Exercises for Discrete Maths
|
|
- Andrew Roderick Day
- 5 years ago
- Views:
Transcription
1 Exercises for Discrete Maths Discrete Maths Rosella Gennari Computer Science Free University of Bozen-Bolzano
2 Disclaimer. The course exercises are meant as complementary material for the students of the course of Discrete Maths and Logic at the Free University of Bozen- Bolzano. The grayed out parts were discussed briefly in class; hence exam exercises are not modeled on them.
3 1 Contents 1. Basic Direct Proofs Basic Notions Divisibility 2 2. Indirect Proofs: Contradiction and Well Ordering for Natural Numbers Prime Numbers The Quotient Remainder Theorem Bases 4 3. Order Relations and Well-founded Induction 6 4. Mathematical and Structural Induction Natural Numbers and Mathematical Induction Invariants and Mathematical Induction Structural Induction Relational Algebras 11 References Basic Notions. 1. Basic Direct Proofs Exercise. If a and b are integers then 2a + 4b + 1 is always odd. Proof. 2a + 4b = 2(a + 2b) = 2c which is, by definition, even. Hence 2c + 1 is, by definition, odd Exercise. The sum of any two even numbers is even. Proof. Such proofs typically start with let m and n be any two objects with a property, in this case being even. What tableau rule does such a statement remind you of? Then such proofs continue with then there exist two integers, n and m with a certain property ; in this case, the property is m = 2m and n = 2n. Is it wise to use m = m or n = n? Why not? What tableaux rule does this remind you of? The proof then continues using the distributivity property of integers: m + n = 2m + 2n = 2(m + n ) and the definition of being even to conclude that m + n is even.
4 Exercise. Do exercises 39 and 40 of [4]. Prove both statements. Proof. Sketch. 39. The error lies in resolving the existentially quantified statements with the same constant k (what does it remind you of?). 40. The error is inverting thesis and hypothesis. A proof of the statement goes as follows, succinctly: let k be any integer greater than 0. Then k 2 + 2k + 1 = (k + 1) 2, which is composite since k Divisibility Exercise. For all integers a, b and c, if a b and a c then a (b + c) Proof. By hypothesis and definition of divisibility, there exist r and s so that ar = b and as = c. Then ar + as = b + c, hence, by distributivity, a(r + s) = b + c, that is, a (b + c) Exercise. Does the vice versa of (1) hold true, that is, if a (b + c) then a b or a c? Proof. The vice-versa is false. To show this, one needs a counter-example. Take a = 2, b = 1 and c = 3. Then a (b + c) and yet a divides neither b nor c. 2. Indirect Proofs: Contradiction and Well Ordering for Natural Numbers In the following section we rely on the well ordering principle for the set N of natural numbers, generalised, in Chapter 5 of [4], to a set S of integers having minimum with respect to the natural order <: Let S be a nonempty subset of N. Then S contains its minimum, that is, there exists an element m of S such that m < b for all natural numbers b of S Prime Numbers Exercise. Every composite natural number n > 1 is divisible by a prime. Proof. Assume not and let n be the minimum composite number that is not divisible by any prime. Now, being composite, n = rs, with 1 < r, s < n. By assumption, r and s cannot be prime and they are both divisible by a prime. Contradiction. For an alternative proof, see Example
5 Exercise. For any natural number N and prime p, if p N then p (N + 1). Proof. Assume not and let p N and p (N + 1). Then there exist r and s with pr = N and ps = N + 1. Then p(s r) = 1, and hence p = 1, which is a contradiction Exercise. There are infinitely many prime natural numbers. Proof. Assume not and let S = {p 1,..., p n } be the finite set of all prime numbers. Consider now N = Π n i=1p i, that is, the product of all the prime numbers in S. Take N + 1. This is divisible by no prime in S by the above Exercise 2.1.2, and hence, by definition of S, it is divisible by no prime. The converse of Exercise implies that N +1 is prime. But N +1 is a prime that does not belong to S, which is defined as the set of all primes. Contradiction The Quotient Remainder Theorem Exercise. Consider the Quotient-Remainder Theorem of [4], p Theorem (Quotient Remainder). For every integer a and integer b > 0, there exist and are unique q and r so that a = q.b + r and 0 r < b. Prove it using the well-ordering principle. Using the proof, define an algorithm for computing the remainder of the division between integers. Proof. Take r 1 = a b 0. If r 1 < b then r = r 1. Else take r 2 = r 1 b 0. If r 2 < b then r = r 2. Repeat until finding r n+1 = r n b 0 with r n+1 < b. The existence of such r n+1 is a consequence of the well-ordering principle applied to the set of natural numbers r n b. Now we have got a b = r 1 with r 1 > b > 0 r 1 b = r 2 with r 2 > b > 0. r n b = r n+1 with b > r n+1 0. giving a (n + 1).b = r n+1, and hence a = (n + 1).b + r n+1, with 0 r 1 < b. Notice that a rigorous proof of this equality requires induction on the property I(n + 1) a = (n + 1).b + r n+1 starting with I(1). Try it yourself as soon as you tackle induction. Uniqueness is proved as follows. If a = q 1 b+r 1 and a = q 2 b+r 2 with 0 r 1, r 2 < b then (q 1 q 2 ).b = (r 2 r 1 ). From this, observe that r 1 = r 2 iff q 1 = q 2, being b > 0. 3
6 4 Let us reason by contradiction and assume that (q 1 q 2 ) 0 (which, given the above observation, is equivalent to (r 1 r 2 ) 0). We now reason by cases. Case 1: (q 1 q 2 ) 1. Take q = (q 1 q 2 ) and r = (r 2 r 1 ). Then qb = r. Since 1 q and b > 0 we have that b q.b. Since r 1 < b and 0 r 2 we have that r < b r 2 < b. The absurdum now follows from b qb = r < b. Case 2: (q 2 q 1 ) 1. Take q = (q 2 q 1 ) and r = (r 1 r 2 ) and reason as in Case 1. As for the algorithm, try it yourself and then check your result against Algorithm 4.8.1, p. 216 of [4]. The Quotient Remainder Theorem allows us to introduce two functions, div b computing the unique quotient and mod b computing the unique rest of the division of any integer by an arbitrarily fixed positive natural number b: for any integer a, a div b is the quotient of the division of a by b; for any integer a, a mod b is the rest of the division of a by b Bases. Consider a natural number b > 1 as a basis (popular bases are 10 and 2) Exercise. Use the Quotient Remainder Theorem to show that, for every natural number a, there exist 1 r 0,..., r k b 1 so that a = r k b k + r k 1 b k r 0 b 0. Proof. The proof is just a sketch and is alternative to the one of Theorem of [4], which is a special case of this exercise. Call the procedure defined in the Quotient Remainder Theorem (QRT, in brief) to compute remainders with mod and quotients with div in the following procedure in pseudo-code: Algorithm 2.1: Representation(a, b) q a div b; r 0 a mod b; i 1; while q 0 do q q div b; r i q mod b; i i + 1; 1 And they are unique.
7 Let q 0 = 0 and q i denote the value of q after the i-th execution of the loop body, with i 1. Observe that q i = q i+1.b + r i+1 (*), with 0 r i+1 < b, and hence q i+1 < q i for all i. The procedure terminates by virtue of the well-ordering principle applied to the set of quotients {q i q i > 0}, with the first q n+1 equal to 0. Given (*) and the fact that q n+1 = 0, we have what follows: a = q 0.b + r 0 = (q 1.b + r 1 ).b + r 0 = q 1.b 2 + r 1.b + r 0 =... = q n.b n+1 + r n.b n + + r 1.b 1 + r 0 = = (q n+1.b + r n+1 ).b n+1 + r n.b n + + r 1.b 1 + r 0 = = r n+1.b n+1 + r n.b n + + r 1.b 1 + r 0 Notice that a rigorous proof of the above rewriting requires induction on the following property, I(i): 5 a = q i.b i+1 + r i.b i + + r 1.b 1 + r 0 ftry it by yourself as soon as you tackle induction Exercise. How many bits (binary digits) are needed to represent a natural number x in binary notation? (Not Part of the Exam) 754 Chapter 11 Analysis of Algorithm Efficiency By definition of the floor function, then, Proof. From Exercise 7 above, we know that log 2 x = k. x = c k.2 k c c 0 b. Recall that the floor of a positive number is its integer part. For instance, 2.82 = 2. Hence property (11.4.2) can be described in words as follows: and hence the number is k + 1. This is equal to the floor of log 2 (x) + 1. The proof of the equality follows from If x is a positive results number in [4], that lies namely: between twotheorem consecutive integer 5.2.3, powers concerning of 2, geometric sequences, seethealso floor of Section the logarithm 4 with below; base 2Example of x is the exponent , of thepp. smaller , power of of 2. which the graphical interpretation is depicted in Fig. 1, and p A graphical interpretation follows: then log 2 x lies in here: k + 1 k y = log 2 x = = 8 2 k 2 k+1 If x lies in here Figure 1. From [4], p. 754 One consequence of property (11.4.2) does not appear particularly interesting in its own right but is frequently needed as a step in the analysis of algorithm efficiency. Example When log 2 (n 1) = log 2 n Prove the following property: For any odd integer n > 1, log 2 (n 1) = log 2 n Solution If n is an odd integer that is greater than 1, then n lies strictly between two successive powers of 2: 2 k < n < 2 k+1 for some integer k > It follows that 2 k n 1because2 k < n and both 2 k and n are integers. Consequently,
8 6 3. Order Relations and Well-founded Induction An order over a set S is a binary relation R over S (that is, R S S) so that: it is irreflexive: for all a S, (a, a) R (an element is never related to itself via the relation); it is antisymmetric: for all a, b S, if (a, b) R then (b, a) R; it is transitive: if (a, b) R and (b, c) R then (a, c) R. Usually, an order relation is denoted by < and, instead of using the set-theoretic notation (a, b) <, we tend to write a < b. Then (S, <) is referred to as ordering. Consider an ordering (S, <). The ordering is total if, for all a and b of S, it is (either) a < b or b < a; else the ordering is partial. An ordering (S, <) is well-founded if the following holds for all non-empty T S: T has a minimal element with respect to <, that is, there exists m T so that, for all n S, if n < m then n T. A well ordering (S, <) is an ordering that is well-founded and total; being the ordering total, an element that is minimal is also the unique minimal element, that is, it is the minimum of S with respect to <. The following theorem, concerning well-founded orderings, goes under the name of well-founded induction: it yields the other forms of induction we will work with, namely, mathematical induction for natural numbers and structural induction for inductively defined sets such as the languages of propositional or first-order logic, or trees. This theorem, alone, could motivate why orderings, and well-founded orderings in particular are important in the study of computer science. Theorem 1 (Well-founded Induction). Consider a well-founded ordering (S, <). Let P be a subset of S, that is, a property of S elements. Assume the following for P and for any y S: for all x S, whenever x < y then x P (i.e., P holds for all x so that x < y); this yields that y P (i.e., P holds for y). Then P = S (that is, P holds for all the elements of S). For a rather comprehensive and yet accessible overview of the topic, see Part I, Chapter 2 of [2]. 4. Mathematical and Structural Induction The following exercises are based on Chapter 5 of [4].
9 4.1. Natural Numbers and Mathematical Induction. Mathematical induction can be derived from well-founded induction once known that (N, <) is a wellfounded ordering. Consider a property P of natural numbers, that is, P N. You suspect that P holds for all natural numbers greater than a given one, say m (this is fixed and depends on P, e.g., it can be 0). A proof by mathematical induction of the statement P holds for all natural numbers greater than or equal to m goes as follows. Idea of a Proof by (Week) Mathematical Induction. Consider a property P N. In order to prove that P = N {i N i < m} (that is, P holds for all natural numbers greater than or equal to m) it is sufficient to prove what follows. induction basis: m P (that is, P holds true for m); induction step: take a generic n N {i N i < m}; as inductive hypothesis, assume that n P (that is, P holds true for n); using the inductive hypothesis, prove that (n + 1) P (that is, P holds true for n + 1). Next, let us see proofs my mathematical induction working with sequences and inequalities. Sequences are functions with domain the set of natural numbers. They can be defined explicitly with a formula, in terms of other known functions, or recursively in terms of themselves. See Section 5.9, p. 290 of [4] Exercise. Reason on how you can explicitly and recursively define a sequence with first terms as follows: (i) with first terms 1, 1, 2, 6, 24, 120; (ii) with first terms 0, 1, 0, 1, 0, 1; (iii) with first terms 0, 2, 4, 6, 8, 10. Proof. (i) Each term s n is equal to n!, and hence a recursive definition of the sequence is s 0 = 1 and, for every n 1, s n = n.s (n 1). (ii) The sequence of terms s n is alternating 0 s and 1 s, returning the remainders of the division of n by 2, and hence a recursive definition of the sequence is s 0 = 0 and, for every n 1, s n = (s (n 1) + 1) mod 2. (iii) The sequence is alternating positive and negative even integers, starting with s 0 = 0 and s 1 = 2, of the form { 2n if n is even s n = 2n otherwise,
10 8 and hence a recursive definition of the sequence, using the sign function sgn defined over integers m as 1 if m < 0 sgn(m) = 0 if m = 0 1 otherwise, is s 0 = 0, s 1 = 2 and, for every n 2, s n = sgn(s n 1 ).( s (n 1) +2). Alternatively, for every n 2, s n = ( 1) n.( s (n 1) +2) Exercise. Reason on how you can recursively define a sequence with first terms as follows: (i) with first terms 1, 1, 2, 3, 5, 8, 13; (ii) with first terms 0, 1, 1, 2, 2, 3, 3, 4. Proof. (i) The given terms can be from the Fibonacci sequence for rabbits reproduction: s 0 = 1 and s 1 = 1; for every n 2, s n = s n 2 + s n 1 ; (ii) The first two terms are 0 and 1; the subsequent terms are obtained from 0 and 1 by recursively adding 1. A recursive definition of a sequence with those terms goes as follows: s 0 = 0 and s 1 = 1; for every n 2, s n+1 = s n Exercise. Find the explicit definition of the sequence that is recursively defined as follows: a 0 =0 a n+1 = 1 2 a n Use induction to prove that your definition is correct. Proof. We start computing the first 6 values and conjecture that, for each n, the explicit definition could be as follows: b n = n n + 1 In order to prove that, for every natural number n, a n = b n, we conduct a proof by mathematical induction. (Induction basis) The explicit definition gives b 0 = 0 which is equal to a 0. (Induction step) Let us assume that b n = a n and prove that b n+1 = a n+1. Let us 1 reason forward on a n+1. This is equal to 2 a n and hence, by induction hypothesis,. Therefore a n+1 = n+1 which is b n+2 n+1. to 1 2 n n+1
11 Exercise. Tackle Exercises 5.2.5, p. 256, and 5.2.H.33, p. 258, from Chapter 5 of [4] Exercise. Let n P be the following statement ( n(n + 1) n 3 = 2 with n N and n 1. Tackle the following questions. (1) What statement is 1 P? Prove that 1 P, which is the basis step of the induction proof. (2) What statement is (n + 1) P? (3) Assume n P as inductive hypothesis. Using this, prove (n + 1) P. In such a manner, you tackle the induction step. ) Exercise. Let n P be the following statement n < 2 n with n N and n 0. Tackle the following questions. (1) What statement is 0 P? Does 0 P hold true? (2) What statement is (n + 1) P? (3) Assume n P as inductive hypothesis. Using this, prove (n + 1) P. In such a manner, you tackle the induction step. Proof. (1) n P is 1 < 2 1, which is true. This is the basis step of an induction proof of the inequality. (2) (n + 1) P is n + 1 < 2 n+1 (3) Let us assume n P, that is, n < 2 n, and let us prove (n + 1) P, that is, n + 1 < 2 n+1. We reason forward from our assumption: since n < 2 n, then 2.n < 2.2 n = 2 n+1 ( ). For all n 1, n + 1 n + n = 2.n ( ). Now, ( ) and ( ) yield that n + 1 < 2 n+1, that is, (n + 1) P holds true Invariants and Mathematical Induction. One of the most important applications of induction is proving that a program or process preserves certain properties from state to state: this property is an invariant. See Section 5.5 of [4] for a definition of (strong) program correctness and of loop invariants.
12 Exercise. Consider a robot that can move only diagonally along an N N grid. More precisely: (0) the robot starts at (0, 0); (I) at step (m, n), the robot moves to (m + k, n + j) where k, j {1, 1}. Can the robot ever reach position (0, 1)? (Taken from [1]) Proof. The answer is negative. This is obtained by proving that the property the sum of the robot coordinates is even is an invariant of the robot moves Exercise. Prove that property I(i) in the proof of Exercise is a loop invariant: for each iteration of the loop, if the predicate is true before the iteration, then it is true after the iteration. Proof. The proof requires induction. Start with proving I(0), that is, that the claimed property holds before the first iteration of the loop. Assume that I(n) holds true at the start of the (n + 1)-th iteration, and that the guard, that is, the while condition, is true: prove that I(n + 1) is true at the end of the iteration Structural Induction. Consider a set U, a subset B of U and functions f 1,..., f n defined from U to U. A subset S of U is defined by structural induction (or generated) from B and the above functions if: (1) P S; (2) for any function f i and s 1,..., s m of S to which f i is applicable, f i (s 1,..., s m ) is in S; (3) nothing else belongs to S. An example of a set defined by structural induction is the set of formulas over a set P of propositions with the connectives defined as functions. For others, see Chapter 5 of [4]. Another is the set of (syntactically correct) RA expressions or queries (see slides of the last set of slides of the course teacher). If S is a set defined by structural induction, then one can prove properties of S as follows. We limit our exposition to the case in which S is defined from P and two functions, f that is binary and g that is unary, for clarity sake. Idea of a Proof by Structural Induction. Consider a set S that is defined by structural induction from a set P, a binary function f and a unary function g. Consider a property P r S. In order to prove that P r = S (that is, P r holds for all elements of S) it is sufficient to prove what follows: induction basis: P P r (that is, P r holds true for all elements of P ); induction step: take generic elements s 1, s 2 and s of S; as inductive hypothesis, assume that s 1 P r, s 2 P r and s P r (that is, P r holds true for s 1, s 2 and s); using the inductive hypothesis, prove that f(s 1, s 2 ) P r and g(s) P r (that is, P r holds true for f(s 1, s 2 ) and g(s)). The fact that a proof by structural induction works for proving properties concerning all elements of S can be derived from well-founded induction, providing S with a suitable order, or directly from the structural induction definition of S.
13 11 A B C C D Table 1. Relations R (to the left) and S (to the right) Next, let us see how to conduct structural induction proofs with inductively defined sets Exercise. Let P be a non-empty set of propositions and L(P ) be the propositional language over P. Using structural induction, prove the following property concerning interpretations over L(P ): if I 1 (p) = I 2 (p) for all p P, then I 1 = φ iff I 2 = φ for all φ L(P ). (This justifies the fact that for a formula with n propositions you only need to consider 2 n interpretations or, differently stated, truth-table rows.) Proof. We reason by structural induction on L(P ). (Induction basis) The basis case follows from the if-condition. (Induction step) We consider three cases (the others are similarly dealt with): φ = ψ, assuming that the property holds for ψ; φ = ψ 1 ψ 2, assuming that the property holds for ψ 1 and ψ 2 ; φ = ψ 1 ψ 2, assuming that the property holds for ψ 1 and ψ 2. Case 1) I 1 = φ = ψ iff I 1 = ψ iff (by induction hypothesis) I 2 = ψ iff I 2 = ψ = φ. Case 2) I 1 = φ = ψ 1 ψ 2 iff I 1 = ψ 1 or I 1 = ψ 2 iff (by induction hypothesis) I 2 = ψ 2 or I 2 = ψ 2 iff I 2 = ψ 1 ψ 2 = φ. Case 3) I 1 = φ = ψ 1 ψ 2 iff I 1 = ψ 1 and I 1 = ψ 2 iff (by induction hypothesis) I 2 = ψ 2 and I 2 = ψ 2 iff I 2 = ψ 1 ψ 2 = φ Relational Algebras. For this part, refer to Sections of [4] and the course slides Exercise. Given the relations R and S in Table 1, compute the following: π A,B (R), π B (S), σ A=1 (R), σ A=1 B=4 (R), R S, σ A=1 B=4 (R S) Exercise. Compute the transitive closure of the binary relations in Fig. 2.
14 1), (1, 2), (2, 2), (3, 3)} 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)} 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), 3)} In Exercises 9 11 determine whether the relation with the e relations on {0, 1, 2, 3} are partial orderne directed graph shown is a partial order. the properties of a partial ordering that a b a b 2), (3, 3)} Exercise. Consider the definition (by structural induction) of I(Q), where Q is a query, as in slides of the course teacher. Prove some of the equivalences 1), (2, 0), (2, 2), (2, 3), (3, 3)} in slides of the course teacher. 1), (1, 2), (2, 2), (3, 1), (3, 3)} a b a b ca d b c d 1), (1, 2), (1, 3), (2, 0), (2, 2), (2, 3), 3)} 11. a b 1), (0, 2), (0, 3), (1, c 0), (1, 1), d(1, 2), c d c d c d 0), (2, 2), (3, 3)} 12. Let (S, R) be a poset. Show that (S,R set if S is the set of all people in the world 1 ) is also a poset, Figure 2. where Digraphs R, where a and b are people, if 1 of isrelations the inverse of R. The poset (S,R 1 ) is called the dual of (S, R). han b? 13. References Find the duals of these posets. ler than b? a) ({0, 1, 2}, ) b) (Z, ) [1] Lehman, E., Leighton, F. T., and Meyer, A. R. Mathematics for Computer is an ancestor of b? c) (P (Z), ) d) (Z + Science. Online, 2010., ) [2] Michal Walicki. Introduction to ve a common friend? 14. Mathematical Which of Logic. these World pairs Scientific, of elements preprint are comparable from http: in the // poset (Z set if S is the set of all people in the world +, )? [3] Rosen, K. Discrete and Its Applications. McGraw Hill, 2011., where a and b are [4] people, SusannaifEpp. Discrete Mathematicsa) with5, Applications. 15 b) BROOKS/COLE, 6, 9 c) 8, d) 7, 7 rter than b? ore than b? is a descendant of b? not have a common friend? e are posets? b) (Z, =) c) (Z, ) d) (Z, ) e are posets? b) (R, <) c) (R, ) d) (R, =) ether the relations represented by these ices are partial orders. b) c) Find two incomparable elements in these posets. a) (P({0, 1, 2}), ) b) ({1, 2, 4, 6, 8}, ) 16. Let S ={1, 2, 3, 4}. With respect to the lexicographic order based on the usual less than relation, a) find all pairs in S S less than (2, 3). b) find all pairs in S S greater than (3, 1). c) draw the Hasse diagram of the poset (S S, ). 17. Find the lexicographic ordering of these n-tuples: a) (1, 1, 2), (1, 2, 1) b) (0, 1, 2, 3), (0, 1, 3, 2) c) (1, 0, 1, 0, 1), (0, 1, 1, 1, 0) 18. Find the lexicographic ordering of these strings of lowercase English letters: a) quack, quick, quicksilver, quicksand, quacking b) open, opener, opera, operand, opened c) zoo, zero, zoom, zoology, zoological 19. Find the lexicographic ordering of the bit strings 0, 01, 11, 001, 010, 011, 0001, and 0101 based on the ordering 0 < Draw the Hasse diagram for the greater than or equal to relation on {0, 1, 2, 3, 4, 5}.
Definition: A binary relation R from a set A to a set B is a subset R A B. Example:
Chapter 9 1 Binary Relations Definition: A binary relation R from a set A to a set B is a subset R A B. Example: Let A = {0,1,2} and B = {a,b} {(0, a), (0, b), (1,a), (2, b)} is a relation from A to B.
More informationSection Summary. Relations and Functions Properties of Relations. Combining Relations
Chapter 9 Chapter Summary Relations and Their Properties n-ary Relations and Their Applications (not currently included in overheads) Representing Relations Closures of Relations (not currently included
More information3 The language of proof
3 The language of proof After working through this section, you should be able to: (a) understand what is asserted by various types of mathematical statements, in particular implications and equivalences;
More informationSets and Motivation for Boolean algebra
SET THEORY Basic concepts Notations Subset Algebra of sets The power set Ordered pairs and Cartesian product Relations on sets Types of relations and their properties Relational matrix and the graph of
More informationWith Question/Answer Animations
Chapter 5 With Question/Answer Animations Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Chapter Summary
More informationRelations Graphical View
Introduction Relations Computer Science & Engineering 235: Discrete Mathematics Christopher M. Bourke cbourke@cse.unl.edu Recall that a relation between elements of two sets is a subset of their Cartesian
More informationWe want to show P (n) is true for all integers
Generalized Induction Proof: Let P (n) be the proposition 1 + 2 + 2 2 + + 2 n = 2 n+1 1. We want to show P (n) is true for all integers n 0. Generalized Induction Example: Use generalized induction to
More informationInduction and Recursion
. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. Induction and Recursion
More informationChapter Summary. Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Recursive Algorithms
1 Chapter Summary Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Recursive Algorithms 2 Section 5.1 3 Section Summary Mathematical Induction Examples of
More informationProof Techniques (Review of Math 271)
Chapter 2 Proof Techniques (Review of Math 271) 2.1 Overview This chapter reviews proof techniques that were probably introduced in Math 271 and that may also have been used in a different way in Phil
More informationNotes. Relations. Introduction. Notes. Relations. Notes. Definition. Example. Slides by Christopher M. Bourke Instructor: Berthe Y.
Relations Slides by Christopher M. Bourke Instructor: Berthe Y. Choueiry Spring 2006 Computer Science & Engineering 235 Introduction to Discrete Mathematics Sections 7.1, 7.3 7.5 of Rosen cse235@cse.unl.edu
More informationInduction and recursion. Chapter 5
Induction and recursion Chapter 5 Chapter Summary Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Recursive Algorithms Mathematical Induction Section 5.1
More informationMathematical Induction
Mathematical Induction MAT231 Transition to Higher Mathematics Fall 2014 MAT231 (Transition to Higher Math) Mathematical Induction Fall 2014 1 / 21 Outline 1 Mathematical Induction 2 Strong Mathematical
More informationMAT115A-21 COMPLETE LECTURE NOTES
MAT115A-21 COMPLETE LECTURE NOTES NATHANIEL GALLUP 1. Introduction Number theory begins as the study of the natural numbers the integers N = {1, 2, 3,...}, Z = { 3, 2, 1, 0, 1, 2, 3,...}, and sometimes
More information1. Consider the conditional E = p q r. Use de Morgan s laws to write simplified versions of the following : The negation of E : 5 points
Introduction to Discrete Mathematics 3450:208 Test 1 1. Consider the conditional E = p q r. Use de Morgan s laws to write simplified versions of the following : The negation of E : The inverse of E : The
More informationThe Real Number System
MATH 337 The Real Number System Sets of Numbers Dr. Neal, WKU A set S is a well-defined collection of objects, with well-defined meaning that there is a specific description from which we can tell precisely
More informationPropositional Logic, Predicates, and Equivalence
Chapter 1 Propositional Logic, Predicates, and Equivalence A statement or a proposition is a sentence that is true (T) or false (F) but not both. The symbol denotes not, denotes and, and denotes or. If
More informationMathematical Induction
Mathematical Induction MAT30 Discrete Mathematics Fall 018 MAT30 (Discrete Math) Mathematical Induction Fall 018 1 / 19 Outline 1 Mathematical Induction Strong Mathematical Induction MAT30 (Discrete Math)
More informationDepartment of Computer Science University at Albany, State University of New York Solutions to Sample Discrete Mathematics Examination II (Fall 2007)
Department of Computer Science University at Albany, State University of New York Solutions to Sample Discrete Mathematics Examination II (Fall 2007) Problem 1: Specify two different predicates P (x) and
More informationDiscrete Structures, Final Exam
Discrete Structures, Final Exam Monday, May 11, 2009 SOLUTIONS 1. (40 pts) Short answer. Put your answer in the box. No partial credit. [ ] 0 1 (a) If A = and B = 1 0 [ ] 0 0 1. 0 1 1 [ 0 1 1 0 0 1 ],
More informationMath 324 Summer 2012 Elementary Number Theory Notes on Mathematical Induction
Math 4 Summer 01 Elementary Number Theory Notes on Mathematical Induction Principle of Mathematical Induction Recall the following axiom for the set of integers. Well-Ordering Axiom for the Integers If
More informationComplete Induction and the Well- Ordering Principle
Complete Induction and the Well- Ordering Principle Complete Induction as a Rule of Inference In mathematical proofs, complete induction (PCI) is a rule of inference of the form P (a) P (a + 1) P (b) k
More informationMATH10040: Chapter 0 Mathematics, Logic and Reasoning
MATH10040: Chapter 0 Mathematics, Logic and Reasoning 1. What is Mathematics? There is no definitive answer to this question. 1 Indeed, the answer given by a 21st-century mathematician would differ greatly
More informationMATH 61-02: PRACTICE PROBLEMS FOR FINAL EXAM
MATH 61-02: PRACTICE PROBLEMS FOR FINAL EXAM (FP1) The exclusive or operation, denoted by and sometimes known as XOR, is defined so that P Q is true iff P is true or Q is true, but not both. Prove (through
More informationICS141: Discrete Mathematics for Computer Science I
ICS141: Discrete Mathematics for Computer Science I Dept. Information & Computer Sci., Originals slides by Dr. Baek and Dr. Still, adapted by J. Stelovsky Based on slides Dr. M. P. Frank and Dr. J.L. Gross
More informationRelations. We have seen several types of abstract, mathematical objects, including propositions, predicates, sets, and ordered pairs and tuples.
Relations We have seen several types of abstract, mathematical objects, including propositions, predicates, sets, and ordered pairs and tuples. Relations use ordered tuples to represent relationships among
More informationCSE 311: Foundations of Computing I Autumn 2014 Practice Final: Section X. Closed book, closed notes, no cell phones, no calculators.
CSE 311: Foundations of Computing I Autumn 014 Practice Final: Section X YY ZZ Name: UW ID: Instructions: Closed book, closed notes, no cell phones, no calculators. You have 110 minutes to complete the
More informationCS1800: Strong Induction. Professor Kevin Gold
CS1800: Strong Induction Professor Kevin Gold Mini-Primer/Refresher on Unrelated Topic: Limits This is meant to be a problem about reasoning about quantifiers, with a little practice of other skills, too
More informationCSE 1400 Applied Discrete Mathematics Proofs
CSE 1400 Applied Discrete Mathematics Proofs Department of Computer Sciences College of Engineering Florida Tech Fall 2011 Axioms 1 Logical Axioms 2 Models 2 Number Theory 3 Graph Theory 4 Set Theory 4
More informationCS1800: Mathematical Induction. Professor Kevin Gold
CS1800: Mathematical Induction Professor Kevin Gold Induction: Used to Prove Patterns Just Keep Going For an algorithm, we may want to prove that it just keeps working, no matter how big the input size
More informationSEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION
CHAPTER 5 SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION Copyright Cengage Learning. All rights reserved. SECTION 5.4 Strong Mathematical Induction and the Well-Ordering Principle for the Integers Copyright
More informationREVIEW QUESTIONS. Chapter 1: Foundations: Sets, Logic, and Algorithms
REVIEW QUESTIONS Chapter 1: Foundations: Sets, Logic, and Algorithms 1. Why can t a Venn diagram be used to prove a statement about sets? 2. Suppose S is a set with n elements. Explain why the power set
More informationHomework #2 Solutions Due: September 5, for all n N n 3 = n2 (n + 1) 2 4
Do the following exercises from the text: Chapter (Section 3):, 1, 17(a)-(b), 3 Prove that 1 3 + 3 + + n 3 n (n + 1) for all n N Proof The proof is by induction on n For n N, let S(n) be the statement
More informationAnnouncements. Read Section 2.1 (Sets), 2.2 (Set Operations) and 5.1 (Mathematical Induction) Existence Proofs. Non-constructive
Announcements Homework 2 Due Homework 3 Posted Due next Monday Quiz 2 on Wednesday Read Section 2.1 (Sets), 2.2 (Set Operations) and 5.1 (Mathematical Induction) Exam 1 in two weeks Monday, February 19
More informationDiscrete Mathematics: Lectures 6 and 7 Sets, Relations, Functions and Counting Instructor: Arijit Bishnu Date: August 4 and 6, 2009
Discrete Mathematics: Lectures 6 and 7 Sets, Relations, Functions and Counting Instructor: Arijit Bishnu Date: August 4 and 6, 2009 Our main goal is here is to do counting using functions. For that, we
More information1 Sequences and Summation
1 Sequences and Summation A sequence is a function whose domain is either all the integers between two given integers or all the integers greater than or equal to a given integer. For example, a m, a m+1,...,
More informationPacket #2: Set Theory & Predicate Calculus. Applied Discrete Mathematics
CSC 224/226 Notes Packet #2: Set Theory & Predicate Calculus Barnes Packet #2: Set Theory & Predicate Calculus Applied Discrete Mathematics Table of Contents Full Adder Information Page 1 Predicate Calculus
More informationAutomata Theory and Formal Grammars: Lecture 1
Automata Theory and Formal Grammars: Lecture 1 Sets, Languages, Logic Automata Theory and Formal Grammars: Lecture 1 p.1/72 Sets, Languages, Logic Today Course Overview Administrivia Sets Theory (Review?)
More informationMATH 114 Fall 2004 Solutions to practice problems for Final Exam
MATH 11 Fall 00 Solutions to practice problems for Final Exam Reminder: the final exam is on Monday, December 13 from 11am - 1am. Office hours: Thursday, December 9 from 1-5pm; Friday, December 10 from
More informationFoundations of Mathematics MATH 220 FALL 2017 Lecture Notes
Foundations of Mathematics MATH 220 FALL 2017 Lecture Notes These notes form a brief summary of what has been covered during the lectures. All the definitions must be memorized and understood. Statements
More informationCSC Discrete Math I, Spring Relations
CSC 125 - Discrete Math I, Spring 2017 Relations Binary Relations Definition: A binary relation R from a set A to a set B is a subset of A B Note that a relation is more general than a function Example:
More information1 Predicates and Quantifiers
1 Predicates and Quantifiers We have seen how to represent properties of objects. For example, B(x) may represent that x is a student at Bryn Mawr College. Here B stands for is a student at Bryn Mawr College
More informationNotes for Math 290 using Introduction to Mathematical Proofs by Charles E. Roberts, Jr.
Notes for Math 290 using Introduction to Mathematical Proofs by Charles E. Roberts, Jr. Chapter : Logic Topics:. Statements, Negation, and Compound Statements.2 Truth Tables and Logical Equivalences.3
More informationmeans is a subset of. So we say A B for sets A and B if x A we have x B holds. BY CONTRAST, a S means that a is a member of S.
1 Notation For those unfamiliar, we have := means equal by definition, N := {0, 1,... } or {1, 2,... } depending on context. (i.e. N is the set or collection of counting numbers.) In addition, means for
More informationLecture Notes on DISCRETE MATHEMATICS. Eusebius Doedel
Lecture Notes on DISCRETE MATHEMATICS Eusebius Doedel c Eusebius J. Doedel, 009 Contents Logic. Introduction............................................................................... Basic logical
More informationRelations. Relations of Sets N-ary Relations Relational Databases Binary Relation Properties Equivalence Relations. Reading (Epp s textbook)
Relations Relations of Sets N-ary Relations Relational Databases Binary Relation Properties Equivalence Relations Reading (Epp s textbook) 8.-8.3. Cartesian Products The symbol (a, b) denotes the ordered
More informationWe are going to discuss what it means for a sequence to converge in three stages: First, we define what it means for a sequence to converge to zero
Chapter Limits of Sequences Calculus Student: lim s n = 0 means the s n are getting closer and closer to zero but never gets there. Instructor: ARGHHHHH! Exercise. Think of a better response for the instructor.
More informationLecture Notes 1 Basic Concepts of Mathematics MATH 352
Lecture Notes 1 Basic Concepts of Mathematics MATH 352 Ivan Avramidi New Mexico Institute of Mining and Technology Socorro, NM 87801 June 3, 2004 Author: Ivan Avramidi; File: absmath.tex; Date: June 11,
More informationCM10196 Topic 2: Sets, Predicates, Boolean algebras
CM10196 Topic 2: Sets, Predicates, oolean algebras Guy McCusker 1W2.1 Sets Most of the things mathematicians talk about are built out of sets. The idea of a set is a simple one: a set is just a collection
More informationMath 13, Spring 2013, Lecture B: Midterm
Math 13, Spring 2013, Lecture B: Midterm Name Signature UCI ID # E-mail address Each numbered problem is worth 12 points, for a total of 84 points. Present your work, especially proofs, as clearly as possible.
More informationCHAPTER 4 SOME METHODS OF PROOF
CHAPTER 4 SOME METHODS OF PROOF In all sciences, general theories usually arise from a number of observations. In the experimental sciences, the validity of the theories can only be tested by carefully
More informationTopics in Logic and Proofs
Chapter 2 Topics in Logic and Proofs Some mathematical statements carry a logical value of being true or false, while some do not. For example, the statement 4 + 5 = 9 is true, whereas the statement 2
More informationChapter 5: Sequences, Mathematic Induction, and Recursion
Chapter 5: Sequences, Mathematic Induction, and Recursion Hao Zheng Department of Computer Science and Engineering University of South Florida Tampa, FL 33620 Email: zheng@cse.usf.edu Phone: (813)974-4757
More informationCIS 375 Intro to Discrete Mathematics Exam 3 (Section M001: Green) 6 December Points Possible
Name: CIS 375 Intro to Discrete Mathematics Exam 3 (Section M001: Green) 6 December 2016 Question Points Possible Points Received 1 12 2 14 3 14 4 12 5 16 6 16 7 16 Total 100 Instructions: 1. This exam
More informationCHAPTER 8: EXPLORING R
CHAPTER 8: EXPLORING R LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN In the previous chapter we discussed the need for a complete ordered field. The field Q is not complete, so we constructed
More informationDiscrete Mathematics. Spring 2017
Discrete Mathematics Spring 2017 Previous Lecture Principle of Mathematical Induction Mathematical Induction: Rule of Inference Mathematical Induction: Conjecturing and Proving Mathematical Induction:
More informationWhat can you prove by induction?
MEI CONFERENCE 013 What can you prove by induction? Martyn Parker M.J.Parker@keele.ac.uk Contents Contents iii 1 Splitting Coins.................................................. 1 Convex Polygons................................................
More informationWriting proofs for MATH 51H Section 2: Set theory, proofs of existential statements, proofs of uniqueness statements, proof by cases
Writing proofs for MATH 51H Section 2: Set theory, proofs of existential statements, proofs of uniqueness statements, proof by cases September 22, 2018 Recall from last week that the purpose of a proof
More informationMath 230 Final Exam, Spring 2008
c IIT Dept. Applied Mathematics, May 15, 2008 1 PRINT Last name: Signature: First name: Student ID: Math 230 Final Exam, Spring 2008 Conditions. 2 hours. No book, notes, calculator, cell phones, etc. Part
More informationLogic, Sets, and Proofs
Logic, Sets, and Proofs David A. Cox and Catherine C. McGeoch Amherst College 1 Logic Logical Operators. A logical statement is a mathematical statement that can be assigned a value either true or false.
More information5. Use a truth table to determine whether the two statements are equivalent. Let t be a tautology and c be a contradiction.
Statements Compounds and Truth Tables. Statements, Negations, Compounds, Conjunctions, Disjunctions, Truth Tables, Logical Equivalence, De Morgan s Law, Tautology, Contradictions, Proofs with Logical Equivalent
More informationMATH 363: Discrete Mathematics
MATH 363: Discrete Mathematics Learning Objectives by topic The levels of learning for this class are classified as follows. 1. Basic Knowledge: To recall and memorize - Assess by direct questions. The
More informationDepartment of Computer Science University at Albany, State University of New York Solutions to Sample Discrete Mathematics Examination I (Spring 2008)
Department of Computer Science University at Albany, State University of New York Solutions to Sample Discrete Mathematics Examination I (Spring 2008) Problem 1: Suppose A, B, C and D are arbitrary sets.
More informationn n P} is a bounded subset Proof. Let A be a nonempty subset of Z, bounded above. Define the set
1 Mathematical Induction We assume that the set Z of integers are well defined, and we are familiar with the addition, subtraction, multiplication, and division. In particular, we assume the following
More informationWeek 4-5: Generating Permutations and Combinations
Week 4-5: Generating Permutations and Combinations February 27, 2017 1 Generating Permutations We have learned that there are n! permutations of {1, 2,...,n}. It is important in many instances to generate
More informationNotation Index. gcd(a, b) (greatest common divisor) NT-16
Notation Index (for all) B A (all functions) B A = B A (all functions) SF-18 (n) k (falling factorial) SF-9 a R b (binary relation) C(n,k) = n! k! (n k)! (binomial coefficient) SF-9 n! (n factorial) SF-9
More informationTree sets. Reinhard Diestel
1 Tree sets Reinhard Diestel Abstract We study an abstract notion of tree structure which generalizes treedecompositions of graphs and matroids. Unlike tree-decompositions, which are too closely linked
More informationa + b = b + a and a b = b a. (a + b) + c = a + (b + c) and (a b) c = a (b c). a (b + c) = a b + a c and (a + b) c = a c + b c.
Properties of the Integers The set of all integers is the set and the subset of Z given by Z = {, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, }, N = {0, 1, 2, 3, 4, }, is the set of nonnegative integers (also called
More informationLecture 7 Feb 4, 14. Sections 1.7 and 1.8 Some problems from Sec 1.8
Lecture 7 Feb 4, 14 Sections 1.7 and 1.8 Some problems from Sec 1.8 Section Summary Proof by Cases Existence Proofs Constructive Nonconstructive Disproof by Counterexample Nonexistence Proofs Uniqueness
More informationProperties of the Integers
Properties of the Integers The set of all integers is the set and the subset of Z given by Z = {, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, }, N = {0, 1, 2, 3, 4, }, is the set of nonnegative integers (also called
More informationPreliminaries. Introduction to EF-games. Inexpressivity results for first-order logic. Normal forms for first-order logic
Introduction to EF-games Inexpressivity results for first-order logic Normal forms for first-order logic Algorithms and complexity for specific classes of structures General complexity bounds Preliminaries
More informationIntroduction to Basic Proof Techniques Mathew A. Johnson
Introduction to Basic Proof Techniques Mathew A. Johnson Throughout this class, you will be asked to rigorously prove various mathematical statements. Since there is no prerequisite of a formal proof class,
More informationBackground for Discrete Mathematics
Background for Discrete Mathematics Huck Bennett Northwestern University These notes give a terse summary of basic notation and definitions related to three topics in discrete mathematics: logic, sets,
More informationThe semantics of propositional logic
The semantics of propositional logic Readings: Sections 1.3 and 1.4 of Huth and Ryan. In this module, we will nail down the formal definition of a logical formula, and describe the semantics of propositional
More informationExam 2. Is g one-to-one? Is g onto? Why? Solution: g is not one-to-one, since for c A, g(b) = g(c) = c. g is not onto, since a / g(a).
Discrete Structures: Exam 2 Solutions to Sample Questions, 1. Let A = B = {a, b, c}. Consider the relation g = {(a, b), (b, c), (c, c)}. Is g one-to-one? Is g onto? Why? Solution: g is not one-to-one,
More informationICS141: Discrete Mathematics for Computer Science I
ICS141: Discrete Mathematics for Computer Science I Dept. Information & Computer Sci., Jan Stelovsky based on slides by Dr. Baek and Dr. Still Originals by Dr. M. P. Frank and Dr. J.L. Gross Provided by
More informationPropositional Logic. What is discrete math? Tautology, equivalence, and inference. Applications
What is discrete math? Propositional Logic The real numbers are continuous in the senses that: between any two real numbers there is a real number The integers do not share this property. In this sense
More informationCIS 375 Intro to Discrete Mathematics Exam 3 (Section M004: Blue) 6 December Points Possible
Name: CIS 375 Intro to Discrete Mathematics Exam 3 (Section M004: Blue) 6 December 2016 Question Points Possible Points Received 1 12 2 14 3 14 4 12 5 16 6 16 7 16 Total 100 Instructions: 1. This exam
More informationCMSC 27130: Honors Discrete Mathematics
CMSC 27130: Honors Discrete Mathematics Lectures by Alexander Razborov Notes by Geelon So, Isaac Friend, Warren Mo University of Chicago, Fall 2016 Lecture 1 (Monday, September 26) 1 Mathematical Induction.................................
More informationWriting proofs for MATH 61CM, 61DM Week 1: basic logic, proof by contradiction, proof by induction
Writing proofs for MATH 61CM, 61DM Week 1: basic logic, proof by contradiction, proof by induction written by Sarah Peluse, revised by Evangelie Zachos and Lisa Sauermann September 27, 2016 1 Introduction
More informationChapter 3. Cartesian Products and Relations. 3.1 Cartesian Products
Chapter 3 Cartesian Products and Relations The material in this chapter is the first real encounter with abstraction. Relations are very general thing they are a special type of subset. After introducing
More informationLogic and Proofs. (A brief summary)
Logic and Proofs (A brief summary) Why Study Logic: To learn to prove claims/statements rigorously To be able to judge better the soundness and consistency of (others ) arguments To gain the foundations
More informationReview 3. Andreas Klappenecker
Review 3 Andreas Klappenecker Final Exam Friday, May 4, 2012, starting at 12:30pm, usual classroom Topics Topic Reading Algorithms and their Complexity Chapter 3 Logic and Proofs Chapter 1 Logic and Proofs
More informationCHAPTER 1. Relations. 1. Relations and Their Properties. Discussion
CHAPTER 1 Relations 1. Relations and Their Properties 1.1. Definition of a Relation. Definition 1.1.1. A binary relation from a set A to a set B is a subset R A B. If (a, b) R we say a is Related to b
More informationChapter 2. Mathematical Reasoning. 2.1 Mathematical Models
Contents Mathematical Reasoning 3.1 Mathematical Models........................... 3. Mathematical Proof............................ 4..1 Structure of Proofs........................ 4.. Direct Method..........................
More informationComplexity Theory VU , SS The Polynomial Hierarchy. Reinhard Pichler
Complexity Theory Complexity Theory VU 181.142, SS 2018 6. The Polynomial Hierarchy Reinhard Pichler Institut für Informationssysteme Arbeitsbereich DBAI Technische Universität Wien 15 May, 2018 Reinhard
More informationCSCE 222 Discrete Structures for Computing. Review for the Final. Hyunyoung Lee
CSCE 222 Discrete Structures for Computing Review for the Final! Hyunyoung Lee! 1 Final Exam Section 501 (regular class time 8:00am) Friday, May 8, starting at 1:00pm in our classroom!! Section 502 (regular
More informationOutline. Complexity Theory EXACT TSP. The Class DP. Definition. Problem EXACT TSP. Complexity of EXACT TSP. Proposition VU 181.
Complexity Theory Complexity Theory Outline Complexity Theory VU 181.142, SS 2018 6. The Polynomial Hierarchy Reinhard Pichler Institut für Informationssysteme Arbeitsbereich DBAI Technische Universität
More informationTECHNISCHE UNIVERSITEIT EINDHOVEN Faculteit Wiskunde en Informatica. Final exam Logic & Set Theory (2IT61) (correction model)
TECHNISCHE UNIVERSITEIT EINDHOVEN Faculteit Wiskunde en Informatica Final exam Logic & Set Theory (2IT61) (correction model) Thursday November 4, 2016, 9:00 12:00 hrs. (2) 1. Determine whether the abstract
More informationInformatics 1 - Computation & Logic: Tutorial 3
Informatics 1 - Computation & Logic: Tutorial 3 Counting Week 5: 16-20 October 2016 Please attempt the entire worksheet in advance of the tutorial, and bring all work with you. Tutorials cannot function
More informationTrinity Christian School Curriculum Guide
Course Title: Calculus Grade Taught: Twelfth Grade Credits: 1 credit Trinity Christian School Curriculum Guide A. Course Goals: 1. To provide students with a familiarity with the properties of linear,
More informationTableau-based decision procedures for the logics of subinterval structures over dense orderings
Tableau-based decision procedures for the logics of subinterval structures over dense orderings Davide Bresolin 1, Valentin Goranko 2, Angelo Montanari 3, and Pietro Sala 3 1 Department of Computer Science,
More informationDo not open this exam until you are told to begin. You will have 75 minutes for the exam.
Math 2603 Midterm 1 Spring 2018 Your Name Student ID # Section Do not open this exam until you are told to begin. You will have 75 minutes for the exam. Check that you have a complete exam. There are 5
More informationProofs. Chapter 2 P P Q Q
Chapter Proofs In this chapter we develop three methods for proving a statement. To start let s suppose the statement is of the form P Q or if P, then Q. Direct: This method typically starts with P. Then,
More informationRelationships between elements of sets occur in many contexts. Every day we deal with
C H A P T E R 9 Relations 9.1 Relations and Their Properties 9.2 n-ary Relations and Their Applications 9.3 Representing Relations 9.4 Closures of Relations 9.5 Equivalence Relations 9.6 Partial Orderings
More informationMATH CSE20 Homework 5 Due Monday November 4
MATH CSE20 Homework 5 Due Monday November 4 Assigned reading: NT Section 1 (1) Prove the statement if true, otherwise find a counterexample. (a) For all natural numbers x and y, x + y is odd if one of
More informationNotes on ordinals and cardinals
Notes on ordinals and cardinals Reed Solomon 1 Background Terminology We will use the following notation for the common number systems: N = {0, 1, 2,...} = the natural numbers Z = {..., 2, 1, 0, 1, 2,...}
More informationMath.3336: Discrete Mathematics. Chapter 9 Relations
Math.3336: Discrete Mathematics Chapter 9 Relations Instructor: Dr. Blerina Xhabli Department of Mathematics, University of Houston https://www.math.uh.edu/ blerina Email: blerina@math.uh.edu Fall 2018
More informationCS 173: Induction. Madhusudan Parthasarathy University of Illinois at Urbana-Champaign. February 7, 2016
CS 173: Induction Madhusudan Parthasarathy University of Illinois at Urbana-Champaign 1 Induction February 7, 016 This chapter covers mathematical induction, and is an alternative resource to the one in
More informationMATH 324 Summer 2011 Elementary Number Theory. Notes on Mathematical Induction. Recall the following axiom for the set of integers.
MATH 4 Summer 011 Elementary Number Theory Notes on Mathematical Induction Principle of Mathematical Induction Recall the following axiom for the set of integers. Well-Ordering Axiom for the Integers If
More information