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

Transcription

1 Functions

2 What is a function?

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

4 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;

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

6 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?

7 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 definition we'll revisit this in a bit.)

8 Dikdik Nubian Ibex Sloth

9

10 but also

11 f(x) = x2 + 3x 15

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

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

14 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 definition: int randomnumber(int numoutcomes) { return rand() % numoutcomes; }

15 f( )= f(137 ) =?

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

17 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 bedefined defined 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.

18 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. double absolutevalueof(double x) { if (x >= 0) { return x; } else { return -x; } }

19 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.

20 The Official 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?

21 Defining 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.

22 Defining Functions Typically, we specify a function by describing a rule that maps every element This of the domain to some element of function the Thisisisthe theceiling ceiling function the the smallest integer greater than smallest integer greater thanor or codomain. equal to x. For example, 1 = 1, equal to x. For example, 1 = 1, Examples: ==2,2,and and π π ==4.4. 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.

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

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

25 ح رم م Is This a Function From A to B? صف ر ربيع الوأ ل ربيع الثاني جمادى الوألى عيد الفطر جمادى الخآرة جب ر عيد الضأحى شع بان مضا ر ن شو ا ل ذوأ القععدة A 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. ذوأ الحجة B

26 Combining Functions

27 f : People Places Cynthia g : Places Prices San Jose Far Too Much Divya San Francisco King's Ransom Hugo Redding A Modest Amount Teresa Fresno A Bit Too Much John People Palo Alto Places h : People Prices h(x) = g(f(x)) Prices

28 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))

29 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 first.

30 Special Types of Functions

31 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

32 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

33 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 different, the outputs are different. ) The following first-order definition 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?

34 Injective Functions Theorem: Let f : ℕ ℕ be defined 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.

35 Injective Functions Theorem: Let f : ℕ ℕ be defined 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.

36 Injective Functions Theorem: Let f : ℕ ℕ be defined 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 startingoff offthis 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.

37 Injective Functions Theorem: Let f : ℕ ℕ be defined 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 Whatdoes doesititmean meanfor forthe thefunction functionffto tobe be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. n₀ n₁ n₀ ℕ. ℕ.means n₁ ℕ. ℕ.((f(n₀) f(n₀)==f(n₁) f(n₁) n₀ n₀==n₁ n₁)) This in turn that n₀ 2n₁, n₁ n₀ ℕ. ℕ. n₁ n₁ ℕ. ℕ.(2n₀ (n₀ n₀= n₁ f(n₀) f(n₀) f(n₁) f(n₁))) so n₀ = n₁, as we'll required. Therefore, pick arbitrary Therefore, we'll pick arbitraryn₀, n₀,n₁ n₁ ℕ ℕwhere where f(n₀) f(n₀)==f(n₁), f(n₁),then thenprove provethat thatn₀ n₀==n₁. n₁.

38 Injective Functions Theorem: Let f : ℕ ℕ be defined 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 Whatdoes doesititmean meanfor forthe thefunction functionffto tobe be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. n₁ n₂ n₁ ℕ. ℕ.means n₂ ℕ. ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) This in turn that n₁ 2n₁, n₂ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.(2n₀ (n₁ n₁= n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as we'll required. Therefore, pick arbitrary Therefore, we'll pick arbitraryn₀, n₀,n₁ n₁ ℕ ℕwhere where f(n₀) f(n₀)==f(n₁), f(n₁),then thenprove provethat thatn₀ n₀==n₁. n₁.

39 Injective Functions Theorem: Let f : ℕ ℕ be defined 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 Whatdoes doesititmean meanfor forthe thefunction functionffto tobe be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. n₁ n₂ n₁ ℕ. ℕ.means n₂ ℕ. ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) This in turn that n₁ 2n₁, n₂ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.(2n₀ (n₁ n₁= n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as we'll required. Therefore, pick arbitrary Therefore, we'll pick arbitraryn₀, n₀,n₁ n₁ ℕ ℕwhere where f(n₀) f(n₀)==f(n₁), f(n₁),then thenprove provethat thatn₀ n₀==n₁. n₁.

40 Injective Functions Theorem: Let f : ℕ ℕ be defined 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 Whatdoes doesititmean meanfor forthe thefunction functionffto tobe be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. n₁ n₂ n₁ ℕ. ℕ.means n₂ ℕ. ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) This in turn that n₁ 2n₁, n₂ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.(2n₀ (n₁ n₁= n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as we'll required. Therefore, pick arbitrary Therefore, we'll pick arbitraryn₁, n₁,n₂ n₂ ℕ ℕwhere where f(n₁) f(n₁)==f(n₂), f(n₂),then thenprove provethat thatn₁ n₁==n₂. n₂.

41 Injective Functions Theorem: Let f : ℕ ℕ be defined 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 Whatdoes doesititmean meanfor forthe thefunction functionffto tobe be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. n₁ n₂ n₁ ℕ. ℕ.means n₂ ℕ. ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) This in turn that n₁ 2n₁, n₂ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.(2n₀ (n₁ n₁= n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as we'll required. Therefore, pick arbitrary Therefore, we'll pick arbitraryn₁, n₁,n₂ n₂ ℕ ℕwhere where f(n₁) f(n₁)==f(n₂), f(n₂),then thenprove provethat thatn₁ n₁==n₂. n₂.

42 Injective Functions Theorem: Let f : ℕ ℕ be defined 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 Whatdoes doesititmean meanfor forthe thefunction functionffto tobe be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. n₁ n₂ n₁ ℕ. ℕ.means n₂ ℕ. ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) This in turn that n₁ 2n₁, n₂ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.(2n₀ (n₁ n₁= n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as we'll required. Therefore, pick arbitrary Therefore, we'll pick arbitraryn₁, n₁,n₂ n₂ ℕ ℕwhere where f(n₁) f(n₁)==f(n₂), f(n₂),then thenprove provethat thatn₁ n₁==n₂. n₂.

43 Injective Functions Theorem: Let f : ℕ ℕ be defined 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 Whatdoes doesititmean meanfor forthe thefunction functionffto tobe be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. n₁ n₂ n₁ ℕ. ℕ.means n₂ ℕ. ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) This in turn that n₁ 2n₁, n₂ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.(2n₀ (n₁ n₁= n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as we'll required. Therefore, pick arbitrary Therefore, we'll pick arbitraryn₁, n₁,n₂ n₂ ℕ ℕwhere where f(n₁) f(n₁)==f(n₂), f(n₂),then thenprove provethat thatn₁ n₁==n₂. n₂.

44 Injective Functions Theorem: Let f : ℕ ℕ be defined 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.

45 Injective Functions Theorem: Let f : ℕ ℕ be defined 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.

46 Injective Functions Theorem: Let f : ℕ ℕ be defined 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.

47 Injective Functions Theorem: Let f : ℕ ℕ be defined 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.

48 Injective Functions Theorem: Let f : ℕ ℕ be defined 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.

49 Injective Functions Theorem: Let f : ℤ ℕ be defined 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.

50 Injective Functions Theorem: Let f : ℤ ℕ be defined 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.

51 Injective Functions Theorem: Let f : ℤ ℕ be defined 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 off this proof? How many of the following are correct ways of starting off 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.

52 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ but f(x₀) = f(x₁). What does it mean for fx₁, to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₀ ℤ. ℤ. x₁ x₀ ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) What is the negation of this statement? = f(-1) = (-1)4 = 1 What is the negation of f(x₀) this statement? 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 x₁4 f(x₀) f(x₀) f(x₁)) f(x₁)) f(1) =x₁ 1 =f(x₀) 1, f(x₁)) x₀ ℤ. (x₀ x₀ ℤ. ℤ. x₁ x₁f(x₁) ℤ.= (x₀ x₁ f(x₀) f(x₁)) x₀ ℤ. ℤ. (x₀ f(x₁))) x₀ ℤ. x₁ x₁ ℤ. (x₀ x₁ x₁ (f(x₀) (f(x₀) 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)) and Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

53 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ but f(x₀) = f(x₁). What does it mean for fx₁, to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. ℤ. x₂ x₁ ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) What is the negation of this statement? = f(-1) = (-1)4 = 1 What is the negation of f(x₀) this statement? 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 x₁4 f(x₀) f(x₀) f(x₁)) f(x₁)) f(1) =x₁ 1 =f(x₀) 1, f(x₁)) x₀ ℤ. (x₀ x₀ ℤ. ℤ. x₁ x₁f(x₁) ℤ.= (x₀ x₁ f(x₀) f(x₁)) x₀ ℤ. ℤ. (x₀ f(x₁))) x₀ ℤ. x₁ x₁ ℤ. (x₀ x₁ x₁ (f(x₀) (f(x₀) 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)) and Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

54 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ but f(x₀) = f(x₁). What does it mean for fx₁, to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. ℤ. x₂ x₁ ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) What is the negation of this statement? = f(-1) = (-1)4 = 1 What is the negation of f(x₀) this statement? 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 x₁4 f(x₀) f(x₀) f(x₁)) f(x₁)) f(1) =x₁ 1 =f(x₀) 1, f(x₁)) x₀ ℤ. (x₀ x₀ ℤ. ℤ. x₁ x₁f(x₁) ℤ.= (x₀ x₁ f(x₀) f(x₁)) x₀ ℤ. ℤ. (x₀ f(x₁))) x₀ ℤ. x₁ x₁ ℤ. (x₀ x₁ x₁ (f(x₀) (f(x₀) 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)) and Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

55 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ but f(x₀) = f(x₁). What does it mean for fx₁, to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. ℤ. x₂ x₁ ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) What is the negation of this statement? = f(-1) = (-1)4 = 1 What is the negation of f(x₀) this statement? x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₀ f(x₀) f(x₁)) x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₁)) 4 = f(1) 1, f(x₁)) x₀ ℤ. = x₁ f(x₀) x₀ ℤ. ℤ. x₁ x₁f(x₁) ℤ. (x₀ (x₀ x₁1 = f(x₀) f(x₁)) x₀ ℤ. f(x₁))) x₀ ℤ. ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ x₁ (f(x₀) (f(x₀) 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)) and Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

56 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ but f(x₀) = f(x₁). What does it mean for fx₁, to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. ℤ. x₂ x₁ ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) What is the negation of this statement? = f(-1) = (-1)4 = 1 What is the negation of f(x₀) this statement? x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₂)) 4 = f(1) 1, f(x₁)) x₀ ℤ. = x₁ f(x₀) x₀ ℤ. ℤ. x₁ x₁f(x₁) ℤ. (x₀ (x₀ x₁1 = f(x₀) f(x₁)) x₀ ℤ. f(x₁))) x₀ ℤ. ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ x₁ (f(x₀) (f(x₀) 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)) and Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

57 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ but f(x₀) = f(x₁). What does it mean for fx₁, to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. ℤ. x₂ x₁ ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) What is the negation of this statement? = f(-1) = (-1)4 = 1 What is the negation of f(x₀) this statement? x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₂)) 4 = f(1) 1, f(x₂)) x₁ ℤ. = x₂ f(x₁) x₁ ℤ. ℤ. x₂ x₂f(x₁) ℤ. (x₁ (x₁ x₂1 = f(x₁) f(x₂)) x₀ ℤ. f(x₁))) x₀ ℤ. ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ x₁ (f(x₀) (f(x₀) 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)) and Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

58 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ but f(x₀) = f(x₁). What does it mean for fx₁, to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. ℤ. x₂ x₁ ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) What is the negation of this statement? = f(-1) = (-1)4 = 1 What is the negation of f(x₀) this statement? x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₂)) 4 = f(1) 1, f(x₂)) x₁ ℤ. = x₂ f(x₁) x₁ ℤ. ℤ. x₂ x₂f(x₁) ℤ. (x₁ (x₁ x₂1 = f(x₁) f(x₂)) x₁ ℤ. f(x₂))) x₁ ℤ. ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) 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)) and Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

59 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ but f(x₀) = f(x₁). What does it mean for fx₁, to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. ℤ. x₂ x₁ ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) What is the negation of this statement? = f(-1) = (-1)4 = 1 What is the negation of f(x₀) this statement? x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₂)) 4 = f(1) 1, f(x₂)) x₁ ℤ. = x₂ f(x₁) x₁ ℤ. ℤ. x₂ x₂f(x₁) ℤ. (x₁ (x₁ x₂1 = f(x₁) f(x₂)) x₁ ℤ. f(x₂))) x₁ ℤ. ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) 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)) and Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

60 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ but f(x₀) = f(x₁). What does it mean for fx₁, to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. ℤ. x₂ x₁ ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) What is the negation of this statement? = f(-1) = (-1)4 = 1 What is the negation of f(x₀) this statement? x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₂)) 4 = f(1) 1, f(x₂)) x₁ ℤ. = x₂ f(x₁) x₁ ℤ. ℤ. x₂ x₂f(x₁) ℤ. (x₁ (x₁ x₂1 = f(x₁) f(x₂)) x₁ ℤ. f(x₂))) x₁ ℤ. ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) 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)) and Therefore, we need to find x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we Therefore, we need to find x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we do that? do that?

61 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ but f(x₀) = f(x₁). What does it mean for fx₁, to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. ℤ. x₂ x₁ ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) What is the negation of this statement? = f(-1) = (-1)4 = 1 What is the negation of f(x₀) this statement? x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₂)) 4 = f(1) 1, f(x₂)) x₁ ℤ. = x₂ f(x₁) x₁ ℤ. ℤ. x₂ x₂f(x₁) ℤ. (x₁ (x₁ x₂1 = f(x₁) f(x₂)) x₁ ℤ. f(x₂))) x₁ ℤ. ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) 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)) and Therefore, we need to find x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we Therefore, we need to find x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we do that? do that?

62 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ but f(x₀) = f(x₁). What does it mean for fx₁, to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. ℤ. x₂ x₁ ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) What is the negation of this statement? = f(-1) = (-1)4 = 1 What is the negation of f(x₀) this statement? x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₂)) 4 = f(1) 1, f(x₂)) x₁ ℤ. = x₂ f(x₁) x₁ ℤ. ℤ. x₂ x₂f(x₁) ℤ. (x₁ (x₁ x₂1 = f(x₁) f(x₂)) x₁ ℤ. f(x₂))) x₁ ℤ. ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) 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)) and Therefore, we need to find x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we Therefore, we need to find x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we do that? do that?

63 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ but f(x₀) = f(x₁). What does it mean for fx₁, to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. ℤ. x₂ x₁ ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) What is the negation of this statement? = f(-1) = (-1)4 = 1 What is the negation of f(x₀) this statement? x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ f(x₁) f(x₂)) x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₂)) 4 = f(1) 1, f(x₂)) x₁ ℤ. = x₂ f(x₁) x₁ ℤ. ℤ. x₂ x₂f(x₁) ℤ. (x₁ (x₁ x₂1 = f(x₁) f(x₂)) x₁ ℤ. f(x₂))) x₁ ℤ. ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) 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)) and Therefore, we need to find x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we Therefore, we need to find x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we do that? do that?

64 Injective Functions Theorem: Let f : ℤ ℕ be defined 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.

65 Injective Functions Theorem: Let f : ℤ ℕ be defined 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.

66 Injective Functions Theorem: Let f : ℤ ℕ be defined 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.

67 Injective Functions Theorem: Let f : ℤ ℕ be defined 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.

68 Injective Functions Theorem: Let f : ℤ ℕ be defined 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.

69 Injective Functions Theorem: Let f : ℤ ℕ be defined 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.

70 Injective Functions Theorem: Let f : ℤ ℕ be defined 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 off this proof? Howmany manyof ofthe thefollowing followingare arecorrect correctways waysof of starting off 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.

71 Injective Functions Theorem: Let f : ℤ ℕ be defined 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.

72 Injections and Composition

73 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 definitions of injectivity and function composition.

74 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.

75 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.

76 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.

77 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.

78 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 two that we use here: Since f isare injective and a₁ofof injectivity a₂, we see f(a₁) f(a₂). There are twodefinitions definitions injectivity thatthat wecan can use here: Then, since g is injective and f(a₁) f(a₂), we see that a₁ A. A. ((g f)(a₁) = g(f(a₁)) g(f(a₂)), required. a₁ A. a₂ a₂ A.as ((g f)(a₁) =(g (g f)(a₂) f)(a₂) a₁ a₁= = a₂) a₂) 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 llchoose choosean anarbitrary arbitrarya₁, a₁,a₂ a₂ AAwhere wherea₁ a₁ a₂, a₂, then thenprove provethat that(g (g f)(a₁) f)(a₁) (g (g f)(a₂). f)(a₂).

79 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 two that we use here: Since f isare injective and a₁ofof injectivity a₂, we see f(a₁) f(a₂). There are twodefinitions definitions injectivity thatthat wecan can use here: Then, since g is injective and f(a₁) f(a₂), we see that a₁ A. A. ((g f)(a₁) = g(f(a₁)) g(f(a₂)), required. a₁ A. a₂ a₂ A.as ((g f)(a₁) =(g (g f)(a₂) f)(a₂) a₁ a₁= = a₂) a₂) 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 llchoose choosean anarbitrary arbitrarya₁, a₁,a₂ a₂ AAwhere wherea₁ a₁ a₂, a₂, then thenprove provethat that(g (g f)(a₁) f)(a₁) (g (g f)(a₂). f)(a₂).

80 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 two that we use here: Since f isare injective and a₁ofof injectivity a₂, we see f(a₁) f(a₂). There are twodefinitions definitions injectivity thatthat wecan can use here: Then, since g is injective and f(a₁) f(a₂), we see that a₁ A. A. ((g f)(a₁) = g(f(a₁)) g(f(a₂)), required. a₁ A. a₂ a₂ A.as ((g f)(a₁) =(g (g f)(a₂) f)(a₂) a₁ a₁= = a₂) a₂) 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 llchoose choosean anarbitrary arbitrarya₁, a₁,a₂ a₂ AAwhere wherea₁ a₁ a₂, a₂, then thenprove provethat that(g (g f)(a₁) f)(a₁) (g (g f)(a₂). f)(a₂).

81 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.

82 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.

83 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 isis(g defined? g(f(a₁)) g(f(a₂)), How as required. How (g f)(x) f)(x) defined? (g (g f)(x) f)(x)==g(f(x)) g(f(x)) So Sowe weneed needto toprove provethat thatg(f(a₁)) g(f(a₁)) g(f(a₂)). g(f(a₂)).

84 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 isis(g defined? g(f(a₁)) g(f(a₂)), How as required. How (g f)(x) f)(x) defined? (g (g f)(x) f)(x)==g(f(x)) g(f(x)) So Sowe weneed needto toprove provethat thatg(f(a₁)) g(f(a₁)) g(f(a₂)). g(f(a₂)).

85 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 isis(g defined? g(f(a₁)) g(f(a₂)), How as required. How (g f)(x) f)(x) defined? (g (g f)(x) f)(x)==g(f(x)) g(f(x)) So Sowe weneed needto toprove provethat thatg(f(a₁)) g(f(a₁)) g(f(a₂)). g(f(a₂)).

86 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 isis(g defined? g(f(a₁)) g(f(a₂)), How as required. How (g f)(x) f)(x) defined? (g (g f)(x) f)(x)==g(f(x)) g(f(x)) So Sowe weneed needto toprove provethat thatg(f(a₁)) g(f(a₁)) g(f(a₂)). g(f(a₂)).

87 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

88 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

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. g(f(a₁)) f(a₁) a₁ a₂ g(f(a₂)) f(a₂) A B C

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. g(f(a₁)) f(a₁) a₁ a₂ g(f(a₂)) f(a₂) A B C

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. g(f(a₁)) f(a₁) a₁ a₂ Great Greatexercise: exercise:repeat Repeatthis this proof proofusing usingthe theother other g(f(a₂)) definition of injectivity. definition of injectivity. f(a₂) A B C

92 Another Class of Functions

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

94 Surjective Functions A function f : A B is called surjective (or onto) if this first-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 first rule of functions?

95 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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.

96 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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.

97 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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. does ititmean for f fto Let What xwhat = 2y. Then we see that does mean for tobe besurjective? surjective? f(x) =ℝ.f(2y) =ℝ. 2yf(x) / 2 ==y.y y x y ℝ. x ℝ. f(x) = y So f(x) = y, as required. Therefore, Therefore,we'll we'llchoose choosean anarbitrary arbitraryyy ℝ, ℝ,then then prove provethat thatthere thereisissome somexx ℝℝwhere wheref(x) f(x)==y. y.

98 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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. does ititmean for f fto Let What xwhat = 2y. Then we see that does mean for tobe besurjective? surjective? f(x) =ℝ.f(2y) =ℝ. 2yf(x) / 2 ==y.y y x y ℝ. x ℝ. f(x) = y So f(x) = y, as required. Therefore, Therefore,we'll we'llchoose choosean anarbitrary arbitraryyy ℝ, ℝ,then then prove provethat thatthere thereisissome somexx ℝℝwhere wheref(x) f(x)==y. y.

99 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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. does ititmean for f fto Let What xwhat = 2y. Then we see that does mean for tobe besurjective? surjective? f(x) =ℝ.f(2y) =ℝ. 2yf(x) / 2 ==y.y y x y ℝ. x ℝ. f(x) = y So f(x) = y, as required. Therefore, Therefore,we'll we'llchoose choosean anarbitrary arbitraryyy ℝ, ℝ,then then prove provethat thatthere thereisissome somexx ℝℝwhere wheref(x) f(x)==y. y.

100 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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. does ititmean for f fto Let What xwhat = 2y. Then we see that does mean for tobe besurjective? surjective? f(x) =ℝ.f(2y) =ℝ. 2yf(x) / 2 ==y.y y x y ℝ. x ℝ. f(x) = y So f(x) = y, as required. Therefore, Therefore,we'll we'llchoose choosean anarbitrary arbitraryyy ℝ, ℝ,then then prove provethat thatthere thereisissome somexx ℝℝwhere wheref(x) f(x)==y. y.

101 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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. does ititmean for f fto Let What xwhat = 2y. Then we see that does mean for tobe besurjective? surjective? f(x) =ℝ.f(2y) =ℝ. 2yf(x) / 2 ==y.y y x y ℝ. x ℝ. f(x) = y So f(x) = y, as required. Therefore, Therefore,we'll we'llchoose choosean anarbitrary arbitraryyy ℝ, ℝ,then then prove provethat thatthere thereisissome somexx ℝℝwhere wheref(x) f(x)==y. y.

102 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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. does ititmean for f fto Let What xwhat = 2y. Then we see that does mean for tobe besurjective? surjective? f(x) =ℝ.f(2y) =ℝ. 2yf(x) / 2 ==y.y y x y ℝ. x ℝ. f(x) = y So f(x) = y, as required. Therefore, Therefore,we'll we'llchoose choosean anarbitrary arbitraryyy ℝ, ℝ,then then prove provethat thatthere thereisissome somexx ℝℝwhere wheref(x) f(x)==y. y.

103 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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.

104 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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.

105 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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.

106 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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.

107 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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.

108 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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.

109 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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.

110 Surjective Functions Theorem: Let f : ℝ ℝ be defined 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 Composing Surjections

112 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.

113 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.

114 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.

115 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.

116 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. What Whatdoes doesititmean meanfor forgg f f: :AA CCto tobe besurjective? surjective? Consider any c C. Since g : B C is surjective, there is some b B such g(b) Similarly, c C. = A. f)(a) = 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 Therefore, we'll arbitrary prove means that there is some a Acsuch that Therefore, we'llchoose choose arbitrary c CCand and provethat thatthere there isissome someaa AAsuch suchthat that(g (g f)(a) f)(a)==c.c. g(f(a)) = g(b) = c, which is what we needed to show.

117 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. What Whatdoes doesititmean meanfor forgg f f: :AA CCto tobe besurjective? surjective? Consider any c C. Since g : B C is surjective, there is some b B such g(b) Similarly, c C. = A. f)(a) = 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 Therefore, we'll arbitrary prove means that there is some a Acsuch that Therefore, we'llchoose choose arbitrary c CCand and provethat thatthere there isissome someaa AAsuch suchthat that(g (g f)(a) f)(a)==c.c. g(f(a)) = g(b) = c, which is what we needed to show.

118 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. What Whatdoes doesititmean meanfor forgg f f: :AA CCto tobe besurjective? surjective? Consider any c C. Since g : B C is surjective, there is some b B such g(b) Similarly, c C. = A. f)(a) = 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 Therefore, we'll arbitrary prove means that there is some a Acsuch that Therefore, we'llchoose choose arbitrary c CCand and provethat thatthere there isissome someaa AAsuch suchthat that(g (g f)(a) f)(a)==c.c. g(f(a)) = g(b) = c, which is what we needed to show.

119 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.

120 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

121 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

122 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

123 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

124 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

125 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

126 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.

127 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?

128 Katniss Everdeen Elsa Hermione Granger

129 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.

130 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.

131 Inverse Functions

132 Katniss Everdeen Elsa Hermione Granger

133 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

134 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

135 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

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

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