dis math techniques1
TRANSCRIPT
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Discrete Mathematics-1
ByProf. Manohar Lal
Director
School of Computer and Information SciencesIndira Gandhi National Open University
New Delhi-110068.
Jan 07, 2012
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Text: Elements of Discrete Mathematics
by C. L. Liu (McGraw Hill, 1985)
References:(1)Discrete Maths and Its Applications
Kenneth H. Rosen (TMH 2003)
(2) Discrete Match Structures with Applicationsto Computer Science,By Trembley & Manohar
(TMH 1975, 1997)
(3) Discrete Mathematics by Richard byJohnsoubaugh (Fifth edition,Pearson, 2001)
(4) Discrete Mathematics with Graph Theory
E.G. Goodaire & M.M. Parmenilor (PH, 2002)
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A computer can not represent real number
system in its entirety, it can essentially
represent only the integers------machine-
epsilon---the smallest positive integer
representable in the machine. Thus,
Computer math is essentially Discrete
mathematics
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The number of natural numbers is essentially
the same as the number of integers, is thesame as the number of rational numbers
Consider the mappings
1 2 3 4 5 6.0 1 -1 2 -2 3..
0, 1/1, 2/1, 1/2, 3/1, 2/2, 1/3, 4/1, 3/2, 2/3,
1/4, 5/1, 4/2, 3/3, 2/4, 1/5, However, the number of real numbers is
strictly more than the number of integers
through diagonalization method
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The number systems (N, +), (N, .), ( I, +), (I, .)(I, +, .),(Q,+), (Q, .), (Q, +, .), (R,+), (R, .), (R, +, .)
(C,+), (C, .), (C, +, .)
(R, +, .) is ordered field
(C, +, .) is not an ordered field
(R, +, .) is complete ordered field
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Set Theory.Set operations.union,
intersection, complementation, symmetric
difference
Finite sets, countably infinite sets, uncountably
infinite sets
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Proof Techniques in Discrete
MathematicsProof by casesConstructive Existence
Non-constructiveMathematical InductionPigeon-Hole Principle
Diagonalization methodto show strictlylarger sets
Principle of inclusion and exclusionSymbolic logical techniques direct, bycontradiction b contra osition
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Proof by cases
(p1 V p2 V p3pn) q
[( p1q) (p2q) .. (pnq)]
Problem: Show there are no solutions for integers x & y
Such that x2 + 3y2 = 8
Case (i) x2
> 8 ,i.e, x 3Case (ii) 3 y2> 8 i.e y 2
Case (iii) x = -2, -1, 0, 1, or 2 &
y = -1, 0, 1
possible values for x2 = 0, 1, 4
possible values for 3y2 = 0, 3
Maximum (x2 + 3y2) = 7
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Constructive Existence Proof
X P(x) is shown by the existence of a such that P(a)
Problem: show that a positive integer can be written as sum of
two cubes in two different way
Solution: 1729 = 103 + 93 = 123 + 13
Non-constructive existence proof
Problem: For some irrationals x and y, xy is rational
Solution: xy
is rational: (2 )2
or ((2 )2
)2
= 2 is rational
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MathematicalInduction
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1. The Principle of Mathematical Induction
Consider the following series
1 = 12
1 + 3 = 22
1 + 3 + 5 = 32
1 + 3 + 5 + 7 = 4
2
Is 1 + 3 + 5 + 7 + . + (2n-1) = n2 ?
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Is it true whenn
= 2800 ?When n = 2800
LHS = 1 + 3 + 5 + 7 +. + (2(100)-1)
= 1 + 3 + 5 + 7 + . + 199
= 10000
RHS = 1002
= 10000The proposition is true for n = 2800
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Apply Mathematical Induction
(M.I.) to prove the propositionA proposition P(n) is true for all positive
integers n ifboth of the following conditions
are satisfied :
Is it true when n = 100000 ?
1. P(1) is true.
2. AssumingP(k) is true for any positive
integer k, it can be proved thatP(k+ 1) is
also true.
1. The Principle of Mathematical Induction
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1. The Principle of Mathematical Induction
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Note :
Mathematical induction cannot be used to
prove whose variables are not positiveintegers.
For instance : it is a serious mistakes to
prove the identity
x3 1 = (x - 1)(x2 +x + 1), for allxR.
1. The Principle of Mathematical Induction
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Prove by mathematical induction that
1 + 3 + 5 + . + (2n 1) = n2 for all positive
integers.
LetP(n) be the proposition 1 + 3 + 5 + 7 + . + (2n 1) = n2
When n = 1, RHS = 12 = 1
LHS = 1
P(1) is true.
AssumeP(k) is true for any positive (+ve) integerk.
i.e. 1 + 3 + 5 + 7 + . + (2k1) = k2
When n = k+ 1, RHS = (k+ 1)2
2. Some Simple Worked Examples
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k2
P(n) is true for n = k+ 1 if n = kistrue .
By M.I.,P(n) is true for all +ve integers n.
LHS = 1+3+5+7+ . +(2k 1) + [2(k+1) 1]
= k2 +2k+ 2 1
= k2 +2k+ 1
= (k+ 1)2
2. Some Simple Worked Examples
1 + 3 + 5 + 7 + . + (2k1)
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Further Examples of
Mathematical Induction
2. Some Simple Worked Examples
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Prove by mathematical induction that
8n 3n is divisible by 5 for all positiveintegers n.
Let P(n) be the proposition 8n 3n is divisible
by 5 .When n = 1, 81 31 = 5 which is divisible by
5.
P(1) is true.
Assume P(k) is true for any positive integerk.
i.e. 8k 3k = 5N, where N is an integer.
2. Some Simple Worked Examples
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When n = k+ 1
8k+1 3k+1=88k 33k
=88k 33k 53k + 53k
=88k 83k + 53k=8(8k 3k) + 53k
=8(5N) + 53k
=5(8N + 3k) which is divisible by 5 P(n) is true for n = k+ 1 if n = kis
true .
2. Some Simple Worked Examples
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3. Variations of the Method of
Induction
(A) 1st type of variation :
LetP(n) be a proposition involving
positive integern.
If (i) P(n) is true for n = 1 and n = 2
and (ii) ifP(n) is true for some positive
integers kand k + 1,then
P(n) is also true forn = k+ 2,
thenP(n) is true for all positive integers n.
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3. Variations of the Method of
Induction
(A) 1st type of variation :
Note :
The principle may be applied
to the proposition of the forman - bn oran + bn.
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3. Variations of the Method of
Induction
(B) 2nd type of variation :
LetP(n) be a proposition involving integern.
If (i)P(n) is true n = ko, where ko is an integer
not necessarily equals 1, and
(ii) ifP(n) is true forn = k(k k0) thenP(n)is
also true forn = k+ 1.
thenP(n) is true for all integers n ko.
3 V i ti f th M th d f
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3. Variations of the Method of
Induction
(B) 2nd type of variation :
.2,5
integerpositiveeveryforthatProve..
2nn
ge
n>
.52..255
,322,5)(
2
2
5
=>
==
===
nfortrueisneiRHS
LHSnWheni
n
3 V i ti f th M th d f
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3. Variations of the Method of
Induction
(C) 3rd type of variation :LetP(n) be a proposition involving integer
n.
If (i)P(n) is true forn = 1 and n = 2,
and (ii) ifP(n) is true for some positive
integerk, thenP(n) is also true forn = k+ 2,
thenP(n) is true for all positive integers n.
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3. Variations of the Method of
Induction
(C) 3rd type of variation :Remarks :
In the above statement, if we only checkP(n) is true forn = 1 (respectively n = 2),
we can only conclude thatP(n) is true for
all positive odd (respectively, even)
integers n.
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3. Variations of the Method of
Induction
(C) 3rd type of variation :
])1(1[4
1
)1(2
1
)(is
2
equationtheofsolutions
integralnegativenonof)(numberthe
,integerpositiveeachforthatProve..
n
nnf
nyx
nf
nge
+++=
=+
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Pigeonhole principle
The pigeonhole principle [Rosen, p.347]: Ifk is placed into kboxes,
then there is at least one a positive integer and k+1 or more objects are
box containing two or more objects.
.
A common way to illustrate this principle is by assuming that k+1 pigeons
fly to kpigeonholes. Then, there must be at least one pigeonhole containing at least
two pigeons.
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The pigeonhole principle is a trivial observation,
which however can be used to prove many
results
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Pigeonhole principle
Example 10.1: Let a, b, c be integers. We canchoose two of them whose sum is even.
Proof:
Apply the pigeonhole principle with odd and evenintegers as the two pigeonholes, and the a, b, c as
the three pigeons.
Two of the a, b, c have the same parity, that is, theyare both even, or both odd.
The sum of two even numbers, or the sum of two
odd numbers is even.
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Pigeonhole principle
2/1
1/2
1/2
Example 10.2: Five points are given on, or inside a unit square.
Show that there is a pair of points whose distance is at most .
Proof: Split the unit square into 4 squares with edge 1/2, see the
figure. By the pigeonhole principle, two of the points are on, or inside
one of the four smaller squares.
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Pigeonhole principle
Their distance is less or equal to the maximum distance between two
points of that square, that is, less or equal the diameter of the square.
By the Pythagorean theorem, the diameter of the small square is
2/14/14/1)2/1()2/1( 22 =+=+
1/2
1/2
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Recursion
Recursively defined functions[Rosen, p.295]:We use two steps to define a function with the
set of nonnegative as its domain:
Basis step: Specify the value of the function atzero.
Recursive step: Give a rule for finding its value at
an integer from its values at smaller integers.
Such a definition is called recursive orinductive
definition.
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Recursion
Exercise 10.3[Rosen, p.295]: Suppose thatfisdefined recursively by
f(0) = 3f(n + 1) = 2f(n) + 3
Findf(1),f(2),f(3), andf(4).
Exercise 10.4[Rosen, p.296]: Give a recursive
definition of the factorial functionf(n)=n!
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Fibonacci numbers
In some recursive definitions of functions, the values of the
function at
the first kpositive integers are specified, and a rule is given
for
determining the value of the function at larger integers from
its values
at some or all of the preceding integers.
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Fibonacci numbers: The Fibonacci numbers,f0,
f1,f2, , are defined recursively by
f0 = 0, f1 = 1
fn =fn - 1 +fn - 2
forn = 2, 3, 4, .
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Fibonacci numbers
Example Letfn denote the n-th Fibonacci number.
Show that
when n is a positive integer.
Proof: For n=1
122
22
1 +=+++ nnn fffff
122
22
1 +=+++ nnn fffff
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Fibonacci numbers
Exercise 10.6[Rosen, p.308, Ex.13]: Letfn denote
the n-th Fibonacci number. Show that
nn ffff 21231 =+++ when n is a positive integer.