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Yet another important mathematical concept
Functions Yet another important mathematical concept
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Functions Most basic representation: Arrow Diagrams B A π
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Functions Most basic representation: Arrow Diagrams
A is called the domain and B is called the co-domain and we say that βf is such that it maps elements from A to Bβ (π:π΄β¦π΅) B A π Linger on this
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Functions Most basic representation: Arrow Diagrams
A is called the domain and B is called the co-domain and we say that βf is such that it maps elements from A to Bβ (π:π΄β¦π΅) B A π \mapsto in LaTeX
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Example 1 Yes No Is this a function? A B π 1 2 4 3
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Example 1 Yes No Is this a function? π A B Domain: {0, 1, 2, 3, 4}
Co-domain: {0, 1, 4} Formula (that we came up with): π π₯ = π₯ 2 % 5 (where % is the MOD operation) A B π 1 2 4 3
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Example 2 Yes No Is this a function? A B 1 2 4 3
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Example 2 Yes No Is this a function? 3 A B
Every element of the domain should map to some co-domain element! A B 1 2 4 3
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Example 3 Yes No Is this a function? π:ββ¦β, and π π₯ = π₯ 2
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Example 3 Yes No Is this a function? π:ββ¦β, and π π₯ = π₯ 2
For any odd selection of π₯ββ, there is no π₯ 2 β β!
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Arrow Diagrams: Pros and Cons
Good for introducing functions as mappings No information on other function properties (continuous, integrable, monotonic, β¦) Help us decouple functions from other mappings in π΄ Γπ΅ Only possible for functions with small, finite domains and co-domains Clarify what the domain and what the co-domain is β¦. Good for teaching us what onto and 1-1 means β¦
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Arrow Diagrams: Pros and Cons
Good for introducing functions as mappings No information on other function properties (continuous, integrable, monotonic, β¦) Help us decouple functions from other mappings in π΄ Γπ΅ Only possible for functions with small, finite domains and co-domains Clarify what the domain and what the co-domain is β¦ Good for teaching us what onto and 1-1 means Those are important consβ¦.
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Graphs We can also represent functions by their graphsβ¦
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(βvertical line testβ)
Graphs (pros,cons) Which has its owns pros and cons Pros Cons Easy to determine 1-1 functions (βhorizontal line testβ) Cannot compute the derivative at any point, only draw it Easy to determine non-functions (βvertical line testβ) Cannot compute the value of the function at any point Can determine monotonicity and continuity easily Cannot compute the indefinite integral of the function Can easily determine if the function intersects the axes Cannot determine where it intersects the axes β¦
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Short quiz
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Short quiz Yes No Are the following valid functions?
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Short quiz Yes No Are the following valid functions?
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Short quiz Yes No Are the following valid functions? Log function
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Short quiz Yes No Are the following valid functions?
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Short quiz Yes No Fails the βvertical lineβ test
Are the following valid functions? Fails the βvertical lineβ test
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Short quiz Yes No Are the following valid functions? B A π
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Short quiz Yes No Are the following valid functions? π B A
It just happens to not be β1-1β (term weβll explain momentarily)!
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Function formulae Yes No Are the following valid functions? B A π
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Function formulae Yes No
Are the following valid functions? B A π Fails the βvertical lineβ test (2 different `yβs mapped to by the same `xβ)
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Surjective functions A function π:πβ¦π is called surjective (or onto) iff βπ¦βπ, βπ₯βπ:π π₯ =π¦
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Surjective functions A function π:πβ¦π is called surjective (or onto) iff βπ¦βπ, βπ₯βπ:π π₯ =π¦ Intuitively: π β² π co-domain is βfullβ with pointers
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Surjective functions A function π:πβ¦π is called surjective (or onto) iff βπ¦βπ, βπ₯βπ:π π₯ =π¦ Intuitively: π β² π co-domain is βfullβ with pointers Mnemonic rule to remember the name βsurjectiveβ: in French, βsurβ means βaboveβ, βonβ, or βontoβ.
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Surjective functions A function π:πβ¦π is called surjective (or onto) iff βπ¦βπ, βπ₯βπ [π π₯ =π¦] Intuitively: π β² π co-domain is βfullβ with pointers Mnemonic rule to remember the name βsurjectiveβ: in French, βsurβ means βaboveβ, βonβ, or βontoβ. βsurβ
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Quiz on surjectivity Yes No Something Else Is π π₯ = π₯ 2 surjective?
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Quiz on surjectivity Yes No Is π π₯ = π₯ 2 surjective?
Something Else Is π π₯ = π₯ 2 surjective? We have not provided you with either the domain or the co-domain!
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Quiz on surjectivity Yes No Is π π₯ = π₯ 2 surjective?
Something Else Is π π₯ = π₯ 2 surjective? We have not provided you with either the domain or the co-domain! π·=β, πΆ=β: Non-Surjective! (e.g -1 is not mapped to)
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Quiz on surjectivity Yes No Is π π₯ = π₯ 2 surjective?
Something Else Is π π₯ = π₯ 2 surjective? We have not provided you with either the domain or the co-domain! π·=β, πΆ=β: Non-Surjective! (e.g -1 is not mapped to) π·=β,πΆ= β β₯0 : Surjective!
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Quiz on surjectivity Yes No Is π π₯ = π₯ 2 surjective?
Something Else Is π π₯ = π₯ 2 surjective? We have not provided you with either the domain or the co-domain! π·=β, πΆ=β: Non-Surjective! (e.g -1 is not mapped to) π·=β,πΆ= β β₯0 : Surjective! π·=β,πΆ=? such that it is surjective?
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Quiz on surjectivity Yes No Is π π₯ = π₯ 2 surjective?
Something Else Is π π₯ = π₯ 2 surjective? We have not provided you with either the domain or the co-domain! π·=β, πΆ=β: Non-Surjective! (e.g -1 is not mapped to) π·=β,πΆ= β β₯0 : Surjective! π·=β,πΆ=ππππ΄π
πΈπ: Surjective!
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Injective functions A function π:πβ¦π is called injective (or 1-1) iff
(β π₯ 1 , π₯ 2 βπ) [π π₯ 1 =π π₯ 2 β π₯ 1 = π₯ 2 ] Intuitively: Every element of the co-domain is mapped to by at most one element of the domain.
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Injective functions A function π:πβ¦π is called injective (or 1-1) iff
(β π₯ 1 , π₯ 2 βπ) [π π₯ 1 =π π₯ 2 β π₯ 1 = π₯ 2 ] Intuitively: Every element of the co-domain is mapped to by at most one element of the domain. Why βat most oneβ and not exactly one?
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Injective functions A function π:πβ¦π is called injective (or 1-1) iff
(β π₯ 1 , π₯ 2 βπ) [π π₯ 1 =π π₯ 2 β π₯ 1 = π₯ 2 ] Intuitively: Every element of the co-domain is mapped to by at most one element of the domain. Why at most one and not exactly one? Because 1-1 but not onto functions are possible! X Y π
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Quiz on injectivity Yes No Something Else Is π π₯ = π₯ 2 injective?
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Quiz on injectivity Yes No Is π π₯ = π₯ 2 injective?
Something Else Is π π₯ = π₯ 2 injective? We need domains and codomains!
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Quiz on injectivity Yes No Is π π₯ = π₯ 2 injective?
Something Else Is π π₯ = π₯ 2 injective? We need domains and codomains! π·=β, πΆ=β: Non-Injective! (e.g β3 2 = 3 2 =9)
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Quiz on injectivity Yes No Is π π₯ = π₯ 2 injective?
Something Else Is π π₯ = π₯ 2 injective? We need domains and codomains! π·=β, πΆ=β: Non-Injective! (e.g β3 2 = 3 2 =9) π·=β,πΆ= β β₯0 : NonβInjective! (same example works)
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Quiz on injectivity Yes No Is π π₯ = π₯ 2 injective?
Something Else Is π π₯ = π₯ 2 injective? We need domains and codomains! π·=β, πΆ=β: Non-Injective! (e.g β3 2 = 3 2 =9) π·=β,πΆ= β β₯0 : NonβInjective! (same example works) π·=β€,πΆ=ππππ΄π
πΈπ: NonβInjective! (same example works)
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Quiz on injectivity Yes No Is π π₯ = π₯ 2 injective?
Something Else Is π π₯ = π₯ 2 injective? We need domains and codomains! π·=β, πΆ=β: Non-Injective! (e.g β3 2 = 3 2 =9) π·=β,πΆ= β β₯0 : NonβInjective! (same example works) π·=β€,πΆ=ππππ΄π
πΈπ: NonβInjective! (same example works) Can this function ever be injective?
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Quiz on injectivity Yes No Is π π₯ = π₯ 2 injective?
Something Else Is π π₯ = π₯ 2 injective? We need domains and codomains! π·=β, πΆ=β: Non-Injective! (e.g β3 2 = 3 2 =9) π·=β,πΆ= β β₯0 : NonβInjective! (same example works) π·=β€,πΆ=ππππ΄π
πΈπ: NonβInjective! (same example works) Can this function ever be injective? Yes. Pick π·=β,πΆ=ππππ΄π
πΈπ
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Making functions onto or 1-1
To make a function onto, we need to make the co-domain smaller. To make a function 1-1 , we need to make the domain smaller.
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Bijective functions A function π:πβ¦π is called bijective (or a bijection, or a 1- 1 correspondence) iff it is both surjective and injective. We will try to avoid using the term β1-1 correspondenceβ (your book uses it) since it can confuse us with the notion of an injective (or 1-1) function.
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Quiz on bijections
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Quiz on bijections Yes No Are the following functions bijections?
(In all examples, πΆ=β)
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Quiz on bijections Yes No Are the following functions bijections?
π π₯ = π₯ , π₯ββ (In all examples, πΆ=β)
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Quiz on bijections Yes No Are the following functions bijections?
(In all examples, πΆ=β) Non-injective!
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Quiz on bijections Yes No Are the following functions bijections?
π π₯ = π₯ , π₯ββ π π₯ =πβ
π₯+π, π,π₯,πββ (In all examples, πΆ=β)
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Quiz on bijections Yes No
Are the following functions bijections? π π₯ = π₯ , π₯ββ No π π₯ =πβ
π₯+π, π,π₯,πββ No (In all examples, πΆ=β) !!!! For a = 0, the graph of the function fails the βhorizontal line testβ!
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Quiz on bijections Yes No Are the following functions bijections?
π π₯ = πβ
π₯ 2 , π,π₯β β, π>0 (In all examples, πΆ=β)
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Quiz on bijections Yes No Are the following functions bijections?
(In all examples, πΆ=β) Non-injective!
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Quiz on bijections Yes No Are the following functions bijections?
β π =4π β1, πββ€ (In all examples, πΆ=β)
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Quiz on bijections Yes No Are the following functions bijections?
(In all examples, πΆ=β) Non-surjective! Set β π =π¦ and solve for π: 4π β 1 = π¦βπ= π¦+1 4 There are infinitely many choices of π¦ for which πββ€!
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Quiz on bijections Yes No Are the following functions bijections?
β π₯ =4π₯ β1, π₯ββ (In all examples, πΆ=β)
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Quiz on bijections Yes No Are the following functions bijections?
(In all examples, πΆ=β) Surjective and injective! Surjective, since, if we set β π =π¦ and solve for π: 4π β 1 = π¦βπ= π¦+1 4 For every real π¦, thereβs always a real solution π. Injective, since itβs of the form of (2) with πβ 0.
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Some interesting functions
We will examine a pair of interesting functions from β to β€. Given a real number π, the floor of π, denoted π is the largest integer π:πβ€π.
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Floor and ceiling We will examine a pair of interesting functions from β to β€. Given a real number π, the floor of π, denoted π is the largest integer π:πβ€π. Given a real number π, the ceiling of π, denoted βπβ is the smallest integer π:πβ₯π.
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Floor and ceiling We will examine a pair of interesting functions from β to β€. Given a real number π, the floor of π, denoted π is the largest integer π:πβ€π. Given a real number π, the ceiling of π, denoted βπβ is the smallest integer π:πβ₯π. β π 1 β β π 1 β β π 2 β β π 2 β -4 -3 -2 -1 1 2 3 4 5 6 7 π 1 π 2
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Examples 1.3 =1
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Examples 1.3 =1 10.3 =10
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Examples 1.3 =1 10.3 =10 0.8 =1
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Examples 1.3 =1 10.3 =10 0.8 =1 β¦1 =1
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Examples 1.3 =1 10.3 =10 0.8 =1 β¦1 =1 β5.6 =β5 At this point, might be worth mentioning that those functions are not injective!
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Examples From these two we can deduce⦠1.3 =1 10.3 =10 0.8 =1
β¦1 =1 β5.6 =β5 From these two we can deduceβ¦ At this point, might be worth mentioning that those functions are not injective!
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Examples From these two we can deduce⦠that floor is non-injective!
1.3 =1 10.3 =10 0.8 =1 β¦1 =1 β5.6 =β6 From these two we can deduceβ¦ that floor is non-injective! (same for ceiling, of course) At this point, might be worth mentioning that those functions are not injective!
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Vote! 3 = 3 2 4 3.5
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Vote! 3 =3 π = 3 2 4 3.5 3 π 4 3.5
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Vote! 3 2 4 3.5 3 =3 π = 4 2 500 Γ = 3 π 4 3.5 2 500 Γ500 2 500 Γ Γ 500 2 2 499 Γ 500 2 Something else
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Vote! 3 2 4 3.5 3 =3 π = 4 2 500 Γ = Γ 500 2 βπββ€, π = π =π 3 π 4 3.5 2 500 Γ500 2 500 Γ Γ 500 2 2 499 Γ 500 2 Something else T πΉ
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Vote! 3 =3 π = 4 βπββ€, π = π =π T βπββ€, π = π =π 3 2 4 3.5 3 π 4 3.5 T
3 =3 π = 4 2 500 Γ = Γ 500 2 βπββ€, π = π =π T βπββ€, π = π =π 3 π 4 3.5 2 500 Γ500 2 500 Γ Γ 500 2 2 499 Γ 500 2 Something else T πΉ T πΉ
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Vote! 3 =3 π = 4 βπββ€, π = π =π T βπββ€, π = π =π T 3 2 4 3.5 3 π 4 3.5
3 =3 π = 4 2 500 Γ = Γ 500 2 βπββ€, π = π =π T βπββ€, π = π =π T 3 π 4 3.5 2 500 Γ500 2 500 Γ Γ 500 2 2 499 Γ 500 2 Something else T πΉ T πΉ
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Relations
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Functions ultra-formally
Consider the function that we talked about before: A B π 1 2 4 3
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Functions ultra-formally
Consider the function that we talked about before: Can view it as the set of ordered pairs { 0,0 , 1,1 , 2,4 , 3,4 , (4, 1)} A B π 1 2 4 3
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Functions ultra-formally
Let π΄, π΅ be sets. A function π:π΄β¦π΅ is any subset π=π΄ Γ π΅ where βπβπ΄ [ βπβπ΅ π, π βπ ] (every element of the domain maps somewhere) βπβπ΄ [ β π 1 βπ΅ π, π 1 βπ β§β π 2 β π 1 π, π 2 βπ ] (every element of the domain maps to exactly one element of the co-domain)
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Functions ultra-formally
Let π΄, π΅ be sets. A function π:π΄β¦π΅ is any subset π=π΄ Γ π΅ where βπβπ΄ [ βπβπ΅ π, π βπ ] (every element of the domain maps somewhere) βπβπ΄ [ β π 1 βπ΅ π, π 1 βπ β§β π 2 β π 1 π, π 2 βπ ] (every element of the domain maps to exactly one element of the co-domain) We generalize the notion of a function to that of a relation! Any subset of π΄ Γ π΅ that doesnβt abide by (1) or (2) we call a relation from A to B.
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Examples (<, βΓβ) {β¦, β1.5, β1.2 , β1.4,β1.2 , 2 , 3 , 2 , 5 ,β¦}
{β¦, β1.5, β1.2 , β1.4,β1.2 , 2 , 3 , 2 , 5 ,β¦} Verbally tell them that is is βless thanβ on R^2
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Examples (<, βΓβ) β€,βΓβ
{β¦, β1.5, β1.2 , β1.4,β1.2 , 2 , 3 , 2 , 5 ,β¦} β€,βΓβ {β¦, 2, 2 , 2, 2.1 , 2 , 3 , 2 , 5 ,β¦}
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Examples (<, βΓβ) β€,βΓβ π΅πππ, βΓβ
{β¦, β1.5, β1.2 , β1.4,β1.2 , 2 , 3 , 2 , 5 ,β¦} β€,βΓβ {β¦, 2, 2 , 2, 2.1 , 2 , 3 , 2 , 5 ,β¦} π΅πππ, βΓβ { π, π | π appears in the decimal expansion of π } E.g: {β¦, π, 1 , π, 7 , , 4 , β¦} We would formally say that all of the above are elements of the relation βBillβ Verbally state the names of the other relations.
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Reflexivity A relation Xβπ΄ Γ π΄ is reflexive if βπβπ΄ [ π, π βπ ]
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Reflexivity A relation Xβπ΄ Γ π΄ is reflexive if βπβπ΄ π, π βπ Examples:
βπβπ΄ π, π βπ Examples: (β€,βΓβ) is reflexive, since βπββ πβ€π (<,βΓβ) is not reflexive, since βΌ βπββ π<π (in fact, there is no such π) (π½ππ ππ,βΓβ) defined as { π₯, π¦ | π₯+π¦β₯100} is not reflexive (e.g 10, 10 βπ½ππ ππ)) Say verbally that one needs only one counter-example (a forall statement can be proven negative with just one witness)
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Symmetry A relation Xβπ΄ Γ π΄ is symmetric if
β π 1 , π 2 βπ΄ π 1 , π 2 βπ β π 2 , π 1 βπ)
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Symmetry A relation Xβπ΄ Γ π΄ is symmetric if
β π 1 , π 2 βπ΄ π 1 , π 2 βπ β π 2 , π 1 βπ) Examples: (β€,βΓβ) is not symmetric since 4β€5 but βΌ(5β€4) (<,βΓβ) is not symmetric (see above) (π½ππ ππ,βΓβ) defined as { π₯, π¦ | π₯+π¦β₯100} is symmetric since π₯+π¦β€100 β(π¦+π₯β€100)
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β π 1 , π 2 , π 3 βπ΄ π 1 , π 2 βπ β§( π 2 , π 3 βπ)β π 1 , π 3 βπ)
Transitivity A relation Xβπ΄ Γ π΄ is transitive if β π 1 , π 2 , π 3 βπ΄ π 1 , π 2 βπ β§( π 2 , π 3 βπ)β π 1 , π 3 βπ)
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β π 1 , π 2 , π 3 βπ΄ π 1 , π 2 βπ β§( π 2 , π 3 βπ)β π 1 , π 3 βπ)
Transitivity A relation Xβπ΄ Γ π΄ is transitive if β π 1 , π 2 , π 3 βπ΄ π 1 , π 2 βπ β§( π 2 , π 3 βπ)β π 1 , π 3 βπ) (β€,βΓβ) is transitive since π₯β€π¦ β§ π¦β€π§ β π₯β€π§ (<,βΓβ) is transitive (see above) (π½ππ ππ,βΓβ) defined as { π₯, π¦ | π₯+π¦β₯100} ???
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Transitivity A relation Xβπ΄ Γ π΄ is transitive if
β π 1 , π 2 , π 3 βπ΄ π 1 , π 2 βπ β§( π 2 , π 3 βπ)β π 1 , π 3 βπ) (β€,βΓβ) is transitive since π₯β€π¦ β§ π¦β€π§ β π₯β€π§ (<,βΓβ) is transitive (see above) (π½ππ ππ,βΓβ) defined as { π₯, π¦ | π₯+π¦β₯100} is not transitive since (counter-example): 1, 100 βπ½ππ ππβ§ 100, 5 βπ½ππ ππ but 1, 5 βπ½ππ ππ
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