1 Copyright © 2011 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 1 Mathematical Tools and Techniques.

1Copyright © 2011 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 1

Mathematical Tools and

Techniques

22

Logic and Proofs

• Logic involves propositions which have truth values, either true or false– “0 = 1” is a proposition whose value is false– “peanut butter is a source of protein” is true

• A proposition containing a free variable can be true or false, depending on the value of the variable– “x - 1 is prime” is true for x = 8, false for x = 10

Introduction to Computation

3

Logic and Proofs (cont’d.)

• Compound propositions are constructed using the

logical connectives ∧, ⋁, ¬, , and ↔

Introduction to Computation 3

4


• The truth value of ¬p is the opposite of the truth value of p

• Truth tables show possible combinations of p and q

p q p ⋀ q p ∨ q p → q p ↔ q

T T T T T T

T F F T F F

F T F T T F

F F F F T T


5


• For a proposition like (p ∨ q) ∧ ¬(p → q), fill in the table in the order shown. Column 3 is the negation of column 2, column 4 is the conjunction of columns 1 and 3

1 4 3 2

p q (p ∨ q) ∧ ¬ (p → q)

T T T F F T

T F T T T F

F T T F F T

F F F F F T


6


• A tautology is a proposition that is always true – Example: p ∨ ¬ p

• A contradiction is a proposition that is always false– Example: p ∧ ¬ p

• Two propositions P and Q are said to be logically equivalent if they always have the same truth value– This is written as P ⇔ Q

• A proposition P is said to logically imply a proposition Q if, whenever P is true, Q is also true – This is written as P ⇒ Q


7


• P → Q and P ⇒ Q look similar; however,– P → Q is a proposition; it has a truth value– P ⇒ Q is a “meta-statement”, an assertion about

the relationship between propositions P and Q– P ⇒ Q means that P → Q is a tautology– Similarly, P ⇔ Q means that P ↔ Q is a tautology


8


• Logical identities can be used to simplify compound propositions:– The commutative laws

p ∨ q q ∨ p p ∧ q q ∧ p

– The associative lawsp ∨ (q ∨ r) (p ∨ q) ∨ r p ∧ (q ∧ r) (p ∧ q) ∧ r

– The distributive lawsp ∨ (q ∧ r) (p ∨ q) ∧ (p ∨ r) p ∧ (q ∨ r) (p ∧ q) ∨ (p ∧ r)


9


• Logical identities (cont’d.)– The De Morgan laws

(p ∨ q) p ∧ q

(p ∧ q) p ∨ q

– An equivalent formulation of the conditional

(p q) (p ∨ q) – The contrapositive of a conditional

(p q) (q p)

– An equivalent formulation of the biconditional

(p q) (p q) ∧ (q p)


10


• The truth value of a proposition like “x - 1 is prime” depends on the value of x– We can use logical quantifiers, (for every) or (for

some) to obtain statements that are no longer statements about specific elements in the domain, but statements about the domain itselfx (x - 1 is prime) x (x - 1 is prime)

– The first proposition is true if x – 1 is prime for every value of x in the domain; the second is true if x – 1 is prime for some (at least one) value of x.


11


• In statements with more than one quantifier, order matters:– x (y (x < y)) and y (x (x < y)) are not logically

equivalent– The first says that for every x there is some larger y

(perhaps depending on x); the second says that there is a single value y that is larger than every x

• Here are two identities involving the negation of a quantified statement:

(x (P(x)) x (P(x))(x (P(x)) x (P(x))


12


• A proof is a series of statements, each of which is derived from: – Initial assumptions– Statements that have been derived previously – Generally accepted facts

• The derivations use principles of logical reasoning


13


• A direct, constructive proof– Prove that for every two integers a and b, if a and b are

odd, then ab is odd– Use the definition of odd to restate this as follows:

If there exist integers i and j so that a = 2i +1 and b = 2j +1, then there exists k such that ab = 2k +1 Proof: ab = (2i + 1)(2j + 1)

= 4ij + 2i + 2j + 1

= 2(2ij + i + j) + 1

If we let k be (2ij + i + j), we have the result we want


14


• An indirect proof – Prove that for every three positive integers i, j, and n, if

ij = n, then i ≤ n or j ≤ n– Prove the contrapositive: assume there exist i, j, and n

such that (i ≤ n j ≤ n)– By De Morgan, this implies (i ≤ n) (j ≤ n), or (i

> n) (j > n)– Therefore i j > n n = n, which implies that i j n– This is a direct proof of the contrapositive statement,

and thus an indirect proof of the original statement


15


• A proof by contradiction that 2 is irrational– Assume that there exist positive integers m and n such

that m/n = 2– Then, by dividing m and n by all common factors, we

get p and q with no common factors such that p/q = 2

– Then all these statements are true: p = q2; p2 = 2q2; p2 is even; p is even; p = 2r; p2 = 4r2; q2 = 2r2; q is even.

– Therefore, p and q have the common factor 2. This is a contradiction, which shows that the assumption is false


16


• A proof by cases– If we can enumerate all of the possible cases, and

prove that the statement is true in each case, then we have proven the statement

– For example, if we want to prove that one proposition P implies another proposition Q, then looking at the truth tables for P and Q gives us one way of enumerating all the possible cases


17

Sets

• A finite set can be described by listing its elementsA = {1, 2, 4, 8}

• Sometimes we use ellipsesB = {0, 3, 6, 9, … } C = {13, 14, 15, … , 71}

E = {0, 3, 5, 6, 8, 9, … } (what set is this? Look 2 slides ahead)

• More generally, we can use a defining propertyB = {x | x is a nonnegative integer multiple of 3}C = {x | x is an integer and 13 ≤ x ≤ 71}B = {3y | y is a nonnegative integer}


18

Sets (cont’d.)

• x A means that x is an element of A– Similarly, x A means that x is not an element of A

• A B means that A is a subset of B– i.e., every element of A is also an element of B

• The empty set is denoted by • The order of elements when we write a set is not

significant: for example, {0, 1} = {1, 0}• Repetition has no effect: {0, 0, 1, 1, 1, 2} = {0, 1, 2}• To show that A = B, we need to show that A B and

that B AIntroduction to Computation 18

19

Sets (cont’d.)

• A few sets will come up frequently:– ℕ is the set of natural numbers, or nonnegative integers– ℤ is the set of all integers– ℝ is the set of all real numbers– ℝ+ is the set of nonnegative real numbers

• Now we can write B and E from above more concisely:

B = {3y | y ℕ}

E = {3i + 5j | i, j ℕ}


20

Sets (cont’d.)

• The union, intersection and difference of two sets are defined as follows:

A ∪ B = {x | x A x B}

A ∩ B = {x | x A x B}

A - B = {x | x A x B}

• Examples:{1, 2, 3, 5} ∪ {2, 4, 6} = {1, 2, 3, 4, 5, 6}

{1, 2, 3, 5} ∩ {2, 4, 6} = {2}

{1, 2, 3, 5} - {2, 4, 6} = {1, 3, 5}


21

Sets (cont’d.)

• The complement of a set– We assume that A is a subset of some universal set U– Then the complement of A, written A’, is U - A– We think of A’ as the set of “everything that is not in A”– What this means, however, can be very different

depending on the choice of U ; for example, what

{0, 1}’ means depends on whether {0,1} is thought of as

a subset of ℕ, ℤ, or ℝ.


22

Sets (cont’d.)

• Many useful set identities are analogous to the logical identities– Union and intersection are commutative and

associative, and distribute across each other

• Two sets are disjoint if their intersection is empty• A collection of sets is pairwise disjoint if every two

distinct sets in the collection are disjoint• A partition of a set S is a collection of pairwise

disjoint sets whose union is S


23

Sets (cont’d.)

• We can describe the union of a set of sets– ∪ {Ai | 0 ≤ i ≤ n} = {x | x Ai for some i with 0 ≤ i ≤ n}• Similarly for intersection:– ∩ {Ai | 0 ≤ i ≤ n} = {x | x Ai for every i with 0 ≤ i ≤ n}• The set of all subsets of a set A is called the power set

of A and is written 2A

– 2A = { X | X A }– Example:

2{a,b,c} = {, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}}– Note that the empty set and the set A itself are in A’s

power set


24

Sets (cont’d.)

• The Cartesian product of two sets A and B, denoted A B, is the set of all ordered pairs with first element from A and second element from B– A B = {(a, b) | a A and b B}– We can generalize this to ordered k-tuples:

A1 A2 … Ak = {(a1, a2, … , ak | ai Ai for each i}


25

Functions and Equivalence Relations

• f : A →B means that f is a function from A to B– To each element of A, one element of B is assigned

• Examples:f : ℕ →R defined by the formula f(x) = xg : 2 ℕ → 2 ℕ defined by g(A) = A ∪ {0}

• A is the domain of the function and B the codomain• f and g are equal if and only if they have the same

domain and codomain and f (x) = g(x) for every x in the domain

• Partial functions from A to B may assign values to only some elements of A


26

Functions and Equivalence Relations (cont’d.)

• The range of a function is the set of elements of the codomain that are actually values of the function– { f (x) | x A} (a subset of the codomain B)


27


• If f : A →B is a bijection then we can define the inverse function f -1 from B to A by these two formulas: for every x A and y B,– f -1( f(x) ) = x– f (f -1(y)) = y

• It is easy to check that f -1 is also a bijection


28


• An n-ary operation on a set A is a function that assigns to every ordered n-tuple of elements of A an element of A

• Unary and binary operations are of most interest– Binary operations on the integers include addition– For every set S, binary operations on 2S include union

and intersection– Unary operations include negation (on the set of

integers, for example) and complementation (on the set 2A)


29


• For a unary or binary operation on a set A– We say that a subset A1 of A is closed under the

operation if the result of applying the operation to elements of A1 is an element of A1

• If A = 2ℕ and A1 is the set of nonempty subsets of A, then A1 is closed under union but not under intersection

• The set of even natural numbers is closed under addition and multiplication

• The set of odd natural numbers is closed under multiplication but not addition


30


• Relations

• We can express relationships several ways: If R is a relation on a set, we can write “a is related to b” as a R b or as (a, b) R


31


• Examples:– The equality relation (a relation on any set)– The relation on A containing all ordered pairs– The relation of congruence mod n on the set ℕ


32


• We may drop the subscript and just say [x] if there is no opportunity for confusion

• Theorem: If R is an equivalence relation on A, the equivalence classes with respect to R form a partition of A, and two elements of A are equivalent if and only if they are elements of the same equivalence class


33

Languages

• An alphabet is a finite set of symbols usually denoted by – Examples: {a,b}, {0,1}, {A, B, C, … , Z}

• A string over is a finite sequence of symbols• |x| stands for the length of the string x

• na(x) is the number of occurrences of a in the string x

• The null string is a string over any alphabet • || = 0


34

Languages (cont’d.)

• The set of all strings over is *• Example: {a,b}* = {, a, b, aa, ab, ba, bb, aaa, aab,…}• A language over is a subset of *• Examples:

– The empty language – {, a, aab}, a finite language– The palindromes over {a,b} (strings like , a, and

baabaab that read the same backwards as forwards)

– {x {a, b}* | na(x) > nb(x)}

– {x {a, b}* | |x| ≥ 2 and x begins and ends with b}Introduction to Computation 34

35


• xy is the concatenation of the two strings x and y; this is the basic operation on strings– If x = ab and y = bab then xy = abbab and yx = babab– For every string x, x = x = x– |xy| = |x| + |y|

• Concatenation is associative, i.e., (xy)z = x(yz), so we can write xyz without worrying about how terms are grouped

• If s = tuv then t is a prefix of s, v is a suffix, and u is a substring. Every string is a prefix (and suffix, and substring) of itself


36


• For languages L1, and L2 over

– L1 ∪ L2, L1 ∩ L2, and L1 − L2 are also languages over

• If L *, the complement of L is a language, * - L

• For languages L1, L2 over

– L1L2 is the language {xy | x L1 and y L2}

• We use exponential notation ak = aaa…a, where there are k occurrences of a

• This also applies to strings (xk = xxx…x) and languages (Lk = LLL…L)

• a0 = x0 = , L0 = {} (for every a , x *, L *)Introduction to Computation 36

37


• If L is a language over , then L* denotes the language of all strings that can be obtained by concatenating zero or more strings in L– This operation is known as the Kleene star

• L* = ∪ {Lk | k ℕ}• L* for every language L, since L0 = {}• Strings are finite, and languages may not be, but to

use a language we need a finite description– L1 = {ab,bab}* ∪ {b} {ba}*{ab}*

– L2 = {x {a,b}* | na(x) > nb(x)}


38

Recursive Definitions

• A recursive definition of a set has a basis statement that specifies at least one member of the set, and a recursive part that specifies how additional members of the set can be generated in terms of given members.

• The prototypical example is ℕ, the set of natural numbers. It can be defined as follows:– Basis statement: 0 ℕ– Recursive part: if n ℕ then n+1 ℕ – Every element of ℕ can be obtained from the first two

statements


39

Recursive Definitions (cont’d.)

• Summary:– The third statement in the definition of ℕ is what says that ℕ

is the smallest set that contains 0 and is closed under the successor operation (addition by 1)

– The statement that the set being defined is the smallest is frequently omitted but always understood

• Example: the subset B = { 2i 5j | i, j ℕ } – 1 B– For every n B, 2n B– For every n B, 5n B


40


• We denote by F the subset of 2{a,b}* defined by: , {}, {a}, {b} F

– if L1, L2 F then L1 ∪ L2 F

– if L1, L2 F then L1L2 F

• F is the smallest set of languages that contains the languages , {}, {a}, and {b} and is closed under union and concatenation

• It is easy to see that F is the set of all finite languages over {a,b}


41


• The set of cities reachable from city s– Suppose that C is a finite set of cities, and the relation

R is defined on C so that c R d means there is a nonstop commercial flight from c to d

– For a city s C, we would like to describe r(s), the set of cities reachable from s by taking zero or more nonstop flights. We can define r(s) this way:

• s r(s)• if c r(s) and c R d then d r(s)


42

Structural Induction

• Consider again the language Expr:– a Expr.– For every x and every y in Expr, x ◦ y and x • y are in Expr

(where x ◦ y and x • y are the strings x+y, x*y)– For every x Expr, ◊(x) Expr (where ◊(x) is (x) )

• Note: this seems like unnecessarily confusing notation (why do we need the operator symbols , •, and ◦ , ◊when we already have the “operators” + and *?)

• The reason is that + and * are operations on numbers, but we’re talking about strings, not numbers. In this discussion +, *, (, and ) are just symbols in the alphabet


43

Structural Induction (cont’d.)

• To prove that every string x Expr satisfies a condition P(x), use structural induction: show that– P(a) is true– For every x and every y in Expr, if P(x) and P(y) are

true, then P(x ◦ y) and P(x • y) are true– For every x Expr, if P(x) is true, then P( (x◊ )) is true

• In other words, show that the set of elements x satisfying the property P contains a and is closed under , •, and ◦ .◊

• This set must then contain every element in the smallest set that contains a and is closed under the operations; i.e., every element in Expr


44


• The recursive definition of Expr consists of – A basis part ( a Expr )– Three recursive parts, the first of the form

If x, y Expr then x ◦ y Expr • The basis step of the proof is to show that P(a) is

true• The induction hypothesis is that x, y Expr and that

P(x) and P(y) are true• The first case of the induction step is to show that

P(x ◦ y) is true. (Similarly for the other 2 operations)Introduction to Computation

45


• For example, suppose P(x) is “x has odd length”• The basis step of the proof is to show that a has odd

length, and this is clearly true: |a| = 1• The induction hypothesis is that x, y Expr and that x

and y have odd length (i.e., |x| and |y| are odd)• The three cases in the induction step are to show that

x ◦ y, x • y, and (◊ x ) have odd length• These are all true, because

| x ◦ y| = | x+y | = |x| + |y| + 1, and odd + odd + 1 = odd | x • y| = | x*y | = |x| + |y| + 1 | (x)| = |(x)| = |x| + 2, and odd + ◊2 = odd


46


• Mathematical induction is simply structural

induction based on the recursive definition of ℕ given earlier

• This is used to prove statements of the form “for every integer n ≥ n0, P(n)”

– Basis step: prove the statement P(n) for n = n0

– Induction hypothesis: k is an integer ≥ n0 and P(k) is true

– Induction step: show using the induction hypothesis that P(k+1) is true


47


• Prove: For every n ℕ, every set A with n elements, 2A has exactly 2n elements– Basis: for every set A with 0 elements, 2A has 20

elements; this is true because only has zero elements, and 2 = {}, which has one element

– Induction hypothesis: k ℕ, and for every set A with k elements, 2A has 2k elements

– Induction step: to show that for every set A with k+1 elements, 2A has 2k+1 elements


48


• Prove: For every n ℕ, and every set A with n elements, 2A has exactly 2n elements (cont’d.)– Proof of induction step

• Let A be a set with k + 1 elements, and let a be any element of A (there is one, since k + 1 ≥ 1)

• Then A - {a} has k elements, and 2A-{a} has 2k elements by the induction hypothesis; therefore, A has 2k subsets that do not contain a and 2k subsets that do contain a, for a total of 2k+1 subsets


49


• Strong induction is another form of induction• Example: To prove that for every n ≥ 2, n is either

prime or a product of two or more primes– Let us strengthen the statement: for n ≥ 2, every

number m such that 2 ≤ m ≤ n is either prime or a product of two or more primes.

– The reason for this, as we’ll see, is that it gives us a stronger induction hypothesis but doesn’t actually require that we prove any more than we would have anyway.


50


• Proof:– Basis step: To show that every m satisfying

2 ≤ m ≤ 2 is either prime or a product of primes. This reduces to showing that 2 is, and 2 is prime

– Induction hypothesis: k ≥ 2, and for every m satisfying 2 ≤ m ≤ k, m is either a prime or a product of primes

– Statement to prove in induction step: For every m satisfying 2 ≤ m ≤ k + 1, m is either prime or a product of primes.


51


• Proof of induction step: For every m with 2 ≤ m ≤ k, we already have the conclusion we want, from the induction hypothesis. The only additional statement we need to prove is that k + 1 is either prime or a product of primes

• If k + 1 is prime, we’re done; if not, k + 1 is the product of two smaller numbers, both bigger than 1. By the (stronger) induction hypothesis, both are prime or the product of primes. Therefore (in either case), k + 1 is the product of primes


52


• The language Balanced was defined as follows: Balanced– If x, y Balanced, then xy Balanced and

(x) Balanced

• Let’s prove that x Balanced if and only if B(x) is true, where B(x) is “x contains equal numbers of left and right parentheses, and no prefix of x contains more right than left”

• The “only if” part is straightforward; use structural induction, based on the definition of Balanced


53


• The basis step is to show that B() is true. This is clear, because has no symbols and no nonnull prefixes

• The induction hypothesis is that x, y Balanced and both B(x) and B(y) are true

• We must show that B(xy) and B( (x) ) are true. Both xy and (x) have equal numbers of left and right parentheses, because x and y do. A prefix of xy is either a prefix of x, or xz for some prefix z of y; a prefix of (x) other than (x) is or (z, for some prefix z of x. In all these cases, the statement is true by the induction hypothesis


54


• For the reverse (the “if” part), use induction on the length of the string

• Prove: For every n ℕ, if x is a string of parentheses so that |x| = n and B(x) is true then x Balanced– Basis: if |x| = 0 then x = , so x Balanced– Induction hypothesis: k ℕ, and for every string x of

parentheses, if |x| ≤ k and B(x), then x Balanced (since we say |x| ≤ k, we’re using strong induction)

– To prove: if |x| = k + 1 and B(x), then x Balanced– Proof: the details get a little involved; see book


55


• Sometimes making a statement stronger makes it easier to prove. Let’s prove that no string in Expr contains ++ as a substring

• The basis step is easy to prove, and so are the induction steps for x*y and (x)

• In the case of x+y, what if x ends with + or y begins with +?

• Strengthen the statement: prove that no string contains ++ as a substring, or begins or ends with +– Now the stronger induction hypothesis makes it

possible to complete the proofIntroduction to Computation 55

56


• Recursive definitions of sets lead naturally to ways of defining functions on those sets

• Example: if we define a function f at 0, and then define f (n+1) assuming that f (n) is defined, then f is

effectively defined over ℕ • The factorial function is defined this way:

f(0) = 1; f(n+1) = (n+1) * f(n)• Definitions like these are particularly well suited to

induction proofs of properties of the corresponding functions


1 Copyright © 2011 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 1 Mathematical Tools and Techniques.

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