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Mathematical Proof Techniques and Fallacies

This chapter explores the concepts of theorems, proofs, axioms, and fallacies in mathematical arguments. Learn about valid arguments, tautologies, direct and indirect proofs, constructive and nonconstructive proofs, mathematical induction, and examples.

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Mathematical Proof Techniques and Fallacies

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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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