Chapter 3. Mathematical Reasoning3.1 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

. tautologya is say that toas same theisIt qq))(p(p

conclusionq

qp

p

hypotheses

Example 1

It is snowing today.

If it snows today, then we will go skiing.

We will go skiing.

Rules of Inference Tautology Name

Addition

Simplification

Conjunction

Modus ponens

(detachment law)

Modus tollens

Hypothetical syllogism

Disjunctive syllogism

qp

p

)( qpp

pqp )(p

qp

qp

q

p

q

qp

p

p

qp

q

rp

rq

qp

q

p

qp

)()( qpqp

qqpp ))((

pqpq ))((

)())()(( rprqqp

qpqp ))((

Table 1 Rules of Inference

Example 2

.9by divisible is

.3by divisible is

.9by divisible is then ,3by divisible is If

2

2

n

n

nn

Example 3

now. rainingor freezing below

either isit Therefore, now. freezing below isIt :argument

following theof basis theis inference of rule which State

qp

p

Solution

Example 4

now. freezing below isit

Therefore, now. raining and freezing below isIt :argument

following theof basis theis inference of rule which State

p

qp

Solution

Example 5

tomorrow.barbecue a

have will we today,rainsit ifTherefore, tomorrow.barbecue

a have will then we today,barbecue a havenot do weIf

today.barbecue a havenot will then werains,it If :argument

following theof basis theis inference of rule which State

rp

rq

qp

Solution

Example 6Show the following hypotheses :

(1) It is not sunny this afternoon and it is colder than yesterday.

(2)We will go swimming only if it is sunny.

(3) If we do not go swimming, then we will take a canoe trip.

(4) If we take a canoe trip, then we will be home by sunset.

lead to the conclusion “We will be home by sunset”.

p: It is sunny this afternoon.q

rs

t

)ponens(6,7 Modus t 8.

Hypotheis t s 7.

(4,5) ponens Modus s 6.

Hypothesis sr5.

,3) tollens(2Modus r 4.

Hypothesis pr 3.

(1)tion Simplifica p 2.

Hypothesis qp 1.

Reason Step

Example 7

Show that the hypotheses

(1) If you send me an e-mail message,

then I will finish writing the program,

(2) If you do not send me an e-mail massage, then I will go to sleep early,

(3) If I go to sleep early, then I will wake up feeling refreshed

lead to the conclusion “If I do not finish

writing program, then I will wake up

feeling refreshed”.

p

q

r

s4,5)syllogism( alHypothetic sq 6.

Hypothsis sr 5.

(2,3) syllogism alHypotheticr q 4.

Hypothsisr p 3.

(1)tion Contraposi pq 2.

Hypothesis qp 1.

Reason Step

true.is )(

n implicatio that theshowing as same theis ,...,

hypotheses thefrom followslogically ,lyConsequent

true.also isn counclutio the

true,are hypotheses whenever if called isargument An

21

21

qppp

p,pp

q

valid

n

n

3.2 Mathematical InductionHow to prove 1+2+…+n=n(n+1)/2 for n=1,2,…?

The Well-Ordering Property

Every nonempty set of nonnegative integers has a least element.

Use mathematical induction!

Mathematical Induction(1)

Mathematical induction is used to prove the form . trueis )(nnP

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 .integer positiveevery for turebe shown to is )1()( kkPkP

).())]1()(()1([ as stated becan techniqueproof The nnPkPkPkP

)2(

)2()1(

)1(

P

PP

P

)3(

)3()2(

)2(

P

PP

P

P(1) )4(

)4()3(

)3(

P

PP

P

)(

)()1(

)1(

kP

kPkP

kP

Why does Mathematical induction work?

q

qp

p

Law of detachment

Example 1

. is integers

positive odd first theof sum that theprove toinduction almathematic Use2n

n

P(n)

true.is 1 then trueis if any for Therefore,

)1()12(

)12()12(531

true.is 1 prove we true,is Assuming :

true.is P(1) Therefore, .11 true.is P(1) prove We:

22

2

)P(kP(k)kkkk

kk

)P(kP(k)stepInductive

Basis step

Example 2

Use Mathematical induction to prove the inequality

. integers positive allfor 2 nn n

P(n)

true.is 1get that we,222121 From

true.is 1 that

show toneed We2 is, that true,is Assume :

.21since true,is1 :

1

1

)P(kk

)P(k

.kP(k)stepInductive

) P(stepBasis

kkkk

k

Example 3 P(n)Use Mathematical induction to prove that

. integers enonnegativ allfor 1222221 132 kkk

.1212 since trueis )0(: 10 PstepBasis: trueis 1)P(k prove we true,is P(k) Assuming: stepInductive

.121222)12(

2)22221(

222221

2111

132

132

kkkk

kk

kk

Mathematical Induction(2)

1. Basis step. The proposition P(m) is shown to be true.

2. Inductive step. The implication .integer positiveevery for truebe shown to is )1()( mkkPkP

Mathematical induction is used to prove the form . trueis )(nnPWhere the universe of discourse is the set of contiguous integers: m,m+1,m+2,….

Example 4 Sums of Geometric Progressions

P(n)

Use Mathematical induction to prove the following formula:

.1 when ,1

12

0

r

r

aararararaar

nn

n

j

j

1 since true,is )0(:

r

aaraPstepBasis

11)(

: trueis 1 prove We true.is :

21

112

12

r

aarar

r

aararararara

arararara

)P(kP(k)AssumestepInductive

kk

kkk

kk

Use Mathematical induction to prove that .2/12

nH n

.2/011 since true,is )0(: 120 HHPstepBasis

.2

1k1

2

1)

2

k(1

)1/2 than lessnot each termsare there(since 2

12)

2

k(1

)hypothesis inductive (by the 2

1

12

1)

2

k(1

2

1

12

1

2

1

12

1

2

1

3

1

2

11

: trueis 1 prove We true.is :

1k1

1

12

12 1

kk

kk

kk

kkk

k

k

H

H

)P(kP(k)AssumestepInductive

P(n)

Example 5 Inequality for Harmonic Numbers.

,....)3,2,1for ,1

3

2

2

11 :numbers (Hormonic i

iH i

Example 6 The numbers of Subsets of a Finite Set

P(n)Use Mathematical induction to prove that subsets. 2has then elements, set with finite a is If n SnS

itself. namely, subsets, 12

exactly hasset empty thesince true,is )0(: 0

PstepBasis

subsets.

222 has therefore,of subsets2 are thereSince

. of subsets theare andsubset each For

: wayfollowing in the obtained becan of subsets The

.' and ofelement one be Let

: trueis 1 that prove We true.is :

1 S S

S{a}X X of S', X

S

S-{a}SS a

)P(kP(k)AssumestepInductive

kkk

X

X

}{aX

3.3 Recursive Definitions

Recursive definitions: defining an object by itself.

,....,,naa

na

nn

nn

210for 2

as defined be alsocan ,...2,1,0for 2 example,For

1

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

defined by 　 f(0)=3, 　 f(n+1)=2f(n)+3 ．

Find f(1),f(2) and f(3).

f(1)=2f(0)+3=9,

f(2)=2f(1)+3=21,

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

integer. enonnegativ a is

andnumber real nonzero a is where of definition recursive a Give

n

aan

,....,,naaa

a

nn 210for

1

Solution

1

0

Example 4 . of definition recursive a Give0

n

kka

.)(

,

Solution

10

1

0

0

0

0

n

n

kk

n

kk

kk

aaa

aa

Example 5

? and numbers Fibonacci theisWhat

.

1

0

:follows asy recursivel defined are

...,, numbers, Fibonacci The

5432

21

1

0

210

f,,f,ff

fff

,f

,f

fff

nnn

.5

,3

,2

,1

345

234

123

012

fff

fff

fff

fff