copy of sets

7/31/2019 Copy of SETS

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DISCRETE

MATHEMATICSSETS


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SETS

Collections of well defined

objects.

Member.

Well defined.


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Representation { a,b, ------} or

{x:x is set of even numbers }

Membership of an element in a

set--pA and qA where A is any

set given as {a,{1,2},p,{q}}.Subset Let A and B be any two

sets. A B iff xA x B


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Equal set (AB & B A)A=B

Proper subsetA is proper subset

of B if A is subset of B but A is not

equal to B.

Universal set(E,U or X)Empty or null set(,{})

Power set ( (A), 2 )

A


5/23

Write the power set of {a,b,c},

N,.

A={1,2,3},B={1,2},C={1,3}and

D={3}.

A B,D A,C A,A = B.


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Index set or Index

Let J = {S1,S2,-------} and A be a family of

sets A = {A ,A ,-------} such that for any

S Jthere corresponds a set A A , and also

A = A iff Si = Sj then A is called index

set, J the index set, and anysubscript such as Si in A is called an index

{A } .

Si

Sj

Si

Si

Si Sj

i

i Ji


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Operations on sets

Intersection, union, disjoint,

complement, relative complement,

absolute complement, symmetric

difference

AB = {x:xA and x B} ={x:(x A)(x B)}.

AUB = {x:xA or x B}


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}6,5,4,3,2,1{BA

}4,3{BA

}6,5,4,3{B},4,3,2,1{A

A={1,2,3,4}, B={3,4,5,6}


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Operations on sets

Disjoint sets A B =

Mutually disjoint- elements of

disjoint collection are said to be

mutually disjoint A = { A }

where A A = .A is a disjoint collection iff

A A = for every i,j J , i j

i Ji

i j

i j


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Relative complement of B in A(difference)

A-B = {x:xA and xB} = {x:(xA) (x

xB)}.Absolute complement

E-A= ~A where E is universal set.

Symmetric difference

AB = (A-B)U(B-A)

A~A=.........,~(~A) =........,AU~ = ......


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A

UB

A B

U

A B

U


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Properties

A,B,C elements of X

AUB = BUA and AB=BA

AU(BUC)=AUBUC & (A B) C =

A (B C)

AU = A and A X= A(Identity laws)

A A= and A U A = X(Complementlaws)

AUA=A&A A=A(law of tautology)

AUX=X & A = (law of absorption)


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Properties

A(BUC)=(AB)U(AC)

AU(B C)=(AU B) (A UC)(Distributive

law)If AUB=X and A B =,then A = A.

(A)=A(law of double complementation)

(AUB)= A B & (A B)= AUB(De

Morgans laws)


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Prove that (AUB)=A B(AUB)AB & AB(AUB)

Let x(AUB)

x(AUB)xA and xB

x A and x B

x AB

(AUB)AB


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Duality

Let E be an equation of set algebra

The dual E* of E is the equation

obtained by replacing each occurance

of U, ,X, in E by , U, ,X,resp.

Dual (X A)U(B A) = A

( UA) (BUA) = A

Principle of duality


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Classes of sets

Subsets of sets

set of sets

class of sets or collection of setsEg- S={1,2,3,4}, A be the class of subsets of S

which contain three elements of S then A =

[{1,2,3},{1,2,4},{1,3,4},{2,3,4}].The elements of

A are {1,2,3},{1,2,4},{1,3,4},{2,3,4}

The power set of any set is the class of all subsets

of that set.


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Partition of sets, Counting

principle

A partition of any set S is a subdivision of S into

nonoverlapping , nonempty subsets.

S = {1,2,3,4,5}, A1 = {2,3}, A2 = {1,4,5}. Then

{A1, A2} is the partition of S.

Finite sets

n(A) or #(A) or card(A).

Theorem- If A & B are finite sets, then AUB &

A B are finite &

n(AUB) = n(A) + n(B) - n(A B).


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Ordered Pairs & n-tuples

Two objects in a given fixed order.

Not set consisting of two elements

or (a,b)

(a,b) = (c,d) iff a = c & b = d

{1,2} = {2,1} = {1,1,2}

but

ordered triple

n-tuple


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Cartesian Products-Let A & B be any two sets. The set of all

ordered pairs such that the first member of the

ordered pair is an element of A & second

member is an element of B.

AB={: ((xA) ( yB)}.

A (B C) = {:(a A) B

C}.

A = {a,b} & B = {1,2,3} find A B, B A,

A A A.

A (B C) (A B) C.


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Properties of Cartesian Product

For the four sets A,B,C and D

)CA()BA()CB(A.4)CB()CA(C)BA.(3

)CB()CA(C)BA.(2

)DB()CA()DC()BA.(1


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Mathematical InductionLet P be a proposition defined on the positive

integer N that is P(n) is either true or false for

each n in N.

Suppose P has the following two properties

(i) P(1) is true

(ii) P(m+1) is true.

Then P is true for every positive integer.

Show that n+2n is divisible by 3


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To show n + 2n is divisible by 3Let n=1

Substituting value of n we get 1+ 21 = 3which is divisible by 3

P(1) is true

Let P(m) be truem +2m is divisible by 3Check for P(m+1)

(m+1)+2(m+1)=m+3m +3m+1+2m+2

= m +2m+3(m+m+1)

It is divisible by 3

P(m+1) is true

Hence P(n) is true for all n.

3

3

3 3

3

2

3

2


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Show that the sum of the first n odd

numbers is .2n

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