
observe that n < N .
1.3
Proof by Mathematical Induction
Mathematical induction is a powerful method for proving theorems about the natural numbers.
Suppose you have a statement P (n) that you want to prove is true for every integer n greater than
or equal to the integer b. How can you prove this statement by mathematical induction? First
you prove that the statement definitely holds for b. Then you have to prove that whenever the
statement holds for an integer n b, then it must hold for the next integer n + 1 as well. In other
words, mathematical induction is a method of proof that works by first proving the statement is
true for a starting value b, which is called the base step. Then one must prove the inductive step,
which shows that the truth of the statement P (n) implies the truth of the statement P (n + 1). If
the base step and the inductive step are both proven, then the statement P (n) is true for all the
natural numbers n b.
Inductive step:
Base step:
······
b
b+1
b+2
······
n
n+1
Figure 1.1: The base step and inductive step force the dominoes to fall, one and all.
38
Chapter 1. The Integers
It may be helpful to think of the domino e↵ect where one is presented with an infinite row
of dominoes, each standing on end, as pictured in Figure 1.1. The base step shows that the first
domino will fall. The inductive step ensures that each domino is perfectly aligned with the one
ahead of it. Thus, we know that the first domino will fall (base step) and whenever the n-th domino
falls, the n + 1 domino will also fall (inductive step). Therefore, all of the dominoes must fall.
Now for the mathematical details
The well-ordering principle implies the following related principle (see Exercise 3 on page 37).
Principle of Mathematical Induction. Let P (n) be a statement about an integer variable n.
Let b be a given integer. Suppose that the following two statements hold:
1. P (b) is true.
2. For all n
b, if P (n) then P (n + 1).
Then for all integers n
b, P (n) is true.
There are many mathematical statements having the form: For every integer n b, “something
about n happens,” where b is some fixed integer. The Principle of Mathematical Induction motivates
the following essential proof strategy called proof by mathematical induction.
Proof Strategy 1.3.1. Let b be a fixed integer. To prove a statement of the form (8n
by mathematical induction, use the diagram
b) P (n)
Prove P (b).
Prove (8n b) [P (n) ! P (n + 1)].
In other words, use the diagram
Base step:
Inductive step:
Prove P (b).
Let n b be arbitrary.
Assume P (n).
Prove P (n + 1).
In a proof of a statement (8n b) P (n) by mathematical induction, b is referred to as the base
value. The proof of P (b) is called the base step and the proof of (8n
b) [P (n) ! P (n + 1)] is
called the inductive step. In the latter proof diagram of proof strategy 1.3.1, the assumption P (n)
is called the induction hypothesis (IH) and the statement to be proven, P (n + 1), is called the
induction conclusion (IC).
In the base step you must show that the statement P (b) is true. To do so, simply replace n by
b everywhere in P (n) and verify that P (b) holds.
The inductive step is more challenging. It requires you to reach the conclusion that P (n + 1) is
true after assuming P (n) is true. To prove that P (n + 1) is true, you should somehow try to rewrite
the statement P (n + 1) in terms that relate to the assumption P (n) (as will soon be illustrated),
for then you will be able to make use of the assumption P (n). Appealing to the assumption P (n)
is referred to as using the induction hypothesis. After establishing that P (n + 1) is true, the proof
will be complete. A proof that uses strategy 1.3.1 is also called an induction proof or proof by
induction.
Our next theorem can be proved in more than one way; however, we shall provide a proof that
uses mathematical induction. We will first present a “proof analysis,” and then we will prove the
theorem by induction.
Theorem 1.3.2. For every natural number n
1, the natural number n2 + n + 2 is even.
1.3 Proof by Mathematical Induction
39
Proof Analysis. When you are looking for a proof by induction, it is helpful to write down the
statement P (n). In this case we have
P (n) :
n2 + n + 2 is even.
In this theorem our base value is b = 1. Next, we construct a proof diagram using the induction
proof strategy 1.3.1:
Base step:
Inductive step:
Prove P (1).
Let n 1 be arbitrary.
Assume P (n).
Prove P (n + 1).
Now, carefully write out the statements P (1) and P (n+1). Replacing n everywhere in the statement
P (n) :
n2 + n + 2 is even
P (1) :
12 + 1 + 2 is even.
with 1, we obtain
By carefully replacing n with n + 1 everywhere in P (n), we obtain
P (n + 1) :
(n + 1)2 + (n + 1) + 2 is even.
Thus, we can rewrite our proof diagram as
Base step:
Inductive step:
Prove 12 + 1 + 2 is even.
Let n 1 be arbitrary.
Assume n2 + n + 2 is even.
Prove (n + 1)2 + (n + 1) + 2 is even.
Now, we write out what it means for these values to be even as follows:
Base step:
Inductive step:
Prove 12 + 1 + 2 = 2k for some k 2 Z.
Let n 1 be arbitrary.
Assume n2 + n + 2 = 2i where i 2 Z.
Prove (n + 1)2 + (n + 1) + 2 = 2j for some j 2 Z.
For the base step we see that 12 + 1 + 2 = 2 · 2, which is clearly even. For the inductive step, we
must use the induction hypothesis,
n2 + n + 2 = 2i where i 2 Z,
(IH)
to deduce the induction conclusion
(n + 1)2 + (n + 1) + 2 = 2j for some j 2 Z.
(IC)
To prove this induction conclusion, we will begin with the left hand side of the equality in (IC) and,
using algebra, will make some changes so that the left hand side of the equation in (IH) appears.
In this case,
(n + 1)2 + (n + 1) + 2 = (n2 + n + 2) + 2(n + 1).
(?)
From the induction hypothesis (IH) we have that n2 + n + 2 = 2i. Upon substituting this into
equation (?), we will be able to prove (IC) and thus conclude that (n + 1)2 + (n + 1) + 2 is even.
40
Chapter 1. The Integers
Our proof diagrams and analysis will now guide the composition of a well-structured proof of
the theorem by mathematical induction.
Proof of Theorem 1.3.2. We prove, by mathematical induction, that n2 +n+2 is even for all n
1.
Base step: For n = 1, we see that 12 + 1 + 2 = 2 · 2 is even.
1 and assume the induction hypothesis that n2 + n + 2 is even, that is,
Inductive step: Let n
n2 + n + 2 = 2i where i 2 Z.
(IH)
We now show that (n + 1)2 + (n + 1) + 2 is even as follows:
(n + 1)2 + (n + 1) + 2 = (n2 + n + 2) + 2(n + 1)
by algebra
= 2i + 2(n + 1)
by induction hyp (IH)
= 2(i + n + 1)
by algebra.
Hence, (n + 1)2 + (n + 1) + 2 = 2j where j = i + n + 1 is an integer. Therefore, (n + 1)2 + (n + 1) + 2
is even and this completes the proof.
1.3.1
Strong Mathematical Induction
A variation of “proof by induction” arises when, in the inductive step, you are having difficulty
relating P (n + 1) to P (n), or when the relationship you discover does not prove fruitful. Suppose,
however, that you can relate P (n) to P (k) for some (or all) k satisfying b  k < n, where b is the
base value. In this case, you would like to use the fact that P (k) is true to conclude that P (n) is
true; but, can you assume that P (k) is true? The answer is yes and, in fact, you can assume all
the statements P (b), P (b + 1), . . . , P (n 1) are true.
It may be helpful, once again, to think of the domino e↵ect as pictured in Figure 1.2. The base
step shows that the first domino will fall. The inductive step ensures that each domino is perfectly
aligned with all of the dominos that come before it. Thus, we can be sure that the first domino
will fall (base step) and whenever all of the dominoes prior to the n-th domino fall, we know that
the n-th domino will also fall (inductive step). Therefore, all of the dominoes must fall.
Inductive step:
Base step:
······
b
b+1
b+2
······
n 1
n
Figure 1.2: Strong induction forces all of the dominoes to fall.
The proof of a statement of the form (8n b)P (n) by strong induction is very much like a proof
by mathematical induction. Strong induction just changes the focus slightly. Rather than proving
P (n + 1), one must prove P (n) by relating P (n) to one (or all) of the statements P (k) where k < n.
Principle of Strong Induction. Let P (n) be a statement about an integer variable n and let b
be a fixed integer. Suppose that
1. P (b) is true, and
2. the statements P (b), P (b + 1), . . . , P (n
Then for all integers n
b, P (n) is true.
1) imply P (n), for all n > b.
1.3 Proof by Mathematical Induction
41
Remark. A more formal way to say that all the statements P (b), P (b + 1), . . . , P (n
say that 8k(b  k < n ! P (k)).
1) hold is to
The principle of strong induction justifies the following proof strategy, which is often used in
mathematics.
Proof Strategy 1.3.3. Let b be a fixed integer. To prove a statement of the form (8n
by strong induction, use the diagram
b) P (n)
Prove P (b).
Prove (8n > b) [(8k(b  k < n ! P (k)) ! P (n)].
In other words, use the diagram
Base step:
Inductive step:
Prove P (b).
Let n > b be arbitrary.
Assume P (k) whenever b  k < n.
Prove P (n).
Exercises 1.3
1. Using mathematical induction, prove that for every natural number n
n2 + 3n + 5 is odd.
2. Prove, by mathematical induction, that for every natural number n
0, the natural number
0, n is even or n is odd.
3. Using Exercise 2, prove that every integer n is either even or odd.
4. Using Exercise 3, prove that for every integer n we have that n(n + 1) is even.
5. For any natural number n 1, let P (n) be the statement: 1 + 2 + · · · + n = n(n+1)
. What is
2
the statement P (3)? What is the statement P (1)? What is the statement P (n + 1)?
6. Prove that 1 + 2 + · · · + n =
n(n+1)
2
for all natural numbers n
7. Prove that for every natural number n
8. Prove that for every natural number n
9. Prove that for every natural number n
10. Prove that for every natural number n
1, 3 | (4n
1, 8 | (9n
1, by mathematical induction,.
1).
1).
2, 2n > n.
6, n! > 2n+2 .
Exercise Notes: For Exercise 2, use the induction proof strategy
Base step:
Inductive step:
Prove 0 is even or odd.
Let n 0 be arbitrary.
Assume n is even or n is odd.
Prove n + 1 is even or n + 1 is odd.
In the inductive step, since you are assuming an ‘or’ statement, use a proof by cases. For Exercise 3,
there are two cases: (1) n 0; (2) n < 0. For Exercise 4, there are two cases: (1) n is odd; (2) n
is even.