# Chapter Three. Functions. In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics.

Save this PDF as:

Size: px
Start display at page:

Download "Chapter Three. Functions. In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics."

## Transcription

1 Chapter Three Functions 3.1 INTRODUCTION In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics. Definition 3.1: Given sets X and Y, a function from X to Y is a subset f of X Y with the property that, for any x X, there exists a unique y Y such that (x, y) f. We denote the fact that f is a function from X to Y by writing f : X Y The set X is called the domain of the function f and is written dom f. The set Y is called the codomain of f. Also associated with f is a subset of Y called the image of f. This is denoted by im f and defined by im f = {y Y there exists x X such that (x, y) f} If X = Y, then we call f a function on X. Example 3.1: Determine which of the following are functions and find the image of each function. (a) The subset f 0 of {1, 2, 3} {1, 2, 3} given by f 0 = {(1, 2), (2, 3)}. (b) The subset f 1 of {1, 2, 3} {1, 2, 3} given by f 1 = {(1, 2), (2, 1), (3, 2)}. (c) The subset f 2 of {1, 2, 3} {1, 2, 3} given by f 2 = {(1, 1), (1, 3), (2,3), (3,1)}. (d) The subset f 3 of Z Z defined by f 3 = {(m, n) n is a multiple of m}. (e) The subset f 4 of Z Z defined by f 4 = {(m, n) n = 2m + 1}.

2 138 Chapter 3 Functions Solution: (a) The subset f 0 is not a function because there is no ordered pair in f 0 of the form (3, y). This violates the condition of the definition that, for each x {1, 2, 3}, there must exist a y {1, 2, 3} such that (x, y) f 0. (b) The subset f 1 is a function on {1, 2, 3}; im f 1 = {1, 2}. (c) The subset f 2 is not a function on {1, 2, 3} because both (1, 1) f 2 and (1, 3) f 2. This violates the condition of the definition that, for each x {1, 2, 3}, there should be a unique y {1, 2, 3} for which (x, y) f 2. (d) The subset f 3 is not a function on Z because, for instance, (2, 4) f 3 and (2, 6) f 3. Again, the definition says that, in order for f 3 to be a function, there should be a unique y Z such that (2, y) f 3, and we see that this is not the case. (e) The subset f 4 is a function on Z; im f 4 is precisely the set of odd integers. Suppose that f is a function from X to Y. Again, the condition of Definition 3.1 is the following: For any x X, there exists a unique y Y such that (x, y) f. We call y the image of x under f and write y = f(x). The notation f(x) is read f of x. Moreover, if y = f(x), then x is said to be a preimage of y under f. Thus, with this notation, we can restate the definition of the image of f as follows: im f = {f(x) x X} It is also common to refer to y = f(x) as the value of f at x and to say that f maps x to y. (A function is sometimes called a mapping.) One commonly defines a function by writing, Define f : X Y by y = f(x), where y = f(x) is a formula that expresses y (uniquely) in terms of x. We call y = f(x) the defining formula of the function f. It can be viewed as a rule specifying how to compute the image of a given x X. Example 3.2: Define f : Q + Q + by f(x) = 1/x. This f is called the reciprocal function (on Q + ) since each positive rational number is mapped to its reciprocal. It is clear that f is a function, since each positive rational number has a unique reciprocal. For example, f(2) = 1/2 and f(2/3) = 1/(2/3) = 3/2. Also, each positive rational number is the reciprocal of its reciprocal, so that im f = Q +. Example 3.3: Define g: R Z by g(x) = x, where x denotes the largest integer that is less than or equal to x. This g is called the greatest integer function or floor function. It is clear that g is a function since the process of rounding down a given real number x to the largest integer m such that m x determines a unique m. For example, g(2) = 2, g(π) = 3, g( 2) = 2, and g( π) = 4. Note that for each integer m, g(m) = m, so that im g = Z. Example 3.4: Define l: R Z by l(x) = x, where x denotes the smallest integer that is greater than or equal to x. This l is called the least integer function or ceiling function. It

3 3.1 Introduction 139 is clear that l is a function since the process of rounding up a given real number x to the smallest integer m such that x m determines a unique m. For example, l(2) = 2, l(π) = 4, l( 2) = 2, and l( π) = 3. Note that for each integer m, l(m) = m, so that im l = Z. Example 3.5: Define n: R Z by n(x) = [x], where [x] denotes the integer that is nearest to x. For example, n(2) = 2, n(π) = 3, n( 2) = 2, and n( π) = 3. This n is called the nearest integer function. But wait, you say (if you ve been reading the section carefully): What, for example, is the value of n(2.5)? Is it the case that n(2.5) = 2 or is n(2.5) = 3? This is a problem! If we want n to be a function with domain R, then n(2.5) must have some value; and, if we want n to be a function, then we can t let n(2.5)have two different values. The common terminology for a problem such as this is to say that the function n is not well-defined. To make n well-defined, we need a rule to insure that every real number has a unique image under n. One rule that is commonly invoked is break ties by rounding up; that is, if there is a tie for the title of the nearest integer to x, then let n(x) = x. With this rule, for example, we have that n(2.5) = 3. Note that this rule can be made more precise as follows: ( ) 2m 1 n = m for any integer m. 2 Example 3.6: Define g: Z 7 Z 7 by g(x) = 4x. Then g(0) = 0, g(1) = 4, g(2) = 1, g(3) = 5, g(4) = 2, g(5) = 6, and g(6) = 3. (A remark: In (Z n, +, ), for m Z n, mx = m x; that is, one gets the same value whether one interprets mx as a multiple of x or as the product m x. So we can interpret mx in whichever way is more convenient.) Note that im g = Z 7. Example 3.7: Define h: Z 6 Z 3 by h(x) = x mod 3. Then h(0) = 0, h(1) = 1, h(2) = 2, h(3) = 3 mod 3 = 0, h(4) = 4 mod 3 = 1, and h(5) = 5 mod 3 = 2. So im h = Z 3. If f is a function whose domain and range are both subsets of the set R of real numbers, then associated with each ordered pair (x, y) f there is a uniquely determined point (x, y) in the xycoordinate plane. In analytic geometry and calculus, the set of all points so determined is called the graph of f. No doubt you have had considerable experience with graphing functions. Example 3.8: Sketch the graphs of the following functions. (a) f : R R; f(x) = 4 2x (b) g: (0, ) (0, ); f(x) = 1/x (c) h: R ( 4, ); h(x) = x 2 2x 3

4 140 Chapter 3 Functions Solution: The graphs are shown in Figure 3.1. Figure 3.1 If X and Y are subsets of R and a function f from X to Y is given, then the property that each x X has a unique image under f can be interpreted geometrically. First of all, note that this property is equivalent to the statement that the following implication holds for any x X and all y 1, y 2 Y : [ (x, y1 ) f and (x, y 2 ) f ] y 1 = y 2 This statement implies that every vertical line intersects the graph of f in at most one point. Thus, if we are given a set of points in the xy-coordinate plane, then we can determine whether the associated set of ordered pairs is a function by applying this vertical line test. Example 3.9: It is easy to apply the vertical line test to each of the graphs in Figure 3.1. Each is seen to determine a function, as noted in the previous example. Figure 3.2 shows the graph of the subset {(x, y) x = y 2 }. Note that any vertical line x = x 0, where x 0 > 0, intersects the graph of this subset in two points, namely, (x 0, x 0 ) and (x 0, x 0 ). Thus, this subset fails the vertical line test, and so it is not the graph of a function.

5 3.1 Introduction 141 Figure 3.2 Exercise Set Let S = {1, 2, 3, 4}. Determine which of the following subsets of S S are functions on S, and determine the images of those that are. (a) f 1 = {(1, 2), (3, 4), (4, 1)} (b) f 2 = {(1, 3), (2, 3), (3, 3),(4, 3)} (c) f 3 = {(1, 1), (2, 2), (3, 3), (3,4), (4, 4)} (d) f 4 = {(1, 3), (2, 4), (3, 1),(4, 2)} 2. Let T denote your maternal family tree (T includes your (biological) mother, your maternal grandmother, your maternal great-grandmother, and so on, and any children of these people). Determine which of the following subsets of T T are functions. (a) g 1 = {(x, y) y is the mother of x} (b) g 2 = {(x, y) x and y are sisters} (c) g 3 = {(x, y) y is an aunt of x} (d) g 4 = {(x, y) y is the eldest daughter of x s maternal grandmother} 3. Graph each of the following subsets of R R. Determine which are functions and find the images of those that are. (a) h 1 = {(x, y) 2x y = 3} (c) h 3 = {(x, y) (x 1) 2 + y 2 = 9} (b) h 2 = {(x, y) x 2 + y = 4} (d) h 4 = {(x, y) y = 8 + 2x x 2 } (e) h 5 = {(x, y) x y = 2} (f) h 6 = {(x, y) (x 2 + 1)y = 1} 4. Each part defines a function and gives a value x in the domain of the function and a value y in the codomain of the function. You are to (1) find the image of x, (2) find the set of preimages of y, and (3) determine the image of the function.

6 142 Chapter 3 Functions (a) f : {1, 2, 3, 4,5} {1, 2, 3, 4}; f(1) = 2, f(2) = 4, f(3) = 1, f(4) = 4, f(5) = 2; x = 3, y = 4 (b) g: {2, 3, 4, 5,...} set P of primes; g(x) =smallest prime factor of x; x = 91, y = 5 (c) h 1 : Z 11 Z 11 ; h 1 (x) = 5x; x = 4, y = 3 (d) h 2 : Z 12 Z 12 ; h 2 (x) = 4x; x = 4, y = 3 (e) r: Z Z 5 ; r(x) = x mod 5; x = 12, y = 3 (f) v: Z Z; v(x) = gcd(x, 12); x = 30, y = ONE-TO-ONE FUNCTIONS AND ONTO FUNCTIONS Let f be a function from the set X to the set Y. Then each x X has a unique image y = f(x) Y. However, it need not be the case that for each y Y there is a unique x X such that f(x) = y; that is, it is not necessarily true that every y Y has exactly one preimage. In fact, it might happen that some y 1 Y has no preimages, or it might happen that some y 2 Y has at least two preimages; that is, there could exist two distinct elements x 1, x 2 X with f(x 1 ) = y 2 = f(x 2 ). For example, consider the function f : Z Z defined by f(m) = 2m It is clear that f(m) is odd for all m Z; so, for instance, there is no m for which f(m) = 0. Moreover, f( 1) = 3 = f(1), and so 3 has both 1 and 1 as preimages. Some functions f : X Y satisfy the property that for each y Y there is at most one x X such that f(x) = y ; that is, each y Y has at most one preimage under f. This condition may be rephrased as follows: For all x 1, x 2 X, if f(x 1 ) = f(x 2 ), then x 1 = x 2. Definition 3.2: A function f : X Y is called one-to-one provided that the following implication holds for all x 1, x 2 X: f(x 1 ) = f(x 2 ) x 1 = x 2 It is sometimes convenient to use the condition in Definition 3.2 in its contrapositive form: For all x 1, x 2 X, x 1 x 2 f(x 1 ) f(x 2 ). When is the function f : X Y not one-to-one? Definition 3.2 yields the following statement: Taking the negation of the condition in f is not one-to-one if and only if, for some x 1, x 2 X, x 1 x 2 and f(x 1 ) = f(x 2 ) Example 3.10: Determine which of the following functions are one-to-one. (a) f 1 : {1, 2, 3} {1, 2, 3, 4}; f 1 (1) = 2, f 1 (2) = 4, f 1 (3) = 2 (b) f 2 : {1, 2, 3} {1, 2, 3, 4}; f 2 (1) = 3, f 2 (2) = 4, f 2 (3) = 1 (c) f : Z Z; f(m) = m 1 (d) g 1 : Z Z; g 1 (m) = 3m + 1 (e) h: Z Z + ; h(m) = m +1 (f) p: Q {1} Q; p(x) = x/(1 x)

7 3.2 One-to-One Functions and Onto Functions 143 Solution: (a) The function f 1 is not one-to-one since f 1 (1) = 2 = f 1 (3). (b) The function f 2 is one-to-one since no two elements of {1, 2, 3} have the same image under f 2. (c) The function f maps each integer to its predecessor, and it is easily seen to be one-to-one. To give a formal proof, note that, for any integers m 1 and m 2, f(m 1 ) = f(m 2 ) m 1 1 = m 2 1 m 1 = m 2 (d) The function g 1 is also one-to-one: For any integers m 1 and m 2, g 1 (m 1 ) = g 1 (m 2 ) 3m = 3m m 1 = 3m 2 m 1 = m 2 (e) Let s see what happens if we try to prove that h is one-to-one. For arbitrary integers m 1 and m 2, h(m 1 ) = h(m 2 ) m 1 +1 = m 2 +1 m 1 = m 2 However, the fact that m 1 = m 2 does not imply that m 1 = m 2, which leads us to suspect that h is not one-to-one. Indeed, if we let m 1 = 1 and m 2 = 1, then we see that h( 1) = 2 = h(1), which shows that h is not one-to-one. (f) For x 1, x 2 Q {1}, we have the following string of implications: This shows that p is a one-to-one function. p(x 1 ) = p(x 2 ) x 1 1 x 1 = x 2 1 x 2 x 1 (1 x 2 ) = x 2 (1 x 1 ) x 1 x 1 x 2 = x 2 x 1 x 2 x 1 = x 2 Suppose we have a function f : X Y, where both the domain X and the codomain Y are subsets of the set R of real numbers. Is there a way to determine from the graph of f whether f is one-to-one? Well, if f is not one-to-one, then there exist distinct elements x 1, x 2 X such that f(x 1 ) = f(x 2 ). Letting y 1 = f(x 1 ), we have the two distinct points (x 1, y 1 ) and (x 2, y 1 ) that are both on the graph of f and that are also both on the horizontal line y = y 1. Conversely, if some horizontal line intersects the graph of f in more than one point, then f is not one-to-one. This yields the horizontal line test, which is stated as follows: f is one-to-one if and only if every horizontal line intersects the graph of f in at most one point Example 3.11: Apply the horizontal line test to determine which of the functions defined in Example 3.8 are one-to-one. (See Figure 3.1.)

8 144 Chapter 3 Functions Solution: (a) The function f : R R defined by f(x) = 4 2x is one-to-one by the horizontal line test. (b) The function g: (0, ) (0, ) defined by g(x) = 1/x is also seen to be one-to-one. (c) The function h: R [ 4, ) defined by h(x) = x 2 2x 3 fails the horizontal line test. For example, the line y = 0 (the x axis) intersects the graph in the points ( 1, 0) and (3, 0). Thus, f is not one-to-one. Please be cautioned that the horizontal line test, as well as the vertical line test mentioned in Section 3.1, apply only when both the domain and codomain of the function f under consideration are subsets of R. Given a function f : X Y, what can be said about im f? Having no specific information about f, all that can be said is that im f is a subset of Y. One extreme possibility is provided by the example g: X Y defined by g(x) = y 0, where y 0 is a fixed element of Y. In this case, im f = {y 0 }, and g is called a constant function (the value of g is constant at y 0 ). The other extreme case is that of a function f : X Y for which im f = Y. Definition 3.3: A function f : X Y is called onto provided im f = Y. Observe that a function f : X Y is onto provided, for each y Y, there exists an x X such that f(x) = y. In other words, f is onto if and only if each y Y has at least one preimage under f. This condition provides a common method for proving that a given function f : X Y is onto. We choose an arbitrary element y Y, set f(x) = y, and then attempt to solve this equation for x in terms of y. If a solution x exists and is in X, then f is onto. On the other hand, if for some y Y there is no solution x X to the equation f(x) = y, then f is not onto. This method is illustrated in the next example. Example 3.12: Determine which of the following functions are onto. (a) h 1 : {1, 2, 3, 4} {1, 2, 3}; h 1 (1) = 2, h 1 (2) = 3, h 1 (3) = 2, h 1 (4) = 3 (b) h 2 : {1, 2, 3, 4} {1, 2, 3}; h 2 (1) = 3, h 2 (2) = 1, h 2 (3) = 2, h 2 (4) = 1 (c) f : Z Z; f(m) = m 1 (d) g 1 : Z Z; g 1 (m) = 3m + 1 (e) h: Z Z + ; h(m) = m +1 (f) p: Q {1} Q; p(x) = x/(1 x) (g) g 2 : Q Q; g 2 (x) = 3x + 1 Solution: (a) The function h 1 is not onto since im h 1 = {2, 3} {1, 2, 3}. (b) The function h 2 is onto since im h 2 = {1, 2, 3}, the codomain of h 2. (c) The function f is onto since, for any m Z, f(m + 1) = (m + 1) 1 = m

9 3.2 One-to-One Functions and Onto Functions 145 (d) Note that im g 1 = {3m + 1 m Z} = {..., 5, 2, 1, 4, 7,...} Since im g 1 Z, g 1 is not onto. (e) For n Z +, h(m) = n m +1 = n m = n 1 m = n 1 or m = 1 n Thus, each n 2 is the image of two integers, namely, n 1 and 1 n. Also, 1 is the image of 0. This shows that h is onto. (f) For y Q, p(x) = y x 1 x = y x = y xy x + xy = y x(1 + y) = y x = y 1 + y Thus, if y 1, then x = y/(1 + y) Q {1} and p(x) = y. However, there does not exist x Q {1} such that p(x) = 1. Therefore, im p = Q { 1}, and the function p just misses being onto. (g) Note that the function g 2 has the same rule as the function g 1 of part (d), but the domain and codomain have been changed from Z to Q. Let s see what happens. Let y Q; we wish to find x Q such that g 2 (x) = y. Now, g 2 (x) = y 3x + 1 = y x = y 1 3 This shows that g 2 is onto; for each rational number y, the image of the rational number x = (y 1)/3 is y. This example illustrates the important point that whether a given function is onto depends not only on the defining formula of the function, but on the domain and codomain as well. We have seen examples of functions that are one-to-one and not onto, and the reverse possibility, functions that are onto but not one-to-one. Under what conditions does the existence of one condition imply the other? One very important case is supplied by the following theorem. Theorem 3.1: Let X and Y be nonempty finite sets and let f be a function from X to Y. 1. If f is one-to-one, then X Y. 2. If f is onto, then X Y. 3. If f is one-to-one and onto, then X = Y. 4. If X = Y, then f is one-to-one if and only if f is onto. (That is, if X = Y, then either f is both one-to-one and onto, or f is neither one-to-one nor onto.)

10 146 Chapter 3 Functions Proof: We prove part 4. You are asked to prove parts 1 and 2 in Exercise 14; note that part 3 follows immediately from parts 1 and 2. For part 4, let X = Y = n; assume X = {x 1, x 2,..., x n }. Under this assumption, we must prove the following two implications: (1) If f is one-to-one, then f is onto. (2) If f is onto, then f is one-to-one. To prove (1), assume f is one-to-one. Note that the image of f is the set im f = {f(x) x X} = {f(x 1 ), f(x 2 ),..., f(x n )} Suppose f(x i ) = f(x j ) for some i and j. Since f is one-to-one, f(x i ) = f(x j ) implies that x i = x j, and hence that i = j. This shows that f(x 1 ), f(x 2 ),..., f(x n ) are distinct elements of Y. Thus, we have both im f Y and im f = Y. We may conclude that im f = Y, thus proving that f is onto. To prove (2), assume that f is onto. Then im f = Y. Thus, {f(x 1 ), f(x 2 ),..., f(x n )} = Y and Y = n, so it follows that f(x 1 ), f(x 2 ),..., f(x n ) are distinct elements. Hence, x i x j implies that f(x i ) f(x j ), which shows that f is one-to-one. Theorem 3.1, part 3, is often applied to prove that two finite sets X and Y have the same cardinality. To show that X = Y, it suffices, by Theorem 3.1, part 3, to construct a one-to-one and onto function from X to Y. This may seem like a rather roundabout way to do things, but it is often quite enlightening. Such a proof is called a bijective proof, and several examples of such proofs are given in Chapter 5. The following example illustrates the application of Theorem 3.1, part 4. Example 3.13: Consider the function f : Z 30 Z 30 defined by f(x) = 7x. Let x 1, x 2 Z 30 ; the following steps show that f is one-to-one: f(x 1 ) = f(x 2 ) 7x 1 = 7x 2 7x 1 mod 30 = 7x 2 mod (7x 1 7x 2 ) (by Theorem 2.6) 30 [ 7(x 1 x 2 ) ] 30 (x 1 x 2 ) (since gcd(7, 30) = 1) x 1 mod 30 = x 2 mod 30 (by Theorem 2.6) x 1 = x 2 We now obtain that f is onto with no additional work; we simply apply Theorem 3.1, part 4! It should be emphasized that to apply Theorem 3.1, part 4, to a function f, the domain and codomain of f must be finite sets with the same cardinality. For example, define f and g on Z by f(m) = 2m and g(m) = m/2. It can then be checked that f is one-to-one but not onto, whereas g is onto but not one-to-one. (Remember, the set Z is an infinite set!) Now let X = {x 1, x 2,..., x n } and suppose that f : X X is a one-to-one function. Then it follows that the n-tuple (f(x 1 ), f(x 2 ),..., f(x n )) is simply an ordered arrangement of the elements

11 3.2 One-to-One Functions and Onto Functions 147 of X. Indeed, in this sense any one-to-one and onto function on a set can be regarded as selecting the elements of the set in some order, or permuting the elements of the set. Definition 3.4: Let X and Y be nonempty sets. A function f : X Y that is both one-to-one and onto is called a bijection from X to Y. If X = Y, then f is called a permutation of X. Example 3.14: Define the function f on {1, 2, 3, 4} by f(1) = 3, f(2) = 2, f(3) = 4, and f(4) = 1. It is easily checked that f is one-to-one and onto, and so f is a permutation of {1, 2, 3, 4}; note that (f(1), f(2), f(3), f(4)) = (3, 2, 4, 1). Consider again the functions in Examples 3.10 and The function f : Z Z defined by f(m) = m 1 is a permutation of Z. In Example 3.12, part (g), we showed that the function g 2 : Q Q defined by g 2 (x) = 3x + 1 is onto. It can also be shown that g 2 is one-to-one. Therefore, g 2 is a permutation of Q. Lastly, consider the function p of part (f); if we modify the function p by changing its codomain to Q { 1}, that is, if we define p: Q {1} Q { 1} by p(x) = x/(1 x), then p is a bijection from Q {1} to Q { 1}. Suppose now that f is a function from X to Y ; recall that im f = {f(x) x X}. It seems natural to write im f = f(x). More generally, for any subset A of X, we define the image of A under f to be the set f(a) = {f(x) x A} Similarly, for any subset B of Y, it is helpful to be able to easily refer to the set of preimages of elements of B. Formally, we define the preimage (or inverse image) of B under f to be the set f 1 (B) = {x X f(x) B} Symbolically, note that, for y Y, y f(a) x A(y = f(x)) and for x X, x f 1 (B) f(x) B Example 3.15: Define f : {1, 2, 3, 4, 5,6,7} {1, 2, 3, 4, 5} by f(1) = 2 = f(3) = f(6), f(2) = 1, f(4) = 5 = f(7), and f(5) = 4. Let Find each of the following: A 1 = {1, 2, 3, 4}, A 2 = {2, 3, 7}, B 1 = {2, 4}, B 2 = {3, 4, 5} (a) f(a 1 A 2 ) (b) f(a 1 ) f(a 2 ) (c) f 1 (B 1 B 2 ) (d) f 1 (B 1 ) f 1 (B 2 ) (e) f(a 1 A 2 ) (f) f(a 1 ) f(a 2 ) (g) f 1 (B 1 B 2 ) (h) f 1 (B 1 ) f 1 (B 2 ) (i) f(a 1 A 2 ) (j) f(a 1 ) f(a 2 ) (k) f 1 (B 1 B 2 ) (l) f 1 (B 1 ) f 1 (B 2 )

12 148 Chapter 3 Functions Solution: For part (a) we have the following: And for part (b) we obtain: f(a 1 A 2 ) = f({1, 2, 3, 4} {2, 3, 7}) = f({1, 2, 3, 4, 7}) = {1, 2, 5} f(a 1 ) f(a 2 ) = f({1, 2, 3, 4}) f({2, 3, 7}) = {1, 2, 5} {1, 2, 5} = {1, 2, 5} Note that f(a 1 A 2 ) = f(a 1 ) f(a 2 ). For parts (c) and (d) we obtain the following: f 1 (B 1 B 2 ) = f 1 ({2, 4} {3, 4, 5}) = f 1 ({2, 3, 4, 5}) = {1, 3, 4, 5, 6,7} f 1 (B 1 ) f 1 (B 2 ) = f 1 ({2, 4}) f 1 ({3, 4, 5}) = {1, 3, 5, 6} {4, 5, 7} = {1, 3, 4, 5, 6,7} Note that f 1 (B 1 B 2 ) = f 1 (B 1 ) f 1 (B 2 ). For parts (e) and (f) we have: f(a 1 A 2 ) = f({1, 2, 3, 4} {2, 3, 7}) = f({2, 3}) = {1, 2} f(a 1 ) f(a 2 ) = f({1, 2, 3, 4}) f({2, 3, 7}) = {1, 2, 5} {1, 2, 5} = {1, 2, 5} Here we see that f(a 1 A 2 ) f(a 1 ) f(a 2 ), although it is the case that f(a 1 A 2 ) is a subset of f(a 1 ) f(a 2 ). Computing the sets for parts (g) and (h) we find that: f 1 (B 1 B 2 ) = f 1 ({2, 4} {3, 4, 5}) = f 1 ({4}) = {5} f 1 (B 1 ) f 1 (B 2 ) = f 1 ({2, 4}) f 1 ({3, 4, 5}) = {1, 3, 5, 6} {4, 5, 7} = {5} Thus, we see that f 1 (B 1 B 2 ) and f 1 (B 1 ) f 1 (B 2 ) are equal in this example. Next, for parts (i) and (j), observe that: f(a 1 A 2 ) = f({1, 2, 3, 4} {2, 3, 7}) = f({1, 4}) = {2, 5} f(a 1 ) f(a 2 ) = f({1, 2, 3, 4}) f({2, 3, 7}) = {1, 2, 5} {1, 2, 5} = So f(a 1 A 2 ) and f(a 1 ) f(a 2 ) are not equal in this example, although it is true that f(a 1 ) f(a 2 ) is a subset of f(a 1 A 2 ). Finally, for parts (k) and (l) we find that: f 1 (B 1 B 2 ) = f 1 ({2, 4} {3, 4, 5}) = f 1 ({2}) = {1, 3, 6} f 1 (B 1 ) f 1 (B 2 ) = f 1 ({2, 4}) f 1 {3, 4, 5} = {1, 3, 5, 6} {4, 5, 7} = {1, 3, 6} So, it turns out that f 1 (B 1 B 2 ) = f 1 (B 1 ) f 1 (B 2 ) in this case. Example 3.16: Let g be the permutation of Q defined by g(x) = 3x + 1. Find the following images and preimages. (a) g(z) (c) g 1 (Z + ) (b) g(2z) (d) g 1 (2Z)

13 3.2 One-to-One Functions and Onto Functions 149 Solution: (a) For m Z, g(m) = 3m + 1. Thus, g(z) = {3m + 1 m Z} = {..., 5, 2, 1, 4, 7,...} that is, g(z) is the set of integers that yield a remainder of 1 when divided by 3, namely, 1 + 3Z. (b) Here we find that y g(2z) y = g(2m) (for some m Z) y = 3(2m) + 1 y = 6m + 1 Hence, g(2z) = {6m + 1 m Z} = {..., 11, 5, 1, 7, 13,...} = 1 + 6Z. (c) For this part we have that It follows that x g 1 (Z + ) g(x) Z + 3x + 1 = n (for some n Z + ) x = n 1 3 g 1 (Z + ) = {(n 1)/3 n Z + } = {0, 1 3, 2 3, 1, 4 3, 5 } 3, 2,... (d) Proceeding in a similar manner for this part, we find that: x g 1 (2Z) g(x) 2Z 3x + 1 = 2m + 1 (for some m Z) x = 2m 3 It follows that g 1 (2Z) = {2m/3 m Z} = {..., 4 3, 2 3, 0, 2 3, 4 } 3,... Example 3.15 illustrates some of the general properties of images and preimages with respect to the set operations of union, intersection, and difference. These properties, along with two others, are listed in the following theorem. Theorem 3.2: Given f : X Y, let A 1 and A 2 be subsets of X and let B 1 and B 2 be subsets of Y. Then the following properties hold: 1. (a) f(a 1 A 2 ) = f(a 1 ) f(a 2 ) (b) f 1 (B 1 B 2 ) = f 1 (B 1 ) f 1 (B 2 ) 2. (a) f(a 1 A 2 ) f(a 1 ) f(a 2 ) (b) f 1 (B 1 B 2 ) = f 1 (B 1 ) f 1 (B 2 ) 3. (a) f(a 1 ) f(a 2 ) f(a 1 A 2 ) (b) f 1 (B 1 ) f 1 (B 2 ) = f 1 (B 1 B 2 ) 4. (a) If A 1 A 2, then f(a 1 ) f(a 2 ). (b) If B 1 B 2, then f 1 (B 1 ) f 1 (B 2 ).

14 150 Chapter 3 Functions Proof: We prove 1(a) and 3(b), and leave the remaining parts for you to prove in Exercise 2. To show that f(a 1 A 2 ) = f(a 1 ) f(a 2 ), we show that each side is a subset of the other. If y f(a 1 A 2 ), then there is some x A 1 A 2 such that y = f(x). This element x is such that x A 1 or x A 2. If x A 1, then y = f(x) f(a 1 ). Similarly, if x A 2, then y f(a 2 ). Hence, y f(a 1 ) or y f(a 2 ), that is, y f(a 1 ) f(a 2 ). This shows that f(a 1 A 2 ) f(a 1 ) f(a 2 ). To show the reverse inclusion, suppose y f(a 1 ) f(a 2 ). Then y f(a 1 ) or y f(a 2 ). This means that y = f(x), where x A 1 or x A 2. Thus, x A 1 A 2, which shows that y f(a 1 A 2 ). Therefore, f(a 1 ) f(a 2 ) f(a 1 A 2 ), and this completes the proof of this part. The proof that f 1 (B 1 ) f 1 (B 2 ) = f 1 (B 1 B 2 ) is easily done using a string of biconditionals as follows: x f 1 (B 1 ) f 1 (B 2 ) x f 1 (B 1 ) and x / f 1 (B 2 ) f(x) B 1 and f(x) / B 2 Therefore, f 1 (B 1 ) f 1 (B 2 ) = f 1 (B 1 B 2 ). f(x) B 1 B 2 x f 1 (B 1 B 2 ) Exercise Set Each part gives a function; determine whether it is one-to-one. (a) f : Z Z + ; f(m) = m (b) g: Q Q; g(x) = x 3 (c) h: R R; h(x) = x 3 x (d) p: Q R; p(x) = 2 x (e) the cardinality function k from P({1, 2,..., n}) (where n is a fixed positive integer) to {0, 1,..., n}; k(x) = X (f) the complement function c on P({1, 2,..., n}) (where n is a fixed positive integer); c(x) = X 2. Prove the remaining parts of Theorem 3.2: (a) part 1(b) (c) part 2(b) (e) part 4(a) (b) part 2(a) (d) part 3(a) (f) part 4(b) 3. For each of the functions in Exercise 1, determine whether it is onto. 4. List all the one-to-one functions from {1, 2} to {1, 2, 3, 4}. (Note: A function with domain {1, 2,..., n} may be specified by giving its images in a list: (f(1), f(2),..., f(n)).) 5. Each part gives sets X and Y and a function from X to Y. Determine whether the function is one-to-one. (a) X = {1, 2, 3, 4}, Y = {1, 2, 3}; f 1 (1) = 2, f 1 (2) = 3, f 1 (3) = 1 = f 1 (4) (b) X = {1, 2, 3}, Y = {1, 2, 3, 4}; f 2 (1) = 3, f 2 (2) = 2, f 2 (3) = 1 (c) X = Y = {1, 2, 3, 4}; f 3 (1) = 2 = f 3 (3), f 3 (2) = 1 = f 3 (4) (d) X = Y = {1, 2, 3, 4}; f 4 (1) = 3, f 4 (2) = 4, f 4 (3) = 1, f 4 (4) = 2 (e) X = Y = Z; f 5 (m) = m

15 3.2 One-to-One Functions and Onto Functions 151 (f) X = Y = Z; (g) X = Y = Z + ; (h) X = Y = Z + ; f 7 (n) = f 6 (m) = f 8 (n) = { 3m if m < 0 2m if m 0 { (n + 1)/2 if n is odd n/2 if n is even { n + 1 if n is odd n 1 if n is even 6. List all the functions from {1, 2, 3, 4} onto {1, 2}. 7. For each of the functions in Exercise 5, determine whether it is onto. 8. Each part gives a set; list the permutations of that set. (a) {1} (b) {1, 2} (c) {1, 2, 3} 9. Each part gives a function; determine whether it is one-to-one. (a) f 1 : Z 10 Z 10 ; f 1 (x) = 3 x (b) f 2 : Z 10 Z 10 ; f 2 (x) = 5 x (c) f 3 : Z 36 Z 36 ; f 3 (x) = 3 x (d) f 4 : Z 36 Z 36 ; f 4 (x) = 5 x (e) f 5 : Z 10 Z 10 ; f 5 (x) = x + 5 (f) f 6 : Z 10 Z 10 ; f 6 (x) = (3 x) + 5 (g) f 7 : Z 12 Z; f 7 (x) = 2 x (h) f 8 : Z 8 Z 12 ; f 8 (x) = 3 x (i) f 9 : Z 6 Z 12 ; f 9 (x) = 2 x (j) f 0 : Z 12 Z 36 ; f 0 (x) = 6 x 10. Let n be a positive integer and let k Z n. Define f : Z n Z n by f(x) = k x. Give a necessary and sufficient condition on k for f to be a permutation of Z n. 11. For each of the functions in Exercise 9, determine whether it is onto. 12. Each part gives a function on Z 12. You are to determine whether the function is a permutation. Also, find the image of A = {1, 5, 7, 11} and the preimage of B = {4, 8}. (a) f 1 (x) = 2 x (b) f 2 (x) = 4 x (c) f 3 (x) = 5 x (d) f 4 (x) = x Give an example of a function on Z + that is: (a) neither one-to-one nor onto (c) onto but not one-to-one (b) one-to-one but not onto (d) both one-to-one and onto 14. Prove Theorem 3.1: (a) part 1 (b) part Give an example of a function on (the closed interval) [ 1, 1 ] that is: (a) neither one-to-one nor onto (c) onto but not one-to-one (b) one-to-one but not onto (d) both one-to-one and onto

16 152 Chapter 3 Functions 16. Let X and Y be nonempty sets and let f be a function from X to Y. Complete each of the following statements by inserting the correct relation:,, or =. (a) f is one-to-one if and only if f 1 ({y}) 1 for every y Y. (b) f is onto if and only if f 1 ({y}) 1 for every y Y. (c) f is a bijection if and only if f 1 ({y}) 1 for every y Y. 17. Each of the following parts refers to the function defined in the corresponding part of Exercise 5. You are given a subset A of X and a subset B of Y ; find the image of A and the preimage of B. (a) A = {1, 2} = B (b) A = {1, 3}, B = {2, 4} (c) A = {1, 3}, B = {1} (d) A = {2} = B (e) A = 2Z, B = Z + (f) A = Z +, B = 2Z (g) A = B = 2Z + (h) A = B = 2Z INVERSE FUNCTIONS AND COMPOSITION Let X and Y be nonempty sets and f be a function from X to Y. Suppose that f is a bijection (that is, f is both one-to-one and onto). Since f is onto, given any y Y, there is an element x X such that f(x) = y. Moreover, since f is one-to-one, this element x is uniquely determined. Thus, for each y Y, there is exactly one x X such that y = f(x). (See Exercise 16 in Exercise Set 3.2.) We can then define a new function g: Y X as follows: For y Y, g(y) = x if and only if f(x) = y In other words, g(y) is that unique element x X for which f(x) = y. Definition 3.5: Let f : X Y be a bijection. The function g: Y X defined by g(y) = x if and only if f(x) = y is called the inverse function of f and is denoted by f 1. The situation of a function f : X Y and its inverse function g = f 1 : Y X is depicted in Figure 3.3, where f(x 0 ) = y 0. In this situation, suppose that B is a subset of Y. At this point in our discussion of functions, we have two possible interpretations for the notation f 1 (B): one is that f 1 (B) denotes the preimage of B under f, and the other is that f 1 (B) denotes the image of B under f 1. In Exercise 2, you are asked to show that these two sets are, in fact, the same, and so there is no problem. It should be pointed out, however, that if f is not a bijection, then f 1 (B) can mean only the preimage of B under f. Theorem 3.3: If f : X Y is a bijection, then the inverse function f 1 : Y X is also a bijection. Proof: We first show that f 1 is one-to-one. Suppose f 1 (y 1 ) = x = f 1 (y 2 ) for some y 1, y 2 Y and x X. Then, by definition, y 1 = f(x) and y 2 = f(x). Since f is a function, x has a unique image under f, and it follows that y 1 = y 2. Thus, f 1 is one-to-one.

17 3.3 Inverse Functions and Composition 153 Figure 3.3 A function and its inverse Next we show that f 1 is onto. Let x X, and let y = f(x). Then, by the definition of f 1, it follows that x = f 1 (y). This shows that any x X has a preimage under f 1, and so f 1 is onto. Corollary 3.4: Let X be a nonempty set. If f is a permutation of X, then f 1 is also a permutation of X. Example 3.17: Define f : {1, 2, 3, 4, 5} {1, 2, 3, 4,5} by f(1) = 3, f(2) = 4, f(3) = 5, f(4) = 1, and f(5) = 2, that is, f is the permutation (3, 4, 5, 1, 2) of {1, 2, 3, 4, 5}. Find f 1. Solution: Since f(1) = 3, we know that f 1 (3) = 1. Similarly, f 1 (4) = 2, f 1 (5) = 3, f 1 (1) = 4, and f 1 (2) = 5. Therefore, f 1 : {1, 2, 3, 4, 5} {1, 2, 3, 4, 5} is defined by f 1 (1) = 4, f 1 (2) = 5, f 1 (3) = 1, f 1 (4) = 2, f 1 (5) = 3 that is, f 1 is the permutation (4, 5, 1, 2, 3) of {1, 2, 3, 4, 5}. If f : X Y is a bijection, how do we find its inverse function? For example, given f : Q Q defined by f(x) = 3x + 1, let s attempt to find f 1. Given y Q, we want to find x Q such that f 1 (y) = x. This means that y = f(x), so that y = 3x + 1. Solving for x we obtain x = (y 1)/3. Thus, f 1 : Q Q is defined by f 1 (y) = y 1 3

18 154 Chapter 3 Functions In general, given a bijection y = f(x), we solve for x in terms of y to obtain x = f 1 (y) Example 3.18: We saw previously that the function p: Q {1} Q { 1}, defined by p(x) = x/(1 x), is a bijection. Find p 1. Solution: If y Q { 1} and p 1 (y) = x, where x Q {1}, then p(x) = y. We then proceed algebraically as follows: x y = y x = y xy x + xy = y x(1 + y) = y x = 1 x 1 + y Therefore, p 1 : Q { 1} Q {1} is given by p 1 (y) = y 1 + y Example 3.19: We saw in Example 3.13 that the function f : Z 30 Z 30 defined by f(x) = 7x is a permutation of Z 30. Find f 1. Solution: If y Z 30 and f 1 (y) = x, where x Z 30, then f(x) = y, that is, 7x = y. We want to solve this equation for x. We do this by multiplying both sides of the equation by the reciprocal of 7 in Z 30 (since gcd(7, 30) = 1, the element 7 has a reciprocal in Z 30 ): 7 x = y 7 1 (7 x) = 7 1 y (7 1 7) x = 7 1 y 1 x = 7 1 y x = 7 1 y Thus, f 1 : Z 30 Z 30 is defined by f 1 (y) = 7 1 y. To complete the problem, we need to find 7 1. Recall that this can be done by using the extended Euclidean algorithm. Using this algorithm, we find that 1 = 7(13) + 30t for some integer t. Therefore, 7 1 = 13, and so f 1 (y) = 13 y. There are various ways in which two functions may be combined to produce a third function. One of the more common and important operations on functions is called composition. Suppose f is a function from X to Y and g is a function from Y to Z. For any x X, there is a unique y Y such that y = f(x). Then, for this element y, there is a unique z Z such that z = g(y) = g(f(x)). Hence, for each x X, there is associated a unique element z Z, namely, z = g(f(x)). This association allows us to define a new function h: X Z by h(x) = g(f(x)). This situation is depicted in Figure 3.4.

19 3.3 Inverse Functions and Composition 155 Figure 3.4 Composition of functions Definition 3.6: Given f : X Y and g: Y Z, the composition of f followed by g (or composite function) is the function g f : X Z defined by (g f)(x) = g(f(x)) Example 3.20: Define f : Z 2Z by f(m) = 2m, and define g: 2Z Z + by g(m) = m / Then g f : Z Z + is given by (g f)(m) = g(f(m)) = g(2m) = 2m = m +1 Example 3.21: Define f : Q {0} Q {1} by f(x) = (x + 1)/x and g: Q {1} Q {2} by g(x) = 3x 1. Then g f : Q {0} Q {2} is given by ( ) x + 1 (g f)(x) = g(f(x)) = g x ( ) x + 1 = 3 1 x 3(x + 1) = x x x = 2x + 3 x

20 156 Chapter 3 Functions Example 3.22: Define f : Z Z by f(m) = m + 3 and g: Z Z by g(m) = m. Then f g: Z Z is given by whereas g f : Z Z is given by (f g)(m) = f(g(m)) = f( m) = m + 3 (g f)(m) = g(f(m)) = g(m + 3) = (m + 3) = m 3 Note that f g g f; also note that f, g, f g, and g f are all permutations of Z. Let f and g be two functions from X to Y. When is it the case that f = g? Since f and g are functions, each is a subset of X Y, and we already know when two sets are equal. Thus, we say that f = g provided the condition f(x) = g(x) holds for every x X. There are several interesting results that involve composition of functions and the properties onto and one-to-one. For example, suppose f : X Y is a bijection; then f 1 : Y X exists. Given x X with f(x) = y, we have that f 1 (y) = x and, hence, f 1 (f(x)) = x. So the composite function f 1 f : X X satisfies the property (f 1 f)(x) = x for all x X. In a similar fashion, we can determine that f f 1 : Y Y satisfies the condition (f f 1 )(y) = y for every y Y. Note that both of f f 1 and f 1 f are functions of the type h: A A, where h(a) = a for all a A. Definition 3.7: For any nonempty set A, the function i A : A A defined by i A (a) = a is called the identity function on A. In view of the preceding discussion, if f : X Y is a bijection, then f 1 f = i X and f f 1 = i Y. A very basic and easily verified property of identity functions is contained in the following theorem, whose proof is left to Exercise 4. Theorem 3.5: Let X and Y be nonempty sets. For any function f : X Y, i Y f = f and f i X = f Theorem 3.6: Given f : X Y and g: Y Z, the following properties hold: 1. If f and g are both one-to-one, then g f is one-to-one. 2. If f and g are both onto, then g f is onto. Proof: We prove part 2; you are asked to prove part 1 in Exercise 6.

21 3.3 Inverse Functions and Composition 157 Assume f and g are both onto. To prove that g f is onto, we begin with an arbitrary element z 0 Z. Since g is onto, there is an element y 0 Y such that g(y 0 ) = z 0. Then, since y 0 Y and f is onto, there is some element x 0 X such that f(x 0 ) = y 0. Thus, (g f)(x 0 ) = g(f(x 0 )) = g(y 0 ) = z 0 and it follows that g f is onto. Corollary 3.7: If f : X Y and g: Y Z are both bijections, then the function g f : X Z is also a bijection. In particular, if X = Y = Z so that f and g are both permutations of X, then g f is a permutation of X. For each of the statements in Theorem 3.6, the converse is false (see Exercises 8 and 10). However, a partial converse does hold. Theorem 3.8: Given f : X Y and g: Y Z, the following properties hold: 1. If g f is one-to-one, then f is one-to-one. 2. If g f is onto, then g is onto. Proof: We prove part 1; you are asked to prove part 2 in Exercise 12. Assume g f is one-to-one, and suppose that f(x 1 ) = f(x 2 ) for some x 1, x 2 X. Then, since f(x 1 ) Y, we have g(f(x 1 )) = g(f(x 2 )), that is, (g f)(x 1 ) = (g f)(x 2 ). Then, since g f is one-to-one, it may be concluded that x 1 = x 2. Therefore, f is one-to-one. Given functions f : A B, g: B C, and h: C D, notice that h g is a function from B to D and that g f is a function from A to C. Thus, (h g) f and h (g f) are both functions from A to D. In fact, they are equal functions. In other words, the associative property holds for composition of functions. Theorem 3.9: Given f : A B, g: B C, and h: C D, the following property holds: (h g) f = h (g f) Proof: Both (h g) f and h (g f) have domain A and codomain D. Hence, to show equality, we must show that the two functions have the same value at each x A. Proceeding, we obtain the following: [ (h g) f ] (x) = (h g) [ f(x) ] = h(g(f(x))) = h [ (g f)(x) ] = [ h (g f) ] (x)

22 158 Chapter 3 Functions Therefore, (h g) f = h (g f). Exercise Set Find the inverse of each of the following functions. (a) f 1 : Q Q; f 1 (x) = 4x + 2 (b) f 2 : Q {1} Q {2}; f 2 (x) = 2x/(x 1) (c) f 3 : Z 12 Z 12 ; f 3 (x) = 5 x (d) f 4 : Z 39 Z 39 ; f 4 (x) = (5 x) + 2 (e) f 5 : Z Z; f 5 (m) = m + 1 { 2m 1 if m > 0 (f) f 6 : Z {0, 1, 2, 3,...}; f 6 (m) = 2m if m 0 (g) f 7 : {1, 2, 3, 4} {1, 2, 3, 4}; f 7 (1) = 4, f 7 (2) = 1, f 7 (3) = 2, f 7 (4) = 3 (h) f 8 : {1, 2, 3, 4} {1, 2, 3, 4}; f 8 (1) = 3, f 8 (2) = 4, f 8 (3) = 1, f 8 (4) = 2 2. Let f : X Y be a bijection and let B be a subset of Y. Let A 1 be the preimage of B under f, and let A 2 be the image of B under f 1. Show that A 1 = A Find g f. (a) f : Z Z + ; f(m) = m +1, g: Z + Q + ; g(n) = 1/n (b) f : R (0, 1); f(x) = 1/(x 2 + 1), g: (0, 1) (0, 1); g(x) = 1 x (c) f : Q {2} Q {0}; f(x) = 1/(x 2), g: Q {0} Q {0}; g(x) = 1/x (d) f : R [1, ); f(x) = x 2 + 1, g: [1, ) [0, ); g(x) = x 1 (e) f : Q {10/3} Q {3}; f(r) = 3r 7, g: Q {3} Q {2}; g(r) = 2r/(r 3) (f) f : Z Z 5 ; f(m) = m mod 5, g: Z 5 Z 5 ; g(m) = m + 1 (g) f : Z 8 Z 12 ; f(m) = 3 m, g: Z 12 Z 6 ; g(m) = 2 m (h) f, g: {1, 2, 3, 4} {1, 2, 3, 4}; 4. Prove Theorem 3.5. f(1) = 4 f(2) = 1 f(3) = 2 f(4) = 3 g(1) = 3 g(2) = 4 g(3) = 1 g(4) = 2 5. Given the permutations f and g, find f 1, g 1, f g, (f g) 1, and g 1 f 1. (a) f : Z Z; f(m) = m + 1, g: Z Z; g(m) = 2 m (b) f : Z 7 Z 7 ; f(m) = m + 3, g: Z 7 Z 7 ; g(m) = 2 m (c) f, g: {1, 2, 3, 4} {1, 2, 3, 4}; (d) f, g: {1, 2, 3, 4} {1, 2, 3, 4}; f(1) = 4 f(2) = 1 f(3) = 2 f(4) = 3 g(1) = 3 g(2) = 4 g(3) = 1 g(4) = 2 f(1) = 2 f(2) = 4 f(3) = 3 f(4) = 1 g(1) = 1 g(2) = 3 g(3) = 4 g(4) = 2 (e) f : Q Q; f(x) = 4x, g: Q Q; g(x) = (x 3)/2 (f) f : Q {1} Q {1}; f(x) = 2x 1, g: Q {1} Q {1}; g(x) = x/(x 1)

23 Chapter Problems Prove Theorem 3.6, part For the permutations f and g given in Exercise 5, find g f and f 1 g Give an example of sets X, Y, and Z, and of functions f : X Y and g: Y Z, such that g f and f are both one-to-one, but g is not one-to-one. 9. Define the functions f and g on your (maternal) family tree by f(x) = the mother of x and g(x) = the eldest child of the mother of x. Describe each of these functions. (a) f f (c) g f (b) f g (d) g g 10. Give an example of sets X, Y, and Z, and of functions f : X Y and g: Y Z, such that g f and g are both onto, but f is not onto. 11. Let f : X Y be a bijection. Prove that (f 1 ) 1 = f. 12. Prove Theorem 3.8, part Let f : X Y and g: Y Z be bijections. Prove that (g f) 1 = f 1 g Let f : X Y and g: Y X be bijections. Prove: If g f = i X (or f g = i Y ), then g = f 1. CHAPTER PROBLEMS 1. Define f : R R by f(x) = x (a) Show that f is a permutation of R. (b) Find f 1. (c) Suppose the domain and codomain of f are changed from R to Q. Is f a permutation of Q? 2. Show that there are infinitely many pairs of distinct functions f and g on Q such that none of f, g, nor f g is the identity function on Q and f g = g f. (Hint: Consider linear functions.) 3. Define f : R R by f(x) = 4x + 1. (a) Show that f is a permutation of R. (b) Find f 1. (c) Suppose the domain and codomain of f are changed from R to Q. Is f a permutation of Q? (d) Suppose the domain and codomain of f are changed from R to Z. Is f a permutation of Z? 4. Let X and Y be nonempty sets and let f : X Y be a function. Prove that, if the condition f(a 1 A 2 ) = f(a 1 ) f(a 2 ) holds for all subsets A 1 and A 2 of X, then f is one-to-one, and conversely. (Hint: For necessity, by Theorem 3.2, part 2(a), it suffices to prove that, if f is one-to-one, then f(a 1 ) f(a 2 ) f(a 1 A 2 ); for sufficiency, prove the contrapositive.)

24 160 Chapter 3 Functions 5. Let f : X Y, where X and Y are subsets of R. The function f is said to be increasing provided the following condition holds for all x 1, x 2 X: x 1 < x 2 f(x 1 ) < f(x 2 ) Similarly, f is said to be decreasing provided the following condition holds for all x 1, x 2 X: x 1 < x 2 f(x 1 ) > f(x 2 ) If f is either increasing or decreasing, then we say that f is monotonic. (a) Prove that a monotonic function is one-to-one. (b) Define f : ( 1, 1) R by f(x) = x/(1 x 2 ). Apply the result of part (a) to show that f is one-to-one. (c) Define g: R R by g(x) = x 3 + x 2. Apply the result of part (a) to show that g is one-to-one. (Hint: Show that g (x) > 0 and apply a result from calculus.) 6. Let X and Y be nonempty sets and let f : X Y be a function. Prove the following results. (a) If the condition f(f 1 (B)) = B holds for every subset B of Y, then f is onto, and conversely. (b) If the condition f 1 (f(a)) = A holds for every subset A of X, then f is one-to-one, and conversely. (Hint: In both parts, prove necessity directly and sufficiency by contrapositive. Also, recall that to prove two sets V and W are equal, it suffices to prove that both V W and W V.) 7. Let f, g, and h be functions on Z defined as follows: f(m) = m + 1, g(m) = 2m, and { 0 if m is even h(m) = 1 if m is odd Determine the following composite functions. (a) f g (c) f h (e) g h (g) g g (b) g f (d) h f (f) h g (h) h f g 8. Let U be a nonempty universal set. For A U, define the function χ A : U {0, 1} by { 0 if x / A χ A (x) = 1 if x A The function χ A is called the characteristic function of A. For A, B P(U), let C = A B, D = A B, and E = A B. Prove that the following relations hold for all x U: (a) χ C (x) = χ A (x) χ B (x) (b) χ D (x) = χ A (x) + χ B (x) [χ A (x) χ B (x)] (c) χ U (x) = 1 (d) χ (x) = 0 (e) χ B (x) = 1 χ B (x) (f) χ E (x) = χ A (x) [1 χ B (x)]

25 Chapter Problems Define f : Z 119 Z 119 by f(x) = 15 x. Show that f is a permutation of Z 119 and find f Let X be a nonempty set, let i denote the identity function on X, and let f be a function on X. Define f 0 = i, f 1 = f, f 2 = f f, f 3 = f f f, and so on. (Recursively, f 0 = i and f n = f f n 1, for n 1.) In particular, take the case X = Z; give an example of a function f : Z Z such that: (a) f i but f 2 = i (b) f 2 i but f 3 = i (Hint: Define f by f(1) = 2, f(2) = 3, f(3) = 1, and f(x) = x for x / {1, 2, 3}.) (c) Generalize parts (a) and (b); for each n > 1, give an example of a function f : Z Z such that f i,..., f n 1 i, but f n = i. 11. Define f : Q {1/4} Q {0} by f(x) = 1 4x and g: Q {0} Q {3/2} by g(x) = (3x 1)/(2x). Determine each of these functions. (a) g f (c) f 1 (e) f 1 g 1 (b) (g f) 1 (d) g Let F denote the set of functions on R, let C = {f F f is continuous}, and let D = {f F f is differentiable}. Consider the function : D F that maps each function f D to its derivative f in F; that is, (f) = f. (a) Is the function one-to-one? (b) Let f D. What is 1 ({f })? (c) Show that C im. 13. Give an example of a function f : (a) neither one-to-one nor onto (c) onto but not one-to-one [ 1, 1 ] [ 0, 4 ] that is: (b) one-to-one but not onto (d) both one-to-one and onto 14. Let m and n be positive integers, and let A = {0, 1,..., m 1}, B = {0, 1,..., n 1}, and C = {0, 1,..., mn 1}. Construct a bijection f : A B C. (Hint: First try a special case, such as m = 2 and n = 3; then try to generalize your construction.) [ ] [ ] 15. Give an example of a function f : 0, 4 1, 1 that is: (a) neither one-to-one nor onto (c) onto but not one-to-one (b) one-to-one but not onto (d) both one-to-one and onto 16. Construct a bijection: (a) from ( 1, 1) to R (b) from Z to Z Let c and d be real numbers with c < d. Construct a bijection from (0, 1) to (c, d). 18. Construct a function f : Q + Z + such that f is one-to-one. 19. Given real numbers a, b, c, and d such that a < b and c < d, construct a bijection from (a, b) to (c, d). 20. For any set X, prove (by contradiction) that there does not exist a bijection from X to P(X).

26 162 Chapter 3 Functions 21. Construct a bijection from Z to 2Z. 22. Let A 1, B 1, A 2, and B 2 be nonempty sets such that A 1 B 1 = = A 2 B 2. Given bijections f : A 1 A 2 and g: B 1 B 2, construct a bijection h: A 1 B 1 A 2 B Construct a bijection from 1 + 2Z to 3Z Let A 1, B 1, A 2, and B 2 be nonempty sets. Given bijections f : A 1 A 2 and g: B 1 B 2, construct a bijection h: A 1 B 1 A 2 B Define the function f : Z Z by f(m) = 5m + 2. (a) Either prove that f is one-to-one or give an example to show that f is not one-to-one. (b) Either prove that f is onto or give an example to show that f is not onto. (c) Find f(2z). (d) Find f 1 (3Z + ). 26. Let X be a nonempty subset of R. Given two functions f and g from X to R, we define functions f + g: X R and f g: X R, called the sum and product of f and g, respectively, as follows: (f + g)(x) = f(x) + g(x) (f g)(x) = f(x)g(x) Each part gives a subset X of R and two functions f and g from X to R; find f + g and f g. (a) X = Z + ; f(x) = x 2 and g(x) = 2x 1 (b) X = Q; f(x) = x/3 and g(x) = 3x + 2 (c) X = R {0}; f(x) = (x 2 + 1)/x and g(x) = x/(x 2 + 1) (d) X = R; f(x) = x 2 2x + 3 and g(x) = x 2 + 2x Define the function f : R [ 1, ) by f(x) = x (a) Either prove that f is one-to-one or give an example to show that f is not one-to-one. (b) Either prove that f is onto or give an example to show that f is not onto. (c) Find f( [ 0, 2 ] ). (d) Find f 1 ( [ 1, 3 ] ). (e) Find f 1 ( [ 2, 5 ] ). 28. Determine f g and g f. (a) f : R R; f(x) = x 2 + x, g: R R; g(x) = 3x + 4 (b) f : (0, ) (0, ); f(x) = x/(x 2 + 1), g: (0, ) (0, ); g(x) = 1/x 29. Define f : R {1} R {2} by f(x) = (x 3 + 7)/4 and g: R {2} R {1} by g(x) = x/(x 2). Determine these functions. (a) g f (c) f g (b) (g f) 1 (d) (f g) 1

### 1 if 1 x 0 1 if 0 x 1

Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or

### 3. Equivalence Relations. Discussion

3. EQUIVALENCE RELATIONS 33 3. Equivalence Relations 3.1. Definition of an Equivalence Relations. Definition 3.1.1. A relation R on a set A is an equivalence relation if and only if R is reflexive, symmetric,

### INTRODUCTORY SET THEORY

M.Sc. program in mathematics INTRODUCTORY SET THEORY Katalin Károlyi Department of Applied Analysis, Eötvös Loránd University H-1088 Budapest, Múzeum krt. 6-8. CONTENTS 1. SETS Set, equal sets, subset,

### Practice with Proofs

Practice with Proofs October 6, 2014 Recall the following Definition 0.1. A function f is increasing if for every x, y in the domain of f, x < y = f(x) < f(y) 1. Prove that h(x) = x 3 is increasing, using

### Cartesian Products and Relations

Cartesian Products and Relations Definition (Cartesian product) If A and B are sets, the Cartesian product of A and B is the set A B = {(a, b) :(a A) and (b B)}. The following points are worth special

### 2.1 Functions. 2.1 J.A.Beachy 1. from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair

2.1 J.A.Beachy 1 2.1 Functions from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 21. The Vertical Line Test from calculus says that a curve in the xy-plane

### Discrete Mathematics: Solutions to Homework (12%) For each of the following sets, determine whether {2} is an element of that set.

Discrete Mathematics: Solutions to Homework 2 1. (12%) For each of the following sets, determine whether {2} is an element of that set. (a) {x R x is an integer greater than 1} (b) {x R x is the square

### 1 The Concept of a Mapping

Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 1 The Concept of a Mapping The concept of a mapping (aka function) is important throughout mathematics. We have been dealing

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

### TOPIC 3: CONTINUITY OF FUNCTIONS

TOPIC 3: CONTINUITY OF FUNCTIONS. Absolute value We work in the field of real numbers, R. For the study of the properties of functions we need the concept of absolute value of a number. Definition.. Let

### H/wk 13, Solutions to selected problems

H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots.

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### 2 Complex Functions and the Cauchy-Riemann Equations

2 Complex Functions and the Cauchy-Riemann Equations 2.1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. Likewise, in complex analysis, we study functions f(z)

### You know from calculus that functions play a fundamental role in mathematics.

CHPTER 12 Functions You know from calculus that functions play a fundamental role in mathematics. You likely view a function as a kind of formula that describes a relationship between two (or more) quantities.

### Mathematics Review for MS Finance Students

Mathematics Review for MS Finance Students Anthony M. Marino Department of Finance and Business Economics Marshall School of Business Lecture 1: Introductory Material Sets The Real Number System Functions,

### 1. Prove that the empty set is a subset of every set.

1. Prove that the empty set is a subset of every set. Basic Topology Written by Men-Gen Tsai email: b89902089@ntu.edu.tw Proof: For any element x of the empty set, x is also an element of every set since

### To define function and introduce operations on the set of functions. To investigate which of the field properties hold in the set of functions

Chapter 7 Functions This unit defines and investigates functions as algebraic objects. First, we define functions and discuss various means of representing them. Then we introduce operations on functions

### Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011

Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011 A. Miller 1. Introduction. The definitions of metric space and topological space were developed in the early 1900 s, largely

### Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

### Chapter 3. Cartesian Products and Relations. 3.1 Cartesian Products

Chapter 3 Cartesian Products and Relations The material in this chapter is the first real encounter with abstraction. Relations are very general thing they are a special type of subset. After introducing

### 3. Mathematical Induction

3. MATHEMATICAL INDUCTION 83 3. Mathematical Induction 3.1. First Principle of Mathematical Induction. Let P (n) be a predicate with domain of discourse (over) the natural numbers N = {0, 1,,...}. If (1)

### Lecture 16 : Relations and Functions DRAFT

CS/Math 240: Introduction to Discrete Mathematics 3/29/2011 Lecture 16 : Relations and Functions Instructor: Dieter van Melkebeek Scribe: Dalibor Zelený DRAFT In Lecture 3, we described a correspondence

### Mathematics for Computer Science/Software Engineering. Notes for the course MSM1F3 Dr. R. A. Wilson

Mathematics for Computer Science/Software Engineering Notes for the course MSM1F3 Dr. R. A. Wilson October 1996 Chapter 1 Logic Lecture no. 1. We introduce the concept of a proposition, which is a statement

### Geometric Transformations

Geometric Transformations Definitions Def: f is a mapping (function) of a set A into a set B if for every element a of A there exists a unique element b of B that is paired with a; this pairing is denoted

### Foundations of Mathematics I Set Theory (only a draft)

Foundations of Mathematics I Set Theory (only a draft) Ali Nesin Mathematics Department Istanbul Bilgi University Kuştepe Şişli Istanbul Turkey anesin@bilgi.edu.tr February 12, 2004 2 Contents I Naive

### Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics

Undergraduate Notes in Mathematics Arkansas Tech University Department of Mathematics An Introductory Single Variable Real Analysis: A Learning Approach through Problem Solving Marcel B. Finan c All Rights

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### Real Roots of Univariate Polynomials with Real Coefficients

Real Roots of Univariate Polynomials with Real Coefficients mostly written by Christina Hewitt March 22, 2012 1 Introduction Polynomial equations are used throughout mathematics. When solving polynomials

### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### CHAPTER 3. Methods of Proofs. 1. Logical Arguments and Formal Proofs

CHAPTER 3 Methods of Proofs 1. Logical Arguments and Formal Proofs 1.1. Basic Terminology. An axiom is a statement that is given to be true. A rule of inference is a logical rule that is used to deduce

### ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS

ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS John A. Beachy Northern Illinois University 2014 ii J.A.Beachy This is a supplement to Abstract Algebra, Third Edition by John A. Beachy and William D. Blair

### it is easy to see that α = a

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore

### This asserts two sets are equal iff they have the same elements, that is, a set is determined by its elements.

3. Axioms of Set theory Before presenting the axioms of set theory, we first make a few basic comments about the relevant first order logic. We will give a somewhat more detailed discussion later, but

### Pythagorean Triples. Chapter 2. a 2 + b 2 = c 2

Chapter Pythagorean Triples The Pythagorean Theorem, that beloved formula of all high school geometry students, says that the sum of the squares of the sides of a right triangle equals the square of the

### PROBLEM SET 6: POLYNOMIALS

PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other

### 6.2 Permutations continued

6.2 Permutations continued Theorem A permutation on a finite set A is either a cycle or can be expressed as a product (composition of disjoint cycles. Proof is by (strong induction on the number, r, of

### Applications of Methods of Proof

CHAPTER 4 Applications of Methods of Proof 1. Set Operations 1.1. Set Operations. The set-theoretic operations, intersection, union, and complementation, defined in Chapter 1.1 Introduction to Sets are

### = 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that

Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without

### Week 5: Binary Relations

1 Binary Relations Week 5: Binary Relations The concept of relation is common in daily life and seems intuitively clear. For instance, let X be the set of all living human females and Y the set of all

### 8 Divisibility and prime numbers

8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express

### Elementary Number Theory and Methods of Proof. CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.

Elementary Number Theory and Methods of Proof CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.edu/~cse215 1 Number theory Properties: 2 Properties of integers (whole

### The last three chapters introduced three major proof techniques: direct,

CHAPTER 7 Proving Non-Conditional Statements The last three chapters introduced three major proof techniques: direct, contrapositive and contradiction. These three techniques are used to prove statements

### Chapter 7. Functions and onto. 7.1 Functions

Chapter 7 Functions and onto This chapter covers functions, including function composition and what it means for a function to be onto. In the process, we ll see what happens when two dissimilar quantifiers

### MATH 4330/5330, Fourier Analysis Section 11, The Discrete Fourier Transform

MATH 433/533, Fourier Analysis Section 11, The Discrete Fourier Transform Now, instead of considering functions defined on a continuous domain, like the interval [, 1) or the whole real line R, we wish

### Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof. Harper Langston New York University

Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof Harper Langston New York University Proof and Counterexample Discovery and proof Even and odd numbers number n from Z is called

### Euclidean Geometry. We start with the idea of an axiomatic system. An axiomatic system has four parts:

Euclidean Geometry Students are often so challenged by the details of Euclidean geometry that they miss the rich structure of the subject. We give an overview of a piece of this structure below. We start

### Quotient Rings and Field Extensions

Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

### Limit processes are the basis of calculus. For example, the derivative. f f (x + h) f (x)

SEC. 4.1 TAYLOR SERIES AND CALCULATION OF FUNCTIONS 187 Taylor Series 4.1 Taylor Series and Calculation of Functions Limit processes are the basis of calculus. For example, the derivative f f (x + h) f

### CHAPTER 5. Number Theory. 1. Integers and Division. Discussion

CHAPTER 5 Number Theory 1. Integers and Division 1.1. Divisibility. Definition 1.1.1. Given two integers a and b we say a divides b if there is an integer c such that b = ac. If a divides b, we write a

### Application. Outline. 3-1 Polynomial Functions 3-2 Finding Rational Zeros of. Polynomial. 3-3 Approximating Real Zeros of.

Polynomial and Rational Functions Outline 3-1 Polynomial Functions 3-2 Finding Rational Zeros of Polynomials 3-3 Approximating Real Zeros of Polynomials 3-4 Rational Functions Chapter 3 Group Activity:

### Understanding Basic Calculus

Understanding Basic Calculus S.K. Chung Dedicated to all the people who have helped me in my life. i Preface This book is a revised and expanded version of the lecture notes for Basic Calculus and other

### Vector Spaces II: Finite Dimensional Linear Algebra 1

John Nachbar September 2, 2014 Vector Spaces II: Finite Dimensional Linear Algebra 1 1 Definitions and Basic Theorems. For basic properties and notation for R N, see the notes Vector Spaces I. Definition

### NOTES ON LINEAR TRANSFORMATIONS

NOTES ON LINEAR TRANSFORMATIONS Definition 1. Let V and W be vector spaces. A function T : V W is a linear transformation from V to W if the following two properties hold. i T v + v = T v + T v for all

### Limits and Continuity

Math 20C Multivariable Calculus Lecture Limits and Continuity Slide Review of Limit. Side limits and squeeze theorem. Continuous functions of 2,3 variables. Review: Limits Slide 2 Definition Given a function

### POWER SETS AND RELATIONS

POWER SETS AND RELATIONS L. MARIZZA A. BAILEY 1. The Power Set Now that we have defined sets as best we can, we can consider a sets of sets. If we were to assume nothing, except the existence of the empty

### Section 2-5 Quadratic Equations and Inequalities

-5 Quadratic Equations and Inequalities 5 a bi 6. (a bi)(c di) 6. c di 63. Show that i k, k a natural number. 6. Show that i k i, k a natural number. 65. Show that i and i are square roots of 3 i. 66.

### Review for Calculus Rational Functions, Logarithms & Exponentials

Definition and Domain of Rational Functions A rational function is defined as the quotient of two polynomial functions. F(x) = P(x) / Q(x) The domain of F is the set of all real numbers except those for

### a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.

Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given

### 1.7 Graphs of Functions

64 Relations and Functions 1.7 Graphs of Functions In Section 1.4 we defined a function as a special type of relation; one in which each x-coordinate was matched with only one y-coordinate. We spent most

### PART I. THE REAL NUMBERS

PART I. THE REAL NUMBERS This material assumes that you are already familiar with the real number system and the representation of the real numbers as points on the real line. I.1. THE NATURAL NUMBERS

### Taylor Polynomials and Taylor Series Math 126

Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will

### We give a basic overview of the mathematical background required for this course.

1 Background We give a basic overview of the mathematical background required for this course. 1.1 Set Theory We introduce some concepts from naive set theory (as opposed to axiomatic set theory). The

### 1.4 Factors and Prime Factorization

1.4 Factors and Prime Factorization Recall from Section 1.2 that the word factor refers to a number which divides into another number. For example, 3 and 6 are factors of 18 since 3 6 = 18. Note also that

### Florida State University Course Notes MAD 2104 Discrete Mathematics I

Florida State University Course Notes MAD 2104 Discrete Mathematics I Florida State University Tallahassee, Florida 32306-4510 Copyright c 2011 Florida State University Written by Dr. John Bryant and Dr.

### Basic Set Theory. Chapter Set Theory. can be written: A set is a Many that allows itself to be thought of as a One.

Chapter Basic Set Theory A set is a Many that allows itself to be thought of as a One. - Georg Cantor This chapter introduces set theory, mathematical induction, and formalizes the notion of mathematical

### Lecture 13 - Basic Number Theory.

Lecture 13 - Basic Number Theory. Boaz Barak March 22, 2010 Divisibility and primes Unless mentioned otherwise throughout this lecture all numbers are non-negative integers. We say that A divides B, denoted

PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient

### Mathematics for Computer Science

Mathematics for Computer Science Lecture 2: Functions and equinumerous sets Areces, Blackburn and Figueira TALARIS team INRIA Nancy Grand Est Contact: patrick.blackburn@loria.fr Course website: http://www.loria.fr/~blackbur/courses/math

### Math 4310 Handout - Quotient Vector Spaces

Math 4310 Handout - Quotient Vector Spaces Dan Collins The textbook defines a subspace of a vector space in Chapter 4, but it avoids ever discussing the notion of a quotient space. This is understandable

### Multiplicity. Chapter 6

Chapter 6 Multiplicity The fundamental theorem of algebra says that any polynomial of degree n 0 has exactly n roots in the complex numbers if we count with multiplicity. The zeros of a polynomial are

### This chapter is all about cardinality of sets. At first this looks like a

CHAPTER Cardinality of Sets This chapter is all about cardinality of sets At first this looks like a very simple concept To find the cardinality of a set, just count its elements If A = { a, b, c, d },

### 8 Primes and Modular Arithmetic

8 Primes and Modular Arithmetic 8.1 Primes and Factors Over two millennia ago already, people all over the world were considering the properties of numbers. One of the simplest concepts is prime numbers.

### Cardinality. The set of all finite strings over the alphabet of lowercase letters is countable. The set of real numbers R is an uncountable set.

Section 2.5 Cardinality (another) Definition: The cardinality of a set A is equal to the cardinality of a set B, denoted A = B, if and only if there is a bijection from A to B. If there is an injection

### 1. LINEAR EQUATIONS. A linear equation in n unknowns x 1, x 2,, x n is an equation of the form

1. LINEAR EQUATIONS A linear equation in n unknowns x 1, x 2,, x n is an equation of the form a 1 x 1 + a 2 x 2 + + a n x n = b, where a 1, a 2,..., a n, b are given real numbers. For example, with x and

### Mathematical Methods of Engineering Analysis

Mathematical Methods of Engineering Analysis Erhan Çinlar Robert J. Vanderbei February 2, 2000 Contents Sets and Functions 1 1 Sets................................... 1 Subsets.............................

### Basic Proof Techniques

Basic Proof Techniques David Ferry dsf43@truman.edu September 13, 010 1 Four Fundamental Proof Techniques When one wishes to prove the statement P Q there are four fundamental approaches. This document

### Number Theory. Proof. Suppose otherwise. Then there would be a finite number n of primes, which we may

Number Theory Divisibility and Primes Definition. If a and b are integers and there is some integer c such that a = b c, then we say that b divides a or is a factor or divisor of a and write b a. Definition

### BANACH AND HILBERT SPACE REVIEW

BANACH AND HILBET SPACE EVIEW CHISTOPHE HEIL These notes will briefly review some basic concepts related to the theory of Banach and Hilbert spaces. We are not trying to give a complete development, but

### SOLUTIONS TO EXERCISES FOR. MATHEMATICS 205A Part 3. Spaces with special properties

SOLUTIONS TO EXERCISES FOR MATHEMATICS 205A Part 3 Fall 2008 III. Spaces with special properties III.1 : Compact spaces I Problems from Munkres, 26, pp. 170 172 3. Show that a finite union of compact subspaces

### Section 2.7 One-to-One Functions and Their Inverses

Section. One-to-One Functions and Their Inverses One-to-One Functions HORIZONTAL LINE TEST: A function is one-to-one if and only if no horizontal line intersects its graph more than once. EXAMPLES: 1.

### 5.1 Commutative rings; Integral Domains

5.1 J.A.Beachy 1 5.1 Commutative rings; Integral Domains from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 23. Let R be a commutative ring. Prove the following

### HOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)!

Math 7 Fall 205 HOMEWORK 5 SOLUTIONS Problem. 2008 B2 Let F 0 x = ln x. For n 0 and x > 0, let F n+ x = 0 F ntdt. Evaluate n!f n lim n ln n. By directly computing F n x for small n s, we obtain the following

### FIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper.

FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is

### MTH4100 Calculus I. Lecture notes for Week 8. Thomas Calculus, Sections 4.1 to 4.4. Rainer Klages

MTH4100 Calculus I Lecture notes for Week 8 Thomas Calculus, Sections 4.1 to 4.4 Rainer Klages School of Mathematical Sciences Queen Mary University of London Autumn 2009 Theorem 1 (First Derivative Theorem

### minimal polyonomial Example

Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We

### God created the integers and the rest is the work of man. (Leopold Kronecker, in an after-dinner speech at a conference, Berlin, 1886)

Chapter 2 Numbers God created the integers and the rest is the work of man. (Leopold Kronecker, in an after-dinner speech at a conference, Berlin, 1886) God created the integers and the rest is the work

### Factoring Polynomials

Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

### 1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style

Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with

### Homework until Test #2

MATH31: Number Theory Homework until Test # Philipp BRAUN Section 3.1 page 43, 1. It has been conjectured that there are infinitely many primes of the form n. Exhibit five such primes. Solution. Five such

### CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12

CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.

### k, then n = p2α 1 1 pα k

Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

### Set theory as a foundation for mathematics

Set theory as a foundation for mathematics Waffle Mathcamp 2011 In school we are taught about numbers, but we never learn what numbers really are. We learn rules of arithmetic, but we never learn why these

### 9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11.

9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n-1 x n-1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role

### Review of Fundamental Mathematics

Review of Fundamental Mathematics As explained in the Preface and in Chapter 1 of your textbook, managerial economics applies microeconomic theory to business decision making. The decision-making tools

### SECTION 10-2 Mathematical Induction

73 0 Sequences and Series 6. Approximate e 0. using the first five terms of the series. Compare this approximation with your calculator evaluation of e 0.. 6. Approximate e 0.5 using the first five terms

### 2.1 Increasing, Decreasing, and Piecewise Functions; Applications

2.1 Increasing, Decreasing, and Piecewise Functions; Applications Graph functions, looking for intervals on which the function is increasing, decreasing, or constant, and estimate relative maxima and minima.