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Chap 3

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.).

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Chap 3

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  1. 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

  2. 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

  3. 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

  4. [(PQ)  Q]  P is not a tautology • fallacy of affirming the conclusion  [(PQ)   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

  5. 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

  6. 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

  7. 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! —  PF is true  P is false , P is true

  8. 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 odd assume 3n+2 is odd and n is even for n is even,we show 3n+2 is even, contradiction!

  9. Proof by cases : If n is an integer not divisible by 3, then n2  1(mod 3) p: n is not divisible by 3 q: 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) ]

  10. Example 21: n is odd  n2 is odd we show pq 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

  11. Prove xp(x) is false find an element a such that p(a) is false xp(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)

  12. every even positive integer greater than 4 is the sum of two primes –Goldbach’s conjecture –no counterexample has been found

  13. 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

  14. 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, n2 Basic step: P(2)

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

  16. Example 14 P(n): postage of n cents can be formed using 4-cent and 5-cent stamps, n12 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

  17. Basic step: P(12), P(13),P(14) and P(15) are true Inductive step: n15, k cents can be formed, 12 k n to form n+1, use n-3 cents and 4-cent

  18. 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)

  19. – 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…

  20. Example 6 ( use fibonacci numbers to prove ) show fn>n-2 , =(1+√5)/2, n3 P(n): fn> n-2 Basic step: P(3) is true: f3=2 >  P(4) is true: f4=3 >(3+√5)/2 =2

  21. Inductive step: assume P(k) is true, 3kn, n4 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

  22. Recursive algorithms Definition 1An algorithm is recursive if it solves a problem by reducing it to an instance of the same problem with smaller input Example 1compute 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)

  23. 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!

  24. 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

  25. 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

  26. 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

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