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 |
|
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 |
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||||||||||||||||||||||||||||||
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 |