26. Suppose that there exist i and j such that si = sj. Prove that there exists k such that sk > A.
27. Prove that 2m+ 5n2= 20 has no solution in positive integers.
28. Prove that m3+ 2n2= 36 has no solution in positive integers.
29. Prove that 2m2+ 4n2− 1 = 2(m + n) has no solution in positive integers.
30. Prove that the product of two consecutive integers is even.
31. Prove that for every n∈ Z, n3+ n is even.
32. Use proof by cases to prove that|xy| = |x||y| for all real numbers x and y.
33. Use proof by cases to prove that|x + y| ≤ |x| + |y| for all real numbers x and y.
34. Define the sign of the real number x, sgn(x), as sgn(x)=
1 if x> 0 0 if x= 0
−1 if x < 0.
Use proof by cases to prove that|x| = sgn(x)x for every real number x.
35. Use proof by cases to prove that sgn(x y)= sgn(x)sgn(y) for all real numbers x and y (sgn is defined in Exercise 34).
36. Use Exercises 34 and 35 to give another proof that|xy| =
|x||y| for all real numbers x and y.
37. Use proof by cases to prove that max{x, y} + min{x, y} = x+ y for all real numbers x and y.
38. Use proof by cases to prove that
max{x, y} = x+ y + |x − y|
2 for all real numbers x and y.
39. Use proof by cases to prove that
min{x, y} = x+ y − |x − y|
2 for all real numbers x and y.
40. Use Exercises 38 and 39 to prove that max{x, y}+min{x, y} = x+ y for all real numbers x and y.
Taking the preceding equation as the definition of ordered pair, prove that (a, b) = (c, d) if and only if a = c and b= d.
47. Prove that the following are equivalent for the integer n:
(a) n is odd. (b) There exists k∈ Z such that n = 2k − 1.
(c) n2+ 1 is even.
48. Prove that the following are equivalent for sets A, B, and C:
(a) A∩ B =∅ (b) B⊆ A (c) A B = A ∪ B, where is the symmetric difference operator (see Exercise 91, Section 1.1).
49. Prove that the following are equivalent for sets A, B, and C:
(a) A∪ B = U (b) A∩ B =∅ (c) A⊆ B, where U is a universal set.
Problem-Solving Corner Proving Some Properties of Real Numbers
Problem
First some definitions:
(a) Let X be a nonempty set of real numbers. An upper bound for X is a real number a having the property that x≤ a for every x ∈ X.
(b) Let a be an upper bound for a set X of real numbers. If every upper bound b for X satisfies b≥ a, we call a a least upper bound for X.
A fundamental property of the real numbers is that ev-ery nonempty subset of real numbers bounded above has a least upper bound.
Answer the following where R serves as a univer-sal set:
1. Give an example of a set X and three distinct upper bounds for X , one of which is a least upper bound for X .
2. Prove that if a and b are least upper bounds for a set X , then a = b. We say that the least upper bound for a set X is unique. If a is the least upper bound of a set X , we sometimes write a= lub X.
3. Let X be a set with least upper bound a. Prove that ifε > 0, then there exists x ∈ X satisfying a− ε < x ≤ a.
4. Let X be a set with least upper bound a, and suppose that t > 0. Prove that ta is the least upper bound of the set{tx | x ∈ X}.
Attacking the Problem
To better understand the definitions, let’s construct ex-amples, write out the definitions in words, look at nega-tions of the defininega-tions, and draw pictures.
We’ll start with definition (a) and construct a sim-ple examsim-ple—taking X to be a small finite set, say
X = {1, 2, 3, 4}.
Now an upper bound a for X satisfies x≤ a for every x in X —here, we must have
1≤ a, 2 ≤ a, 3 ≤ a, 4 ≤ a.
Examples of upper bounds for X are 4, 6.9, 3π, 9072.
In words, definition (a) says that a is an upper bound for a set X if every element in X is less than or equal to a. We see that upper bounds e, f , and g for a set X (shown in color) look like
e f g
X
What would it mean that a is not an upper bound for a set X ? We would have to negate definition (a):
¬∀x(x ≤ a) or, equivalently, ∃x¬(x ≤ a) or ∃x(x >
a). In words, a is not an upper bound for a set X if there exists x in X such that x> a. Looking at the preceding picture, we see that any number less than e is not an upper bound for X .
Let’s turn to definition (b), which says, in words, that a is a least upper bound for a set X if, among all upper bounds for X , a is smallest. Looking ahead, prob-lem 2 is to show that there is only one (distinct) upper bound for a set X ; thus, we usually say the least up-per bound rather than a least upup-per bound. The least upper bound of our previous set
X= {1, 2, 3, 4}
is 4. We have already noted that 4 is an upper bound for X . If a is any upper bound for X , since 4 ∈ X, 4≤ a. Therefore, 4 is the least upper bound for X. In the preceding figure, e is a least upper bound for the set X .
Finding a Solution
Now we consider the problems.
[Problem 1.] Our previous example, X = {1, 2, 3, 4}, will suffice. We have noted that 4 is the
least upper bound for X . Any values greater than 4 serve as additional upper bounds.
[Problem 2.] One way to prove that two numbers a and b are equal is to show that a ≤ b and b ≤ a.
We’ll try this first. Another possibility is proof by con-tradiction and assume that a = b.
[Problem 3.] Here we can use the fact that a− ε is not an upper bound (since it’s less than the least upper bound) and, as discussed previously, what it means for a value to not be an upper bound.
[Problem 4.] Here we are given a value, ta, and asked to prove that it is the least upper bound of the given set, which here we denote as t X . Going directly to the definitions, we must show that
(a) z ≤ ta for every z ∈ t X (i.e., ta is an upper particular, a is an upper bound for X so
x≤ a for all x ∈ X.
How do we deduce the first inequality from the sec-ond? Multiply by t! We hope that the proof of part (b) proceeds in a similar way.
The least upper bound for X is 4. We have already noted that 4 is an upper bound for X . If a is any upper bound for X , since 4∈ X, 4 ≤ a. Therefore, 4 is the least upper bound for X .
[Problem 2.] Since a is a least upper bound for X and b is an upper bound for X , a ≤ b. Since b is a least upper bound for X and a is an upper bound for X , b≤ a. Therefore, a = b.
[Problem 3.] Letε > 0. Since a is the least upper bound for X and a−ε < a, a −ε is not an upper bound for X . Therefore, by definition (a) there exists x ∈ X such that a− ε < x. Since a is an upper bound for X, x ≤ a. We have shown that there exists x ∈ X such that a− ε < x ≤ a.
Problem-Solving Corner: Proving Some Properties of Real Numbers
89
[Problem 4.] Let t X denote the set {tx | x ∈ X}.
Next we prove part (b). Let b be an upper bound for t X . Then t x ≤ b for every x ∈ X (since an arbi-trary element in t X is of the form t x for some x∈ X).
Dividing by t and noting that t> 0, we have x ≤ b/t for every x ∈ X. Therefore b/t is an upper bound for X . Since a is the least upper bound for X , b/t ≥ a.
Multiplying by t and noting again that t> 0, we have b≥ ta. Therefore ta is the least upper bound for t X.
The proof is complete.
Summary of Problem-Solving Techniques
■ Before beginning a proof, familiarize yourself with relevant definitions, theorems, examples, and so on.
■ Construct additional examples—especially small examples (e.g., for sets look at some small finite sets).
■ Write out some of the technical statements in words.
■ Look at negations of statements.
■ Draw pictures.
■ If one proof technique seems not to be working, try another. For example, if a direct proof seems unpromising, try a proof by contradiction.
■ Review the Problem-Solving Tips sections in this chapter and the previous chapter.
Comments
The fact that every nonempty set of real numbers that is bounded above has a least upper bound is called the completeness property of the real numbers. The real numbers are complete in the sense that there are
no “holes” in the number line. Informally, if there was a hole in the line, the set of numbers to the left of the hole, although bounded above, would not have a least upper bound:
The set of rational numbers is not complete. The subset of rational numbers less than √
2 is bounded above, but does not have a rational least upper bound.
(The least upper bound of the subset of rational num-bers less than√
2 is the irrational number√ 2.)
Exercises
1. What is the least upper bound of a nonempty finite set of real numbers?
2. What is the least upper bound of the set {1 − 1/n | n is a positive integer}?
Prove your answer.
3. Let X and Y be nonempty sets of real numbers such that X ⊆ Y and Y is bounded above. Prove that X is bounded above and lub X ≤ lub Y .
4. Let X be a nonempty set. What is the least upper bound of the set{tx | x ∈ X} if t = 0?
5. Let X be a set with least upper bound a, and let Y be a set with least upper bound b. Prove that the set
{x + y | x ∈ X and y ∈ Y }
is bounded above and its least upper bound is a+ b.
Let X be a nonempty set of real numbers. A lower bound for X is a real number a having the property that x≥ a for every x ∈ X. Let a be a lower bound for a set X of real numbers. If every lower bound b for X satisfies b ≤ a, we call a a greatest lower bound for X .
6. Prove that if a and b are greatest lower bounds for a set X , then a= b.
7. Prove that every nonempty subset of real numbers bounded below has a greatest lower bound. Hint: If X is a nonempty set of real numbers bounded below, let Y denote the set of lower bounds. Prove that Y has a least upper bound, say a. Prove that a is the greatest lower bound for X .
8. Let X be a set with greatest lower bound a. Prove that ifε > 0, then there exists x ∈ X satisfying a+ ε > x ≥ a.
9. Let X be a set with least upper bound a, and let t < 0. Prove that ta is the greatest lower bound of the set{tx | x ∈ X}.