Transcript Chapter 3. Mathematical Reasoning 3.1 Methods of proof • A theorem is a statement that can be shown to be true. • A.

Chapter 3. Mathematical Reasoning
3.1 Methods of proof
• A theorem is a statement that can be shown to be true.
• A proof is to demonstrate that a theorem is true with a sequence of
statements that form an argument.
• An axiom or postulate is the underlying assumption about
mathematical structures, the hypothesis of the theorem to be proved,
and previously proved theorems.
• The rules of inference are the means used to draw conclusion from
other assertions which tie together the steps of a proof.
• A lemma is a simple theorem used in the proof of other theorems.
• A corollary is a proposition that can be established directly from a
theorem that has been proved.
• A conjecture is a statement whose truth is unknown. When its proof
is found, it becomes a theorem.
• Rules of Inference
The rules of inference provide the justification of the steps used to
show that a conclusion follows logically from a set of hypotheses.
• Basis of the rule of inference: the law of detachment
p
pq
q
hypotheses
conclusion
It is thesame as tosay that (p  (p  q))  q is a tautology.
Example 1
It is snowing today.
If it snows today, then we will go skiing.
We will go skiing.
Table 1 Rules of Inference
Rules of Inference
p
pq
pq
p
p
q
pq
p
pq
q
q
pq
 p
pq
qr
pr
pq
p
q
Tautology
Name
p  ( p  q)
Addition
( p  q)  p
Simplification
( p  q)  ( p  q)
Conjunction
( p  ( p  q))  q
Modus ponens
(detachment law)
(q  ( p  q))  p
Modus tollens
Hypothetical
(( p  q)  (q  r ))  ( p  r )
syllogism
(( p  q)  p)  q
Disjunctive
syllogism
Example 2
If n is divisible by 3, thenn 2 is divisible by 9.
n is divisible by 3.
 n 2 is divisible by 9.
Example 3
State which rule of inferenceis thebasis of thefollowing
argument: It is below freezingnow. T herefore,it is either
below freezingor rainingnow.
Solution
p
pq
Example 4
State which rule of inferenceis thebasis of thefollowing
argument: It is below freezingand rainingnow. T herefore,
it is below freezingnow.
Solution
pq
p
Example 5
State which rule of inferenceis thebasis of thefollowing
argument: If it rains, then wewill not havea barbecue today.
If we do not havea barbecue today,then wewill havea
Solution
pq
qr
pr
barbecue tomorrow.T herefore,if it rains today,we will have
a barbecue tomorrow.
Step
p: It is sunny this afternoon.
Example 6
1. p  q
q
Show the following hypotheses :
2. p
(1) It is not sunny this afternoon and it is colder than
3. r  p
yesterday.
r
4. r
(2)We will go swimming only if it is sunny.
s
5.r  s
(3) If we do not go swimming, then we will take a
6. s
canoe trip.
t
7. s  t
(4) If we take a canoe trip, then we will be home by
8. t
sunset.
lead to the conclusion “We will be home by sunset”.
Reason
Hypothesis
Simplification (1)
Hypothesis
Modus tollens(2,3)
Hypothesis
Modus ponens(4,5)
Hypotheis
Modus ponens(6,7)
Example 7
Step
Reason
1. p  q
Hypothesis
Show that the hypotheses
2. q  p Contraposition (1)
q
(1) If you send me an e-mail message,
3. p  r
Hypothsis
Hypothet ic
al syllogism (2,3)
then I will finish writing the program, 4. q  r
r
5. r  s
Hypothsis
(2) If you do not send me an e-mail massage,
6. q  s
Hypothet ic
al syllogism(4,5)
then I will go to sleep early,
s
(3) If I go to sleep early, then I will wake up
feeling refreshed
p
lead to the conclusion “If I do not finish
writing program, then I will wake up
An argumentis called valid if wheneverhypothesesare true,
feeling refreshed”.
thecounclut ion is also true.
Consequently, q logically follows from thehypotheses
p1,p2 ,..., pn is thesame as showing that theimplication
( p1  p2    pn )  q is true.
3.2 Mathematical Induction
How to prove 1+2+…+n=n(n+1)/2 for n=1,2,…?
Use mathematical induction!
The Well-Ordering Property
Every nonempty set of nonnegative integers has a least element.
Mathematical Induction(1)
Mathematical induction is used to prove the form nP(n) is true.
Where the universe of discourse is the set of positive integers.
1. Basis step. The proposition P(1) is shown to be true.
2. Inductive step. The implication
P(k )  P(k  1) is shown tobe turefor everypositiveintegerk.
T heproof techniquecan be statedas [ P(1)  k ( P(k )  P(k  1))]  nP(n).
p
pq
q
Why does Mathematical induction work?
Law of detachment
P (1)
P (1)  P (2)
P(1)
 P ( 2)
P(2)
P(2)  P(3)
 P(3)
P(3)
P(3)  P(4)
 P(4)
P(k  1)
P(k  1)  P(k )
 P( k )
P(n)
Example 1
Use mathematic
al inductionto provethat thesum of thefirst n odd positive
integersis n 2 .
1 = 12. Therefore, P(1) is true.
Inductive step : Assuming P(k) is true, we prove P(k  1 ) is true.
1  3  5    (2k  1)  (2k  1)
Basis step : We prove P(1) is true.
= k2  (2k  1) = ( k  1) 2
Therefore, for any k if P(k) is true then P(k  1 ) is true.
Example 2
P(n)
Use Mathematical induction to prove the inequality
n  2n for all positiveintegersn.
Basis step : P(1 ) is true,since 1  21.
Inductivestep : Assume P(k)is true, thatis, k  2 k . We need to show
thatP(k  1 ) is true.
From k  1  2 k  1  2 k  2 k = 2 k 1 , we get thatP(k  1 ) is true.
Mathematical Induction(2)
Mathematical induction is used to prove the form nP(n) is true.
Where the universe of discourse is the set of contiguous integers:
m,m+1,m+2,….
1. Basis step. The proposition P(m) is shown to be true.
2. Inductive step. The implication
P(k )  P(k  1) is shown tobe truefor everypositiveintegerk  m.
Example 3
P(n)
Use Mathematical induction to prove that
1  2  22  23   2k = 2k 1 1 for all nonnegative integersk.
Basis step : P(0) is truesince 20 = 1 = 21 1.
Inductivestep : Assuming P(k)is true, we proveP(k  1) is true:
1  2  2 2  23    2 k  2 k 1
= (1  2  2 2  23    2 k )  2 k 1
= (2 k 1  1)  2 k 1 = 2  2 k 1  1 = 2 k  2  1.
Example 4 Sums of Geometric Progressions
Use Mathematical induction to prove the following formula:
n 1
n
ar
a
j
2
n
P(n)
ar
=
a

ar

ar


ar
=
,
when
r

1
.

r 1
j =0
ar  a
Basis step : P(0) is true, since a =
r 1
Inductivestep : Assum eP(k)is true.We proveP(k  1 ) is true:
a  ar  ar2   ark  ar k 1
k 1
k 2
ar

a
ar
a
2
k
k 1
k 1
= (a  ar  ar   ar )  ar =
 ar =
r 1
r 1
Example 5 Inequality for Harmonic Numbers.
1 2
1
(Hormonic numbers : H i = 1      , for i = 1,2,3,....)
2 3
i
Use Mathematical induction to prove that H n  1  n / 2.
P(n)
2
Basis step : P(0) is true,since H 20 = H1 = 1  1  0 / 2.
Inductivestep : Assum eP(k)is true.We proveP(k  1 ) is true:
1 1
1
1
1
H 2k 1 = 1      k  k
   k 1
2 3
2
2 1
2
1
1
= H 2k  k
   k 1
2 1
2
k
1
1
 (1 )  k
   k 1 (by theinductivehypothesis)
2 2 1
2
k
1
k
 (1 )  2  k 1 (since thereare termseach not less than1/2k 1 )
2
2
k 1
k 1
 (1 )  = 1 
.
2 2
2
Example 6 The numbers of Subsets of a Finite Set
Use Mathematical induction to prove that
If S is a finiteset withn elements,thenS has 2n subsets.
P(n)
Basis step : P(0) is true,since theemptyset has exactly
20 = 1 subsets, namely,itself.
Inductivestep : Assum eP(k)is true.We provethat P(k  1 ) is true:
Let a be one elementof S and S ' = S-{a}.
T hesubsets of S can be obtainedin thefollowing way :
For each subset X of S', Xand X  {a} are thesubsets of S .
Since thereare 2 k subsets of S , thereforeS has 2  2 k = 2 k 1
subsets.
X  {a}
X
X
3.3 Recursive Definitions
Recursive definitions: defining an object by itself.
For example,an = 2n for n = 0,1,2,...can also be defined as
an1 = 2an for n = 0,1,2,....
Recursively defined functions
To define a function with these of nonnegative integers as its domain,
1.
Specify the value of the function at zero.
2.
Give a rule for finding its value at an integer from its values at
smaller integers.
Such a definition is called a recursive or inductive definition.
Example 1 Function f is recursively
f(1)=2f(0)+3=9,
defined by f(0)=3, f(n+1)=2f(n)+3．
f(2)=2f(1)+3=21,
Find f(1),f(2) and f(3).
f(3)=2f(3)+3=93.
Example 2
Give an inductive definition of the factorial function F(n)=n!.
Solution
F(0)=1
F(n+1)=(n+1)F(n)
Example 3
Give a recursivedefinitionof a n where a is a nonzeroreal number and
n is a nonnegative integer.
Solution
a0 = 1
an 1 = a  an for n = 0,1,2,....
n
Example 4 Give a recursivedefinitionof  ak .
k =0
Solution
0
a
k =0
k
n 1
a
k =0
= a0 ,
n
k
= ( a k )  an 1.
k =0
Example 5
T heFibonaccinumbers, f 0 , f1 , f 2 ...
f 2 = f1  f 0 = 1,
are defined recursively as follows:
f 3 = f 2  f1 = 2,
f 0 = 0,
f1 = 1,
f n = f n 1  f n  2 .
Whatis theFibonaccinumbers f 2 ,f3 ,f4 , and f 5 ?
f 4 = f 3  f 2 = 3,
f 5 = f 4  f 3 = 5.