The Max/Min Methods
15.10 THE MAX/MIN METHODS
When a statement indicates that the smallest (largest) element of a set of real numbers is less (greater) than or equal to a particular real number, say z, then you should use a max/min method. Doing so involves rewriting the statement in an equivalent form using the quantifier “for all” or “there is,”
whichever is appropriate. Once in this form, you can then apply the choose, construction, or specialization method.
How to Read and Do Proofs
A final summary of how to read and do proofs is in order.
How to Read a Proof
A written proof is nothing more than a sequence of applications of the indi-vidual techniques you have now learned. However, due to the way in which they are written, there are three reasons why proofs are challenging to read:
1. The author does not always refer to the techniques by name.
2. Several steps are combined in a single sentence with little or no justifi-cation.
3. The steps of a proof are not necessarily presented in the order in which they were performed when the proof was done.
To read a proof, you have to reconstruct the author’s thought processes. Doing so requires that you identify which techniques are used and how they apply to the particular problem. Begin by trying to determine what technique is used to start the proof. Then try to follow the methodology associated with that technique. Watch for quantifiers to appear, for then the author is likely to use the corresponding choose, induction, construction, and/or specialization methods. The inability to follow a particular step of a written proof is often due to the lack of sufficient detail. To fill in the gaps, learn to ask yourself how you would proceed to do the proof. Then try to see if the written proof matches your thought process.
The following is a final example of how to read a proof, where it is shown that, when you divide an integer b by an integer a≥ 1, you get a unique whole number and a unique remainder. The subsequent analysis-of-proof provides an explanation of which techniques are used and how they are applied to the specific problem.
Proposition 27 If a and b are integers with a ≥ 1, then there are unique integers q and r such that b = aq + r, where 0≤ r < a.
Proof of Proposition 27. (For reference purposes, each sentence of the proof is written on a separate line.)
S1: The values of q and r are obtained from the least element of the following set of integers: M = {integers w ≥ 0 : M has a least element by the Least-Integer Principle and so, in either case, let r be the least element of M .
S10: It remains to show that these values of q and r are unique, so, suppose that m and n are also integers for which b = am + n and 0≤ n < a.
Analysis of Proof. An interpretation of statements S1 through S13 follows.
Interpretation of S1: The values of q and r are obtained from the least element of the following set of integers: M = {integers w ≥ 0 : there is an integer k such that w = b− ak }.
The author recognizes the keyword “unique” in the conclusion and is there-fore using a backward uniqueness method (is it the direct or indirect method?).
As such, the first step is to construct the integers q and r, which the author states are going to come from the least element of the set M . Thus, the author is using previous knowledge of the Least Integer Principle (see page 158).
Interpretation of S2: Now if b≥ 0, then for k = 0, you have w = b ∈ M.
To use the Least Integer Principle it must be shown that M 6= ∅, which is what the author is doing in S1, when b≥ 0 and in S2, when b < 0. Specifically, because either b≥ 0 or b < 0, the author uses a proof by cases to show that M 6= ∅. In the first case, when b ≥ 0, the author constructs k = 0 in S1 to show that w = b≥ 0 satisfies the defining property of M.
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Interpretation of S3: While if b < 0, then for k = b, you have w = b−ak = b− ab = b(1 − a) ≥ 0 as b < 0 and 1 − a ≤ 0, so w ∈ M.
The author is completing the proof by cases. Specifically, for the case when b < 0, the author shows that w = b− ab satisfies the defining property of M.
To do so, the author constructs k = b and shows that w = b− ak = b − ab = b(1− a) ≥ 0, the last inequality being true because b < 0 and a ≥ 1 from the hypothesis. Thus, from S1 and S2, the author has shown that M6= ∅.
Interpretation of S4: Now if 0 ∈ M, then 0 is the least element of M;
otherwise, M has a least element by the Least-Integer Principle and so, in either case, let r be the least element of M .
Note that to use the Least Integer Principle requires that the set consist of strictly positive integers. The author mentally observes that the set M consists of nonnegative integers and therefore works forward to note that either 0∈ M or 0 6∈ M. Proceeding with a proof by cases, in the first case, when 0∈ M, the author states that 0 is the least integer of M, which is true because all elements of M are ≥ 0. In the second case, when 0 6∈ M, the author uses the Least Integer Principle to claim that M has a least element, which is true because, in this case, M consists of strictly positive integers and it has already been shown in S2 and S3 that M 6= ∅. Thus, in either case, M has a smallest element, r. Here is where the author has finally constructed the integer r. It remains to construct the integer q, which the author does next.
Interpretation of S5: Because r∈ M, (1) r ≥ 0 and (2) there is an integer q such that r = b− aq, or equivalently, b = aq + r.
The author uses the defining property of M and the fact that r ∈ M to claim that there is an integer q such that r = b−aq, or equivalently, b = aq+r.
The author has now constructed the integers r in S4 and q in S5. According to the construction method, the author must show that these values of r and q are correct—that is, that 0 ≤ r < a. The first inequality is true by the defining property of M , as stated in S5. The fact that r < a is shown next.
Interpretation of S6: To see that r < a, assume that r≥ a.
The author is using the contradiction method to show that r < a and, accordingly, assumes that this is not true—that is, that r≥ a.
Interpretation of S7: Then w = b− a(q + 1) is a smaller element of M than r.
The author is making a claim that, if true, is in fact a contradiction. Specif-ically, from S4, the author has shown that r is the smallest element of M . However, if the author’s claim in S7 is true, then w would be a smaller ele-ment of M than r, which cannot happen. The author must therefore show that w = b− a(q + 1) ∈ M (which is done in S8) and that w < r (which is done in S9).
Interpretation of S8: You can see that w∈ M because, for k = q + 1, you have w = b− ak = b − a(q + 1) = b − aq − a = r − a ≥ 0.
The author is showing that w ∈ M by using the defining property of M.
Thus, the author constructs the integer k = q + 1 and shows, by algebra and the fact that r = b− aq from S5, that w = b − ak ≥ 0.
Interpretation of S9: Also, w < r because w = b− a(q + 1) = b − aq − a <
b− aq = r.
The author is showing that w < r by algebra and the fact that r = b− aq from S5. The author has now shown that w ∈ M in S8 and w < r in S9.
This establishes the contradiction that r is not the smallest element of M , as stated in S4. This contradiction means that r < a and so the author has now completed the construction method by establishing that the constructed values of q and r satisfy the desired properties that b = aq + r (see S5) and 0≤ r < a (see S5 and S6).
Interpretation of S10: It remains to show that these values of q and r are unique, so, suppose that m and n are also integers for which b = am + n and 0≤ n < a.
The author is now completing the proof by using the direct uniqueness method to show that q and r are unique. Thus, the author assumes that m and n are also integers that satisfy the same properties as q and r, that is, that b = am + n and 0≤ n < a. According to the direct uniqueness method, the author must show that m and n are the same as q and r, that is, that m = q and n = r. This is accomplished in S11, S12, and S13.
Interpretation of S11: But then, a(q− m) = n − r and −a < n − r < a.
The author applies algebra to the properties of these integers. Specifically, subtracting b = am + n from b = aq + r and bringing the term r− n to the other side leads to a(q− m) = n − r. To see that −a < n − r, the author uses the fact that 0≤ n and r < a. Specifically, 0 ≤ n and −a < −r, so adding these two inequalities results in −a < n − r. Similarly, because 0 ≤ r and n < a, you have−r ≤ 0 and n < a, so adding these two inequalities results in n− r < a.
Interpretation of S12: From this you have −a < a(q − m) < a and so q− m = 0.
In S11, the author replaces the expression n− r in the inequalities with a(q− m) to obtain −a < a(q − m) < a. The author then mentally (1) divides the foregoing inequality through by a > 0 to obtain −1 < q − m < 1 and (2) notes that the only integer strictly between −1 and 1 is 0 and hence concludes that q− m = 0.
Interpretation of S13: Finally, substituting m = q in a(q− m) = n − r yields n− r = 0, that is, n = r.
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The author uses the fact that q− m = 0 from S12 to state that m = q and then substitutes that into a(q− m) = n − r (see S11) to obtain n − r = 0, which means that n = r. This together with the fact that m = q obtained in S12 completes the direct uniqueness method and hence the proof.
How to Do a Proof
When trying to prove that “A implies B,” let the form of A and B guide you as much as possible. For example, you should scan the statement B for certain keywords, which often indicate how to proceed. If you come across the quantifier “there is,” then consider the construction method, whereas the quantifier “for all” suggests using the choose or induction method. When the statement B contains the word “no” or “not,” you will probably want to use the contrapositive or contradiction method. Other keywords to look for are “unique,” “either/or,” and “maximum” and “minimum,” for then the corresponding uniqueness, either/or, and max/min methods are appropriate.
If no keywords appear in the hypothesis or conclusion, then you should proceed with the forward-backward method. Here is a final example of doing a proof in which the following definitions are used.
Definition 18 A function f of one real variable is a convex function if and only if, for all real numbers x, y, and t with 0≤ t ≤ 1, it follows that f(tx + (1− t)y) ≤ tf(x) + (1 − t)f(y).
Definition 19 A function f of one variable is increasing if and only if, for all real numbers x and y with x≤ y, f(x) ≤ f(y).
Proposition 28 If g is a convex function and f is an increasing convex func-tion, then the function f ◦ g is a convex function. [Recall that f ◦ g is the function defined for all real numbers x by (f ◦ g)(x) = f(g(x)).]
Analysis of Proof. The forward-backward method is used to begin the proof because the hypothesis A and the conclusion B do not contain keywords (such as “for all” or “there is”). Working backward, you are led to the key question, “How can I show that a function (namely, f◦ g) is convex?” Using Definition 18, you must show that
B1: For all real numbers x, y, and t with 0≤ t ≤ 1, it follows that (f◦ g)(tx + (1 − t)y) ≤ t(f ◦ g)(x) + (1 − t)(f ◦ g)(y).
Because the backward statement B1 contains the quantifier “for all,” you should now proceed with the choose method. Accordingly, you should choose
A1: Real numbers x, y, and t with 0≤ t ≤ 1, for which you must show that
B2: (f◦ g)(tx + (1 − t)y) ≤ t(f ◦ g)(x) + (1 − t)(f ◦ g)(y).
This is accomplished by working forward. Specifically, from the hypothesis that g is convex, by definition, this means that
A2: For all real numbers u, v, and s with 0≤ s ≤ 1, it follows that g(su + (1− s)v) ≤ sg(u) + (1 − s)g(v).
Recognizing the keywords “for all” in the forward statement A2, you should now use specialization. Specifically, specializing A2 with u = x, v = y, and s = t and noting from A1 that 0≤ t ≤ 1, the result is that
A3: g(tx + (1− t)y) ≤ tg(x) + (1 − t)g(y).
Working forward from the hypothesis that f is an increasing function, by Definition 19, this means that
A4: For all real numbers u and v with u≤ v, f(u) ≤ f(v).
Recognizing the keywords “for all” in the forward statement A4, you should now use specialization. Specifically, specializing A4 with u = g(tx + (1− t)y) (the left side of the inequality in A3) and v = tg(x) + (1− t)g(y) (the right side of the inequality in A3), and noting from A3 that, for these values of u and v, u≤ v, the result is that
A5: f(g(tx + (1− t)y)) ≤ f(tg(x) + (1 − t)g(y)).
Now work forward from the hypothesis that f is a convex function. Accord-ingly, by definition, this means that
A6: For all real numbers u, v, and s with 0≤ s ≤ 1, it follows that f(su + (1− s)v) ≤ sf(u) + (1 − s)f(v).
Recognizing the keywords “for all” in the forward statement A6, you should now use specialization. Specifically, specializing A6 with u = g(x), v = g(y), and s = t and noting from A1 that 0≤ t ≤ 1, the result is that
A7: f(tg(x) + (1− t)g(y)) ≤ tf(g(x)) + (1 − t)f(g(y)).
Combining the inequalities in A5 and A7, you have A8: f(g(tx + (1− t)y)) ≤ tf(g(x)) + (1 − t)f(g(y)), or equivalently, by the definition of f◦ g,
A9: (f◦ g)(tx + (1 − t)y) ≤ t(f ◦ g)(x) + (1 − t)(f ◦ g)(y).
The proof is now complete because A9 is the same as B2.
Proof of Proposition 28. To see that f◦ g is convex, let x, y, and t be real numbers, with 0≤ t ≤ 1, for which it must be shown that
(f◦ g)(tx + (1 − t)y) ≤ t(f ◦ g)(x) + (1 − t)(f ◦ g)(y). (15.1) Now because g is convex, by definition,
CHAPTER 15: EXERCISES 173
g(tx + (1− t)y) ≤ tg(x) + (1 − t)g(y). (15.2) Applying f to both sides of (15.2) and using the hypothesis that f is an increasing function yields
f(g(tx + (1− t)y)) ≤ f(tg(x) + (1 − t)g(y)). (15.3) Also, because f is convex, by definition, the right side of (15.3) satisfies
f(tg(x) + (1− t)g(y)) ≤ tf(g(x)) + (1 − t)f(g(y)). (15.4) Combining the inequalities in (15.3) and (15.4), you have
f(g(tx + (1− t)y)) ≤ tf(g(x)) + (1 − t)f(g(y)). (15.5) The proof is now complete because (15.5) is the same as (15.1).
You now have all of the techniques you need to do a proof, but keep in mind that this is a creative endeavor. For example, you know when to use the construction method, but actually constructing the object can require a great deal of creativity. Nevertheless, learning to do proofs is like learning a language—the more you practice, the easier it gets. If all of these techniques fail, you may wish to stick to Greek—after all, it’s all Greek to me.
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.
W15.1 For each of the following statements, indicate which technique you would use to begin the proof and explain why.
a. If p and q are odd integers, then the equation x2+ 2px + 2q = 0 has no rational solution for x.
b. For every integer n≥ 4, n! > n2.
c. If f and g are convex functions, then f + g is a convex function.
d. If a, b, and c are real variables, then the maximum value of ab + bc + ac subject to the condition that a2+ b2+ c2= 1 is less than or equal to 1.
e. In a plane, there is one and only one line perpendicular to a given line L through a point P on the line.
15.2 For each of the following statements, indicate which technique you would use to begin the proof and explain why.
a. If p > 1 is an integer that is not prime, then there is an integer m with 1 < m≤√p such that m|p.
b. If f and g are continuous functions at the point x, then so is the function f + g.
c. If a and b are integers for which a|b and b|a, then a = ±b.
d. If f and g are functions such that (1) for all real numbers x, f(x)≤ g(x) and (2) there is no real number M such that, for all x, f(x) ≤ M, then there is no real number M > 0 such that, for all real numbers x, g(x)≤ M.
e. If f and g are continuous functions at the point x, then, for every real number > 0, there is a real number δ > 0 such that, for all real numbers y with |x − y| < δ, |f(x) + g(x) − (f(y) + g(y))| < .
W15.3 For each of the problems in Exercise 15.1, state how the technique you chose to begin the proof would be applied to that problem. Indicate what you would assume, what you would conclude, and how you go about doing it.
15.4 For each of the problems in Exercise 15.2, state how the technique you chose to begin the proof would be applied to that problem. Indicate what you would assume, what you would conclude, and how you go about doing it.
W15.5 Describe how you would use each of the following techniques to prove that, “For every integer n≥ 4, n! > n2.” State what you would assume and what you would conclude.
a. Induction method. b. Choose method.
c. Forward-Backward method. d. Contradiction method.
15.6 Suppose that S and T are sets of real numbers with S⊆ T . Describe how you would use each of the following techniques to prove that, “If > 0 is a real number such that, for every element x ∈ T , x ≤ , then for every element x ∈ S, x ≤ .” State what you would assume and what you would conclude.
a. Choose method. b. Specialization method.
c. Contradiction method. d. Contrapositive method.
15.7 Reword the following proposition in such a way that it would be ap-propriate, based on keywords, to use the given technique to begin the proof:
“If X = { 1 +n1
n
: n > 0 is an integer}, then, for every element x ∈ X, x≤ 3.”
CHAPTER 15: EXERCISES 175
a. Max/Min method.
b. Contradiction method.
c. Induction method.
15.8 Suppose the forward-backward method is used to start each of the fol-lowing proofs. List all of the techniques that are likely to be used subsequently in the proof.
a. (C AND D) implies (E OR F ).
b. (C OR D) implies (E AND F ).
c. If X is an object such that, for all objects Y with a certain property, something happens, then there is an object Z with a certain property such that something else happens.
d. If for all objects X with a certain property, something happens, then there is an object Y with a certain property such that, for all objects Z with a certain property, something else happens.
15.9 Repeat the previous exercise assuming that the contrapositive method is used to start each proof.
15.10 Identify all of the techniques that are used in the condensed proof in Exercise 13.13 on page 153.
15.11 Identify all of the techniques that are used in the condensed proof in Exercise 10.24 on page 123.
∗15.12 Identify all of the techniques that are used in the condensed proof in Exercise 6.25 on page 80.
15.13 Prove that, if ABC is a right triangle with sides of integer length a and b and hypotenuse of integer length c, then 12ax2+ cx + b has a rational root.
15.14 Prove that, if S is a set of real numbers and x = max{s : s ∈ S}, then x is the only element of S such that, for every element s∈ S, s ≤ x.
∗15.15 Prove that, if p is a polynomial and q is a polynomial, then p + q is a polynomial [where for every real number x, (p + q)(x) = p(x) + q(x)]. (Recall that a polynomial is a function f of one real variable for which there is an integer n≥ 0 and real numbers a0, a1, . . . , ansuch that, for every real number x, f(x) = a0+ a1x1+· · · + anxn.)
15.16 Let f be a function of one real variable. Prove that, if x1< x2< x3
are real numbers for which f(x1) < f(x2) and f(x2) > f(x3), then f is not linear. (Recall that a function f is linear if and only if there are real numbers m and b such that, for every real number x, f(x) = mx + b.)
15.17 Answer the given questions pertaining to the following condensed proof that, for a function f of one variable, if x1< x2< x3 are real numbers for which f(x1) < f(x2) and f(x2) > f(x3), then f is not a convex function.
Proof. Suppose, to the contrary, that f is a convex function.
Then consider the real number t = xx33−x−x21. It is not hard to show that 0 < t < 1 and that x2= tx1+ (1− t)x3. Therefore, by the fact that f is a convex function, it follows that
f(x2) = f(tx1+ (1− t)x3) ≤ tf(x1) + (1− t)f(x3)
< tf(x2) + (1− t)f(x2) = f(x2).
The foregoing contradiction completes the proof.
a. The first sentence indicates that the author is using the contradiction method. What is the contradiction?
b. Why is the author creating a value for t? What proof technique is this value of t being used for?
c. Justify the third sentence by showing that 0 < t < 1 and also that x2= tx1+ (1− t)x3.
d. Justify the strict inequality that
tf(x1) + (1− t)f(x3) < tf(x2) + (1− t)f(x2).