Functions, High-School Edition

Transcription

1 Functions

2 What is a function?

3 Functions, High-School Edition

4 f(x) = x4 5x2 + 4 source:

5 source:

6 Functions, High-School Edition In high school, functions are usually given as objects of the form x3 +3 x2 +15 x+7 f (x) = 1 x137 What does a function do? It takes in as input a real number. It outputs a real number except when there are vertical asymptotes or other discontinuities, in which case the function doesn't output anything.

7 Functions, CS Edition

8 int int flipuntil(int flipuntil(int n) n) {{ int int numheads numheads == 0; 0; int int numtries numtries == 0; 0; while while (numheads (numheads << n) n) {{ if if (randomboolean()) (randomboolean()) numheads++; numheads++; }} }} numtries++; numtries++; return return numtries; numtries;

9 Functions, CS Edition In programming, functions might take in inputs, might return values, might have side efects, might never return anything, might crash, and might return diferent values when called multiple times.

10 What's Common? Although high-school math functions and CS functions are pretty diferent, they have two key aspects in common: They take in inputs. They produce outputs. In math, we like to keep things easy, so that's pretty much how we're going to defne a function.

11 Rough Idea of a Function: A function is an object f that takes in an input and produces exactly one output. x f f(x) (This is not a complete defnition we'll revisit this in a bit.)

12 High School versus CS Functions In high school, functions usually were given by a rule: f(x) = 4x + 15 In CS, functions are usually given by code: int factorial(int n) { int result = 1; for (int i = 1; i <= n; i++) { result *= i; } return result; } What sorts of functions are we going to allow from a mathematical perspective?

13 Dikdik Nubian Ibex Sloth

14

15 but also

16 f(x) = x2 + 3x 15

17 n/2 if n is even f (n)= (n+1)/2 otherwise { Functions Functionslike likethese these are arecalled calledpiecewise piecewise functions. functions.

18 To defne a function, you will typically either draw a picture, or give a rule for determining the output.

19 In mathematics, functions are deterministic. That is, given the same input, a function must always produce the same output. The following is a perfectly valid piece of C++ code, but it s not a valid function under our defnition: int randomnumber(int numoutcomes) { return rand() % numoutcomes; }

20 One Challenge

21 f(x) = x2 + 2x + 5 f( 3 ) = = 20 f( 0 ) = = 5 f( 3 ) =?

22 f(x) = x2 + 2x + 5 f( 3 ) = = 20 f( 0 ) = = 5 f( 3 ) =?

23 f(x) = x2 + 2x + 5 f( 3 ) = = 20 f( 0 ) = = 5 f( 3 ) =?

24 f(x) = x2 + 2x + 5 f( 3 ) = = 20 f( 0 ) = = 5 f( 3 ) =?

25 f( )= f(137 ) =?

26 We need to make sure we can't apply functions to meaningless inputs.

27 Domains and Codomains Every function f has two sets associated with it: its domain and its codomain. A function f can only be applied to elements of its domain. For any x in the domain, f(x) belongs to the codomain. The Thefunction function must be must bedefned defned for every element for every element ofofthe thedomain. domain. Domain Codomain The Theoutput outputofofthe the function must function must always alwaysbe beininthe the codomain, but codomain, but not notall allelements elements ofofthe codomain the codomain must mustbe be produced producedas as outputs. outputs.

28 Domains and Codomains Every function f has two sets associated with it: its domain and its codomain. A function f can only be applied to elements of its domain. For any x in the domain, f(x) belongs to the codomain. The Thecodomain codomainofofthis thisfunction functionisis ℝ. ℝ.Everything Everythingproduced producedisisaareal real number, but not all real numbers number, but not all real numbers can canbe beproduced. produced. The Thedomain domainofofthis thisfunction function isisℝ. Any real number ℝ. Any real numbercan canbe be provided as input. provided as input. private double absolutevalueof(double x) { if (x >= 0) { return x; } else { return -x; } }

29 Domains and Codomains If f is a function whose domain is A and whose codomain is B, we write f : A B. This notation just says what the domain and codomain of the function are. It doesn't say how the function is evaluated. Think of it like a function prototype in C or C++. The notation f : ArgType RetType is like writing RetType f(argtype argument); We know that f takes in an ArgType and returns a RetType, but we don't know exactly which RetType it's going to return for a given ArgType.

30 The Oficial Rules for Functions Formally speaking, we say that f : A B if the following two rules hold. First, f must be obey its domain/codomain rules: a A. b B. f(a) = b ( Every input in A maps to some output in B. ) Second, f must be deterministic: a₁ A. a₂ A. (a₁ = a₂ f(a₁) = f(a₂)) ( Equal inputs produce equal outputs. ) If you re ever curious about whether something is a function, look back at these rules and check! For example: Can a function have an empty domain? Can a function with a nonempty domain have an empty codomain?

31 Defning Functions Typically, we specify a function by describing a rule that maps every element of the domain to some element of the codomain. Examples: f(n) = n + 1, where f : ℤ ℤ f(x) = sin x, where f : ℝ ℝ f(x) = x, where f : ℝ ℤ Notice that we're giving both a rule and the domain/codomain.

32 Defning Functions Typically, we specify a function by describing a rule that maps every element This function of the domain to some element of the This isis the the ceiling ceiling function the codomain. the smallest smallest integer integer Examples: greater greater than than or or equal equal to to x. x. For For example, example, 1 1 == 1,1, == 2, 2, and and == f(n) = n + 1, where f : ℤ ℤ f(x) = sin x, where f : ℝ ℝ f(x) = x, where f : ℝ ℤ Notice that we're giving both a rule and the domain/codomain.

33 Is This a Function From A to B? Stanford Cardinal Berkeley Blue Michigan Gold Arkansas White A B

34 Is This a Function From A to B? California Dover New York Sacramento Delaware Albany Washington DC A B

35 م ح رم Is This a Function From A to B? ص فر الأل ربيع ربيع الثاني جمادى الألى عيد الفطر جمادى الخآرة ر جب عيد الضحى ش عبان A ر مضان شوال ذأ القعدة Answer Answerat atpollev.com/cs103 PollEv.com/cs103or or text CS103 to once to join, then text CS103 to once to join, thenyyor orn. N. ذأ الحجة B

36 Combining Functions

37 f : People Places Cynthia g : Places Prices Mountain View Far Too Much Guy San Francisco King's Ransom Dana Redding, CA A Modest Amount Chioma Barrow, AK Pocket Change Diego People Palo Alto Places h : People Prices h(x) = g(f(x)) Prices

38 Function Composition Suppose that we have two functions f : A B and g : B C. Notice that the codomain of f is the domain of g. This means that we can use outputs from f as inputs to g. x f f(x) g g(f(x))

39 Function Composition Suppose that we have two functions f : A B and g : B C. The composition of f and g, denoted g f, is a function where g f : A C, and (g f)(x) = g(f(x)). The Thename nameofofthe thefunction functionisisgg f.f. When Whenwe weapply applyititto toan aninput inputx,x, we wewrite write(g (g f)(x). f)(x).i Idon't don'tknow know why, but that's what we do. why, but that's what we do. A few things to notice: The domain of g f is the domain of f. Its codomain is the codomain of g. Even though the composition is written g f, when evaluating (g f)(x), the function f is evaluated frst.

40 Special Types of Functions

41 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

42 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

43 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

44 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

45 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

46 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

47 Injective Functions A function f : A B is called injective (or one-to-one) if the following statement is true about f: a₁ A. a₂ A. (a₁ a₂ f(a₁) f(a₂)) ( If the inputs are diferent, the outputs are diferent. ) The following frst-order defnition is equivalent and is often useful in proofs. a₁ A. a₂ A. (f(a₁) = f(a₂) a₁ = a₂) ( If the outputs are the same, the inputs are the same. ) A function with this property is called an injection. How does this compare to our second rule for functions?

48 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. Since f(n₀) = f(n₁), we see that 2n₀ + 7 = 2n₁ + 7. This in turn means that 2n₀ = 2n₁, so n₀ = n₁, as required.

49 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. Since f(n₀) = f(n₁), we see that 2n₀ + 7 = 2n₁ + 7. This in turn means that 2n₀ = 2n₁, so n₀ = n₁, as required.

50 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. Since f(n₀) = f(n₁), we see that How Howmany manyof ofthe thefollowing followingare arecorrect correctways waysof ofstarting startingof ofthis thisproof? proof? 2n₀ + 7 = 2n₁ + 7. Consider any n₁, n₂ ℕ where n₁ ==n₂. We will prove that f(n₁) ==f(n₂). Consider any n₁, n₂ ℕ where n₁ n₂. We will prove that f(n₁) f(n₂). This in turn means that Consider Considerany anyn₁, n₁,n₂ n₂ ℕℕwhere wheren₁ n₁ n₂. n₂.we Wewill willprove provethat thatf(n₁) f(n₁) f(n₂). f(n₂). Consider Considerany anyn₁, n₁,n₂ n₂ ℕℕwhere wheref(n₁) f(n₁)= =f(n₂). f(n₂).we Wewill willprove provethat thatn₁ n₁==n₂. n₂. 2n₀ = 2n₁, Consider any n₁, n₂ ℕ where f(n₁) f(n₂). We will prove that n₁ n₂. Consider any n₁, n₂ ℕ where f(n₁) f(n₂). We will prove that n₁ n₂. so n₀ = n₁, as required. Answer Answerat atpollev.com/cs103 PollEv.com/cs103or or text CS103 to once to join, then a number text CS103 to once to join, then a numberbetween between00and and4. 4.

51 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does it mean for the function ff to What does it mean for the function to Since f(n₀) = f(n₁), we see that be be injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₀ n₀ ℕ. ℕ. n₁ n₁ ℕ. ℕ.((f(n₀) f(n₀)==f(n₁) f(n₁) n₀ n₀==n₁ n₁)) 2n₀ = 2n₁, n₀ n₀ ℕ. ℕ. n₁ n₁ ℕ. ℕ.((n₀ n₀ n₁ n₁ f(n₀) f(n₀) f(n₁) f(n₁))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₀, n₀,n₁ n₁ ℕ ℕ where where f(n₀) f(n₀)==f(n₁), f(n₁), then then prove prove that that n₀ n₀==n₁. n₁.

52 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does it mean for the function ff to What does it mean for the function to Since f(n₀) = f(n₁), we see that be be injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) 2n₀ = 2n₁, n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((n₁ n₁ n₂ n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₀, n₀,n₁ n₁ ℕ ℕ where where f(n₀) f(n₀)==f(n₁), f(n₁), then then prove prove that that n₀ n₀==n₁. n₁.

53 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does it mean for the function ff to What does it mean for the function to Since f(n₀) = f(n₁), we see that be be injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) 2n₀ = 2n₁, n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((n₁ n₁ n₂ n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₀, n₀,n₁ n₁ ℕ ℕ where where f(n₀) f(n₀)==f(n₁), f(n₁), then then prove prove that that n₀ n₀==n₁. n₁.

54 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does it mean for the function ff to What does it mean for the function to Since f(n₀) = f(n₁), we see that be be injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) 2n₀ = 2n₁, n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((n₁ n₁ n₂ n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₁, n₁,n₂ n₂ ℕ ℕ where where f(n₁) f(n₁)==f(n₂), f(n₂), then then prove prove that that n₁ n₁==n₂. n₂.

55 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does it mean for the function ff to What does it mean for the function to Since f(n₀) = f(n₁), we see that be be injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) 2n₀ = 2n₁, n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((n₁ n₁ n₂ n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₁, n₁,n₂ n₂ ℕ ℕ where where f(n₁) f(n₁)==f(n₂), f(n₂), then then prove prove that that n₁ n₁==n₂. n₂.

56 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does it mean for the function ff to What does it mean for the function to Since f(n₀) = f(n₁), we see that be be injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) 2n₀ = 2n₁, n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((n₁ n₁ n₂ n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₁, n₁,n₂ n₂ ℕ ℕ where where f(n₁) f(n₁)==f(n₂), f(n₂), then then prove prove that that n₁ n₁==n₂. n₂.

57 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does it mean for the function ff to What does it mean for the function to Since f(n₀) = f(n₁), we see that be be injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) 2n₀ = 2n₁, n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((n₁ n₁ n₂ n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₁, n₁,n₂ n₂ ℕ ℕ where where f(n₁) f(n₁)==f(n₂), f(n₂), then then prove prove that that n₁ n₁==n₂. n₂.

58 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₁, n₂ ℕ where f(n₁) = f(n₂). We will prove that n₁ = n₂. Since f(n₀) = f(n₁), we see that 2n₀ + 7 = 2n₁ + 7. This in turn means that 2n₀ = 2n₁, so n₀ = n₁, as required.

59 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₁, n₂ ℕ where f(n₁) = f(n₂). We will prove that n₁ = n₂. Since f(n₁) = f(n₂), we see that 2n₁ + 7 = 2n₂ + 7. This in turn means that 2n₀ = 2n₁, so n₀ = n₁, as required.

60 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₁, n₂ ℕ where f(n₁) = f(n₂). We will prove that n₁ = n₂. Since f(n₁) = f(n₂), we see that 2n₁ + 7 = 2n₂ + 7. This in turn means that 2n₁ = 2n₂ so n₀ = n₁, as required.

61 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₁, n₂ ℕ where f(n₁) = f(n₂). We will prove that n₁ = n₂. Since f(n₁) = f(n₂), we see that 2n₁ + 7 = 2n₂ + 7. This in turn means that 2n₁ = 2n₂ so n₁ = n₂, as required.

62 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₁, n₂ ℕ where f(n₁) = f(n₂). We will prove that n₁ = n₂. Since f(n₁) = f(n₂), we see that 2n₁ + 7 = 2n₂ + 7. This in turn means that 2n₁ = 2n₂ so n₁ = n₂, as required.

63 Injective Functions Theorem: Let f : ℕ ℕ be defned as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₁, n₂ ℕ where f(n₁) = f(n₂). We will prove that n₁ = n₂. Since f(n₁) = f(n₂), we see that How Howmany manyof ofthe thefollowing followingare arecorrect correctways waysof ofstarting startingof ofthis thisproof? proof? 2n₁n₁+=7n₂. =We 2n₂will + prove 7. Consider any n₁, n₂ ℕ where Consider any n₁, n₂ ℕ where n₁ = n₂. We will provethat thatf(n₁) f(n₁)==f(n₂). f(n₂). This any in turn means that Consider n₁, n₂ ℕ where n₁ n₂. We will prove that f(n₁) f(n₂). Consider any n₁, n₂ ℕ where n₁ n₂. We will prove that f(n₁) f(n₂). Consider Considerany anyn₁, n₁,n₂ n₂ ℕℕwhere wheref(n₁) f(n₁)= =f(n₂). f(n₂).we Wewill willprove provethat thatn₁ n₁==n₂. n₂. 2n₁ = 2n₂ Consider Considerany anyn₁, n₁,n₂ n₂ ℕℕwhere wheref(n₁) f(n₁) f(n₂). f(n₂).we Wewill willprove provethat thatn₁ n₁ n₂. n₂. so n₁ = n₂, as required. Good Good exercise: exercise: Repeat Repeat this this proof proof using using the the other other defnition defnition of of injectivity! injectivity!

64 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). Let x₀ = -1 and x₁ = +1. Then f(x₀) = f(-1) = (-1)4 = 1 and f(x₁) = f(1) = 14 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

65 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). Let x₀ = -1 and x₁ = +1. Then f(x₀) = f(-1) = (-1)4 = 1 and f(x₁) = f(1) = 14 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

66 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). How many of the following are correct ways of starting of this proof? How many of the following are correct ways of starting of this proof? Let for x₀ the = -1sake and x₁ = +1. Then Assume of Assumefor thesake ofcontradiction contradictionthat thatf fisisnot notinjective. injective. 4 are integers x₁ and x₂ Assume there f(x₀) = f(-1)that = (-1) =are 1 integers x₁ and x₂ Assumefor forthe thesake sakeof ofcontradiction contradiction that there where wheref(x₁) f(x₁)==f(x₂) f(x₂)but butx₁ x₁ x₂. x₂. and arbitrary integers x₁ and x₂ where x₁ x₂. We will prove Consider Consider arbitrary integers x₁ and x₂ where x₁ x₂. We will prove that f(x₁) = f(1) = 14 = 1, thatf(x₁) f(x₁)==f(x₂). f(x₂). Consider arbitrary integers x₁ and x₂ where f(x₁) ==f(x₂). We will prove Consider arbitrary integers x₁ and x₂ where f(x₁) f(x₂). We will prove so f(x₀) = f(x₁) even though x₀ x₁, as required. that thatx₁ x₁ x₂. x₂. Answer Answerat atpollev.com/cs103 PollEv.com/cs103or or text CS103 to once to join, then a number text CS103 to once to join, then a numberbetween between00and and4. 4.

67 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₀ ℤ. x₁ ℤ. (x₀ x₀ ℤ. x₁ ℤ. (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) 4 = f(-1) = (-1) =1 What is the negation f(x₀) of this statement? What is the negation of this statement? and x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) x₀ ℤ. x₀ ℤ. ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ f(x₀) f(x₁)) f(x₁) = f(1) =x₁ 14 =f(x₀) 1, f(x₁)) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ. (x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ (f(x₀) (f(x₀) f(x₁))) f(x₁))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

68 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 = f(-1) = (-1) =1 What is the negation f(x₀) of this statement? What is the negation of this statement? and x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) x₀ ℤ. x₀ ℤ. ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ f(x₀) f(x₁)) f(x₁) = f(1) =x₁ 14 =f(x₀) 1, f(x₁)) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ. (x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ (f(x₀) (f(x₀) f(x₁))) f(x₁))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

69 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 = f(-1) = (-1) =1 What is the negation f(x₀) of this statement? What is the negation of this statement? and x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) x₀ ℤ. x₀ ℤ. ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ f(x₀) f(x₁)) f(x₁) = f(1) =x₁ 14 =f(x₀) 1, f(x₁)) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ. (x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ (f(x₀) (f(x₀) f(x₁))) f(x₁))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

70 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 = f(-1) = (-1) =1 What is the negation f(x₀) of this statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₀ ℤ. = x₁ 4 f(x₀) f(x₁)) x₀ ℤ. ℤ. x₁ x₁ ℤ. (x₀ x₁1 =f(x₀) f(x₁) =(x₀ f(1) 1, f(x₁)) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ. (x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ (f(x₀) (f(x₀) f(x₁))) f(x₁))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

71 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 = f(-1) = (-1) =1 What is the negation f(x₀) of this statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 =f(x₁) f(x₁) =(x₁ f(1) 1, f(x₂)) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ. (x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ (f(x₀) (f(x₀) f(x₁))) f(x₁))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

72 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 = f(-1) = (-1) =1 What is the negation f(x₀) of this statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 =f(x₁) f(x₁) =(x₁ f(1) 1, f(x₂)) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ (f(x₀) (f(x₀) f(x₁))) f(x₁))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

73 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 = f(-1) = (-1) =1 What is the negation f(x₀) of this statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 =f(x₁) f(x₁) =(x₁ f(1) 1, f(x₂)) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) f(x₂))) f(x₂))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

74 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 = f(-1) = (-1) =1 What is the negation f(x₀) of this statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 =f(x₁) f(x₁) =(x₁ f(1) 1, f(x₂)) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) f(x₂))) f(x₂))) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁)==f(x₂)) f(x₂)) Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to fnd x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

75 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 = f(-1) = (-1) =1 What is the negation f(x₀) of this statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 =f(x₁) f(x₁) =(x₁ f(1) 1, f(x₂)) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) f(x₂))) f(x₂))) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁)==f(x₂)) f(x₂)) Therefore, we need to fnd x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = Therefore, we need to fnd x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we do that? f(x₂). Can we do that?

76 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 = f(-1) = (-1) =1 What is the negation f(x₀) of this statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 =f(x₁) f(x₁) =(x₁ f(1) 1, f(x₂)) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) f(x₂))) f(x₂))) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁)==f(x₂)) f(x₂)) Therefore, we need to fnd x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = Therefore, we need to fnd x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we do that? f(x₂). Can we do that?

77 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 = f(-1) = (-1) =1 What is the negation f(x₀) of this statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 =f(x₁) f(x₁) =(x₁ f(1) 1, f(x₂)) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) f(x₂))) f(x₂))) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁)==f(x₂)) f(x₂)) Therefore, we need to fnd x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = Therefore, we need to fnd x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we do that? f(x₂). Can we do that?

78 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 = f(-1) = (-1) =1 What is the negation f(x₀) of this statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 =f(x₁) f(x₁) =(x₁ f(1) 1, f(x₂)) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) f(x₂))) f(x₂))) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁)==f(x₂)) f(x₂)) Therefore, we need to fnd x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = Therefore, we need to fnd x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we do that? f(x₂). Can we do that?

79 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₀ = -1 and x₁ = +1. Then f(x₀) = f(-1) = (-1)4 = 1 and f(x₁) = f(1) = 14 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

80 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₁ = -1 and x₂ = +1. Then f(x₀) = f(-1) = (-1)4 = 1 and f(x₁) = f(1) = 14 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

81 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₁ = -1 and x₂ = +1. f(x₁) = f(-1) = (-1)4 = 1 and f(x₁) = f(1) = 14 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

82 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₁ = -1 and x₂ = +1. f(x₁) = f(-1) = (-1)4 = 1 and f(x₂) = f(1) = 14 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

83 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₁ = -1 and x₂ = +1. f(x₁) = f(-1) = (-1)4 = 1 and f(x₂) = f(1) = 14 = 1, so f(x₁) = f(x₂) even though x₁ x₂, as required.

84 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₁ = -1 and x₂ = +1. f(x₁) = f(-1) = (-1)4 = 1 and f(x₂) = f(1) = 14 = 1, so f(x₁) = f(x₂) even though x₁ x₂, as required.

85 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₀ = -1 and x₁ = +1. Then How starting of this proof? Howmany manyof ofthe thefollowing followingare arecorrect correctways waysof of starting of this proof? 4 f(x₀) = f(-1)that = (-1) = injective. 1 Assume for the sake of contradiction f is not Assume for the sake of contradiction that f is not injective. Assume for and Assume forthe thesake sakeof ofcontradiction contradictionthat thatthere thereare areintegers integersx₁ x₁and andx₂ x₂ where f(x₁) = f(x₂) but x₁ x₂. where f(x₁) = f(x₂) but x₁ x₂. 4 f(x₁) = f(1) = 1 = 1, Consider Considerarbitrary arbitraryintegers integersx₁ x₁and andx₂ x₂where wherex₁ x₁ x₂. x₂.we Wewill willprove prove that so f(x₁) f(x₀) =f(x₂). f(x₁) even though x₀ x₁, as required. that f(x₁)== f(x₂). Consider Considerarbitrary arbitraryintegers integersx₁ x₁and andx₂ x₂where wheref(x₁) f(x₁)==f(x₂). f(x₂).we Wewill willprove prove that thatx₁ x₁ x₂. x₂.

86 Injective Functions Theorem: Let f : ℤ ℕ be defned as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₁ = -1 and x₂ = +1. f(x₁) = f(-1) = (-1)4 = 1 and f(x₂) = f(1) = 14 = 1, so f(x₁) = f(x₂) even though x₁ x₂, as required.

87 Injections and Composition

88 Injections and Composition Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is an injection. Our goal will be to prove this result. To do so, we're going to have to call back to the formal defnitions of injectivity and function composition.

89 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required.

90 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required.

91 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required.

92 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required.

93 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). There There are are two two defnitions defnitions of of injectivity injectivity that that we we can can use use here: here: Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that a₁ A. A. ((g f)(a₁) = a₁ A. a₂ a₂ A.as ((g f)(a₁) =(g (g f)(a₂) f)(a₂) a₁ a₁= = a₂) a₂) g(f(a₁)) g(f(a₂)), required. a₁ a₁ A. A. a₂ a₂ A. A.(a₁ (a₁ a₂ a₂ (g (g f)(a₁) f)(a₁) (g (g f)(a₂)) f)(a₂)) Therefore, Therefore, we ll we ll choose choose an an arbitrary arbitrary a₁, a₁,a₂ a₂ AA where where a₁ a₁ a₂, a₂, then then prove prove that that (g (g f)(a₁) f)(a₁) (g (g f)(a₂). f)(a₂).

94 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). There There are are two two defnitions defnitions of of injectivity injectivity that that we we can can use use here: here: Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that a₁ A. A. ((g f)(a₁) = a₁ A. a₂ a₂ A.as ((g f)(a₁) =(g (g f)(a₂) f)(a₂) a₁ a₁= = a₂) a₂) g(f(a₁)) g(f(a₂)), required. a₁ a₁ A. A. a₂ a₂ A. A.(a₁ (a₁ a₂ a₂ (g (g f)(a₁) f)(a₁) (g (g f)(a₂)) f)(a₂)) Therefore, Therefore, we ll we ll choose choose an an arbitrary arbitrary a₁, a₁,a₂ a₂ AA where where a₁ a₁ a₂, a₂, then then prove prove that that (g (g f)(a₁) f)(a₁) (g (g f)(a₂). f)(a₂).

95 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). There There are are two two defnitions defnitions of of injectivity injectivity that that we we can can use use here: here: Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that a₁ A. A. ((g f)(a₁) = a₁ A. a₂ a₂ A.as ((g f)(a₁) =(g (g f)(a₂) f)(a₂) a₁ a₁= = a₂) a₂) g(f(a₁)) g(f(a₂)), required. a₁ a₁ A. A. a₂ a₂ A. A.(a₁ (a₁ a₂ a₂ (g (g f)(a₁) f)(a₁) (g (g f)(a₂)) f)(a₂)) Therefore, Therefore, we ll we ll choose choose an an arbitrary arbitrary a₁, a₁,a₂ a₂ AA where where a₁ a₁ a₂, a₂, then then prove prove that that (g (g f)(a₁) f)(a₁) (g (g f)(a₂). f)(a₂).

96 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required.

97 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required.

98 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁)defned? f(a₂), we see that How is (g f)(x) Howasisrequired. (g f)(x) g(f(a₁)) g(f(a₂)), defned? (g (g f)(x) f)(x)==g(f(x)) g(f(x)) So So we we need need to to prove prove that that g(f(a₁)) g(f(a₁)) g(f(a₂)). g(f(a₂)).

99 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁)defned? f(a₂), we see that How is (g f)(x) Howasisrequired. (g f)(x) g(f(a₁)) g(f(a₂)), defned? (g (g f)(x) f)(x)==g(f(x)) g(f(x)) So So we we need need to to prove prove that that g(f(a₁)) g(f(a₁)) g(f(a₂)). g(f(a₂)).

100 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁)defned? f(a₂), we see that How is (g f)(x) Howasisrequired. (g f)(x) g(f(a₁)) g(f(a₂)), defned? (g (g f)(x) f)(x)==g(f(x)) g(f(x)) So So we we need need to to prove prove that that g(f(a₁)) g(f(a₁)) g(f(a₂)). g(f(a₂)).

101 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁)defned? f(a₂), we see that How is (g f)(x) Howasisrequired. (g f)(x) g(f(a₁)) g(f(a₂)), defned? (g (g f)(x) f)(x)==g(f(x)) g(f(x)) So So we we need need to to prove prove that that g(f(a₁)) g(f(a₁)) g(f(a₂)). g(f(a₂)).

102 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required. g(f(a₁)) f(a₁) a₁ a₂ g(f(a₂)) f(a₂) A B C

103 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required. g(f(a₁)) f(a₁) a₁ a₂ g(f(a₂)) f(a₂) A B C

104 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required. g(f(a₁)) f(a₁) a₁ a₂ g(f(a₂)) f(a₂) A B C

105 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required. g(f(a₁)) f(a₁) a₁ a₂ g(f(a₂)) f(a₂) A B C

106 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required. g(f(a₁)) f(a₁) a₁ a₂ Great Great exercise: exercise: Repeat Repeat this this proof proof using using the the g(f(a₂)) f(a₂) A B other other defnition defnition of of injectivity. C injectivity.

107 Another Class of Functions

108 Mt. Lassen California Mt. Hood Washington Mt. St. Helens Oregon Mt. Shasta

109 Surjective Functions A function f : A B is called surjective (or onto) if this frst-order logic statement is true about f: b B. a A. f(a) = b ( For every possible output, there's at least one possible input that produces it ) A function with this property is called a surjection. How does this compare to our frst rule of functions?

110 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

111 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

112 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. for Let What xwhat = 2y.does Then it we see that does it mean mean for ff to to be be surjective? surjective?f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required. y y ℝ. ℝ. x x ℝ. ℝ.f(x) f(x)==yy Therefore, Therefore, we'll we'll choose choose an an arbitrary arbitrary yy ℝ, ℝ, then then prove prove that that there there isis some some xx ℝℝ where where f(x) f(x)==y. y.

113 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. for Let What xwhat = 2y.does Then it we see that does it mean mean for ff to to be be surjective? surjective?f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required. y y ℝ. ℝ. x x ℝ. ℝ.f(x) f(x)==yy Therefore, Therefore, we'll we'll choose choose an an arbitrary arbitrary yy ℝ, ℝ, then then prove prove that that there there isis some some xx ℝℝ where where f(x) f(x)==y. y.

114 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. for Let What xwhat = 2y.does Then it we see that does it mean mean for ff to to be be surjective? surjective?f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required. y y ℝ. ℝ. x x ℝ. ℝ.f(x) f(x)==yy Therefore, Therefore, we'll we'll choose choose an an arbitrary arbitrary yy ℝ, ℝ, then then prove prove that that there there isis some some xx ℝℝ where where f(x) f(x)==y. y.

115 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. for Let What xwhat = 2y.does Then it we see that does it mean mean for ff to to be be surjective? surjective?f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required. y y ℝ. ℝ. x x ℝ. ℝ.f(x) f(x)==yy Therefore, Therefore, we'll we'll choose choose an an arbitrary arbitrary yy ℝ, ℝ, then then prove prove that that there there isis some some xx ℝℝ where where f(x) f(x)==y. y.

116 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. for Let What xwhat = 2y.does Then it we see that does it mean mean for ff to to be be surjective? surjective?f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required. y y ℝ. ℝ. x x ℝ. ℝ.f(x) f(x)==yy Therefore, Therefore, we'll we'll choose choose an an arbitrary arbitrary yy ℝ, ℝ, then then prove prove that that there there isis some some xx ℝℝ where where f(x) f(x)==y. y.

117 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. for Let What xwhat = 2y.does Then it we see that does it mean mean for ff to to be be surjective? surjective?f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required. y y ℝ. ℝ. x x ℝ. ℝ.f(x) f(x)==yy Therefore, Therefore, we'll we'll choose choose an an arbitrary arbitrary yy ℝ, ℝ, then then prove prove that that there there isis some some xx ℝℝ where where f(x) f(x)==y. y.

118 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

119 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

120 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

121 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

122 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

123 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

124 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

125 Surjective Functions Theorem: Let f : ℝ ℝ be defned as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

126 Composing Surjections

127 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

128 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

129 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

130 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

131 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. itit mean What What does does mean for for gg ff :: AA CC to to be be surjective? surjective? Consider any c C. Since g : B C is surjective, there is c C. = A. f)(a) = some b B such g(b) Similarly, c that C. a a A.c.(g (g f)(a)since = cc f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such Therefore, we'll choose arbitrary c Cthat and prove that there Therefore, we'll choose arbitrary c C and prove that there isis some AA such that g(b)(g = c, some aa g(f(a)) such= that (g f)(a) f)(a) == c. c. which is what we needed to show.

132 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. itit mean What What does does mean for for gg ff :: AA CC to to be be surjective? surjective? Consider any c C. Since g : B C is surjective, there is c C. = A. f)(a) = some b B such g(b) Similarly, c that C. a a A.c.(g (g f)(a)since = cc f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such Therefore, we'll choose arbitrary c Cthat and prove that there Therefore, we'll choose arbitrary c C and prove that there isis some AA such that g(b)(g = c, some aa g(f(a)) such= that (g f)(a) f)(a) == c. c. which is what we needed to show.

133 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. itit mean What What does does mean for for gg ff :: AA CC to to be be surjective? surjective? Consider any c C. Since g : B C is surjective, there is c C. = A. f)(a) = some b B such g(b) Similarly, c that C. a a A.c.(g (g f)(a)since = cc f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such Therefore, we'll choose arbitrary c Cthat and prove that there Therefore, we'll choose arbitrary c C and prove that there isis some AA such that g(b)(g = c, some aa g(f(a)) such= that (g f)(a) f)(a) == c. c. which is what we needed to show.

134 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

135 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, c which is what we needed to show. a A B b C

136 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, c which is what we needed to show. a A B b C

137 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, c which is what we needed to show. a A B b C

138 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, c which is what we needed to show. a A B b C

139 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, c which is what we needed to show. a A B b C

140 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, c which is what we needed to show. a A B b C

141 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

142 Injections and Surjections An injective function associates at most one element of the domain with each element of the codomain. A surjective function associates at least one element of the domain with each element of the codomain. What about functions that associate exactly one element of the domain with each element of the codomain?

143 Katniss Everdeen Elsa Hermione Granger

144 Bijections A function that associates each element of the codomain with a unique element of the domain is called bijective. Such a function is a bijection. Formally, a bijection is a function that is both injective and surjective. Bijections are sometimes called one-toone correspondences. Not to be confused with one-to-one functions.

145 Bijections and Composition Suppose that f : A B and g : B C are bijections. Is g f necessarily a bijection? Yes! Since both f and g are injective, we know that g f is injective. Since both f and g are surjective, we know that g f is surjective. Therefore, g f is a bijection.

146 Inverse Functions

147 Katniss Everdeen Elsa Hermione Granger

148 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

149 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

150 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

151 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

152 Mt. Lassen California Mt. Hood Washington Mt. St. Helens Oregon Mt. Shasta

153 Mt. Lassen California Mt. Hood Washington Mt. St. Helens Oregon Mt. Shasta