Discrete Structures

(CS-104)
Lecture 07
Previous Lectures Summary
• Rational Number
• Irrational numbers
• Absolute values
• Triangular inequality
• Modular Arithmetic
• Floor and Ceiling
Today’s Lectures Summary
• Floor and Ceiling
• Properties of Floor and Ceiling
Functions
• Methods of Proof
• Direct Proofs
Today's Lecture

 Why study set Theory

 Sets

 Operations on sets

 Memberships

 Notations

 Venn diagrams
Examples
Compute  x  and  x  for each of the following values of x:
a. 25/4
b. 0.999
c. −2.01
Solution:
a. 25/4 = 6.25 and 6 < 6.25 < 7; hence 25/ 4  6 and 25/4  7

b. 0 < 0.999 < 1; hence 0.999  0 and 0.999  1.

c. −3 < −2.01 < −2; hence 2.01  3 and -2.01  2

Example
The 1,370 students at a college are given the opportunity
to take buses to an out-of-town game. Each bus holds
a maximum of 40 passengers.
a. For reasons of economy, the athletic director will send
only full buses. What is the maximum number of buses
the athletic director will send?
b. If the athletic director is willing to send one partially
filled bus, how many buses will be needed to allow all
the students to take the trip?

a). 1370 / 40   34.25  34

b). 1370/40   34.25  35
General Values of Floor
If k is an integer, what are k  and k  1/ 2 ? Why?
Solution:
Suppose k is an integer. Then
k   k becuase k is an integer and k  k  k 1
and
 1 1
k+
 2   k becuase k is an integer and k  k  2  k 1
Cont…
Is the following statement true or false? For all real
Numbers x and y,  x  y    x    y 

Solution: The statement is false, take x = y = 1/2, then

1 1
 x  y       1  1
2 2
where as
1 1
 x    y         0  0  0
2 2
Hence
 x  y    x    y 
Properties of Floor
Theorem: For all numbers x and all integers m,
 x  m    x   m
Proof: Suppose a real number x and an integer m are
given. Let n =  x  that unique integer n such that,
n ≤ x < n + 1. Add m to all sides to obtain n + m ≤ x + m <
n + m + 1. Now n + m is an integer and so, by definition of
floor,  x  m   n  m
But n =  x  . Hence by substitution
 x  m    x   m
Cont….
Theorem: F o r a n y in te g e r n
 n
 ; if n is e v e n
 n   2
 2   
 n  1 ; if n is o d d
 2

Proof: Suppose n is a integer. Then either n is odd or n is even.

Case 1: in this case n = 2k+1 for some integers k.
 n   2k  1  2k 1 
 2    2    2  2 
 1
 k  k
2 
because k is an integer and k  k  1 / 2  k  1. But since
n 1
n  2k  1  n  1  2k  k 
2
Cont…
also. Since both the left-hand and right-hand sides equal k,
n
they are equal to each other. That is,    k
2
Case 2: In this case, n = 2k for some integer k.

 n   2k 
 2    2    k   k

Since k=n/2 by the definition of even number. So

n n
 2   k  2
Methods of Proof
Proof
A proof is a logically structured argument which
demonstrates that a certain proposition is true. When
the proof is complete, the resulting proposition becomes
a theorem, or if it is rather simple, a lemma.
Direct Proofs
Theorem

Theorem: The sum of any two even integers is even integer.

Proof: m,n  Z, if m,n are even then m+n is even.

Three basic steps in a direct proof:

Deconstruct Axioms
Mathematical Insights
Reconstruct Conclusion
Cont…
1) Deconstruct Axioms
Take the hypothesis and turn it into a usable form.
Usually this amounts to just applying the definition.
e.g., Let m, n be two even integers, then m= 2r, n=2s.
2) Mathematical Insights
Use your human intellect and get at “real reason”
behind theorem. e.g., looking at what we’re trying to
prove, we see that we’d really like to prove m+n is even.
Cont….
So let’s take sum of m and n, m+n=2r+2s.
From here, we’ll have to use some algebra to get the
formula into a form usable by the final step.

3) Reconstruct Conclusion
This is the reverse of step 1. At the end of step 2 we
should have a simple form that could be derived by
applying the definition of the conclusion.
e.g., m+n=2(r+s)
Theorem
Disproving by Counterexample
Disproving by Counterexample

Theorem: For all integers a and b, if a | b and

b | a then a = b.
Counterexample: Let a = 2 and b = −2. Then
a | b since 2 | (−2) and b | a since (−2) | 2, but a
= b since 2 = −2.
Therefore, the statement is false.
Proof by Contradiction
Theorem
Theorem
Proof by contradiction
Theorem: The square of an even number is even
Proof: Suppose k is an even number but k 2 is not even.
 So k 2 is odd.  n k 2 = 2n + 1
 n k 2 - 1 = 2n n (k - 1)(k + 1) = 2n
 2 | (k - 1)(k + 1)
 2 | (k - 1) or 2 | (k + 1) since 2 is prime
 a k - 1 = 2a or  b k+1 = 2b
 a k = 2a + 1 or  b k = 2b – 1
In both cases k is odd
So k is not even, which is a contradiction
Indirect Proof
In addition to direct proofs, there are two other standard
methods for proving k P(k)  Q(k)

1. Indirect Proof : For any k assume: Q(k) and

derive: P(k), Uses the contra positive logical
equivalence: P Q  Q  P
2. Proof by Contradiction: For any k assume:
P(k)  Q(k) and derive: P(k)  Q(k)
Uses the logical equivalence:
P Q  P  Q  P  Q  P  Q
 (P  Q )  (P  Q )  (P  Q )  (P  Q )
 (P  Q )  (P  Q )
Intuitively: Assume claim is false (so P must be true and
Q false). Show that assumption was absurd (so P
false or Q true) so claim true!
Example
Theorem : There is no greatest integer.

Proof: Suppose not. That is, suppose there is a

greatest integer N. [We must deduce a contradiction.]
Then N ≥ n for every integer n. Let M = N + 1. Now M is
an integer since it is a sum of integers. Also M > N since
M = N + 1. Thus M is an integer that is greater than N.
So N is the greatest integer and N is not the greatest
integer, which is a contradiction.
Proof by contradiction
PROVE: The square of an even number is even
Proof: Suppose k is an even number but k 2 is not even.
 So k 2 is odd.  n k 2 = 2n + 1
 n k 2 - 1 = 2n n (k - 1)(k + 1) = 2n
 2 | (k - 1)(k + 1)
 2 | (k - 1) or 2 | (k + 1) since 2 is prime
 a k - 1 = 2a or  b k+1 = 2b
 a k = 2a + 1 or  b k = 2b – 1
In both cases k is odd
So k is not even, which is a contradiction
Easier Characterization
Recall the set of rational numbers Q = the set of numbers
with decimal expansion which is periodic past some point
(I.e. repeatinginginginginging…)
Easier characterization
Q = { p/q | p∙q are integers with q  0 }

Theorem: that the sum of any irrational number with a

rational number is irrational:
Disproving claims is often much easier than proving them.
Claims are usually of the form k P(k).
Cont….
Thus to disprove, enough to find one k –called a
counterexample– which makes P(k) false.
Proof: The product of irrational numbers is irrational.

1
1. Let x  2 and y  . Both are irrational.
2
1
2. Their product xy  2   1 is rational.
2
Method of Proof by Contra Positive
1. Express the statement to be proved in the form
∀x in D, if P(x) then Q(x).
2. Rewrite this statement in the contra positive form
∀x in D, if Q(x) is false then P(x) is false.
3. Prove the contra positive by a direct proof.
a. Suppose x is a (particular but arbitrarily chosen)
element of D such that Q(x) is false.
b. Show that P(x) is false.
Example
Proposition: For all integers n, if n2 is even then n is
even.
Contra positive: For all integers n, if n is not even then n2
is not even.
Proof: Suppose n is any odd integer. [We must show that
n2 is odd.] By definition of odd, n = 2k + 1 for some
integer k. By substitution and algebra,
n2 = (2k+1)22 = 4 k2 + 4k + 1 = 2(2 k2 + 2k) + 1.
But 2k2 + 2k is an integer because products and sums of
integers are integers. So n2 = 2·(an integer) + 1, and
thus, by definition of odd, n2 is odd.
Relation ship between Contra positive and
Contradiction Proofs
In a proof by contraposition, the statement
∀x in D, if P(x) then Q(x)
is proved by giving a direct proof of the equivalent
statement
∀x in D, if ∼Q(x) then ∼P(x).
To do this, you suppose you are given an arbitrary element
x of D such that ∼Q(x). You then show that ∼P(x). This is
illustrated in Figure
Cont….
To rewrite the proof as a proof by contradiction, you
suppose there is an x in D such that P(x) and ∼Q(x). You
then follow the steps of the proof by contraposition to
deduce the statement ∼P(x). But ∼P(x) is a contradiction to
the supposition that P(x) and ∼Q(x). (Because to contradict
a conjunction of two statements, it is only necessary to
contradict one of them.) This process is illustrated in Figure
Two Classical Theorems
T h e o r e m 1 : I r r a t i o n a li t y o f 2 .

Proof:
Suppose not, suppose 2 is a rational number. Then there are integers m and n with no
m m2
common factors such that 2  , Squaring both side of the equation gives 2= 2
n n
or equivalently, m2  2n2  m2 is even. it fallows that m  2k, for some integer k.
so m2  (2k)2  4k 2  2n2 , dividing both sides of the right most equation by 2 gives
n2  2k 2. Consequently, n2 is even, and so n is even. Hence both m and n have a common
factor of 2. But this contradicts the supposition that m and n have no common
factors.
 Cont…
T h e o re m 2 : P ro v e b y c o n tra d ic tio n s 1 + 3 2 i s i r r a t i o n a l.
Methods of Proof and Number Theory
Mod Functions
Compute following
1. 113 mod 24
2. -29 mod 7
4
1. 113 mod 24:
24 113 113 div 24
96
17
2. -29 mod 7: 5
7  29 -29 div 7
 35
6
Mod and div
Definitions
Divisibility and Floor
Theorem
Cont…
Computing div and mod
Use the floor notation to compute 3850 div 17

and 3850 mod 17.

Sol:
Mod Congruence's
Let a, b be integers and n be a positive integer. We say
that a is congruent to b modulo n (i.e. a  b(mod n) )
iff n | (b-a), implies that there exist some integer k such
that b-a = n·k.

Note: a mod n = b mod n

Which of the following are true?
1. 3  3 (mod 17)
2. 3  -3 (mod 17)
3. 172  177 (mod 5)
4. -13  13 (mod 26)
Cont…
1. 3  3 (mod 17) True: any number is congruent to
itself (3-3 = 0, divisible by all)
2. 3  -3 (mod 17) False: (-3-3) = 6 isn’t divisible by
17.
3. 172  177 (mod 5) True: 177-172 = 5 is a multiple
of 5
4. -13  13 (mod 26) True: 13-(-13) = 26 divisible by
26.
Congruence's Identities
Let n > 1 be fixed and a, b, c, d be arbitrary integers.
Then the following properties holds:
a) (Reflexive Property ) a  a (mod n).
b) (Symmetric Property) If a b(mod n) then b a(mod
n).
c) ( Transitive Property) If a  b(mod n) and b  c (mod
n) then a c(mod n).
d) If a  b(mod n) and c d (mod n) then a + c  (b
+ d ) (mod n) and a·c  b·d(mod n).
e) If a  b(mod n) then a + c b+c(mod n) and a·c 
b·c(mod n).
f) If a  b(mod n) then a k  b k (mod n) for any positive
integer k.
Theorem
If k is any integer such that k  1 (mod 3),
then k3  1 (mod 9).
Proof: k  Z, k 1(mod 3)  k 31(mod 9)
k 1(mod 3)
  n, k-1 = 3n   n, k = 3n + 1
  n, k 3 = (3n + 1)3
  n, k 3 = 27n 3 + 27n 2 + 9n + 1
  n, k 3-1 = 27n 3 + 27n 2 + 9n
  n, k 3-1 = (3n 3 + 3n 2 + n)·9
  m, k 3-1 = m·9 where m = 3n 3 + 3n 2 + n
 k 31(mod 9)
Indirect Proofs
Method of Proof by Contra-Positive
1. Express the statement to be proved in the form
∀x in D, if P(x) then Q(x).
2. Rewrite this statement in the contra positive form
∀x in D, if Q(x) is false then P(x) is false.
3. Prove the contra-positive by a direct proof.
a. Suppose x is a (particular but arbitrarily chosen)
element of D such that Q(x) is false (or ¬Q(x) is true).
b. Show that P(x) is false (or ¬P(x) is true).
Proof by Contra-positive
Proposition: For all integers n, if n2 is even then n is
even.
Contra positive: For all integers n, if n is not even then n2
is not even.
Proof: Suppose n is any odd integer. [We must show that
n2 is odd.] By definition of odd, n = 2k + 1 for some
integer k. By substitution and algebra,
n2 = (2k+1)2 = 4 k2 + 4k + 1 = 2(2 k2 + 2k) + 1.
But 2k2 + 2k is an integer because products and sums of
integers are integers. So n2 = 2·(an integer) + 1, and
thus, by definition of odd, n2 is odd.
Relation ship between Contra-positive and
Contradiction Proofs
In a proof by contraposition, the statement
∀x in D, if P(x) then Q(x)
is proved by giving a direct proof of the equivalent
statement
∀x in D, if ∼Q(x) then ∼P(x).
To do this, you suppose you are given an arbitrary
element x of D such that ∼Q(x). You then show that
∼P(x). This is illustrated in Figure
Cont….
To rewrite the proof as a proof by contradiction, you suppose
there is an x in D such that P(x) and ¬Q(x). You then follow
the steps of the proof by contraposition to deduce the
statement ¬P(x). But ¬P(x) is a contradiction to the
supposition that P(x) and ¬Q(x). (Because to contradict a
conjunction of two statements, it is only necessary to
contradict one of them.) This process is illustrated in Figure
Proof by Contradiction
Proposition: For all integers n, if n2 is even then n is
even.
Proof: Suppose n is not even integer. Then n is odd
integer. By definition of odd, n = 2k + 1 for some integer
k. By substitution and algebra,
n2 = (2k+1)2 = 4 k2 + 4k + 1 = 2(2 k2 + 2k) + 1.
But 2k2 + 2k is an integer because products and sums of
integers are integers. So n2 = 2·(an integer) + 1, and
thus, by definition of odd, n2 is odd.
But n2 is even in hypothesis. Which is a contradiction
because any integer cannot be both even and odd. Thus
our supposition was wrong. Hence n is even.
When to use which method…???

In the absence of obvious clues suggesting indirect

argument, Try first to prove a statement directly. Then,
if that does not succeed, look for a counterexample. If
the search for a counterexample is unsuccessful, look
for a proof by contradiction or contraposition.
Two Classical Theorems
Theorem 1: is an irrational number.
Proof:
Cont…
m = 2k for some integer k.
m2 = (2k)2 = 4k2 = 2n2.
n2=2k2
Consequently, n2 is even, and so n is even. But we also
know that m is even. Hence both m and n have a
common factor of 2. But this contradicts the supposition
that m and n have no common factors. [Hence the
supposition is false and so the theorem is true.]
Theorem 2
T h e o r e m 2 : P ro v e b y c o n tra d ic tio n s 1 + 3 2 is ir ra tio n a l.
Introductions to Set Theory I
Why study Set Theory?

Understanding set theory helps people to …

1. See things in terms of systems.

2. Organize things into groups.

3. Begin to understand logic.

Sets
A set is a gathering together into a whole of definite,
distinct objects of our perception and of our thought
which are called elements of the set.
The elements or members of a set can be anything:
numbers, people, letters of the alphabet, other sets,
and so on. Sets are conventionally denoted with capital
letters.
Note: A set should be well defined and distinct.
Examples
(1) A = {tiger, lion, puma, cheetah, leopard, cougar, ocelot}
(this is a set of large species of cats).
(2) A = {a, b, c, ..., z} (this is a set consisting of the
lowercase letters of the alphabet)
(3) A = {-1, -2, -3, ...} (this is a set of the negative
numbers)
In all above examples each element of the sets is distinct
and well defined.
Operations on sets
Union
Two sets can be "added" together. The union of A and B,
denoted by A ∪ B, is the set of all things which are
members of either A or B.
Examples:
• {1, 2} ∪ {red, white} ={1, 2, red, white}.
• {1, 2, green} ∪ {red, white, green} ={1, 2, red, white,
green}.
• {1, 2} ∪ {1, 2} = {1, 2}.
Intersection
A new set can also be constructed by determining which
members two sets have "in common".
Cont….
The intersection of A and B, denoted by A ∩ B, is the set
of all things which are members of both A and B. If A ∩ B
= ∅, then A and B are said to be disjoint.
Examples
{1, 2} ∩ {red, white} = ∅.
{1, 2, green} ∩ {red, white, green} = {green}.
{1, 2} ∩ {1, 2} = {1, 2}.
Compliments
Two sets can also be "subtracted". The relative
complement of B in A (also called the set-theoretic difference of A and B),
Cont….
denoted by A \ B (or A − B), is the set of all elements
which are members of A but not members of B.
Note: That it is valid to "subtract" members of a set that
are not in the set, such as removing the element green
from the set {1, 2, 3}; doing so has no effect.
Example:
 {1, 2} \ {red, white} = {1, 2}.
 {1, 2, green} \ {red, white, green} = {1, 2}.
 {1, 2} \ {1, 2} = ∅.
 If U is the set of integers, E is the set of even
integers, and O is the set of odd integers, then E′ = O.
Cont…
Cartesian products
A new set can be constructed by associating every
element of one set with every element of another set.
The Cartesian product of two sets A and B, denoted by
A × B is the set of all Ordered pairs (a, b) such that a is
a member of A and b is a member of B.
Examples
 {1, 2} × {red, white} = {(1, red), (1, white), (2, red), (2,
white)}.
 {1, 2, green} × {red, white, green} = {(1, red), (1,
white), (1, green), (2, red), (2, white), (2, green), (green,
red), (green, white), (green, green)}.
 {1, 2} × {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)}.
Special Sets
There are some sets which hold great mathematical
importance and are referred to with such regularity that
they have acquired special names and notational
conventions to identify them. One of these is the empty
set, denoted { } or ∅. Another is the singleton set {x}
which contains exactly one element, namely x.
• P or ℙ, denoting the set of all primes P = {2, 3, 5, 7,11,
13, 17, ...}.
• N or ℕ, denoting the set of all natural numbers: N = {1,
2, 3, . . .}.
• Z or ℤ, denoting the set of all integers (whether
positive, negative or zero): Z = {..., −2, −1, 0, 1, 2, ...}.
• Q or ℚ, Q = {a/b : a, b ∈ Z, b ≠ 0}. For example, 1/4 ∈
Q and 11/6 ∈ Q.
Cont….
• R or ℝ, denoting the set of all real numbers. This set
includes all rational numbers, together with all
irrational numbers (that is, numbers which cannot be
rewritten as fractions, such as π, e, and √2, as well
as numbers that cannot be defined).
• C or ℂ, denoting the set of all complex numbers : C =
{a + bi : a, b ∈ R}. For example, 1 + 2i ∈ C.
Finite and Infinite Sets
Finite Set: A set is finite if it contains a specific (finite)
number of elements, i.e., If we can count the element
in a set, such sets are called finite sets.
Example: Some finite numbers in a set: the number of
digits on your hand, the number of seats on a bus, and
the number of people on earth.
Infinite Set: If we can not count the elements in a set
such sets are called infinite sets.
Example: Set of Natural numbers. Set of Whole
numbers.
Cardinality: refers to the number of elements in a set.
Finite and Infinite Set Cardinality
Set Definition Cardinality

A = {x | x is a lower case letter} |A| = 26

B = {2, 3, 4, 5, 6, 7} |B| = 6

C = {x | x is an even number  10} |C|= 4

A = {1, 2, 3, …} |A| = 0

B = {x | x is a point on a line} |B| = 0

C = {x| x is a point in a plane} |C| = 0

Memberships
The key relation between sets is membership when one
set is an element of another. If a is a member of B, this is
denoted a ∈ B, while if c is not a member of B then c ∉ B.
For example, With respect to the sets A = {1,2,3,4} and B
= {blue, white, red}, 4 ∈ A and green ∉ B.
 Universal Sets: The universal set is the set of all
things relevant to a given discussion and is designated
by the symbol U. i.e. it contains every set.
 Subsets: If every member of set A is also a member of
set B, then A is said to be a subset of B, written A ⊆ B
(also pronounced A is contained in B). The relationship
between sets established by ⊆ is called inclusion or
containment.
Cont…

Examples
A = {1,2,3,4}, B = {1,2,3,4,5,7}, and C = {7,9,3}, and
the universal set U = {1,2,3,4,5,6,7,8,9}.
Cont…
 Super Set: if we can write B ⊇ A, read as B is a
superset of A, B includes A, or B contains A.
 Proper Subset: If A is a subset of, but not equal to, B,
then A is called a proper subset of B, written A ⊂ B (A
is a proper subset of B) or B ⊃ A (B is a proper
superset of A).
Examples
 The set of all men is a proper subset of the set of all
people.
 {1, 3} ⊂ {1, 2, 3, 4}.
 {1, 2, 3, 4} ⊆ {1, 2, 3, 4}.
Cont…
An obvious but useful identity, which can often be used to
show that two seemingly different sets are equal:
A = B if and only if A ⊆ B and B ⊆ A.

Power set: The power set of a set A is the set

containing all possible subsets of A including the empty
subset. It contains 2n elements where n is the number of
elements in A. It is typically denoted by P(A) or 2A. For
example, the power set of the set A={a,b,c,d} is the set
P(A)={{}, {a}, {b}, {c}, {d}, {a,b}, {a,c}, {a,d}, {b,c}, {b,d},
{c,d}, {a,b,c}, {a,b,d}, {a,c,d}, {b,c,d}, {a,b,c,d}}.
Subset Relationships
A = {x | x is a positive integer  8}
set A contains: 1, 2, 3, 4, 5, 6, 7, 8
B = {x | x is a positive even integer  10}
set B contains: 2, 4, 6, 8
C = {2, 4, 6, 8, 10}
set C contains: 2, 4, 6, 8, 10
The universal set U = {1,2,3,4,5,6,7,8,9,10}.
Subset Relationships
AA AB AC
BA BB BC
CA CB CC
Set Equality
Two sets are equal if and only if they contain precisely
the same elements. The order in which the elements
are listed is un important. Elements may be repeated in
set definitions without increasing the size of the sets.
Examples
 A = {1, 2, 3, 4} B = {1, 4, 2, 3}
A  B and B  A; therefore, A = B and B = A.
 A = {1, 2, 2, 3, 4, 1, 2} B = {1, 2, 3, 4}
A  B and B  A; therefore, A = B and B = A.
Notations
Symbol Meaning
Upper case designates set name
Lower case designates set elements
{ } enclose elements in set
 (or ∉ ) is (or is not) an element of
 is a subset of (includes equal sets)
 is a proper subset of
 is not a subset of
 is a superset of
| or : such that (if a condition is true)
| | the cardinality of a set
Venn Diagrams
Venn diagrams or set diagrams are diagrams that show all
possible logical relations between a finite collection of sets. Venn
diagrams were conceived around 1880 by John Venn.
Venn diagrams show relationships between sets and their
elements.
Sets A & B

Universal Set
Examples
Set Definition Elements
A = {x | x  Z+ and x  8} {1, 2 ,3, 4, 5, 6 ,7, 8}
B = {x | x  Z+, x is even and  10} {2, 4, 6, 8, 10}

AB
BA
Cont…
Set Definition Elements
A = {x | x  Z+ and x  9} {1, 2 ,3, 4, 5, 6 ,7, 8,9}
B = {x | x  Z+ ; x is even and  8} {2, 4, 6, 8,}

AB
BA
AB
Cont…
Set Definition Elements
A = {x | x  Z+ x is even and x  10} { 2 , 4, 6, 8, 9}
B = {x | x  Z+ ; x is odd and  10} {1, 3, 5, 7, 9}

AB
BA
Set operations and Venn diagram
Set operations and Venn diagram

A∩B

A ∩ B = { x | x  A and x  B }
Set operations and Venn diagram

A ∪ B = { x | x  A or x  B }
Set operations and Venn diagram

B \ A = { x | x ∉ A and x  B }
Set operations and Venn diagram

Ac = U-A = { x | x ∉ A and x  U }
Sets and Universal Set
A = {1,2,3,4}, B = {1,3,5,7}, and C = {7,9,3}, and the
universal set U = {1,2,3,4,5,6,7,8,9}. Locate all this
information appropriately in a Venn diagram.
Sets and Universal Set
A = {1,2,3,4}, B = {1,3,5,7}, and C = {7,9,3}, and the
universal set U={1,2,3,4,5,6,7,8,9}. Locate all this
information appropriately in a Venn diagram.
Sketching Regions representing Sets
Sketch the region corresponding to the set
(A ∪ Bc) ∩ C
Sketching Regions representing Sets
Sketch the region corresponding to the set (A ∪ Bc) ∩ C
Sketching Regions representing Sets
Sketch the region corresponding to the set (A ∪ Bc) ∩ C

(A ∪ Bc) ∩ C