Yet another important mathematical concept Functions Yet another important mathematical concept

Functions Most basic representation: Arrow Diagrams B A 𝑓

Functions Most basic representation: Arrow Diagrams A is called the domain and B is called the co-domain and we say that “f is such that it maps elements from A to B” (𝑓:𝐴↦𝐵) B A 𝑓 Linger on this

Functions Most basic representation: Arrow Diagrams A is called the domain and B is called the co-domain and we say that “f is such that it maps elements from A to B” (𝑓:𝐴↦𝐵) B A 𝑓 \mapsto in LaTeX

Example 1 Yes No Is this a function? A B 𝑓 1 2 4 3

Example 1 Yes No Is this a function? 𝑓 A B Domain: {0, 1, 2, 3, 4} Co-domain: {0, 1, 4} Formula (that we came up with): 𝑓 𝑥 = 𝑥 2 % 5 (where % is the MOD operation) A B 𝑓 1 2 4 3

Example 2 Yes No Is this a function? A B 1 2 4 3

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

Example 3 Yes No Is this a function? 𝑓:ℕ↦ℕ, and 𝑓 𝑥 = 𝑥 2

Example 3 Yes No Is this a function? 𝑓:ℕ↦ℕ, and 𝑓 𝑥 = 𝑥 2 For any odd selection of 𝑥∈ℕ, there is no 𝑥 2 ∈ ℕ!

Arrow Diagrams: Pros and Cons Good for introducing functions as mappings No information on other function properties (continuous, integrable, monotonic, …) Help us decouple functions from other mappings in 𝐴 ×𝐵 Only possible for functions with small, finite domains and co-domains Clarify what the domain and what the co-domain is …. Good for teaching us what onto and 1-1 means …

Arrow Diagrams: Pros and Cons Good for introducing functions as mappings No information on other function properties (continuous, integrable, monotonic, …) Help us decouple functions from other mappings in 𝐴 ×𝐵 Only possible for functions with small, finite domains and co-domains Clarify what the domain and what the co-domain is … Good for teaching us what onto and 1-1 means Those are important cons….

Graphs We can also represent functions by their graphs…

(“vertical line test”) Graphs (pros,cons) Which has its owns pros and cons Pros Cons Easy to determine 1-1 functions (“horizontal line test”) Cannot compute the derivative at any point, only draw it Easy to determine non-functions (“vertical line test”) 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 …

Short quiz

Short quiz Yes No Are the following valid functions?

Short quiz Yes No Are the following valid functions?

Short quiz Yes No Are the following valid functions? Log function

Short quiz Yes No Are the following valid functions?

Short quiz Yes No Fails the “vertical line” test Are the following valid functions? Fails the “vertical line” test

Short quiz Yes No Are the following valid functions? B A 𝑓

Short quiz Yes No Are the following valid functions? 𝑓 B A It just happens to not be “1-1” (term we’ll explain momentarily)!

Function formulae Yes No Are the following valid functions? B A 𝑓

Function formulae Yes No Are the following valid functions? B A 𝑓 Fails the “vertical line” test (2 different `y’s mapped to by the same `x’)

Surjective functions A function 𝑓:𝑋↦𝑌 is called surjective (or onto) iff ∀𝑦∈𝑌, ∃𝑥∈𝑋:𝑓 𝑥 =𝑦

Surjective functions A function 𝑓:𝑋↦𝑌 is called surjective (or onto) iff ∀𝑦∈𝑌, ∃𝑥∈𝑋:𝑓 𝑥 =𝑦 Intuitively: 𝑓 ′ 𝑠 co-domain is “full” with pointers

Surjective functions A function 𝑓:𝑋↦𝑌 is called surjective (or onto) iff ∀𝑦∈𝑌, ∃𝑥∈𝑋:𝑓 𝑥 =𝑦 Intuitively: 𝑓 ′ 𝑠 co-domain is “full” with pointers Mnemonic rule to remember the name “surjective”: in French, “sur” means “above”, “on”, or “onto”.

Surjective functions A function 𝑓:𝑋↦𝑌 is called surjective (or onto) iff ∀𝑦∈𝑌, ∃𝑥∈𝑋 [𝑓 𝑥 =𝑦] Intuitively: 𝑓 ′ 𝑠 co-domain is “full” with pointers Mnemonic rule to remember the name “surjective”: in French, “sur” means “above”, “on”, or “onto”. “sur”

Quiz on surjectivity Yes No Something Else Is 𝑓 𝑥 = 𝑥 2 surjective?

Quiz on surjectivity Yes No Is 𝑓 𝑥 = 𝑥 2 surjective? Something Else Is 𝑓 𝑥 = 𝑥 2 surjective? We have not provided you with either the domain or the co-domain!

Quiz on surjectivity Yes No Is 𝑓 𝑥 = 𝑥 2 surjective? Something Else Is 𝑓 𝑥 = 𝑥 2 surjective? We have not provided you with either the domain or the co-domain! 𝐷=ℝ, 𝐶=ℝ: Non-Surjective! (e.g -1 is not mapped to)

Quiz on surjectivity Yes No Is 𝑓 𝑥 = 𝑥 2 surjective? Something Else Is 𝑓 𝑥 = 𝑥 2 surjective? We have not provided you with either the domain or the co-domain! 𝐷=ℝ, 𝐶=ℝ: Non-Surjective! (e.g -1 is not mapped to) 𝐷=ℝ,𝐶= ℝ ≥0 : Surjective!

Quiz on surjectivity Yes No Is 𝑓 𝑥 = 𝑥 2 surjective? Something Else Is 𝑓 𝑥 = 𝑥 2 surjective? We have not provided you with either the domain or the co-domain! 𝐷=ℝ, 𝐶=ℝ: Non-Surjective! (e.g -1 is not mapped to) 𝐷=ℝ,𝐶= ℝ ≥0 : Surjective! 𝐷=ℕ,𝐶=? such that it is surjective?

Quiz on surjectivity Yes No Is 𝑓 𝑥 = 𝑥 2 surjective? Something Else Is 𝑓 𝑥 = 𝑥 2 surjective? We have not provided you with either the domain or the co-domain! 𝐷=ℝ, 𝐶=ℝ: Non-Surjective! (e.g -1 is not mapped to) 𝐷=ℝ,𝐶= ℝ ≥0 : Surjective! 𝐷=ℕ,𝐶=𝑆𝑄𝑈𝐴𝑅𝐸𝑆: Surjective!

Injective functions A function 𝑓:𝑋↦𝑌 is called injective (or 1-1) iff (∀ 𝑥 1 , 𝑥 2 ∈𝑋) [𝑓 𝑥 1 =𝑓 𝑥 2 ⇒ 𝑥 1 = 𝑥 2 ] Intuitively: Every element of the co-domain is mapped to by at most one element of the domain.

Injective functions A function 𝑓:𝑋↦𝑌 is called injective (or 1-1) iff (∀ 𝑥 1 , 𝑥 2 ∈𝑋) [𝑓 𝑥 1 =𝑓 𝑥 2 ⇒ 𝑥 1 = 𝑥 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?

Injective functions A function 𝑓:𝑋↦𝑌 is called injective (or 1-1) iff (∀ 𝑥 1 , 𝑥 2 ∈𝑋) [𝑓 𝑥 1 =𝑓 𝑥 2 ⇒ 𝑥 1 = 𝑥 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 𝑓

Quiz on injectivity Yes No Something Else Is 𝑓 𝑥 = 𝑥 2 injective?

Quiz on injectivity Yes No Is 𝑓 𝑥 = 𝑥 2 injective? Something Else Is 𝑓 𝑥 = 𝑥 2 injective? We need domains and codomains!

Quiz on injectivity Yes No Is 𝑓 𝑥 = 𝑥 2 injective? Something Else Is 𝑓 𝑥 = 𝑥 2 injective? We need domains and codomains! 𝐷=ℝ, 𝐶=ℝ: Non-Injective! (e.g −3 2 = 3 2 =9)

Quiz on injectivity Yes No Is 𝑓 𝑥 = 𝑥 2 injective? Something Else Is 𝑓 𝑥 = 𝑥 2 injective? We need domains and codomains! 𝐷=ℝ, 𝐶=ℝ: Non-Injective! (e.g −3 2 = 3 2 =9) 𝐷=ℝ,𝐶= ℝ ≥0 : Non−Injective! (same example works)

Quiz on injectivity Yes No Is 𝑓 𝑥 = 𝑥 2 injective? Something Else Is 𝑓 𝑥 = 𝑥 2 injective? We need domains and codomains! 𝐷=ℝ, 𝐶=ℝ: Non-Injective! (e.g −3 2 = 3 2 =9) 𝐷=ℝ,𝐶= ℝ ≥0 : Non−Injective! (same example works) 𝐷=ℤ,𝐶=𝑆𝑄𝑈𝐴𝑅𝐸𝑆: Non−Injective! (same example works)

Quiz on injectivity Yes No Is 𝑓 𝑥 = 𝑥 2 injective? Something Else Is 𝑓 𝑥 = 𝑥 2 injective? We need domains and codomains! 𝐷=ℝ, 𝐶=ℝ: Non-Injective! (e.g −3 2 = 3 2 =9) 𝐷=ℝ,𝐶= ℝ ≥0 : Non−Injective! (same example works) 𝐷=ℤ,𝐶=𝑆𝑄𝑈𝐴𝑅𝐸𝑆: Non−Injective! (same example works) Can this function ever be injective?

Quiz on injectivity Yes No Is 𝑓 𝑥 = 𝑥 2 injective? Something Else Is 𝑓 𝑥 = 𝑥 2 injective? We need domains and codomains! 𝐷=ℝ, 𝐶=ℝ: Non-Injective! (e.g −3 2 = 3 2 =9) 𝐷=ℝ,𝐶= ℝ ≥0 : Non−Injective! (same example works) 𝐷=ℤ,𝐶=𝑆𝑄𝑈𝐴𝑅𝐸𝑆: Non−Injective! (same example works) Can this function ever be injective? Yes. Pick 𝐷=ℕ,𝐶=𝑆𝑄𝑈𝐴𝑅𝐸𝑆

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.

Bijective functions A function 𝑓:𝑋↦𝑌 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 “1-1 correspondence” (your book uses it) since it can confuse us with the notion of an injective (or 1-1) function.

Quiz on bijections

Quiz on bijections Yes No Are the following functions bijections? (In all examples, 𝐶=ℝ)

Quiz on bijections Yes No Are the following functions bijections? 𝑓 𝑥 = 𝑥 , 𝑥∈ℝ (In all examples, 𝐶=ℝ)

Quiz on bijections Yes No Are the following functions bijections? (In all examples, 𝐶=ℝ) Non-injective!

Quiz on bijections Yes No Are the following functions bijections? 𝑓 𝑥 = 𝑥 , 𝑥∈ℝ 𝑓 𝑥 =𝑎⋅𝑥+𝑏, 𝑎,𝑥,𝑏∈ℝ (In all examples, 𝐶=ℝ)

Quiz on bijections Yes No Are the following functions bijections? 𝑓 𝑥 = 𝑥 , 𝑥∈ℝ No 𝑓 𝑥 =𝑎⋅𝑥+𝑏, 𝑎,𝑥,𝑏∈ℝ No (In all examples, 𝐶=ℝ) !!!! For a = 0, the graph of the function fails the “horizontal line test”!

Quiz on bijections Yes No Are the following functions bijections? 𝑔 𝑥 = 𝑎⋅𝑥 2 , 𝑎,𝑥∈ ℝ, 𝑎>0 (In all examples, 𝐶=ℝ)

Quiz on bijections Yes No Are the following functions bijections? (In all examples, 𝐶=ℝ) Non-injective!

Quiz on bijections Yes No Are the following functions bijections? ℎ 𝑛 =4𝑛 −1, 𝑛∈ℤ (In all examples, 𝐶=ℝ)

Quiz on bijections Yes No Are the following functions bijections? (In all examples, 𝐶=ℝ) Non-surjective! Set ℎ 𝑛 =𝑦 and solve for 𝑛: 4𝑛 – 1 = 𝑦⇒𝑛= 𝑦+1 4 There are infinitely many choices of 𝑦 for which 𝑛∉ℤ!

Quiz on bijections Yes No Are the following functions bijections? ℎ 𝑥 =4𝑥 −1, 𝑥∈ℝ (In all examples, 𝐶=ℝ)

Quiz on bijections Yes No Are the following functions bijections? (In all examples, 𝐶=ℝ) Surjective and injective! Surjective, since, if we set ℎ 𝑛 =𝑦 and solve for 𝑛: 4𝑛 – 1 = 𝑦⇒𝑛= 𝑦+1 4 For every real 𝑦, there’s always a real solution 𝑛. Injective, since it’s of the form of (2) with 𝑎≠0.

Some interesting functions We will examine a pair of interesting functions from ℝ to ℤ. Given a real number 𝑟, the floor of 𝑟, denoted 𝑟 is the largest integer 𝑛:𝑛≤𝑟.

Floor and ceiling We will examine a pair of interesting functions from ℝ to ℤ. Given a real number 𝑟, the floor of 𝑟, denoted 𝑟 is the largest integer 𝑛:𝑛≤𝑟. Given a real number 𝑟, the ceiling of 𝑟, denoted ⌈𝑟⌉ is the smallest integer 𝑛:𝑛≥𝑟.

Floor and ceiling We will examine a pair of interesting functions from ℝ to ℤ. Given a real number 𝑟, the floor of 𝑟, denoted 𝑟 is the largest integer 𝑛:𝑛≤𝑟. Given a real number 𝑟, the ceiling of 𝑟, denoted ⌈𝑟⌉ is the smallest integer 𝑛:𝑛≥𝑟. ⌊ 𝑟 1 ⌋ ⌈ 𝑟 1 ⌉ ⌊ 𝑟 2 ⌋ ⌈ 𝑟 2 ⌉ -4 -3 -2 -1 1 2 3 4 5 6 7 𝑟 1 𝑟 2

Examples 1.3 =1

Examples 1.3 =1 10.3 =10

Examples 1.3 =1 10.3 =10 0.8 =1

Examples 1.3 =1 10.3 =10 0.8 =1 0.0000000000…1 =1

Examples 1.3 =1 10.3 =10 0.8 =1 0.0000000000…1 =1 −5.6 =−5 At this point, might be worth mentioning that those functions are not injective!

Examples From these two we can deduce… 1.3 =1 10.3 =10 0.8 =1 0.0000000000…1 =1 −5.6 =−5 From these two we can deduce… At this point, might be worth mentioning that those functions are not injective!

Examples From these two we can deduce… that floor is non-injective! 1.3 =1 10.3 =10 0.8 =1 0.0000000000…1 =1 −5.6 =−6 From these two we can deduce… that floor is non-injective! (same for ceiling, of course) At this point, might be worth mentioning that those functions are not injective!

Vote! 3 = 3 2 4 3.5

Vote! 3 =3 𝜋 = 3 2 4 3.5 3 𝜋 4 3.5

Vote! 3 2 4 3.5 3 =3 𝜋 = 4 2 500 × 500 2 = 3 𝜋 4 3.5 2 500 ×500 2 500 × 500 2 2 500 × 500 2 2 499 × 500 2 Something else

Vote! 3 2 4 3.5 3 =3 𝜋 = 4 2 500 × 500 2 = 2 500 × 500 2 ∀𝑛∈ℤ, 𝑛 = 𝑛 =𝑛 3 𝜋 4 3.5 2 500 ×500 2 500 × 500 2 2 500 × 500 2 2 499 × 500 2 Something else T 𝐹

Vote! 3 =3 𝜋 = 4 ∀𝑛∈ℤ, 𝑛 = 𝑛 =𝑛 T ∀𝑛∈ℤ, 𝑛 = 𝑛 =𝑛 3 2 4 3.5 3 𝜋 4 3.5 T 3 =3 𝜋 = 4 2 500 × 500 2 = 2 500 × 500 2 ∀𝑛∈ℤ, 𝑛 = 𝑛 =𝑛 T ∀𝑛∈ℤ, 𝑛 = 𝑛 =𝑛 3 𝜋 4 3.5 2 500 ×500 2 500 × 500 2 2 500 × 500 2 2 499 × 500 2 Something else T 𝐹 T 𝐹

Vote! 3 =3 𝜋 = 4 ∀𝑛∈ℤ, 𝑛 = 𝑛 =𝑛 T ∀𝑛∈ℤ, 𝑛 = 𝑛 =𝑛 T 3 2 4 3.5 3 𝜋 4 3.5 3 =3 𝜋 = 4 2 500 × 500 2 = 2 500 × 500 2 ∀𝑛∈ℤ, 𝑛 = 𝑛 =𝑛 T ∀𝑛∈ℤ, 𝑛 = 𝑛 =𝑛 T 3 𝜋 4 3.5 2 500 ×500 2 500 × 500 2 2 500 × 500 2 2 499 × 500 2 Something else T 𝐹 T 𝐹

Relations

Functions ultra-formally Consider the function that we talked about before: A B 𝑓 1 2 4 3

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 𝑓 1 2 4 3

Functions ultra-formally Let 𝐴, 𝐵 be sets. A function 𝑓:𝐴↦𝐵 is any subset 𝑆=𝐴 × 𝐵 where ∀𝑎∈𝐴 [ ∃𝑏∈𝐵 𝑎, 𝑏 ∈𝑆 ] (every element of the domain maps somewhere) ∀𝑎∈𝐴 [ ∃ 𝑏 1 ∈𝐵 𝑎, 𝑏 1 ∈𝑆 ∧∀ 𝑏 2 ≠ 𝑏 1 𝑎, 𝑏 2 ∉𝑆 ] (every element of the domain maps to exactly one element of the co-domain)

Functions ultra-formally Let 𝐴, 𝐵 be sets. A function 𝑓:𝐴↦𝐵 is any subset 𝑆=𝐴 × 𝐵 where ∀𝑎∈𝐴 [ ∃𝑏∈𝐵 𝑎, 𝑏 ∈𝑆 ] (every element of the domain maps somewhere) ∀𝑎∈𝐴 [ ∃ 𝑏 1 ∈𝐵 𝑎, 𝑏 1 ∈𝑆 ∧∀ 𝑏 2 ≠ 𝑏 1 𝑎, 𝑏 2 ∉𝑆 ] (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 𝐴 × 𝐵 that doesn’t abide by (1) or (2) we call a relation from A to B.

Examples (<, ℝ×ℝ) {…, −1.5, −1.2 , −1.4,−1.2 , 2 , 3 , 2 , 5 ,…} {…, −1.5, −1.2 , −1.4,−1.2 , 2 , 3 , 2 , 5 ,…} Verbally tell them that is is “less than” on R^2

Examples (<, ℝ×ℝ) ≤,ℝ×ℝ {…, −1.5, −1.2 , −1.4,−1.2 , 2 , 3 , 2 , 5 ,…} ≤,ℝ×ℝ {…, 2, 2 , 2, 2.1 , 2 , 3 , 2 , 5 ,…}

Examples (<, ℝ×ℝ) ≤,ℝ×ℝ 𝐵𝑖𝑙𝑙, ℝ×ℕ {…, −1.5, −1.2 , −1.4,−1.2 , 2 , 3 , 2 , 5 ,…} ≤,ℝ×ℝ {…, 2, 2 , 2, 2.1 , 2 , 3 , 2 , 5 ,…} 𝐵𝑖𝑙𝑙, ℝ×ℕ { 𝑟, 𝑛 | 𝑛 appears in the decimal expansion of 𝑟 } E.g: {…, 𝜋, 1 , 𝑒, 7 , 1 7 , 4 , …} We would formally say that all of the above are elements of the relation “Bill” Verbally state the names of the other relations.

Reflexivity A relation X⊆𝐴 × 𝐴 is reflexive if ∀𝑎∈𝐴 [ 𝑎, 𝑎 ∈𝑋 ]

Reflexivity A relation X⊆𝐴 × 𝐴 is reflexive if ∀𝑎∈𝐴 𝑎, 𝑎 ∈𝑋 Examples: ∀𝑎∈𝐴 𝑎, 𝑎 ∈𝑋 Examples: (≤,ℕ×ℕ) is reflexive, since ∀𝑛∈ℕ 𝑛≤𝑛 (<,ℕ×ℕ) is not reflexive, since ∼ ∀𝑛∈ℕ 𝑛<𝑛 (in fact, there is no such 𝑛) (𝐽𝑎𝑠𝑜𝑛,ℕ×ℕ) defined as { 𝑥, 𝑦 | 𝑥+𝑦≥100} is not reflexive (e.g 10, 10 ∉𝐽𝑎𝑠𝑜𝑛)) Say verbally that one needs only one counter-example (a forall statement can be proven negative with just one witness)

Symmetry A relation X⊆𝐴 × 𝐴 is symmetric if ∀ 𝑎 1 , 𝑎 2 ∈𝐴 𝑎 1 , 𝑎 2 ∈𝑋 ⇒ 𝑎 2 , 𝑎 1 ∈𝑋)

Symmetry A relation X⊆𝐴 × 𝐴 is symmetric if ∀ 𝑎 1 , 𝑎 2 ∈𝐴 𝑎 1 , 𝑎 2 ∈𝑋 ⇒ 𝑎 2 , 𝑎 1 ∈𝑋) Examples: (≤,ℕ×ℕ) is not symmetric since 4≤5 but ∼(5≤4) (<,ℕ×ℕ) is not symmetric (see above) (𝐽𝑎𝑠𝑜𝑛,ℕ×ℕ) defined as { 𝑥, 𝑦 | 𝑥+𝑦≥100} is symmetric since 𝑥+𝑦≤100 ⇒(𝑦+𝑥≤100)

∀ 𝑎 1 , 𝑎 2 , 𝑎 3 ∈𝐴 𝑎 1 , 𝑎 2 ∈𝑋 ∧( 𝑎 2 , 𝑎 3 ∈𝑋)⇒ 𝑎 1 , 𝑎 3 ∈𝑋) Transitivity A relation X⊆𝐴 × 𝐴 is transitive if ∀ 𝑎 1 , 𝑎 2 , 𝑎 3 ∈𝐴 𝑎 1 , 𝑎 2 ∈𝑋 ∧( 𝑎 2 , 𝑎 3 ∈𝑋)⇒ 𝑎 1 , 𝑎 3 ∈𝑋)

∀ 𝑎 1 , 𝑎 2 , 𝑎 3 ∈𝐴 𝑎 1 , 𝑎 2 ∈𝑋 ∧( 𝑎 2 , 𝑎 3 ∈𝑋)⇒ 𝑎 1 , 𝑎 3 ∈𝑋) Transitivity A relation X⊆𝐴 × 𝐴 is transitive if ∀ 𝑎 1 , 𝑎 2 , 𝑎 3 ∈𝐴 𝑎 1 , 𝑎 2 ∈𝑋 ∧( 𝑎 2 , 𝑎 3 ∈𝑋)⇒ 𝑎 1 , 𝑎 3 ∈𝑋) (≤,ℕ×ℕ) is transitive since 𝑥≤𝑦 ∧ 𝑦≤𝑧 ⇒ 𝑥≤𝑧 (<,ℕ×ℕ) is transitive (see above) (𝐽𝑎𝑠𝑜𝑛,ℕ×ℕ) defined as { 𝑥, 𝑦 | 𝑥+𝑦≥100} ???

Transitivity A relation X⊆𝐴 × 𝐴 is transitive if ∀ 𝑎 1 , 𝑎 2 , 𝑎 3 ∈𝐴 𝑎 1 , 𝑎 2 ∈𝑋 ∧( 𝑎 2 , 𝑎 3 ∈𝑋)⇒ 𝑎 1 , 𝑎 3 ∈𝑋) (≤,ℕ×ℕ) is transitive since 𝑥≤𝑦 ∧ 𝑦≤𝑧 ⇒ 𝑥≤𝑧 (<,ℕ×ℕ) is transitive (see above) (𝐽𝑎𝑠𝑜𝑛,ℕ×ℕ) defined as { 𝑥, 𝑦 | 𝑥+𝑦≥100} is not transitive since (counter-example): 1, 100 ∈𝐽𝑎𝑠𝑜𝑛∧ 100, 5 ∈𝐽𝑎𝑠𝑜𝑛 but 1, 5 ∉𝐽𝑎𝑠𝑜𝑛