Onetone Blog

Math 541 – 2/5

$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 ̅$

Math 541 – 2/2

$Homework 1 (a) • Let A and B be nonempty sets, and let f:A→B be a function. • Suppose f is injective, prove that f has a left inverse • Since f is injective, ∀b∈im(f),∃!a∈A s.t. f(a)=b • Define g:B→A in the following way ○ Choose a_0∈A ○ If b∈im(f) § Choose a∈A s.t. f(a)=b § Define g(b)=a ○ If b∉im(f) § Define g(b)=a_0 • Check that g is a left inverse ○ If a∈A, (g∘f)(a)=g(f(a))=a ○ Thus, g∘f=id_A Example of The Euclidean Algorithm • Let a=97, b=20 • Use the Euclidean Algorithm to find (a,b) ○ 97=20×4+17 ○ 20=17×1+3 ○ 17=3×5+2 ○ 3=2×1+1 ○ Therefore (a,b)=1 • And then find x,y∈Z s.t. (a,b)=ax+by ○ (a,b)=1 ○ =3−2×1 ○ =3−(17−3×5)×1 ○ =3×6−17×1 ○ =(20−17×1)×6−17 ○ =20×6−17×7 ○ =20×6−(97−20×4)×7 ○ =97×(−7)+20×34 ○ We can take x=−7, y=34 Equivalence Class • Let X be a set, and let ~ be an equivalence relation on X • If x∈X, then the equivalence class represented by x is the set • [x]={x^′∈X│x~x^′ }⊆X Proposition 8 • Statement ○ Let X be a set with equivalence relationship ~ ○ If x,x^′∈X, then [x] and [x′] are either equal or disjoint • Proof ○ Suppose ∃y∈[x]∩[x^′ ] ○ It suffices to show if z∈X, then x~z⟺x^′~z ○ x~z⇒x^′~z § Suppose x~z § ⇒z~x (Symmetry) § ⇒z~y (Transitivity) § ⇒y~z (Symmetry) § ⇒x^′~z (Transitivity) ○ x~z⇐x^′~z § Suppose x′~z § ⇒z~x′ (Symmetry) § ⇒z~y (Transitivity) § ⇒y~z (Symmetry) § ⇒x~z (Transitivity) Integers Modulo n • Let n∈Z( 0) • The relation on Z given by a~b⟺n|(a−b) is an equivalence relation • The set of equivalence classes under ~ is denoted as Z\/nZ • We call this set integers modulo n (or integers mod n) • We can check that there are n elements in Z\/nZ • Use a ̅ to denote the equivalence class in Z\/nZ • Then Z\/nZ={0 ̅,1 ̅,2 ̅,…,(n−1) ̅ } Group • Definition ○ If G is a set equipped with a binary operation § G×G→G § (g,h↦g⋅h ○ that satisfy § Associativity: ∀g,h,k∈G, g⋅(hk)=(g⋅h⋅k § Identity: ∃1∈G s.t. ∀g∈G,1⋅g=g⋅1=g § Inverses: ∀g∈G, ∃g^(−1)∈G s.t. gg^(−1)=g^(−1) g=1 ○ Then we say G is a group under this operation • Z,Q,R,ℂ are groups with operation + ○ If a,b∈Z, then a+b∈Z (Similarly for Q,R,ℂ) ○ + is certainly associative in all 4 sets ○ 0 is the identity in each case ○ If a∈Z (or QRℂ), then the inverse of a is −a$

Math 541 – 1/31

Proposition 6 • Statement ○ If a,b∈Z, then (a,b) exists • Proof ○ By Proposition 5, we may assume b≠0 ○ Choose q,r∈Z s.t. a=bq+r, where 0≤r |b| ○ We argue by induction on r ○ Inductive hypothesis § If a′,b′∈Z s.t. b^′≠0, and a^′=b^′ q^′+r^′, where 0≤r^′ r § Then (a′,b′) exists ○ Base case § Suppose r=0, then a=bq § We have ├ |b|┤|├ a┤ and ├ |b|┤|├ b┤ § If e∈Z s.t. e|a and e|b, then ├ e┤|├ |b|┤ § Therefore (a,b) exists, and is |b| ○ Inductive step § Suppose r 0 § Choose q^′,r^′∈Z s.t. b=q^′ r+r′, where 0≤r^′ r § By induction, (b,r) exists § By Proposition 4, (a,b) exists, and euqals (b,r) The Euclidean Algorithm • Input ○ a,b∈Z with |b|≤|a| • Output ○ (a,b) • Algorithm (0) If b=0, output |a| Else, proceed to step (1) (1) If b≠0, find q,r∈Z s.t. a=bq+r, where 0≤r |b| (2) If r=0, output |b| Otherwise, repeat step (1) with b and r playing the roles of a and b • Note ○ The algorithm terminates ○ Since the remainder decreases at each application of step (1) ○ By Proposition 4, the output will be (a,b) • Example: use the Euclidean Algorithm to compute (4148, 2057) ○ Take a=4148, b=2057 ○ ⏟4148┬a=⏟2057┬b×⏟2┬q+⏟34┬r ○ ⏟2057┬a=⏟34┬b×⏟60┬q+⏟17┬r ○ ⏟34┬a=⏟17┬b×⏟2┬q+⏟0┬r ○ Here r=0, so the algorithm terminates ○ Thus, (4148, 2057)=17 Proposition 7 • Statement ○ If a,b∈Z, then ∃x,y∈Z s.t. (a,b)=ax+by • Note ○ x,y need not to be unique • Proof ○ If a=b=0 § We may take x=y=0 § In fact, any pair of (x,y) works ○ If a=0 or b=0 § Without loss of generality, assume b=0 § Then (a,b)=|a|=±a+b § Take x=±1, y=1 ○ If a≠0 and b≠0 § Without loss of generality, assume |a|≥|b| § Choose q,r∈Z s.t. a=qb+r, where 0≤r |b| § We argue by induction on r § Base case □ Suppose r=0, then (a,b)=|b|=0⋅a+(±1)⋅b □ So take x=0, y=±1 § Inductive step □ Suppose r 0 □ Choose q^′,r^′∈Z s.t. b=q^′ r+r^′, where 0≤r^′ r □ By induction, ∃x^′,y^′∈Z s.t. (b,r)=bx^′+ry^′ □ Thus, by Proposition 4 □ (a,b)=(b,r)=bx^′+(a−bq) y^′=ay^′+b(x^′−qy′) □ Take x=y′ and y=x^′−qy^′ • Example: express (4148, 2057) as 4148x+2057y where x,y∈Z ○ Recall when we computed (4148, 2057), we had § 4148=2057×2+34 § 2057=34×60+17 § 34=17×2+0 ○ Let s now find x,y∈Z s.t. (4148, 2057)=17=4148x+2057y ○ Start with the second to last equation, and "back-fill" § 17=2057−34×60 § =2057−(4148−2×2057)×60 § =4148×(−60)+2057×121 ○ Therefore x=−60, y=121

Math 521 – 1/31

$Upper Bound and Lower Bound • Suppose S is an ordered set and E⊂S • If there exists β∈S such that x≤β, ∀x∈E • We say that x is bounded above and call β an upper bound for E • If there exists β∈S such that x≥β, ∀x∈E • We say that x is bonded below by β, and β is a lower bound for E Least Upper Bound and Greatest Lower Bound • Definition ○ Suppose S is an ordered set and E⊂S is bounded above. ○ Suppose there exists α∈S s.t. § α is an upper bond of E § If γ α, then γ is not an upper bound of E ○ Then we call α the least upper bound (or lub or sup or supremium) of E ○ Suppose there exists α∈S s.t. § α is an lower bond of E § If γ α, then γ is not an lower bound of E ○ Then we call α the greastst lower bound (or glb or inf or infimum) of E • Examples ○ Recall § A={q∈Qq^2 2} has no sup in Q § B={q∈Qq^2 2} has no inf in Q ○ If α=sup⁡E exists, α may or may not be in E § E_1≔{r∈Qr 0} □ inf⁡〖E_1 〗 doesn t exist □ sup⁡〖E_1 〗=0∉E_1 § E_2≔{r∈Qr≤0} □ inf⁡〖E_2 〗 doesn t exist □ sup⁡〖E_2 〗=0∈E_2 § E≔{1/n│n∈N={1, 1/2,1/3,1/4,⋯} □ inf⁡E=0∉E □ sup⁡E=1∈E • Least-upper-bound property ○ We say that a ordered set S has least-upper-bound property provided that ○ if E∈S s.t. E≠∅ and E is bounded above, then sup⁡E exists and sup⁡E∈S Theorem 1.11 • Statement ○ Suppose S is an ordered set with the least-upper-bound property ○ Suppose B⊂S, B≠∅ and B is bounded below ○ Let L be the set of lower bounds of B ○ Then α=sup⁡L exists in S and α=inf⁡B • Proof ○ L≠∅ § B is bounded below, so L is not empty ○ L is bounded above § Given b∈B and l∈L, we have l≤b by definition of L § Therefore, L is bounded above ○ sup⁡L exists in S § L≠∅, L is bounded above § And S has least upper bound property § So sup⁡L exists § Let α=sup⁡L∈S ○ α is a lower bound for B (i.e. α∈L) § If γ α, then γ is not an upper bound for L, so γ∉B § So α≤x for all x∈B § Thus, α is a lower bound for B § i.e. α∈L ○ α=inf⁡B § If β α is another lower bound for B § Then β∉L since α is an upper bound for L § So, α∈L, but β∉L if β α § Therefore α is the least upper bound of B § i.e. α=inf⁡B ○ Therefore α=sup⁡L=inf⁡B∈S Ordered Field • Definition ○ An ordered field is a field F which is also an ordered set, such that § x+y x+z if x,y,z∈F and y z § xy 0 if x,y∈F, x 0 and y 0 ○ If x 0, we call x positive ○ If x 0, we call x negative • Examples ○ N,Z,Q, R • Note ○ R is an ordered field with least-upper-bound property$

1.4 Predicates and Quantifiers

Propositional Logic Not Enough • If we have: ○ “All men are mortal.” ○ “Socrates is a man.” • Does it follow that “Socrates is mortal?” • Can’t be represented in propositional logic. • Need a language that talks about objects, their properties, and their relations. • Later we’ll see how to draw inferences. Introducing Predicate Logic • Predicate logic uses the following new features: ○ Variables: x, y, z ○ Predicates: P(x), M(x) ○ Quantifiers: exists and for all • Propositional functions are a generalization of propositions. ○ They contain variables and a predicate, e.g., P(x) ○ Variables can be replaced by elements from their domain. Propositional Functions • Propositional functions become propositions (and have truth values) when their variables are each replaced by a value from the domain (or bound by a quantifier). • The statement P(x) is said to be the value of the propositional function P at x. • For example, let P(x) denote “x0” and the domain be the integers. Then: ○ P(−3) is false. ○ P(0) is false. ○ P(3) is true. • Often the domain is denoted by U. So in this example U is the integers. Examples of Propositional Functions • Let “x+y=z” be denoted by R(x, y, z) and U be the integers. • Find these truth values: ○ R(2,−1,5) =F ○ R(3,4,7)=T ○ R(x, 3, z)⇒ Not a Proposition • Now let “x is the least number” be denoted by Q(x), with U={0,1,2,3,5}. • Find these truth values: ○ Q(0)=T ○ Q(5)=F ○ Q(6)⇒ undefined • What is Q(0) is U is the integers? Q(0)=F Compound Expressions • Connectives from propositional logic carry over to predicate logic. • If P(x) denotes “x0,” find these truth values: ○ P(3)∨ P(−1)=T∨F=T ○ P(3)∨ P(−1)=T∧F=F ○ P(3)→P(−1)=T→F=F ○ P(3)→¬P(−1)=T→T=T • Expressions with variables are not propositions and therefore do not have truth values. For example, ○ P(3)∧P(y) ○ P(x)→P(y) • When used with quantifiers (to be introduced next), these expressions (propositional functions) become propositions. Quantifiers • We need quantifiers to express the meaning of English words including all and some: ○ “All men are Mortal.” ○ “Some cats do not have fur.” • The two most important quantifiers are: ○ Universal Quantifier, “For all,” symbol: ∀ ○ Existential Quantifier, “There exists,” symbol: ∃ • We write as in ∀x P(x) and ∃x P(x). ○ ∀x P(x) asserts P(x) is true for every x in the domain. ○ ∃x P(x) asserts P(x) is true for some x in the domain. • The quantifiers are said to bind the variable x in these expressions. Universal Quantifier • ∀x P(x) is read as “For all x, P(x)” or “For every x, P(x)” • If P(x) denotes “x0” and U is the integers, then ∀x P(x) is false. • If P(x) denotes “x0” and U is the positive integers, then ∀x P(x) is true. • If P(x) denotes “x is even” and U is the integers, then ∀x P(x) is false. Existential Quantifier • ∃x P(x) is read as “For some x, P(x)”, or as “There is an x such that P(x),” or “For at least one x, P(x).” • If P(x) denotes “x0” and U is the integers, then ∃x P(x) is true. It is also true if U is the positive integers. • If P(x) denotes “x0” and U is the positive integers, then ∃x P(x) is false. • If P(x) denotes “x is even” and U is the integers, then ∃x P(x) is true. Thinking about Quantifiers • When the domain of discourse is finite, we can think of quantification as looping through the elements of the domain. • To evaluate ∀x P(x) loop through all x in the domain. • If at every step P(x) is true, then ∀x P(x) is true. • If at a step P(x) is false, then ∀x P(x) is false and the loop terminates. • To evaluate ∃x P(x) loop through all x in the domain. • If at some step, P(x) is true, then ∃x P(x) is true and the loop terminates. • If the loop ends without finding an x for which P(x) is true, then ∃x P(x) is false. • Even if the domains are infinite, we can still think of the quantifiers this fashion, but the loops will not terminate in some cases. Thinking about Quantifiers as Conjunctions and Disjunctions • If the domain is finite, a universally quantified proposition is equivalent to a conjunction of propositions without quantifiers and an existentially quantified proposition is equivalent to a disjunction of propositions without quantifiers. • If U consists of the integers 1,2, and 3: ○ ∀x P(x)≡P(1)∧P(2)∧P(3) ○ ∃x P(x)≡P(1)∨P(2)∨P(3) • Even if the domains are infinite, you can still think of the quantifiers in this fashion, but the equivalent expressions without quantifiers will be infinitely long. Precedence of Quantifiers • The quantifiers ∀ and ∃ have higher precedence than all the logical operators. • For example, ∀x P(x)∨Q(x) means (∀x P(x))∨Q(x) • ∀x (P(x)∨Q(x)) means something different. Translating from English to Logic • Every student in this class has taken a course in Java. ○ Solution 1 § If U=every student in the class § Let J(x)≔x has taken a course in Java § ∀x J(x) ○ Solution 2 § If U= every student § Let C(x)≔x is a student in the class § Let J(x)≔x has taken a course in Java § Let ∀x (C(x)→J(x)) • Some but not all students in this class has taken a course in Java. ○ Let C(x)≔x is a student in the class ○ Let J(x)≔x has taken a course in Java ○ Some but not all § ∃x J(x)∧¬∀x J(x) § ≡∃x J(x)∧ ∃x ¬J(x) ○ Solution § ∃x(C(x)∧J(x))∧¬∀x(C(x)→J(x)) § ≡∃x(C(x)∧J(x))∧¬∀x (C(x)→J(x)) § ≡∃x(C(x)∧J(x))∧∀x ¬(C(x)→J(x)) § ≡∃x(C(x)∧J(x))∧ ∃x (C(x)∧¬J(x)) Equivalences in Predicate Logic • Statements involving predicates and quantifiers are logically equivalent if and only if they have the same truth value ○ for every predicate substituted into these statements and ○ for every domain of discourse used for the variables in the expressions. • The notation S≡T indicates that S and T are logically equivalent. • Example: ∀x ¬¬S(x) ≡ ∀x S(x) Negating Quantified Expressions • Consider ∀x J(x) ○ “Every student in your class has taken a course in Java.” ○ Here J(x) is “x has taken a course in Java” and ○ the domain is students in your class. ○ Negating the original statement gives “It is not the case that every student in your class has taken Java.” ○ This implies that “There is a student in your class who has not taken Java.” ○ Symbolically ¬∀x J(x) and ∃x ¬J(x) are equivalent • Consider ∃x J(x) ○ “There is a student in this class who has taken a course in Java.” ○ Where J(x) is “x has taken a course in Java.” ○ Negating the original statement gives “It is not the case that there is a student in this class who has taken Java.” ○ This implies that “Every student in this class has not taken Java” ○ Symbolically ¬∃x J(x) and ∀x ¬J(x) are equivalent Equivalent Statements • ∀x(P(x)∧Q(x))≡∀x P(x)∧∀x Q(x) • ∀x(P(x)∨Q(x))≠∀x P(x)∨∀x Q(x) ○ Let U=N={0,1,2,3…} ○ Let P(x)≔x is even ○ Let Q(x)≔x is odd ○ ∀x(P(x)∨Q(x)): every natural number is even or odd ○ ∀x P(x)∨∀x Q(x): every natural number is even or every natural number is odd • ∀x P(x)≡∀z P(z) Lewis Carroll Example • The first two are called premises and the third is called the conclusion. 1. “All lions are fierce.” 2. “Some lions do not drink coffee.” 3. “Some fierce creatures do not drink coffee.” • Define ○ U≔ all creatures ○ L(x)≔x is a lion ○ F(x)≔x is fierce ○ C(x)≔x drinks coffee • Translation ○ ∀x (L(x)→F(x)) ○ ∃x (L(x)∧¬C(x)) ○ ∃x(F(x)∧¬C(x)) Some Predicate Calculus Definitions • An assertion involving predicates and quantifiers is valid if it is true ○ for all domains ○ every propositional function substituted for the predicates in the assertion. • Example: ∀x ¬S(x)⟷¬∃x S(x) • An assertion involving predicates is satisfiable if it is true ○ for some domains ○ some propositional functions that can be substituted for the predicates in the assertion. • Otherwise it is unsatisfiable. • Example: ∀x (F(x)⟷T(x)) not valid but satisfiable • Example: ∀x(F(x)∧¬F(x)) unsatisfiable

Shawn Zhong

Shawn Zhong

Math 541 – 2/5

Math 541 – 2/2

Math 541 – 1/31

Math 521 – 1/31

1.4 Predicates and Quantifiers

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