Chap 3

– A theorem is a statement that can be shown to be true

– A proof is a sequence of statements to show that a theorem is true

– Axioms: statements which are assumptions, hypotheses or previously proved theorem

– Pales of inference: draw conclusion from other assertions

Chap 3 (cont.)

– Lemma: simple theorems used to prove other theorems

– Corollary: established from a theorem– Modus ponens P P Q ∴ Q

– Table 1: rules of inference

Chap 3 (cont.)

– An argument is valid if whenever all the hypotheses are true, the conclusion is also true– if all propositions used in a valid argument are true,

if leads to a correct conclusion

– “if | o | is divisible by 3, than | 0 |2 is divisible by 9. | o | is divisible by 3. Consequently, | 0 |2 is divisible by 9.” is a valid argument; however, the conclusion is false

– Example 6 & 7

[(PQ) Q] P is not a tautology – fallacy of affirming the conclusion

[(PQ) Q] Q is not a tautology – fallacy of denying the hypothesis

n is an even integer whenever n2 is an even integer suppose n2 is even, then n2 is = 2k For some integer k. Let n=2l for some integer l . This show n is ever .

– fallacy if begging the question Table 2 rule of inference for quantified statements Example 13

direct proof: If n is odd, then n2 is odd n=2k+1, n2 =(2k+1)2 = 4k2 +4h+1 = 2(2k2 +2h)+1

n2 is odd indirect proof: If 3n+2 is odd, n is odd assume n is ever, n=2k 3n+2 = 3(2h)+2 = 2(3h+1) 3n+2 is even

P Q Q P

trivial proof: P(n): “If a and b are positive , a b

then an an bn “,show P(o) is true

If a b, then a0 b0

since 1 1, P(o) is true

- Q is true, then P Q is true

Proof by contradiction: √2 is irrational Let P: √2 is irrational

Suppose that P is true, √2 is rational √2 = a / b, a and b have no common factors 2 = a2 /b2

a 2 is even , a is even , a=2c

2b2=4c2

b2=2c2

b2 is even , b is even contradiction! — PF is true P is false , P is true

Rewrite an indirect proof by a proof by contradiction q p is true then p q is true q is true and show p must also true Suppose p and p are true(proof by contradiction) use

direct proof q p to show p is also true,contradiction

Example 19 : If 3n+2 is odd, n is oddassume 3n+2 is odd and n is evenfor n is even,we show 3n+2 is even,

contradiction!

Proof by cases : If n is an integer not divisible by 3, then n2 1(mod 3)

p: n is not divisible by 3q: n2 1(mod 3)

p is equivalent to p1V p2,p1:n 1,p2 :n 2

[(p1V p2 V. ...pn) q] [(p1 q)(p2 q) … (pn q)](p1V p2) q

p q

(p q) [(p q) (q p) ]

Example 21: n is odd n2 is odd we show pq and q p are true[ p1 p2 … pn ] [ (p1 p2) … (pn-1 pn)

pn p1) ]Constructive existence proof find an element a such that p(a) is true for proving x p(x)Nonconstructive existence proof proof by constructive

Prove xp(x) is false

find an element a such that p(a) is false

xp(x) is true, xp(x) is true, xp(x) is false

counterexample

Example 25

Example 26 (the truth of a statement cannot be

established by one or more examples)

every even positive integer greater than 4 is

the sum of two primes

–Goldbach’s conjecture

–no counterexample has been found

Mathematical Induction

– The sum of the first n positive cold integers,n2?

– P(n) is true for every positive integer n:

Basic step: P(n) is true

Inductive step: P(n) P(n-1) is true for every

positive integer n –[P(1)n(P(n) P(n+1)] n P(n)– Example 2,3,5,6,7,8,11,12

Second Principle of Mathematical Induction

– P(n) is true for every positive integer n: Basic step: P(1) is true Inductive step: P(1) P(2) … P(m) P(m+1) is

true– Example 13 P(n): n can be written as product of primes, n2 Basic step: P(2)

Second Principle of Mathematical Induction, cont.

Inductive step: assume P(k) is true for all positive integers k, kn i ) n+1 is prime ii) n+1 is composite n+1= a*b, 2 ab n+1 by inductive hypothesis, both a and b can be written as product of primes difficult to prove using principle of math. Induction!

Example 14 P(n): postage of n cents can be formed using 4-cent and 5-cent stamps, n12 Basic step: P(n) is true Inductive step: P(n) is true i) one 4-cent stamp is used replace it with a 5-cent stamp ii) no 4-cent stamps were used n 12, at least three 5-cent were used replace three 5-cent with for 4-cent

Basic step: P(12), P(13),P(14) and P(15) are true

Inductive step: n15, k cents can be formed,

12 k n to form n+1, use n-3

cents and 4-cent

Application of Mathematical Induction

An=2n, n=0,1,2,… a0 =1 an+1 =2an, n=0,1,2,…– recessive / inductive definitions

Example 1 f(0)=3 f(n+1)=2f(n)+3

Example 2 F(n)=n! F(0)=1 F(n+1)=(n+1)F(n)

– Some recessive definitions of functions are

based on the second principle of mathematical induction

Example 5 The Fibonacci numbers

f0=0, f1=1

fn=fn-1+fn-2, n=2,3,4…

Example 6 ( use fibonacci numbers to prove )

show fn>n-2 , =(1+√5)/2, n3

P(n): fn > n-2

Basic step: P(3) is true: f3=2 > P(4) is true: f4=3 >(3+√5)/2

=2

Inductive step: assume P(k) is true, 3kn, n4

2 = +1, n-1= 2 × n-3

= × n-3+ n-3

= n-2+ n-3

fn-1>n-3, fn > n-2

∴ fn+1= fn +fn-1 > n-2+ n-3 =n-1

P(n+1) is true

Recursive algorithms

Definition 1 An algorithm is recursive if it solves a problem by reducing it to an instance of the same problem with smaller input Example 1 compute an where a is non ero and n 0 procedure power (a:nonzero, n:nonnegative ) if n=0 than power (a, n):=1 else power (a,n):= a×power(a,n-1)

Example 5 compute n! procedure factorial ( n:positive ) if n=1 than factorial(n) : = 1 else factorial (n) : = n × factorial(n-1)– a corresponding iterative procedure procedure iterative factorial ( n:positive ) x : = 1 for i : =1 to n x : =i × x x is n!

Example 7 found the nth Fabonacci number

procedure fibonacci (n:nonnegative)

if n=0 then fibonacci(0):=0

else if n=0 then fibonacci(1):=1

else fabonacci(n):=fabonacci(n-1)+

f4 fabonacci(n-2)

f3 f2

f2 f1 f1 f0

f1 f0

procedure iterative fibonacci (n: nonnegative) if n=0 than y:= 0 else begin x:=0 y:=1 for i:=1 to n-1 begin z : = x+y x : = y y : = z end end y is the nth fibonacci number

• Require n-1 addition to find fn

• Require far less computation

• A recursive procedure is sometimes preferable – eases to be implemented– Machine designed to handle recursion