Shawn Zhong

Shawn Zhong

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Shawn Zhong

钟万祥
  • Tutorials
  • Mathematics
    • Math 240
    • Math 375
    • Math 431
    • Math 514
    • Math 521
    • Math 541
    • Math 632
    • Abstract Algebra
    • Linear Algebra
    • Category Theory
  • Computer Sciences
    • CS/ECE 252
    • CS/ECE 352
    • Learn Haskell
  • AP Notes
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Mathematics

Home / Notes / Mathematics / Page 32

1.6 Rules of Inference

  • Feb 06, 2018
  • Shawn
  • Math 240
  • No comments yet
The Socrates Example • We have the two premises: ○ “All men are mortal.” ○ “Socrates is a man.” • And the conclusion: ○ “Socrates is mortal.” • How do we get the conclusion from the premises? The Argument • We can express the premises (above the line) and the conclusion (below the line) in predicate logic as an argument: • We will see shortly that this is a valid argument Arguments in Propositional Logic • An argument in propositional logic is a sequence of propositions. • All but the final proposition are called premises. • The last statement is the conclusion. • The argument is valid if the premises imply the conclusion. • An argument form is an argument that is valid no matter what propositions are substituted into its propositional variables. • If the premises are p_1, p_2, …,p_n and the conclusion is q then ○ (p_1∧p_2∧…∧p_n)→q is a tautology. • Inference rules are all argument simple argument forms that will be used to construct more complex argument forms. Rules of Inference for Propositional Logic: • Modus Ponens ○ Equation ○ Corresponding Tautology: § (p∧(p→q))→q ○ Example: § Let p be “It is snowing.” § Let q be “I will study discrete math. § “If it is snowing, then I will study discrete math.” § “It is snowing.” § “Therefore , I will study discrete math.” • Modus Tollens ○ Equation ○ Corresponding Tautology: § (¬q∧(p→q))→¬p ○ Example: § Let p be “it is snowing.” § Let q be “I will study discrete math.” § “If it is snowing, then I will study discrete math.” § “I will not study discrete math.” § “Therefore , it is not snowing.” • Hypothetical Syllogism ○ Equation ○ Corresponding Tautology: § ((p→q)∧(q→r))→(p→r) ○ Example: § Let p be “it snows.” § Let q be “I will study discrete math.” § Let r be “I will get an A.” § “If it snows, then I will study discrete math.” § “If I study discrete math, I will get an A.” § “Therefore , If it snows, I will get an A.” • Disjunctive Syllogism ○ Equation ○ Corresponding Tautology: § (¬p∧(p ∨q))→q ○ Example: § Let p be “I will study discrete math.” § Let q be “I will study English literature.” § “I will study discrete math or I will study English literature.” § “I will not study discrete math.” § “Therefore , I will study English literature.” • Addition ○ Equation ○ Corresponding Tautology: § p→(p ∨q) ○ Example: § Let p be “I will study discrete math.” § Let q be “I will visit Las Vegas.” § “I will study discrete math.” § “Therefore, I will study discrete math or I will visit Las Vegas.” • Simplification ○ Equation ○ Corresponding Tautology: § (p∧q)→p ○ Example: § Let p be “I will study discrete math.” § Let q be “I will study English literature.” § “I will study discrete math and English literature” § “Therefore, I will study discrete math.” • Conjunction ○ Equation ○ Corresponding Tautology: § ((p)∧(q))→(p∧q) ○ Example: § Let p be “I will study discrete math.” § Let q be “I will study English literature.” § “I will study discrete math.” § “I will study English literature.” § “Therefore, I will study discrete math and I will study English literature.” • Resolution ○ Equation ○ Corresponding Tautology: § ((¬p∨r)∧(p∨q))→(q ∨ r) ○ Example: § Let p be “I will study discrete math.” § Let r be “I will study English literature.” § Let q be “I will study databases.” § “I will not study discrete math or I will study English literature.” § “I will study discrete math or I will study databases.” § “Therefore, I will study databases or I will study English literature.” Using the Rules of Inference to Build Valid Arguments • A valid argument is a sequence of statements. • Each statement is either a premise or follows from previous statements by rules of inference. • The last statement is called conclusion. Valid Arguments Example 1 ○ From the single proposition p∧(p→q) ○ Show that q is a conclusion. Example 2 ○ With these hypotheses: § “It is not sunny this afternoon and it is colder than yesterday.” § “We will go swimming only if it is sunny.” § “If we do not go swimming, then we will take a canoe trip.” § “If we take a canoe trip, then we will be home by sunset.” ○ Using the inference rules, construct a valid argument for the conclusion: § “We will be home by sunset.” ○ Choose propositional variables: § p: “It is sunny this afternoon.” § r: “We will go swimming.” § t: “We will be home by sunset.” § q: “It is colder than yesterday.” § s: “We will take a canoe trip.” ○ Translation into propositional logic: § Hypotheses: ¬p∧q, r→p,¬r→s,s→t § Conclusion: t ○ Argument Handling Quantified Statements • Universal Instantiation (UI) ○ Example: § Our domain consists of all dogs and Fido is a dog. § “All dogs are cuddly.” § “Therefore, Fido is cuddly.” • Universal Generalization (UG) ○ Used often implicitly in Mathematical Proofs. • Existential Instantiation (EI) ○ Example: § “There is someone who got an A in the course.” § “Let’s call her a and say that a got an A” • Existential Generalization (EG) ○ Example: § “Michelle got an A in the class.” § “Therefore, someone got an A in the class.” Using Rules of Inference • Example 1 ○ Using the rules of inference, construct a valid argument to show that § “John Smith has two legs” ○ is a consequence of the premises § “Every man has two legs.” § “John Smith is a man.” ○ Notation and domain § Let M(x) denote “x is a man” § L(x) “x has two legs” § Let John Smith be a member of the domain. ○ Argument • Example 2 ○ Use the rules of inference to construct a valid argument showing that the conclusion § “Someone who passed the first exam has not read the book.” follows from the premises “A student in this class has not read the book.” “Everyone in this class passed the first exam.” ○ Notation § Let C(x) denote “x is in this class.” § B(x) denote “x has read the book.” § P(x) denote “x passed the first exam.” ○ First we translate the premises and conclusion into symbolic form. ○ Argument Returning to the Socrates Example • Premises and conclusion • Argument The Barber Example • Show that from the statements •  Every barber in Jonesville shaves those and only those who don t shave themselves.  and  There is a barber in Jonesville  • We can derive a contradiction ○ ∀x(B(x)→∀y(S(x,y)⟷¬S(y,y))) ○ ∃x B(x) ○ ∀c B(c) ○ B(c)→∀y(S(c,y)⟷¬S(y,y)) ○ ∀y (S(c,y)⟷¬S(y,y)) ○ S(c,c)⟷¬S(c,c) ○ Thus, we have a contradiction Lewis Carroll • The first three are called premises and the third is called the conclusion ○ “All hummingbirds are richly colored.” ○ “No large birds live on honey.” ○ “Birds that do not live on honey are dull in color.” ○ “Hummingbirds are small.” • Notation ○ H(x)≔x is a hummingbird ○ C(x)≔x is richly colored ○ L(x)≔x is large ○ Ho(x)≔x lives on honey • Here is one way to translate these statements to predicate logic ○ ∀x (H(x)→C(x)) ○ ∀x (L(x)→¬Ho(x)) ○ ∀x (¬Ho(x)→¬C(x)) ○ ∀x (H(x)→¬L(x)) • Let c be an arbilirary element of the universe (1) ∀x (H(x)→C(x)) (2) ∀x (L(x)→¬Ho(x)) (3) ∀x (¬Ho(x)→¬C(x)) (4) H(c)→C(c)≡¬H(c)∨C(c) (5) L(c)→¬Ho(c)≡¬L(c)∨¬Ho(c) (6) ¬Ho(c)→¬C(c)≡Ho(c)∨¬C(c) (7) By resolution of (4) and (6), ¬H(c)∨Ho(c) (8) By resolution of (5) and (7), ¬H(c)∨¬L(c)≡H(c)→¬L(c) (9) By (8), ∀x(H(x)→¬L(x))
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1.5 Nested Quantifiers

  • Feb 05, 2018
  • Shawn
  • Math 240
  • No comments yet
Nested Quantifiers • Nested quantifiers are often necessary to express the meaning of sentences in English as well as important concepts in computer science and mathematics. • Example ○ “Every real number has an inverse” is ○ ∀x ∃y(x+y=0) ○ where the domains of x and y are the real numbers. • We can also think of nested propositional functions: ○ ∀x ∃y(x+y=0) can be viewed as ∀x Q(x) ○ where Q(x) is ∃y P(x, y) where P(x, y) is (x+y=0) Thinking of Nested Quantification • Nested Loops • To see if ∀x∀y P(x,y) is true, loop through the values of x: • At each step, loop through the values for y. • If for some pair of x and y, P(x,y) is false, then ∀x∀y P(x,y) is false and both the outer and inner loop terminate. • ∀x∀y P(x,y) is true if the outer loop ends after stepping through each x. • To see if ∀x∃y P(x,y) is true, loop through the values of x: • At each step, loop through the values for y. • The inner loop ends when a pair x and y is found such that P(x, y) is true. • If no y is found such that P(x, y) is true • the outer loop terminates as ∀x∃y P(x,y) has been shown to be false. • ∀x∃y P(x,y) is true if the outer loop ends after stepping through each x. • If the domains of the variables are infinite, • then this process cannot actually be carried out. Order of Quantifiers • Let P(x,y) be the statement “x+y=y+x.” • Assume that U is the real numbers. • Then ∀x∀y P(x,y) and ∀y∀x P(x,y) have the same truth value. • Let Q(x,y) be the statement “x+y=0.” • Assume that U is the real numbers. • Then ∀x∃y Q(x,y) is true, but ∃y∀x Q(x,y) is false. Questions on Order of Quantifiers • Example 1 ○ Let U be the real numbers, ○ Define P(x,y): x∙y=0 ○ ∀x∀yP(x,y)=F ○ ∀x∃yP(x,y)=T ○ ∃x∀yP(x,y)=T ○ ∃x∃yP(x,y)=T Example 2 ○ Let U be the positive real numbers, ○ Define P(x,y): x/y=1 ○ ∀x∀yP(x,y)=F ○ ∀x∃yP(x,y)=T ○ ∃x∀yP(x,y)=F ○ ∃x∃yP(x,y)=T Translating Nested Quantifiers into English • Example 1 ○ Translate the statement ∀x(C(x)∨∃y(C(y)∧F(x, y))) § where C(x) is “x has a computer,” § and F(x,y) is “x and y are friends,” § and the domain for both x and y consists of all students in your school. ○ Solution § Every student in your school has a computer or has a friend who has a computer. • Example 2 ○ Translate the statement § ∃x∀y∀z ((F(x, y)∧ F(x,z)∧(y ≠z))→¬F(y,z)) ○ Solution § There is a student none of whose friends are also friends with each other. • Example 3 ○ Translate the statement ∀x (B(x)∨∃y (B(y)∧S(y,x))) § Where B(x) is
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Math 521 – 2/5

  • Feb 05, 2018
  • Shawn
  • Math 521
  • No comments yet
Theorem 1.21 • Notation ○ For a positive integer n § x^n≔⏟(x⋅x⋅x⋯x)┬(n times) ○ For a negative integer n § x^n≔⏟((1/x)⋅(1/x)⋅(1/x)⋯(1/x) )┬(−n times) • Statement ○ For every real x 0, and positive integer n ○ There is one and only one positive real number y s.t. y^n=x ○ In this case, we write y=x^(1/n) • Intuition ○ Try this for n=2 and x=2, so y=√2 • Proof (Uniqueness) ○ If there were y_1 and y_2 s.t. ○ y_1^n=x,y_2^n=x, but y_1≠y_2 ○ Without loss of generality, assume y_1 y_2 ○ Then y_1^n y_2^n, so they can t both equal x ○ So, there is at most one such y • Lemma ○ If n is a positive integer, then § b^n−a^n=(b−a)(b^(n−1)+ab^(n−2)+…+a^(n−2) b+a^(n−1) ) ○ Moreover, if b a 0, then § b^n−a^n (b−a) ⏟((b^(n−1)+b^(n−1)+…+b^(n−1)+b^(n−1) ) )┬(n terms) § b^n−a^n (b−a)nb^(n−1) § Also, (b^n−a^n)/(nb^(n−1) ) b−a b • Proof (Existence) ○ Let E≔{t∈Rt 0 and t^n x} ○ E is not empty § Let t≔x/(x+1), then 0 t 1 and t x § So, 0 t^n t x § Thus, t∈E § Therefore E is not empty ○ E is bounded above § Let t∈R s.t. t 1+x § Therefore t^n t 1+x x § So t∉E and E is bounded above by 1+x § By least upper bound property, sup⁡E exists § Let y≔sup⁡E ○ We now show that y^n≮x and y^n≯x ○ Assume y^n x § Choose h∈R s.t. § 0 h 1 and h (x−y^n)/(n(y+1)^(n−1) ) § Then hn(y+1)^(n−1) y^n § Use the lemma b^n−a^n (b−a)nb^(n−1) § Set a≔y,b≔y+h § (y+h^n−y^n (y+hy)n(y+h^(n−1) § (y+h^n−y^n hn(y+1)^(n−1) § (y+h^n−y^n y^n § (y+h^n x § Since y+h h and y+h∈E § y is not an upper bound of E § This contradicts y=sup⁡E § Thus, y^n≮x ○ Assume y^n x § Let k≔(y^n−x)/(ny^(n−1) ) 0 § Use the lemma (b^n−a^n)/(nb^(n−1) ) b § Set b≔y,a^n≔a, then k=(y^n−x)/(ny^(n−1) ) y § Thus,0 k y § Let t∈R s.t. t≥y−k, then § y^n−t^n≤y^n−(y−k)^n § Use the lemma b^n−a^n (b−a)nb^(n−1) § Set a≔y,b≔y−k, then § y^n−t^n≤y^n−(y−k)^n kny^(n−1)=y^n−x § Therefore, t^n x § By definition of E={t∈Rt 0 and t^n x} § t∉E and t is greater than everything in E § Also t≥y−k, so y−k is an upper bound for E § But y−k y, which contradicts y=sup⁡E § Thus, y^n≯x ○ Therefore y^n=x • Corollary: If a,b∈R+, and n∈Z+, then a^(1/n)⋅b^(1/n)=(ab)^(1/n) ○ Let α=a^(1/n), β=b^(1/n), then ○ α^n β^n=ab ○ (αβ)^n=ab ○ So αβ=(ab)^(1/n) Complex Numbers ℂ • Definition ○ The complex numbers ℂ can be thought of ○ as numbers of the form a+bi, where a,b∈R, and i^2=−1 • Addition, multiplication, subtraction, and division ○ If a+bi, c+di∈ℂ, then ○ (a+bi)+(c+di)=(a+c)+(b+d)i ○ (a+bi)−(c+di)=(a−c)+(b−d)i ○ (a+bi)⋅(c+di)=(ac−bd)+(ad+bc)i ○ (a+bi)/(c+di)=((a+bi)/(c+di))((c−di)/(c−di))=(a+bi)(c−di)/(c^2+d^2 )
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Math 521 – 2/2

  • Feb 05, 2018
  • Shawn
  • Math 521
  • No comments yet
Proposition 1.18 • Let F be an ordered filed, for x,y,z∈F (1) If x 0 then −x0, and vice versa § x 0 § x+(−x) 0+(−x) § 0 −x∎ (2) If x 0 and yz then xyxz § x 0, z−y 0 § x(z−y) 0 § xz−xy 0 § xyxz∎ (3) If x0 and yz then xy xz § x0 § By (1), −x 0 § By (2), (−x)y(−x)z § 0(−x)(z−y) § By (1), x(z−y)0 § xzxy∎ (4) If x≠0 then x^2 0. In particular 1 0 § If x 0, by (2), x^2 0⋅x=0 § If x0, by (3), x^2 0⋅x=0 § 1=1^2=1×1 0 § So 1 0∎ (5) If 0xy, then 01/y1/x § If y 0, then 1/y⋅y=1 0=0⋅1/y by (4) § So, 1/y must have been positive by (2) § Similarly, 1/x 0 § Therefore (1/x)(1/y) 0 § Multiply both sides of xy by (1/x)(1/y) § We get 1/y1/x § Therefore 01/y1/x∎ Theorem 1.19 • There exists an ordered filed with the least upper bond property called R • Moreover R has Q as a subfield • Proof: See appendix Theorem 1.20 • The Archimedean property of R ○ Given x,y∈R, and x 0 ○ There is a positive integer n such that nx y • Proof: The Archimedean property of R ○ Let A={nx│n is a positive integer} ○ Assume the Archimedean property is false ○ Then A has an upper bound ○ i.e. sup⁡A exists ○ Let α=sup⁡A ○ x 0, so α−xα ○ And α−x is not an upper bound for A ○ By definition of A={nx│n is a positive integer} ○ α−xmx for some positive integer m ○ So, αmx+x=(m+1)x∈A ○ This contradicts α=sup⁡A ○ Therefore the Archimedean property is true • Corollary ○ Given x 0 ○ Let y=1, then ○ ∃n∈Z+ s.t. nx 1 ○ Therefore given x 0, ∃n∈Z+ s.t. 1/nx • Q is dense in R ○ If x,y∈R, and xy, then there exists a p∈Q s.t. xpy ○ We can always find a rational number between two real numbers • Proof: Q is dense in R ○ Let x,y∈R, and xy ○ So y−x 0 ○ By the Archimedean property of R § There exists a positive integer n s.t. § n(y−x) 1 § ⇒ny−nx 1 § ⇒ny nx+1 ○ By the Archimedean property of R again § There are positive integers m_1,m_2 s.t. § m_1 nx, m_2 −nx § i.e. −m_2nxm_1 § So there is an integer m s.t. § −m_2≤m≤m_1 § And more importantly, m−1≤nxm ○ Combining two parts together, we have § nxm≤1+nxny § In particular, nxmny § Since n 0, we can multiply by 1/n and get § 1/n (nx)1/n (m)1/n (ny) § Therefore xqy, where q=m/n∈Q Theorem 1.21 • Notation ○ For a positive integer n § x^n≔⏟(x⋅x⋅x⋯x)┬(n times) ○ For a negative integer n § x^n≔⏟((1/x)⋅(1/x)⋅(1/x)⋯(1/x) )┬(−n times) • Statement ○ For every real x 0, and positive integer n ○ There is one and only one positive real number y s.t. y^n=x ○ In this case, we write y=x^(1/n) • Intuition ○ Try this for n=2 and x=2, so y=√2 • Proof (Uniqueness) ○ If there were y_1 and y_2 s.t. ○ y_1^n=x, y_2^n=x, but y_1≠y_2 ○ Without loss of generality, assume y_1y_2 ○ Then y_1^ny_2^n, so they can t both equal x ○ So, there is at most one such y • Proof (Existence) ○ (To be continued)
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Math 541 – 2/5

  • Feb 05, 2018
  • Shawn
  • Math 541
  • No comments yet
Examples of Groups • Is Z a group under multiplication? ○ No, because there is no inverses for 2 ○ Let x∈Z∖{±1} ○ The multiplicative inverse of x (i.e. 1/x) is not an integer • What about Q,R, and ℂ? ○ No, because 0 still has no multiplicative inverse • Multiplicative group of Q,R,ℂ ○ Let Q∗=Q∖{0} and R∗, ℂ^∗ similarly ○ Then Q∗,R∗,ℂ^∗ are groups with operation given by multiplication ○ We argue this for Q∗; the same proof works for R∗ and ℂ^∗ ○ Check: multiplication is an operation on Q∗ § If a,b∈Q∗, then ab∈Q∗ ○ Associativity § This is clear ○ Identity § 1∈Q∗ ○ Inverses § If a∈Q∗, then 1/a is the inverse of a, and 1/a∈Q∗ • Is Z a group with operation given by subtration? ○ No, because subtraction is not associative ○ (1−2)−3=−4 ○ 1−(2−3)=2 • GL_n (R under matrix multiplication ○ Let n∈Z( 0) ○ GL_n (R≔{invertible n×n matrices with entries in R ○ GL_n (R is a group under matrix multiplication ○ Check: matrix multiplication is an operation on GL_n (R § If A,B∈GL_n (R § AB∈GL_n (R, since (AB)^(−1)=B^(−1) A^(−1) ○ Associativity § This is clear ○ Identity § I_n, the n×n identity matrix ○ Inverses § If A∈GL_n (R, its inverse is A^(−1) ○ Notice: When n 1, the operation in GL_n (R is not commutative Abelian Group • We say a group G is abelian, if ∀a,b∈G, a⋅b=b⋅a Proposition 9 • Statement ○ Let n∈Z( 0), and let a_1,a_2,b_1,b_2∈Z ○ If (a_1 ) ̅=(b_1 ) ̅, and (a_2 ) ̅=(b_2 ) ̅ in Z\/nZ ○ Then (a_1+a_2 ) ̅=(b_1+b_2 ) ̅, and (a_1 a_2 ) ̅=(b_1 b_2 ) ̅ • Proof: (a_1+a_2 ) ̅=(b_1+b_2 ) ̅ ○ Choose c_1,c_2∈Z s.t. c_1 n=a_1−b_1 and c_2 n=a_2−b_2 ○ (c_1+c_2 )n ○ =a_1−b_1+a_2−b_2 ○ =(a_1+a_2 )−(b_1+b_2 ) ○ Thus, ├ n┤|├ ((a_1+a_2 )−(b_1+b_2 ))┤ ○ So, (a_1+a_2 ) ̅=(b_1+b_2 ) ̅ ∎ • Proof: (a_1 a_2 ) ̅=(b_1 b_2 ) ̅ ○ Choose c_1,c_2∈Z s.t. c_1 n=a_1−b_1 and c_2 n=a_2−b_2 ○ a_1 a_2−b_1 b_2 ○ =a_1 a_2+a_1 b_2−a_1 b_2−b_1 b_2 ○ =a_1 (a_2−b_2 )+(a_1−b_1 ) b_2 ○ =a_1 c_2 n+b_2 c_1 n ○ =(a_1 c_2+b_2 c_1 )n ○ Thus, n|(a_1 c_2+b_2 c_1 ) ○ So, (a_1 a_2 ) ̅=(b_1 b_2 ) ̅ ∎ Well-definedness • Example ○ Say we want to "define" a map § f:Z\/2Z→Z § f(a ̅ )=a ○ f is not a function § 1 ̅=3 ̅ in Z\/2Z § But f(1 ̅ )=1≠f(3 ̅ )=3 ○ So we say that f is not well defined • How to check well-definedness ○ To check that a purported function f:A→B is well-defined ○ (i.e. to check f is a function) ○ one needs to check that a=a^′⇒f(a)=f(a′) Corollary 10 (Integers Modulo n) • Statement ○ If n∈Z( 0), Z\/nZ is a group under the operation § Z\/nZ×Z\/nZ→Z\/nZ § (a ̅,b ̅ )↦(a+b) ̅ ○ We will denote this operation by + ○ So a ̅+b ̅=(a+b) ̅ • Proof ○ Well-definedness § By proposition 9, the operation a ̅+b ̅=(a+b) ̅ is well-defined ○ Associative § Associativity is inherited from associativity of addition for Z ○ Identity § The identity is 0 ̅ § ∀a ̅∈Z\/nZ, a ̅+0 ̅=(a+0) ̅=a ̅=(0+a) ̅=0 ̅+a ̅ ○ Inverses § ∀a ̅∈Z\/nZ, the inverse of a ̅ is (−a) ̅ § a ̅+(−a) ̅=(a−a) ̅=0 ̅=(−a+a) ̅=(−a) ̅+a ̅
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