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REFLEXIVE RELATION:IRREFLEXIVE RELATION, ANTISYMMETRIC RELATION

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MTH001 ­ Elementary Mathematics
2
1
1
2
4
4
3
3
R2 is symmetric
R1 is symmetric
1
2
1
2
4
3
4
3
R3 is not symmetric since there
R4 is not symmetric since
are arrows from 2 to 3 and from
there is an arrow from 4 to 3
3 to 4 but not conversely
but no arrow from 3 to 4
MATRIX REPRESENTATION OF A SYMMETRIC RELATION
Let
A = {a1, a2, ..., an}.
A relation R on A is symmetric if and only if for all
ai, aj A, if (ai, aj) R then (aj, ai)R.
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MTH001 ­ Elementary Mathematics
Accordingly, R is
symmetric if the
elements in the ith row are the same as the elements in the ith column of the
t
matrix M representing R. More precisely, M is a symmetric matrix.i.e. M = M
EXAMPLE
The relation R = {(1,3), (2,2), (3,1), (3,3)}
on A = {1,2,3} represented by the following matrix M is symmetric.
123
1 0 0 1
M = 2 0 1 0
3 1 0 1
TRANSITIVE RELATION
Let R be a relation on a set A.R is transitive if and only if for all a, b, c A,
if (a, b) R and (b, c) R then (a, c) R.
That is, if aRb and bRc then aRc.
In words, if any one  element is related to a second
and that second element is related to a third, then the first is related to the third.
Note that the "first", "second" and "third" elements need not to be distinct.
REMARK
R is not transitive iff there are elements a, b, c in A such that
If (a,b) R and (b,c) R but (a,c) R.
EXAMPLE
Let A = {1, 2, 3, 4} and define relations R1, R2 and R3 on A
as follows:
R1 = {(1, 1), (1, 2), (1, 3), (2, 3)}
R2 = {(1, 2), (1, 4), (2, 3), (3, 4)}
R3 = {(2, 1), (2, 4), (2, 3), (3,4)}
Then R1 is transitive because (1, 1), (1, 2) are in R then to be transitive relation
(1,2) must be there and it belongs to R Similarly for other order pairs.
R2 is not transitive since (1,2) and (2,3) R2 but (1,3) R2.
R3 is transitive.
DIRECTED GRAPH OF A TRANSITIVE RELATION
For a transitive directed graph, whenever there is an arrow going from one point to
the second, and from the second to the third, there is an arrow going
directly from the
first to the
third.
EXAMPLE
Let A = {1, 2, 3, 4} and define relations R1, R2 and R3 on A by the
directed graphs:
R1 = {(1, 1), (1, 2), (1, 3), (2, 3)}
R2 = {(1, 2), (1, 4), (2, 3), (3, 4)}
R3 = {(2, 1), (2, 4), (2, 3), (3,4)}
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MTH001 ­ Elementary Mathematics
1
1
2
2
4
3
4
3
R1 is transitive
R2 is not transitive since there is
an arrow from 1 to 2 and from 2
to 3 but no arrow from 1 to 3
2
1
directly
4
3
R3 is transitive
EXERCISE:
Let A = {1, 2, 3, 4} and define the null relation φ and universal
relation A ×A on A. Test these relations for reflexive, symmetric and
transitive
properties.
SOLUTION:
Reflexive:
is not reflexive since (1,1), (2,2), (3,3), (4,4) ∉ ∅.
(i)
A × A is reflexive since (a,a) A × A for all a A.
(ii)
Symmetric
For the null relation on A to be symmetric, it must
(i)
satisfy the implication:
if (a,b) ∈ ∅ then (a, b) ∈ ∅.
Since (a, b) ∈ ∅ is never true, the implication is vacuously true or true by default.
Hence is symmetric.
The universal relation A × A is symmetric, for it contains
(ii)
all
ordered pairs of elements of A. Thus,
if (a, b) A × A then (b, a) A × A for all a, b in A.
Transitive
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MTH001 ­ Elementary Mathematics
The null relation on A is transitive, because the
(i)
implication.
if (a, b) ∈ ∅ and (b, c) ∈ ∅ then (a, c) ∈ ∅ is true by default,
since the condition (a, b) ∈ ∅ is always false.
The universal relation A × A is transitive for it contains all ordered pairs of
(i)
elements of A.
Accordingly, if (a, b) A × A and (b, c) A × A then (a, c) A × A as well.
EXERCISE:
Let
A = {0, 1, 2} and
R = {(0,2), (1,1), (2,0)} be a relation on A.
1. Is R reflexive? Symmetric? Transitive?
2. Which ordered pairs are needed in R to make it a reflexive and transitive
relation.
SOLUTION:
1. R is not reflexive, since 0 A but (0, 0) R and also 2 A but (2, 2)
R.
R is clearly symmetric.
R is not transitive, since (0, 2) & (2, 0) R but (0, 0) R.
2.
For R to be reflexive, it must contain ordered pairs (0,0) and (2,2).
For R to be transitive,
we note (0,2) and (2,0) but (0,0) R.
Also (2,0) and (0,2) R but (2,2)R.
Hence (0,0) and (2,2). Are needed in R to make it a transitive relation.
EXERCISE:
Define a relation L on the set of real numbers R be defined as follows:
for all x, y R, x L y x < y.
a. Is L reflexive?
b. Is L symmetric?
c. Is L transitive?
SOLUTION:
a. L is not reflexive, because x < x for any real number x.
(e.g. 1 < 1)
L is not symmetric, because for all x, y R, if
b.
x < y then y < x
(e.g. 0 < 1 but 1 < 0)
L is transitive, because for all, x, y, z R, if x < y
c.
and y < z, then x < z.
(by transitive law of order of real numbers).
EXERCISE:
+
Define a relation R on the set of positive integers Z as follows:
for all a, b Z+, a R b iff a × b is odd.
Determine whether the relation is
a. reflexive
b. symmetric c. transitive
SOLUTION:
Firstly, recall that the product of two positive integers is
odd if and only if both of them are odd.
a. reflexive
R is not reflexive, because 2 Z+ but 2 R 2
for 2 × 2 = 4 which is not odd.
b. symmetric
R is symmetric, because
if a R b then a × b is odd or equivalently b × a is odd
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MTH001 ­ Elementary Mathematics
( b × a = a × b) b R a.
c. transitive
R is transitive, because if a R b then a × b is
odd
both "a" and "b" are odd. Also bRc means b × c is odd
both "b" and "c" are odd.
Now if aRb and bRc, then all of a, b, c are odd and so a × c
is odd.
Consequently aRc.
EXERCISE:
Let "D" be the "divides" relation on Z defined as:
for all m, n Z, m D nm|n
Determine whether D is reflexive, symmetric or transitive. Justify your answer.
SOLUTION:
Reflexive
Let m Z, since every integer divides
itself  so
m|m m Z therefore m D m m Z
Accordingly D is reflexive
Symmetric
Let m, n Z and suppose m D n.
By definition of D, this means m|n (i.e.= an integer)
Clearly, then it is not necessary that = an integer.
Accordingly, if m D n then n D m, m, n Z
Hence D is not symmetric.
Transitive
Let m, n, p Z and suppose m D n and n D p.
Now m D n m|n  = an integer.
Also n D p n|p  = an integer.
*  ⎛ p (an in) n an int)
=p
We note
=
t *(
⎜  ⎟
⎜  ⎟
m = an itn
n
m
m|p and so mDp
Thus if mDn and nDp then mDp m, n, p Z
Hence D is transitive.
EXERCISE:
Let A be the set of people living in the world today. A
binary relation R is defined on A as follows:
for all p, q A, pRq p has the same first name as q.
Determine whether the relation R is reflexive, symmetric and/or transitive.
SOLUTION:
a. Reflexive
Since every person has the same first name as his/her self.
Hence for all p A, pRp. Thus, R is reflexive.
b. Symmetric:
Let p, q A and suppose pRq.
p has the same first name as q.
q has the same first name as p.
qRp
Thus if pRq then qRp p,q A.
R is symmetric.
a. Transitive
Let p, q, s A and suppose p R q and qRr.
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MTH001 ­ Elementary Mathematics
Now pRq p has the same first name as q
and qRr q has the same first name as r.
Consequently, p has the same first name as r.
pRr
Thus, if pRq and qRs then pRr, p, q, r A.
Hence R is transitive.
EQUIVALENCE RELATION:
Let A be a non-empty set and R a binary relation on A. R is an equivalence
relation if, and only if, R is reflexive, symmetric, and transitive.
EXAMPLE:
Let A = {1, 2, 3, 4} and
R = {(1,1), (2,2), (2,4), (3,3), (4,2), (4,4)}
be a binary relation on A.
Note that R is reflexive, symmetric and transitive, hence an
equivalence
relation.
CONGRUENCES:
Let m and n be integers and d be a positive integer. The notation
m n (mod d) means that
d | (m ­ n) {d divides m minus n}.There exists an integer k such that
(m ­ n) = d k
EXAMPLE:
c. Is 22 1(mod 3)?
Is ­5 +10 (mod 3)?
b.
d. Is 7 7 (mod 3)?
Is 14 4 (mod 3)?
d.
SOLUTION
a. Since 22-1 = 21 = 3×7.
Hence 3|(22-1), and so 22 1 (mod 3)
b. Since ­ 5 ­ 10 = - 15 = 3 × (-5),
Hence 3|((-5)-10), and so  - 5 10 (mod 3)
c. Since 7 ­ 7 = 0 = 3 × 0
Hence 3|(7-7), and so 7 7 (mod 3)
d. Since 14 ­ 4 = 10, and 3 / 10 because 10 3k for any integer
k. Hence 14 4 (mod 3).
EXERCISE:
Define a relation R on the set of all integers Z as follows:
for all integers m and n, m R n m n (mod 3)
Prove that R is an equivalence relation.
SOLUTION:
1. R is reflexive.
R is reflexive iff for all m Z, m R m.
By definition of R, this means that
For all m Z, m m (mod 3)
Since m ­ m = 0 = 3 ×0.
Hence 3|(m-m), and so m m (mod 3)
mRm
R is reflexive.
2. R is symmetric.
R is symmetric iff for all m, n Z
if m R n then n R m.
mn (mod 3)
Now mRn
3|(m-n)
m-n = 3k, for some integer k.
n ­ m = 3(-k), -k Z
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MTH001 ­ Elementary Mathematics
3|(n-m)
n m (mod 3)
nRm
Hence R is symmetric.
1. R is transitive
R is transitive iff for all m, n, p Z,
if mRn and nRp then mRp
Now mRn and nRp means m n (mod 3) and n p (mod 3)
3|(m-n)
and
3|(n-p)
(m-n) = 3r
for some r, s Z
and
(n-p) = 3s
Adding these two equations, we get,
(m ­ n) + (n ­ p) = 3 r + 3 s
m ­ p = 3 (r + s),where r + s Z
3|(m ­ p)
m p (mod 3) m Rp
Hence R is transitive. R being reflexive, symmetric and transitive, is an
equivalence relation.
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MTH001 ­ Elementary Mathematics
LECTURE # 10
EXERCISE:
Suppose R and S are binary relations on a set A.
a. If R and S are reflexive, is R S reflexive?
b. If R and S are symmetric, is R S symmetric?
c. If R and S are transitive, is R S transitive?
SOLUTION:
a. R S is reflexive:
Suppose R and S are reflexive.
Then by definition of reflexive relation
a A (a,a) R and (a,a) S
a A (a,a) R S
(by definition of intersection)
Accordingly, R S is reflexive.
b. R S is symmetric.
Suppose R and S are symmetric.
To prove R S is symmetric we need to show that
a, b A, if (a,b) R S then (b,a) R S.
Suppose (a,b) R S.
(a,b) R and (a,b) S
( by the definition of Intersection of two sets )
Since R is symmetric, therefore if (a,b) R then
(b,a) R. Similarly S is symmetric, so if (a,b) S then (b,a) S.
Thus (b,a) R and (b,a) S
(b,a) R S
(by definition of intersection)
Accordingly, R S is symmetric.
c. RS is transitive.
Suppose R and S are transitive.
To prove RS is transitive we must show that
a,b,c, A, if (a,b) RS and (b,c) RS
then (a,c) RS.
Suppose (a,b) RS and (b,c) RS
(a,b) R and (a,b) S and (b,c) R and (b,c) S
Since R is transitive, therefore
if (a,b) R and (b,c) R then (a,c) R.
Also S is transitive, so (a,c) S
Hence we conclude that (a,c) R and (a,c) S
and so (a,c) RS  (by definition of intersection)
Accordingly, RS is transitive.
EXAMPLE:
Let A = {1,2,3,4}
and let R and S be transitive binary relations on A defined as:
R = {(1,2), (1,3), (2,2), (3,3), (4,2), (4,3)}
and
S = {(2,1), (2,4),(3,3)}
Then R S = {(1,2), (1,3), (2,1), (2,2), (2,4), (3,3), (4,2), (4,3)}
We note (1,2) and (2,1) RS, but (1,1) RS
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MTH001 ­ Elementary Mathematics
Hence RS is not transitive.
IRREFLEXIVE RELATION:
Let R be a binary relation on a set A. R is irreflexive iff for all aA,(a,a) R.
That is, R is irreflexive if no element in A is related to itself by R.
REMARK:
R is not irreflexive iff there is an element aA such that (a,a) R.
EXAMPLE:
Let A = {1,2,3,4} and define the following
relations on A:
R1 = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)}
R2 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)}
R3 = {(1,2), (2,3), (3,3), (3,4)}
Then R1 is irreflexive since no element of A is related to itself in R1. i.e.
(1,1)R1, (2,2) R1, (3,3) R1,(4,4) R1
R2 is not irreflexive, since all elements of A are related to  themselves in R2
R3 is not irreflexive since (3,3) R3. Note that R3 is not reflexive.
NOTE:
A relation may be neither reflexive nor irreflexive.
DIRECTED GRAPH OF AN IRREFLEXIVE RELATION:
Let R be an irreflexive relation on a set A. Then by
definition, no element of
A is
related to itself by R. Accordingly, there is no loop at each point of A in the
directed graph of R.
EXAMPLE:
Let A = {1,2,3}
and R = {(1,3), (2,1), (2,3), (3,2)} be represented by the
directed graph.
1
2
MATRIX REPRESENTATION OF AN IRREFLEXIVE RELATION
Let R be an irreflexive relation on a set A. Then by definition, no element of A is
related to itself by R.
Since the self related elements are represented by 1's on the main diagonal of the
matrix representation of the relation, so for irreflexive relation R, the matrix will
contain all 0's in its  main diagonal.
It means that a relation is irreflexive if in its matrix  representation the diagonal
elements are all zero, if one of them is not zero the we will say that the
relation is
not
irreflexive.
EXAMPLE:
Let A = {1,2,3} and R = {(1,3), (2,1), (2,3), (3,2)}
be represented
by the matrix
1 2 3
1 0 0 1
M = 2 1 0 1
3 0 1 0
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MTH001 ­ Elementary Mathematics
Then R is irreflexive, since all elements in the main diagonal are 0's.
EXERCISE:
Let R be the relation on the set of integers Z
defined as:
for all a,b Z, (a,b) R a > b.
Is R irreflexive?
SOLUTION:
R is irreflexive if for all a Z, (a,a) R.
Now by the definition of given relation R,
for all a Z, (a,a) R since a > a.
Hence R is irreflexive.
ANTISYMMETRIC RELATION:
Let R be a binary relation on a set A.R is anti-symmetric iff
a, b A if (a,b) R and (b,a) R then a = b.
REMARK:
1. R is not anti-symmetric iff there are elements a and b in A such
that (a,b) R and (b,a) R but a b.
2. The properties of being symmetric and being
anti-symmetric are not negative of each other.
EXAMPLE:
Let A = {1,2,3,4} and define the following relations on A.
R1 = {(1,1),(2,2),(3,3)}
R2 = {(1,2),(2,2), (2,3), (3,4), (4,1)}
R3={(1,3),(2,2), (2,4), (3,1), (4,2)}
R4={(1,3),(2,4), (3,1), (4,3)}
R1 is anti-symmetric and symmetric .
R2 is anti-symmetric but not symmetric because (1,2) R2but (2,1) R2.
R3 is not anti-symmetric since (1,3) & (3,1) R3 but 1 3.
Note that R3 is symmetric.
(1,3) & (3,1) R4 but 1 3 nor
R4is neither anti-symmetric because
because (2,4) R4 but (4,2) R4
symmetric
DIRECTED GRAPH OF AN ANTISYMMETRIC RELATION:
Let R be an anti-symmetric relation on a set A. Then by definition, no two distinct
elements of A are related to each other.
Accordingly, there is no pair of arrows between two distinct elements of A in the
directed graph of R.
EXAMPLE:
Let A = {1,2,3} And R be the relation defined on
A is
R ={(1,1), (1,2), (2,3), (3,1)}.Thus R is represented by the directed graph as
1
2
3
R is anti-symmetric, since there is no pair of arrows between two distinct points in
A.
MATRIX REPRESENTATION OF AN ANTISYMMETRIC RELATION:
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MTH001 ­ Elementary Mathematics
Let R be an anti-symmetric relation on a set
if (ai, aj) R for i j then (ai, aj) R.
A = {a1, a2, ..., an}. Then
Thus in the matrix representation of R there is a 1 in the ith row and jth column iff
the jth row and ith column contains 0 vice versa.
EXAMPLE:
Let A = {1,2,3} and a relation
R = {(1,1), (1,2), (2,3), (3,1)}on A be represented by the matrix.
1 2 3
1 1 1 0
M = 2 0 0 1
3 1 0 0
Then R is anti-symmetric as clear by the form of matrix M
PARTIAL ORDER RELATION:
Let R be a binary relation defined on a set A. R is a partial order relation,if and
only if, R is reflexive, antisymmetric, and transitive. The set A together with a
partial ordering R is  called a partially ordered set or poset.
EXAMPLE:
Let R be the set of real numbers and define the"less than or
equal to" , on R as follows:
for all real numbers x and y in R.x y x < y or x = y
Show that is a partial order relation.
SOLUTION:
is reflexive
For to be reflexive means that x x for all x R
But x x means that x < x or x = x and x = x is always true.
Hence under this relation every element is related to itself.
is anti-symmetric.
For to be anti-symmetric means that
x,y R, if x y and y x, then x = y.
This follows from the definition of and the trichotomy
property, which says
that
"given any real numbers x and y, exactly one of thefollowing holds:
x < y or x = y or x > y"
is transitive
For to be transitive means that
x,y,z R, if xy and y z then x z.
This follows from the definition of and the transitive property of order of real
numbers, which says that "given any real numbers x, y and z,
if x < y and y < z then x < z"
Thus being reflexive, anti-symmetric and transitive is a  partial order relation on
R.
EXERCISE:
Let A be a non-empty set and P(A) the power set of A.
Define the "subset" relation, , as follows:
for all X,Y P(A), X Y ⇔ ∀ x, iff x X then x Y.
Show that is a partial order relation.
SOLUTION:
1. is reflexive
Let X P(A). Since every set is a subset of itself, therefore
X X, X P(A).
Accordingly is reflexive.
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MTH001 ­ Elementary Mathematics
2. is anti-symmetric
Let X, Y P(A) and suppose X Y and Y X.Then by definition of equality of
two
sets it follows that X = Y.
Accordingly, is anti-symmetric.
3. is transitive
Let X, Y, Z P(A) and suppose X Y and Y Z.  Then by the transitive
property of subsets "if U V and V W then U W"it follows X Z.
Accordingly is transitive.
EXERCISE:
Let "|" be the "divides" relation on a set A of positive
integers.
That is, for all a, b A, a|b b = k a for
some integer k.
Prove that | is a partial order relation on A.
SOLUTION:
1. "|" is reflexive. [We must show that, a A, a|a]
Suppose a A. Then a = 1a and so a|a by definition of
divisibility.
2. "|" is anti-symmetric
[We must show that for all a, b A, if a|b and b|a then a=b]
Suppose a|b and b|a
By definition of divides there are integers k1, and k2 such that
b = k1 a
a = k2 b
and
Now  b = k1 a
= k1(k2 b)
(by substitution)
= (k1 k2) b
Dividing both sides by b gives
1 = k1 k2
Since a, b A, where A is the set of positive integers, so the
equations
b = k1a
a = k2 b
and
implies that k1 and k2 are both positive integers. Now the
equation
k1k2 =1
can hold only when k1 = k2 = 1
Thus a = k2b=1 b=b
i.e., a = b
3. "|" is transitive
[We must show that a,b,cA if a|b and b|c than a|c]
Suppose a|b and b|c
By definition of divides, there are integers k1 and k2 such that
b = k1 a
c = k2 b
and
Now  c = k2 b
= k2 (k1 a) (by substitution)
= (k2 k1) a (by associative law under
multiplication)
= k3 a
where k3= k2k1 is an integer
a|c
by definition of divides
Thus "|" is a partial order relation on A.
EXERCISE:
Let "R" be the relation defined on the set of integers Z as  follows:
r
for all a, b Z, aRb iff b=a for some positive integer r.
Show that R is a partial order on Z.
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MTH001 ­ Elementary Mathematics
SOLUTION:
Let a, b Z and suppose aRb and bRa. Then there are positive
integers r and s such that
r
s
b = a  and
a=b
s
Now,
a=b
r s
= (a )
by substitution
rs
=a
 rs =1
Since r and s are positive integers, so this equation can hold if, and only if,
r
=1 and s = 1
1
and then a = bs = b = b
i.e., a = b
Thus R is anti-symmetric.
3. Let a, b, c Z and suppose aRb and bRc.
Then there are positive integers r and s such that
r
s
b = a  and
c=b
r
Now  c = b
r s
= (a )
(by substitution)
rs
t
=a =a
(where t = rs is also a positive integer)
Hence by definition of R, aRc. Therefore, R is transitive.
Accordingly, R is a partial order relation on Z.
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Table of Contents:
  1. Recommended Books:Set of Integers, SYMBOLIC REPRESENTATION
  2. Truth Tables for:DE MORGAN’S LAWS, TAUTOLOGY
  3. APPLYING LAWS OF LOGIC:TRANSLATING ENGLISH SENTENCES TO SYMBOLS
  4. BICONDITIONAL:LOGICAL EQUIVALENCE INVOLVING BICONDITIONAL
  5. BICONDITIONAL:ARGUMENT, VALID AND INVALID ARGUMENT
  6. BICONDITIONAL:TABULAR FORM, SUBSET, EQUAL SETS
  7. BICONDITIONAL:UNION, VENN DIAGRAM FOR UNION
  8. ORDERED PAIR:BINARY RELATION, BINARY RELATION
  9. REFLEXIVE RELATION:SYMMETRIC RELATION, TRANSITIVE RELATION
  10. REFLEXIVE RELATION:IRREFLEXIVE RELATION, ANTISYMMETRIC RELATION
  11. RELATIONS AND FUNCTIONS:FUNCTIONS AND NONFUNCTIONS
  12. INJECTIVE FUNCTION or ONE-TO-ONE FUNCTION:FUNCTION NOT ONTO
  13. SEQUENCE:ARITHMETIC SEQUENCE, GEOMETRIC SEQUENCE:
  14. SERIES:SUMMATION NOTATION, COMPUTING SUMMATIONS:
  15. Applications of Basic Mathematics Part 1:BASIC ARITHMETIC OPERATIONS
  16. Applications of Basic Mathematics Part 4:PERCENTAGE CHANGE
  17. Applications of Basic Mathematics Part 5:DECREASE IN RATE
  18. Applications of Basic Mathematics:NOTATIONS, ACCUMULATED VALUE
  19. Matrix and its dimension Types of matrix:TYPICAL APPLICATIONS
  20. MATRICES:Matrix Representation, ADDITION AND SUBTRACTION OF MATRICES
  21. RATIO AND PROPORTION MERCHANDISING:Punch recipe, PROPORTION
  22. WHAT IS STATISTICS?:CHARACTERISTICS OF THE SCIENCE OF STATISTICS
  23. WHAT IS STATISTICS?:COMPONENT BAR CHAR, MULTIPLE BAR CHART
  24. WHAT IS STATISTICS?:DESIRABLE PROPERTIES OF THE MODE, THE ARITHMETIC MEAN
  25. Median in Case of a Frequency Distribution of a Continuous Variable
  26. GEOMETRIC MEAN:HARMONIC MEAN, MID-QUARTILE RANGE
  27. GEOMETRIC MEAN:Number of Pupils, QUARTILE DEVIATION:
  28. GEOMETRIC MEAN:MEAN DEVIATION FOR GROUPED DATA
  29. COUNTING RULES:RULE OF PERMUTATION, RULE OF COMBINATION
  30. Definitions of Probability:MUTUALLY EXCLUSIVE EVENTS, Venn Diagram
  31. THE RELATIVE FREQUENCY DEFINITION OF PROBABILITY:ADDITION LAW
  32. THE RELATIVE FREQUENCY DEFINITION OF PROBABILITY:INDEPENDENT EVENTS