The Uniqueness Methods
11.2 THE BACKWARD UNIQUENESS METHOD
the hypothesis that there is a unique solution, the forward uniqueness method allows you to state that
A2: ¯x = ¯y.
The conclusion is obtained by working forward from A1 and A2, as follows:
¯
x = −b+√2ab2−4ac = −b−√2ab2−4ac = ¯y (from A1 and A2)
b2− 4ac = 0 (algebra).
The proof is now complete.
Proof of Proposition 17. Because the equation ax2+ bx + c = 0 has a real solution, you know from the quadratic formula that the solutions are
¯
However, because the hypothesis states that the solution to the equation is unique, it follows (by the forward uniqueness method) that ¯x = ¯y, and so, by algebra, b2− 4ac = 0, thus completing the proof.
11.2 THE BACKWARD UNIQUENESS METHOD
When the keyword “unique” appears in the backward process, you must show not only that there is an object with a certain property such that something happens, but also that there is only one such object. That is, with the back-ward uniqueness method you must prove two statements: (a) there is an object with the certain property such that the something happens and (b) there is only one such object. As you already know, the first task is done either by the construction or the contradiction method. The second task is accomplished in one of the following two standard ways.
11.2 THE BACKWARD UNIQUENESS METHOD 127
The Direct Uniqueness Method
With the direct uniqueness method, you assume that there are two objects having the certain property and for which the something happens. If there really is only one such object then, using the two objects with their certain properties, the something that happens, and the information in A, you must conclude that the two objects are one and the same—that is, that they are equal. The forward-backward method is usually the best way to prove that the two objects are equal. This process is illustrated now.
Proposition 18 If a, b, c, d, e, and f are real numbers with the property that ad− bc 6= 0, then there are unique real numbers x and y such that ax + by = e and cx + dy = f.
Analysis of Proof. Recognizing the keyword “unique” in the conclusion, you should use the backward uniqueness method. Accordingly, it is first neces-sary to construct real numbers x and y for which ax + by = e and cx + dy = f.
This is done in Proposition 4 on page 44 with the construction method. It remains to ensure that there is only one pair of numbers that satisfy the two equations. This is now established by the direct uniqueness method. Accord-ingly, you assume that (x1, y1) and (x2, y2) are two objects with the certain property and for which the something happens. In this case, that means that
A1:
A2:
ax1+ by1= e and cx1+ dy1= f and ax2+ by2= e and cx2+ dy2= f.
With these four equations and the hypothesis A, the forward-backward method is used to show that the two objects are the same; that is, that
B1: (x1, y1) = (x2, y2).
A key question associated with B1 is, “How can I show that two ordered pairs of real numbers are equal?” Using Definition 4 on page 26 for equality of ordered pairs, one answer is to show that
B2: x1= x2and y1= y2.
Both of these statements are obtained from the forward process by applying algebraic manipulations to the four equations in A1 and A2 and by using the hypothesis that ad−bc 6= 0, as indicated in the condensed proof that follows.
Proof of Proposition 18. The existence of the real numbers x and y satisfying the two equations is established in Proposition 4 on page 44. To see that there is only one such solution, assume that (x1, y1) and (x2, y2) are two pairs of real numbers satisfying
(1) ax1+ by1= e (2) cx1+ dy1= f (3) ax2+ by2= e (4) cx2+ dy2= f.
Subtracting (3) from (1) and (4) from (2) results in (5) a(x1− x2) + b(y1− y2) = 0 (6) c(x1− x2) + d(y1− y2) = 0.
Multiplying (5) by d and (6) by b and then subtracting (6) from (5) yields (7) (ad− bc)(x1− x2) = 0.
Because, by hypothesis, ad− bc 6= 0, one has x1− x2= 0 and hence x1= x2. A similar sequence of algebraic manipulations establishes that y1 = y2 and thus the uniqueness is proved.
The Indirect Uniqueness Method
With the indirect uniqueness method for showing that there is only one object satisfying a certain property and for which something happens, you assume that there are two different objects having the certain property and for which the something happens. Now supposedly this cannot happen, so, by using the certain property, the something that happens, the information in A, and especially the fact that the objects are different, you must reach a contradiction. This process is demonstrated now.
Proposition 19 If r > 0, then there is a unique real number x such that x3= r.
Analysis of Proof. The appearance of the keyword “unique” in the con-clusion suggests starting with the backward uniqueness method. Accordingly, the first step is to use the construction method to produce a real number x such that x3 = r. This part of the proof is omitted so as to focus on how the indirect uniqueness method is subsequently used to establish that there is only one such real number. To that end, suppose that
A1: x and y are two different real numbers (that is, x 6= y) such that x3= r and y3= r.
Using this information, and especially the fact that x6= y, it is shown that r = 0, which contradicts the hypothesis that r > 0.
To show that r = 0, work forward from A1. Specifically, because x3 = r and y3= r, it follows that
A2: x3= y3 or x3− y3= 0.
On factoring, you have that
A3: (x− y)(x2+ xy + y2) = 0.
11.2 THE BACKWARD UNIQUENESS METHOD 129
Here is where you can use the fact that x6= y to divide both sides of A3 by x− y 6= 0, obtaining
Because x is real and the foregoing formula requires taking the square root of
−3y2, it must be that A6: y = 0,
and if y = 0, then a contradiction to r > 0 is reached from A1 because A7: r = y3= 0.
Proof of Proposition 19. The proof that there is a real number x such that x3= r is omitted. To see that x is the only such number, assume that y6= x contra-dicting the hypothesis that r > 0, and so the proof is complete.
Summary
In this chapter, you have learned the various uniqueness methods. Use a uniqueness method when you come across a statement in the form, “there is a unique object (or one and only one object, or exactly one object) with a certain property such that something happens.”
When this statement occurs in the forward process, with the forward uniqueness method you
1. Look for two objects, say X and Y , with that certain property and for which that something happens.
2. You can then write, as a new statement in the forward process, that X and Y are the same; that is, that X = Y . This statement should then help you establish that the conclusion B is true.
Use the backward uniqueness method when, in the backward process, you need to show that “there is a unique object with a certain property such that
something happens.” Doing so requires two steps: first showing that there is one such object, say X, and then showing that there is only one such object.
While the first task is accomplished with the construction or contradiction method, you can accomplish the second task in one of two ways. With the direct uniqueness method you
1. Assume that, in addition to the object X, Y is also an object with the certain property and for which the something happens.
2. Use the certain property and something that happens for X and Y together with the hypothesis A to show that X and Y are the same (that is, that X = Y ).
With the indirect uniqueness method, you
1. Assume that Y is a different object from X with the certain property and for which the something happens.
2. Use the properties of X and Y , the fact that they are different, and the hypothesis A to reach a contradiction.
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.
W11.1 Suppose that each of the following statements arises in the forward process. When subsequently applying the forward uniqueness method with the given objects, indicate (i) what you would have to show about those objects and (ii) what you would then be able to conclude as a new statement in the forward process.
a. Statement: There is one and only one line through two given points (x1, y1) and (x2, y2) in the plane.
Given objects: Lines y = mx + b and y = cx + d.
b. Statement: There is exactly one solution to the system of equa-tions ax + by = 0 and cx + dy = 0, where a, b, c, and d are given real numbers.
Given objects: Solutions (x1, y1) and (x2, y2).
c. Statement: There is a unique complex number c + di for which (a + bi)(c + di) = 1, where a and b are given real numbers and i =√
−1.
Given objects: Complex numbers r + si and t + ui.
CHAPTER 11: EXERCISES 131
11.2 Suppose that each of the following statements arises in the forward process. When subsequently applying the forward uniqueness method with the given objects, indicate (i) what you would have to show about those objects and (ii) what you would then be able to conclude as a new statement in the forward process.
a. Statement: There is a unique maximizer of the function ax2+bx + c, where a, b, and c are given real numbers with a < 0.
Given objects: Real numbers x∗ and y∗.
b. Statement: For a given function f of one real variable, there is one and only one function g such that, for every real number x, f(g(x)) = g(f(x)) = x.
Given objects: Functions F and G.
c. Statement: For a given integer n, there is a unique integer a > 0 such that a|n and for every integer b > 0 such that b|n, b|a.
Given objects: Integers p and q.
W11.3 Suppose that each of the statements in Exercise 11.1 appears in the backward process. How would you proceed to prove these statements using (i) the direct uniqueness method and (ii) the indirect uniqueness method?
11.4 Suppose that each of the statements in Exercise 11.2 appears in the backward process. How would you proceed to prove these statements using (i) the direct uniqueness method and (ii) the indirect uniqueness method?
∗11.5 Answer the given questions about the following proof that, “If u is an upper bound for a set S of real numbers and u ∈ S, then u is the unique element of S with the property that, for every element x∈ S, x ≤ u.”
Proof. Suppose that v∈ S also satisfies the property that, for every element x∈ S, x ≤ v. Then, because u ∈ S, it must be that u≤ v. Likewise, because for every element x ∈ S, x ≤ u and v∈ S, it must be that v ≤ u. It follows that u = v and so the proof is complete.
a. In this backward uniqueness method, why does the author not first establish that there is an object with the certain property such that something happens?
b. Does this proof use the direct or indirect uniqueness method? Explain.
c. What proof technique is used in the second sentence?
d. Why is the author justified in claiming that the proof is complete in the last sentence?
11.6 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 a and c are real numbers with a < 0, then x = 0 is the unique maximizer of the function f(x) = ax2+ c.
Proof. It was already shown in Exercise 5.15 that x = 0 is a maximizer of f. Suppose now that y is also a maximizer of f.
Then, for every real number z, ay2+ c≥ az2+ c. In particular, ay2+ c≥ a(0)2+ c = c. But then it follows by algebra that y2≤ 0. Thus y = 0 and so the proof is complete.
11.7 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 a, b, c, and d are real numbers for which the equations ax + by = 0 and cx + dy = 0 have a unique solution, then ad− bc 6= 0.
Proof. Assume that ad− bc = 0. Note that, because the two equations have a unique solution, it cannot be that both c and d are 0. It then follows that x = d and y =−c is a solution to the two equations, as is x = 0 and y = 0. By the uniqueness, it must be that d = 0 and c = 0. This contradiction completes the proof.
W11.8 Prove that, if x is a real number > 2, then there is a unique real number y < 0 such that x = 2y/(1 + y).
11.9 Prove that, if a and b are integers with a6= 0 such that a|b, then there is a unique integer k such that b = ka. (See Definition 1 on page 26.)
W11.10 Prove, by the indirect uniqueness method, that, if m and b are real numbers with m6= 0, then there is a unique number x such that mx + b = 0.
11.11 Prove, by the indirect uniqueness method, that there is a unique integer n for which 2n2− 3n − 2 = 0.
W11.12 Prove that, if a and b are real numbers, at least one of which is not 0, and i =√
−1, then there is a unique complex number, say c + di, such that (a + bi)(c + di) = 1.
∗11.13 Prove that, if f is a function of one real variable such that for every real number y, there is a unique real number x such that f(x) = y, then the function f is one-to-one.