Review I Rosen 1.1-1.5, 3.1 Know your definitions!

Review I

Rosen 1.1-1.5, 3.1

Know your definitions!

Definition 1. Negation of pLet p be a proposition. The statement “It is not the case that p” is also a proposition, called the “negation of p” or ¬p (read “not p”)

Table 1. The Truth Table for theNegation of a Proposition

p ¬p

T FF T

p = The sky is blue.

p = It is not the case that the sky is blue.

p = The sky is not blue.

Definition 2. Conjunction of p and q

Let p and q be propositions. The proposition “p and q,” denoted by pq is true when both p and q are true and is false otherwise. This is called the conjunction of p and q.

Table 2. The Truth Table for the Conjunction of two propositions

p q pq

T T T T F FF T FF F F

Definition 3. Disjunction of p and q

Let p and q be propositions. The proposition “p or q,” denoted by pq, is the proposition that is false when p and q are both false and true otherwise.

Table 3. The Truth Table for the Disjunction of two propositions

p q pq

T T T T F TF T TF F F

Definition 4. Exclusive or of p and q

Let p and q be propositions. The exclusive or of p and q, denoted by pq, is the proposition that is true when exactly one of p and q is true and is false otherwise.

Table 4. The Truth Table for the Exclusive OR of two propositions

p q pq

T T F T F TF T TF F F

Definition 5. Implication pq

Let p and q be propositions. The implication pq is the proposition that is false when p is true and q is false, and true otherwise. In this implication p is called the hypothesis (or antecedent or premise) and q is called the conclusion (or consequence).

Table 5. The Truth Table for the Implication of pq.

p q pq

T T T T F FF T TF F T

Implications

• If p, then q• p implies q• if p,q• p only if q• p is sufficient for q• q if p• q whenever p• q is necessary for p

• Not the same as the if-then construct used in programming languages such as If p then S

ImplicationsHow can both p and q be false, and pq be true?•Think of p as a “contract” and q as its “obligation” that is only carried out if the contract is valid.•Example: “If you make more than $25,000, then you must file a tax return.” This says nothing about someone who makes less than $25,000. So the implication is true no matter what someone making less than $25,000 does.•Another example:

p: Bill Gates is poor.q: Pigs can fly.

pq is always true because Bill Gates is not poor. Another way of saying the implication is“Pigs can fly whenever Bill Gates is poor” which is true since neither p nor q is true.

Related Implications

Converse of p q is q p

Contrapositive of p q is the proposition q p

Definition 6. Biconditional

Let p and q be propositions. The biconditional pq is the proposition that is true when p and q have the same truth values and is false otherwise. “p if and only if q, p is necessary and sufficient for q”

Table 6. The Truth Table for the biconditional pq.

p q pq

T T T T F FF T FF F T

Logical Equivalence• An important technique in proofs is to replace a

statement with another statement that is “logically equivalent.”

• Tautology: compound proposition that is always true regardless of the truth values of the propositions in it.

• Contradiction: Compound proposition that is always false regardless of the truth values of the propositions in it.

Logically Equivalent

• Compound propositions P and Q are logically equivalent if PQ is a tautology. In other words, P and Q have the same truth values for all combinations of truth values of simple propositions.

• This is denoted: PQ (or by P Q)

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Example: DeMorgans

• Prove that (pq) (p q)p q (pq) (pq) p q (p q)

T T

T F

F T

F F

T F F T F

T F F F F

T F T F F F T T T T

List of Logical EquivalencespT p; pF p Identity Laws

pT T; pF F Domination Laws

pp p; pp p Idempotent Laws

(p) p Double Negation Law

pq qp; pq qp Commutative Laws

(pq) r p (qr); (pq) r p (qr) Associative Laws

List of Equivalencesp(qr) (pq)(pr) Distribution Lawsp(qr) (pq)(pr)

(pq)(p q) De Morgan’s Laws(pq)(p q)

Misc. , Table 6p p T Or Tautologyp p F And Contradiction(pq) (p q) Implication Equivalence

pq(pq) (qp) Biconditional Equivalence

Prove: (pq) q pq (pq) q Left-Hand Statement

q (pq) Commutative

(qp) (q q) Distributive

(qp) T Or Tautology (Misc. T6)

qp Identity

pq Commutative

Begin with exactly the left-hand side statementEnd with exactly what is on the right

Justify EVERY step with a logical equivalence

Prove: (pq) q pq (pq) q Left-Hand Statement

q (pq) Commutative(qp) (q q) Distributive

Why did we need this step?

Our logical equivalence specified that is distributive on the right. This does not guarantee distribution on the left!

Ex.: Matrix multiplication is not always commutative

(Note that whether or not is distributive on the left is not the point here.)

Prove or Disprovep q p q ???To prove that something is not true it is enough to

provide one counter-example. (Something that is true must be true in every case.)

p q pq pqF T T FThe statements are not logically equivalent

Method to construct DNF• Construct a truth table for the proposition.

• Use the rows of the truth table where the proposition is True to construct minterms– If the variable is true, use the propositional variable in the minterm

– If a variable is false, use the negation of the variable in the minterm

• Connect the minterms with ’s.

How to find the DNF of (p q)r p q r (p q)r (p q)rT T T T F FT T F T T TT F T T F FT F F T T TF T T T F FF T F T T TF F T F F TF F F F T TThere are five sets of input that make the statement true.

Therefore there are five minterms.

p q r (p q)r (p q)rT T T T F FT T F T T TT F T T F FT F F T T TF T T T F FF T F T T TF F T F F TF F F F T TFrom the truth table we can set up the DNF(p q)r (pqr) (pqr) (pqr) (pqr)

(pqr)

Quantifiers

Universe of Discourse, U: The domain of a variable in a propositional function.

Universal Quantification of P(x) is the proposition:“P(x) is true for all values of x in U.”

Existential Quantification of P(x) is the proposition: “There exists an element, x, in U such that P(x) is true.”

Universal Quantification of P(x)xP(x)“for all x P(x)”“for every x P(x)”Defined as:P(x0) P(x1) P(x2) P(x3) . . . for all xi in U

Example:Let P(x) denote x2 xIf U is x such that 0 < x < 1 then xP(x) is false.If U is x such that 1 < x then xP(x) is true.

Existential Quantification of P(x)xP(x)“there is an x such that P(x)”“there is at least one x such that P(x)”“there exists at least one x such that P(x)”Defined as:P(x0) P(x1) P(x2) P(x3) . . . for all xi in U

Example:Let P(x) denote x2 xIf U is x such that 0 < x 1 then xP(x) is true.If U is x such that x < 1 then xP(x) is true.

Quantifiers

xP(x)•True when P(x) is true for every x.•False if there is an x for which P(x) is false.

xP(x)•True if there exists an x for which P(x) is true.•False if P(x) is false for every x.

Negation (it is not the case)

xP(x) equivalent to xP(x)•True when P(x) is false for every x•False if there is an x for which P(x) is true.

xP(x) is equivalent to xP(x) •True if there exists an x for which P(x) is false.•False if P(x) is true for every x.

Quantification of Two Variables(read left to right)

xyP(x,y) or yxP(x,y)•True when P(x,y) is true for every pair x,y.•False if there is a pair x,y for which P(x,y) is false.

xyP(x,y) or yxP(x,y)True if there is a pair x,y for which P(x,y) is true.False if P(x,y) is false for every pair x,y.

Quantification of Two VariablesxyP(x,y)•True when for every x there is a y for which P(x,y) is true.

(in this case y can depend on x)•False if there is an x such that P(x,y) is false for every y.

yxP(x,y)•True if there is a y for which P(x,y) is true for every x.

(i.e., true for a particular y regardless (or independent) of x)•False if for every y there is an x for which P(x,y) is false.

Note that order matters hereIn particular, if yxP(x,y) is true, then xyP(x,y) is true.However, if xyP(x,y) is true, it is not necessary that yxP(x,y) is true.

Basic Number Theory Definitionsfrom Chapters 1.6, 2

• Z = Set of all Integers

• Z+ = Set of all Positive Integers

• N = Set of Natural Numbers (Z+ and Zero)

• R = Set of Real Numbers

• Addition and multiplication on integers produce integers. (a,b Z) [(a+b) Z] [(ab) Z]

Number Theory Defs (cont.)

• n is even is defined as k Z n = 2k• n is odd is defined as k Z n = 2k+1• x is rational is defined as a,b Z x = a/b, b0• x is irrational is defined as a,b Z x = a/b,

b0 or a,b Z, x a/b, b0• p Z+ is prime means that the only positive

factors of p are p and 1. If p is not prime we say it is composite.

= “such that”

Methods of Proof p q (Example: if n is even, then n2 is even)

• Direct proof: Assume p is true and use a series of previously proven statements to show that q is true.

• Indirect proof: Show q p is true (contrapositive), using any proof technique (usually direct proof).

• Proof by contradiction: Assume negation of what you are trying to prove (pq). Show that this leads to a contradiction.

Direct ProofProve: nZ, Z, if n is even, then n2 is even.Tabular-style proof:n is even hypothesisn=2k for some kZ definition of evenn2 = 4k2 algebran2 = 2(2k2) which is algebra and mult of 2*(an integer) integers gives integersn2 is even definition of even

Same Direct ProofProve: nZ, Z, if n is even, then n2 is even.

Sentence-style proof:

Assume that n is even. Thus, we know that n = 2k for some integer k. It follows that n2 = 4k2 = 2(2k2). Therefore n2 is even since it is 2 times 2k2, which is an integer.

Prove: nZ, Z, if n is even, then n2 is even.

Proof:

Assume that n is even. Thus, we know that n = 2k

for some integer k. It follows that n2 = 4k2 =

2(2k2). Therefore n2 is even since it is 2 times 2k2

which is an integer.

Structure of a Direct Proof

Example of an Indirect ProofProve: If n3 is even, then n is even.

Proof: The contrapositive of “If n3 is even, then n is even” is “If n is odd, then n3 is odd.” If the contrapositive is true then the original statement must be true.

Assume n is odd. Then kZ n = 2k+1. It follows that n3 = (2k+1)3 = 8k3+8k2+4k+1 = 2(4k3+4k2+2k)+1. (4k3+4k2+2k) is an integer. Therefore n3 is 1 plus an even integer. Therefore n3 is odd.

Assumption, Definition, Arithmetic, Conclusion

Discussion of Indirect Proof

Could we do a direct proof of If n3 is even, then n is even?

Assume n3 is even . . . then what?

We don’t have a rule about how to take n3 apart!

Example: Proof by ContradictionProve: The sum of an irrational number and a

rational number is irrational.Proof: Let q be an irrational number and r be a

rational number. Assume that their sum is rational, i.e., q+r=s where s is a rational number. Then q = s-r. But by our previous proof the sum of two rational numbers must be rational, so we have an irrational number on the left equal to a rational number on the right. This is a contradiction. Therefore q+r can’t be rational and must be irrational.

Structure of Proof by Contradiction• Basic idea is to assume that the opposite of what you are

trying to prove is true and show that it results in a violation of one of your initial assumptions.

• In the previous proof we showed that assuming that the sum of a rational number and an irrational number is rational and showed that it resulted in the impossible conclusion that a number could be rational and irrational at the same time. (It can be put in a form that implies n n is true, which is a contradiction.)

Using Cases

Prove: n Z, n3 + n is even.Separate into cases based on whether n is even or odd. Prove each separately using direct proof.Proof: We can divide this problem into two cases. n can be even or n can be odd.

Case 1: n is even. Then kZ n = 2k.n3+n = 8k3 + 2k = 2(4k3+k) which is even since 4k3+k must be an integer.

Cases (cont.)

Case 2: n is odd. Then kZ n = 2k+1.

n3 + n = (8k3 +12k2 + 6k + 1) + (2k + 1) = 2(4k3 + 6k2 + 4k + 1) which is even since 4k3 + 6k2 + 4k + 1 must be an integer.

Therefore n Z, n3 + n is even

Proof?Prove if n3 is even then n is even.Proof: Assume n3 is even. Then kZ n3 = 8k3 for some integer k. It follows

that n = 38k3 = 2k. Therefore n is even.Statement is true but argument is false. Argument assumes that n is even in making the

claim n3=8k3, rather than n3 = 2k. This is circular reasoning.

Prove or Disprove• If m and n are even integers, then mn is divisible by 4.

• The sum of two odd integers is odd.

• The sum of two odd integers is even.

• If n is a positive integer, then n is even iff 3n2+8 is even.

• n2 + n + 1 is a prime number whenever n is a positive integer.

• n2 + n + 1 is a prime number whenever n is a prime number.

• |x| + |y| |x + y| when x,y R. 3 is irrational.