The Choose Method
6.1 HOW TO USE SPECIALIZATION
Quantifiers III:
Specialization
The previous two chapters illustrate how to proceed when a quantifier appears in the statement B. A method is introduced in this chapter for working forward from a statement that contains the universal quantifier “for all.”
6.1 HOW TO USE SPECIALIZATION
When the statement A contains the quantifier “for all” in the standard form:
A: For all “objects” with a “certain property,”
“something happens,”
one typical method emerges for working forward from A—specialization.
In general terms, specialization works as follows. As a result of assuming A is true, you know that, for all objects with the certain property, something happens. If, at some point, you were to come across one of these objects that does have the certain property, then you can use the information in A by being able to conclude that, for this particular object, the something does indeed happen. That fact should help you to conclude that B is true. In other words, you will have specialized the statement A to one particular object having the certain property.
To illustrate the idea of specialization in a more tangible way, suppose you know that
A: All cars with 4 cylinders get good gas mileage.
69
In this statement, you can identify the following three items:
Objects:
Certain property:
Something happens:
cars.
with 4 cylinders.
get good gas mileage.
Suppose that you are interested in buying a car that gets good gas mileage, so your objective is
B: To buy a car that gets good gas mileage.
You can work forward from the information in A by specialization to establish B as follows. Suppose you are in a dealer’s lot one day and you see a particular car that you like. Talking with the salesperson, suppose you verify that the car has 4 cylinders. Recalling that the foregoing statement A is assumed to be true, you can use this information to conclude that
A1: This particular car gets good gas mileage.
In other words, you have specialized the for-all statement in A to one particular object with the certain property.
If you analyze this example in detail, you can identify the following steps associated with applying specialization to a forward statement of the form:
A: For all “objects” with a “certain property,”
“something happens.”
Steps for Using Specialization
1. Identify, in the for-all statement, the object and its type, the certain property, and the something that happens.
2. Look for one particular object with the certain property that you can apply specialization to.
3. Conclude, by writing a new statement in the forward process, that the something happens for this particular object.
The following example demonstrates the proper use of specialization. To that end, suppose you know that
A: For all real numbers x, y≥ 0, x + y ≥ 2√xy.
In the foregoing statement, you can identify the objects (real numbers x and y), the certain property (being ≥ 0), and the something that happens (x + y ≥ 2√xy). You can therefore specialize this statement to any two real numbers that are≥ 0 (that is, objects with the certain property). For example, the result of specializing A to x = 3 and y = 27 is
A1: 3 + 27≥ 2p
3(27), or equivalently, 30≥ 18.
6.1 HOW TO USE SPECIALIZATION 71
Alternatively, for given real numbers a and b, the result of specializing A to x = a2≥ 0 and y = b2≥ 0 is
A2: a2+ b2≥ 2√ ab.
Specialization is now used in a complete proof.
Definition 14 A real number u is an upper bound for a set of real numbers T if and only if for all elements t∈ T , t ≤ u.
Proposition 8 If R is a subset of a set S of real numbers and u is an upper bound for S, then u is an upper bound for R.
Analysis of Proof. The forward-backward method gives rise to the key question, “How can I show that a real number (namely, u) is an upper bound for a set of real numbers (namely, R)?” Definition 14 is used to answer the question. Thus it must be shown that
B1: For all elements r∈ R, r ≤ u.
The appearance of the quantifier “for all” in the backward process suggests proceeding with the choose method, whereby one chooses
A1: An element, say r, in R, for which it must be shown that
B2: r≤ u.
(Here, the symbol r is used both for the chosen object in A1 and the general object in the for-all statement in B1, though they have different meanings.)
Turning now to the forward process, you will see how specialization is used to reach the conclusion that r ≤ u in B2. From the hypothesis that R is a subset of S, and by Definition 12 on page 55, you know that
A2: For each element x∈ R, x ∈ S.
Recognizing the keywords “for each” in the forward process, you should con-sider using specialization. According to the discussion preceding Proposition 8, the first step in doing so is to identify, in A2, the object with its type (element x∈ R), the certain property (there is none) and the something that happens (x ∈ S). Next, you must look for one particular object with the certain property with which to specialize. Recall call that, as a result of the backward process, you chose the particular element r ∈ R (see A1). The final step of specialization is to conclude, by writing a new statement in the forward process, that the something in A2 happens for the particular object r∈ R. In this case, specialization of A2 allows you to conclude that
A3: r∈ S.
The proof is not yet complete because the last statement in the backward process (B2) has not yet been reached in the forward process. To do so, continue to work forward. For example, from the hypothesis you know that u is an upper bound for S. By Definition 14 this means that
A4: For every element s∈ S, s ≤ u.
Once again, the appearance of the quantifier “for every” in the forward process suggests using specialization. Accordingly, identify, in A4, the object with its type (element s∈ S), the certain property (there is none), and the something that happens (s ≤ u). Now look for one particular object with the certain property with which to apply specialization. The same element r chosen in A1 serves the purpose noting, from A3, that r∈ S. Specialization then allows you to conclude that, for this particular object r∈ S, the something in A4 happens, so
A5: r≤ u.
The proof is now complete because A5 is the last statement obtained in the backward process (see B2).
In the condensed proof that follows, note the lack of reference to the forward-backward, choose, and specialization methods.
Proof of Proposition 8. To show that u is an upper bound for R, let r∈ R. (The word “let” here indicates that the choose method is used.) By hypothesis, R ⊆ S and so r ∈ S. (Here is where specialization is used.) Furthermore, by hypothesis, u is an upper bound for S, thus, every element in S is≤ u. In particular, r ∈ S, so r ≤ u. (Again specialization is used.)
When using specialization, be careful to keep your notation and symbols in order. Doing so involves a correct “matching up of notation,” similar to what you learned in Chapter 3 when using definitions. To illustrate, suppose you are going to apply specialization to a statement of the form:
A: For all objects X with a certain property, something happens.
When looking for a particular object, say Y , with which to specialize, it is necessary to verify that Y satisfies the certain property in A. To do so, replace X with Y everywhere in the certain property in A and see if the resulting condition is true. Similarly, when concluding that the particular object Y satisfies the something that happens in A, again replace X everywhere with Y in the something that happens to obtain the correct statement in the forward process. (This is done when writing statements A3 and A5 in the foregoing analysis of the proof of Proposition 8.) Be careful of overlapping notation, for example, when the particular object you have identified has precisely the same symbol as the one in the for-all statement you are specializing.
6.2 READING A PROOF 73
6.2 READING A PROOF
The process of reading and understanding a proof that uses specialization is demonstrated with the following proposition.
Definition 15 A real number u is a least upper bound for a set S of real numbers if and only if (1) u is an upper bound for S and (2) for every upper bound v for S, u≤ v.
Proposition 9 If v∗and w∗are least upper bounds for a set T , then v∗= w∗. Proof of Proposition 9. (For reference purposes, each sentence of the proof is written on a separate line.)
S1: From the hypothesis, both v∗and w∗are upper bounds for T . S2: Because v∗is a least upper bound for T , v∗≤ u, for any upper
bound u for T .
S3: In particular, w∗is an upper bound for T , so v∗≤ w∗. S4: Similarly, w∗ is a least upper bound for T and, because v∗ is
an upper bound for T , w∗≤ v∗. S5: It now follows that v∗= w∗. The proof is now complete.
Analysis of Proof. An interpretation of statements S1 through S5 follows.
Interpretation of S1: From the hypothesis, both v∗and w∗are upper bounds for T .
The author is working forward from the hypothesis using part (1) of the definition of a least upper bound to claim that
A1: v∗ and w∗ are upper bounds for T .
When reading a proof, it is advisable to determine where the author is heading. To do so, work backward from B yourself. In this case, you are led to the key question, “How can I show that two real numbers (namely, v∗ and w∗) are equal?” Read forward in the proof to see how the author answers this question. From S3 and S4, the answer in this case is to show that
B1: v∗≤ w∗ and w∗≤ v∗.
Interpretation of S2: Because v∗ is a least upper bound for T , v∗≤ u, for any upper bound u for T .
The author is continuing to work forward by stating part (2) of the defini-tion of a least upper bound applied to v∗; that is,
A2: For every upper bound u for T , v∗≤ u.
Interpretation of S3: In particular, w∗is an upper bound for T , so v∗≤ w∗.
It is here that specialization is applied to the for-all statement in A2, indi-cated by the words “in particular.” Specifically, A2 is specialized to the value u = w∗, which is an upper bound for T (see A1). The result of specialization, as the author claims is S3, is
A3: v∗≤ w∗.
Interpretation of S4: Similarly, w∗is a least upper bound for T and, because v∗ is an upper bound for T , w∗≤ v∗.
The author is using the same analysis as in S3 but, this time, applied to the least upper bound w∗ and the upper bound v∗ for T . The result of this specialization is that
A4: w∗≤ v∗.
Interpretation of S5: It now follows that v∗ = w∗, and so the proof is complete.
The author is working forward by combining v∗≤ w∗from A3 and w∗≤ v∗ from A4 to claim correctly that v∗ = w∗. Finally, the author states that the proof is complete, which is true because the conclusion B has been established.
Summary
You now have various techniques for dealing with quantifiers that can appear in either A or B. As always, let the form of the statement guide you. When B contains the quantifier “there is,” the construction method is used to produce the desired object. The choose method is associated with the quantifier “for all” in the backward process. Finally, if the quantifier “for all” appears in the forward process, use specialization. To do so, follow these steps:
1. Identify, in the for-all statement, the object with its type, the certain property, and the something that happens.
2. Look for one particular object with the certain property that you can apply specialization to. (This object often arises as a result of the back-ward process, especially when the choose method is used.)
3. Conclude, by writing a new statement in the forward process, that the something happens for this one particular object.
It is common to confuse the choose method with the specialization method.
Use the choose method when you encounter the keywords “for all” in the backward process; use specialization when the keywords “for all” arise in the forward process. Another way to say this is to use the choose method when you want to show that “for all objects with a certain property, something happens”; use specialization when you know that “for all objects with a certain property, something happens.”
CHAPTER 6: EXERCISES 75
All of the statements thus far have contained only one quantifier. In the next chapter you will learn what to do when statements contain more than one quantifier.
Exercises
Note: Solutions to those exercises marked with a W are located on the web at http://www.wiley.com/college/solow/.
Note: All proofs should contain an analysis of proof and a condensed version.
Definitions for all mathematical terms are provided in the glossary at the end of the book.
W6.1 Suppose you want to specialize the statement that “for every object with a certain property, something happens” to the particular object Y . Explain why you need to show that Y has the certain property before you can do so.
W6.2 Suppose you are working backward and want to show that, for a partic-ular object Y with a certain property P , S happens. When and how can you reach this conclusion by working forward from the statement that, for every object X with a certain property Q, T happens? State your answer in terms of the objects, the certain properties, and the somethings that happen.
W6.3 For each definition in Exercise 5.1 on page 63, explain how you would work forward from the associated for-all statement. For example, to work forward from the for-all statement in Exercise 5.1(b), (1) look for a specific element, say y, with which to apply specialization, (2) show that y∈ S, and (3) conclude that g(y)≥ f(y) as a new statement in the forward process.
6.4 For each definition in Exercise 5.2 on page 63, explain how you would work forward from the associated for-all statement.
6.5 Would you use specialization to prove each of the following statements?
Why or why not? Explain.
a. If a, b, and c are real numbers for which there is a real number x 6= 0 such that ax2+ bx + c = 0, then cx2+ bx + a has a rational root.
b. If a and b are real numbers with a < 0 and y = −b/(2a), then for all real numbers x, ax2+ bx≤ ay2+ by.
c. If a6= 0 and b are real numbers and y = −b/(2a) satisfies the property that, for all real numbers x, ax2+ bx≤ ay2+ by, then a < 0.
d. If f is a real-valued function such that, for all real numbers x, y, and t with 0 ≤ t ≤ 1, f(tx + (1 − t)y) ≤ tf(x) + (1 − t)f(y), then the set C ={real numbers x : f(x) ≤ 0} is a convex set.
6.6 For each of the following for-all statements, what properties must the given object satisfy so that you can apply specialization? Given that the object does satisfy those properties, what can you conclude about the object?
a. Statement: For all prime numbers p, p + 7 is composite.
Given object: An integer m.
b. Statement: For all elements x > 0 in a set S of real numbers, x is a root of the polynomial p(x).
Given object: A real number y.
c. Statement: Every triangle ABC with sides of length a = BC , b = CA, and c = AB satisfies c2= a2+b2−2ab cos(C).
Given object: The isosceles right triangle ABC whose legs a = BC and b = CA are both equal to m.
d. Statement: For all pairs of equilateral triangles ABC and DEF , if one side of triangle ABC is parallel to one side of trian-gle DEF , then the other two sides of triantrian-gle ABC are parallel to the corresponding sides of triangle DEF . Given object: The triangle CDE whose side DE is parallel to side
DA of triangle F DA.
W6.7 To what specific object could you specialize each of the following for-all statements so that the result of specialization leads to the desired conclusion?
Verify that the object to which you are applying specialization satisfies the certain property in the for-all statement so that you can apply specialization.
a. For-all statement: For all angles α and β, sin(α + β) = sin(α) cos(β) + cos(α) sin(β).
Desired conclusion: For a particular angle X, sin(2X) = 2 sin(X) cos(X).
b. For-all statement: For any sets S and T , (S∪T )c= Sc∩Tc(where Xc is the complement of the set X).
Desired conclusion: For two sets A and B, (A∩ B)c= Ac∪ Bc.
∗6.8 To what specific object could you specialize each of the following for-all statements so that the result of specialization leads to the desired conclusion?
Verify that the object to which you are applying specialization satisfies the certain property in the for-all statement so that you can apply specialization.
a. For-all statement: f is a function of one variable such that, for all real numbers x, y, and t with 0 ≤ t ≤ 1, f(tx + (1− t)y) ≤ tf(x) + (1 − t)f(y).
Desired conclusion: the function f satisfies f(1/2)≤ (f(0) + f(1))/2.
CHAPTER 6: EXERCISES 77
b. For-all statement: For all real numbers c and d for which c√ 2≥ d2, c2− d2≤ c.
Desired conclusion: For the real numbers a, b≥ 0,√
ab≤ (a+b)/2.
∗6.9 Answer the given questions about the following proof that, “If f is a convex function and y is a real number, then C ={real numbers x : f(x) ≤ y}
is a convex set (see the definitions in Exercise 5.2(b) and (c) on page 63).
Proof. Let a, b∈ C, and t be a real number with 0 ≤ t ≤ 1.
Because f is convex, f(ta + (1− t)b) ≤ tf(a) + (1 − t)f(b).
Furthermore, f(a)≤ y and f(b) ≤ y and hence it follows that tf(a)+(1−t)f(b) ≤ ty+(1−t)y = y. Thus, f(ta+(1−t)b) ≤ y and so ta + (1− t)b ∈ C.
a. Explain what the author is doing in the first sentence of the proof. What techniques have been used?
b. Where and how is specialization used?
c. Why is the author justified in saying that f(a)≤ y and f(b) ≤ y?
W6.10 What, if anything, is wrong with the following proof of the statement,
“If R is a nonempty subset of a set S of real numbers and R is convex (see the definition in Exercise 5.2(b) on page 63), then S is convex?”
Proof. To show that S is a convex set, let x, y∈ S, and t be a real number with 0≤ t ≤ 1. Because R is a convex set, by definition, for any two elements u and v in R, and for any real number s with 0≤ s ≤ 1, su + (1 − s)v ∈ R. In particular, for the specific elements x and y, and for the real number t, it follows that tx + (1− t)y ∈ R. Because R ⊆ S, it follows that tx + (1− t)y ∈ S and so S is convex.
∗6.11 What, if anything, is wrong with the following proof of the statement,
“If a, b, and c are real numbers with a < 0 and x∗ is a maximizer of the function f(x) = ax2+ bx + c (see the definition in Exercise 5.1(a) on page 63), then for every real number > 0, ≤ (2ax∗+ b)/a?”
Proof. Let > 0. It will be shown that ≤ (2ax∗+ b)/a.
Because x∗ is a maximum of f, by definition, for every real number x, f(x∗)≥ f(x). Thus, for x = x∗− , you have that
a(x∗)2+ bx∗+ c ≥ a(x∗− )2+ b(x∗− ) + c
= a(x∗)2+ bx∗+ c− (b + 2ax∗) + a2.
Subtracting a(x∗)2+ bx∗+ c from both sides and dividing by
> 0, it follows that
a− (b + 2ax∗)≤ 0.
The result that ≤ (2ax∗+ b)/a follows by adding b + 2ax∗to both sides of the foregoing inequality and dividing by a.
∗6.12 Identify two errors in the following proof of the statement, “If f is a convex function (see the definition in Exercise 5.2(c) on page 63), then for all real numbers a1, a2, a3 with a1+ a2+ a3 = 1 and 1− a3 > 0, it follows that
Multiplying both sides of the foregoing inequality by 1−a3> 0 and combining the result with the previous inequality yields the desired conclusion that f(a1x1+ a2x2+ a3x3)≤ a1f(x1) + a2f(x2) + a3f(x3).
W6.13 For sets R, S, and T , prove that, if R⊆ S and S ⊆ T , then R ⊆ T . 6.14 Prove that, if a and b are real numbers such that for every integer n > 0, a≤ b +1n, then for all real numbers > 0, a≤ b + .
6.15 Prove that, if for all real numbers x and y, |x + y| ≤ |x| + |y|, then for all real numbers x, y, and z, |x − z| ≤ |x − y| + |y − z|. (Be careful of overlapping notation.)
W6.16 For functions f, g, and h, prove that, if f ≥ g on a set S of real numbers (see the definition in Exercise 5.1(b) on page 63) and g ≥ h on S, then f≥ h on S.
CHAPTER 6: EXERCISES 79
6.17 Prove that, if the real numbers u and v are upper bounds, respectively, for the set S of real numbers and the set−S = {−x : x ∈ S} and v ≥ u, then for every element x∈ S, |x| ≤ v.
6.18 For real numbers u and v, prove that, if u is an upper bound for a set S of real numbers and u≤ v, then v is an upper bound for S.
W6.19 Prove that, if S and T are convex sets (see the definition in Exercise 5.2(b) on page 63), then S∩ T is a convex set.
W6.20 Prove that, if f is a convex function (see the definition in Exercise 5.2(c) on page 63), then for all real numbers s≥ 0, the function sf is convex [where the value of the function sf at any point x is sf(x)].
6.21 For functions f and g of one variable, prove that, if g≥ f on the set of real numbers and x∗ is a maximizer of g (see the definitions in Exercise 5.1(a) and (b) on page 63), then for every real number x, f(x)≤ g(x∗) .
∗6.22 Suppose that a, b, and c are real numbers with a6= 0. Prove that, if x∗=−b/(2a) is a maximizer of the function f(x) = ax2+ bx + c (see Exercise 5.1(a) on page 63), then a < 0. (Hint: Specialize x to x∗+ , where > 0.)
W6.23 Write an analysis of proof that corresponds to the condensed proof given below. Indicate which techniques are used and how they are applied.
Fill in the details of any missing steps, where appropriate.
Proposition. If R is subset of a set S of real numbers and if f and g are functions for which g≥ f on S (see the definition in Exercise 5.1(b) on page 63), then g≥ f on R.
Proof. To show that g ≥ f on R, let x ∈ R. Because R is a subset of S, every element r ∈ R is in S. In particular, x∈ R, so x ∈ S. Also, because g ≥ f on S, it follows that, for every element s∈ S, g(s) ≥ f(s). In particular, x ∈ S, so
Proof. To show that g ≥ f on R, let x ∈ R. Because R is a subset of S, every element r ∈ R is in S. In particular, x∈ R, so x ∈ S. Also, because g ≥ f on S, it follows that, for every element s∈ S, g(s) ≥ f(s). In particular, x ∈ S, so