3.7 Modular Arithmetic
3.7.1 Introduction:
Let us
look at the Calendar of May 2006 which is given on the right hand side. 1st
date of the month is Monday. What day is 29th of that month? It is Monday
again. How did we arrive at that? If 1st
is Monday then next Mondays are 1,8,15,22,29
Why is this so? This is
because days repeat after every 7 days. Mathematically we say 81(modulo 7) which implies that 7 exactly divides (8-1) 15 1(mod 7) which implies that 7 exactly divides (15-1) 22 1(mod 7) which implies that 7 exactly divides (22-1) In
general when xy(mod m ) we mean that m exactly divides (x-y). The above
statement can also be represented by If x-y 0(mod m) then (x-y)/m = is an integer. ‘’ is pronounced as congruent |
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We know that when we divide
any positive integer by another positive integer say ‘m’, the reminders has to
be among the set {0,1,2,3,4…..(m-1)}.
The numbers 0,1,2,….(m-1) are called residues of mod m.
The set Zm= {0,1,2,3,4…..(m-1)}is called the set of mod m.
Definition
: The ‘residue set’ of mod m is the possible set of reminders when a number
is divided by m and is represented by
Zm= {0,1,2,3,4…..(m-1)}
3.7.1 Example 1: Write the residues when a positive integer is
divided by 10
Solution :
The possible reminders when
10 is a divisor are
0,1,2,3….9
Therefore Z10=
{0,1,2,3,4,5,6,7,8,9}
Will it be possible to
define addition and multiplication operations under the
modulo system?
Definition: If mod is m then
The symbol for addition
is with m as subscript
to the symbol.
The symbol for
multiplication is with m as subscript to the symbol.
We define
1. a +m b r (=reminder of
(a+b)/m)
2. a m b r (=reminder of
(a*b)/m)
3.7.1 Example 2 :
Find 10 +12 2 +12 3
Solution :
10 +12 2 +12
3
=(10
+12 2) +12 3
=0+12 3(reminder of (10+2)/12 = 0)
= 3(reminder of 3/12 is 3)
3.7.1 Example 3 :
Find 4 11 3 11 7
Solution :
4 11 3 11 7
=(4
11 3) 11 7
= 1 11 7(reminder of (4*3)/11 = 1)
=7 (reminder of (1*7)/11 = 7)
3.7.1 Example 4 : What is
the value of y if yy 1(mod 8)
Solution :
Since yy 1(mod 8)
8 should divide y*y with
reminder = 1
(y2-1)/8
= 0.
This is possible only if
y=3.
Verification:
38 3 =1(reminder of (3*3)/8 = 1) which is as given in the problem.
3.7.2 Caley’s Table:
The Caley’s table is a
representation of modular arithmetic operation.
The results of modulo
operations (addition and multiplication) on a given set, when represented in a
table format is called Caley’s table.
Let us represent Caley’s
table for addition operation (ab) for mod 4.
Since
the residues of mod 4 can only be 0 or 1 or 2 or 3, we calculate a+4b only for a =0,1,2,3 and b=
0,1,2,3.The set on which Caley’s table is arrived at is Z4= {0,1,2,3} |
The
results of the operation [ab mod 4] (denoted
by +4) |
||||||||||||||||||||||||||||||
0 +4 0 0 0 +4 1 1
0 +4 2 2 0 +4 3 3 1 +4
0 1 1 +4 1 2 1
+4 2 3 1 +4 3 0 2
+4 0 2 2 +4 1 3 2 +4
2 0 2 +4 3 1 3 +4 0 3 3 +4 1 0 3 +4
2 1 3 +4 3 2 |
|
||||||||||||||||||||||||||||||
Observe that the results in the table (Green Color Numbers) also belong to the set Z4= {0,1,2,3} |
Since the
residues of mod 4 can only be 0 or 1 or 2 or 3, we calculate a X4b
only for a =0,1,2,3 and b=
0,1,2,3.The set on which Caley’s table is arrived at is Z4= {0,1,2,3} |
The
results of the operation [ab mod4] (denoted by X4) |
||||||||||||||||||||||||||||||
0 X4 0 0 0 X4 1 0
0 X4 2 0 0 X4 3 0 1 X4
0 0 1 X4 1 1 1
X4 2 2 1 X4 3 3 2
X4 0 0 2 X4 1 2 2 X4
2 0 2 X4 3 2 3 X4 0 0 3 X4 1 3 3 X4
2 2 3 X4 3 1 |
|
||||||||||||||||||||||||||||||
Observe that the results in the table (Green Color Numbers) also belong to the set Z4= {0,1,2,3} |
3.7.2 Problem 1
: Construct
Caley’s Table for Q ={0,2,4,6,8} under mod 10
Solution
:
We are required to arrive
at values of ab mod 10 when a, b to the set Q= {0,2,4,6,8}
Since 0 is reminder of
(6+4)/10, 2 is reminder of (6+6)/10 and
4 is reminder of (8+6)/10
We get
6
+10 4 0 6 +10 6 2 8 +10
6 4
Similarly we can arrive at
the remainders for other values of a and b.
bà |
0 |
2 |
4 |
6 |
8 |
a |
a +10 b= |
||||
0 |
0 |
2 |
4 |
6 |
8 |
2 |
2 |
4 |
6 |
8 |
0 |
4 |
4 |
6 |
8 |
0 |
2 |
6 |
6 |
8 |
0 |
2 |
4 |
8 |
8 |
0 |
2 |
4 |
6 |
3.7.2 Problem 2
: Construct
Caley’s Table for A = {1,5,7,11} under mod 12
We are required to arrive
at values of ab mod 12 when a, b to the set A ={1,5,7,11}
Since 1 is reminder of
(7*7)/12 , 5 is reminder of (7*11)/12 and 1 is
reminder of (11*11)/12
We get
7
127 1 7 12 11 5 11 12 11 1
Similarly we can arrive at
the remainders for other values of a and b.
bà |
1 |
5 |
7 |
11 |
a |
a 12 b = |
|||
1 |
1 |
5 |
7 |
11 |
5 |
5 |
1 |
11 |
7 |
7 |
7 |
11 |
1 |
5 |
11 |
11 |
7 |
5 |
1 |
3.7 Summary of learning
No. |
Points learnt |
1 |
Definition
of Modulo operations |
2 |
Modulo
addition and multiplication |
3 |
Caley’s
Table |