chap 3 –a theorem is a statement that can be shown to be true –a proof is a sequence of...
TRANSCRIPT
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