2. LEARNING
OBJECTIVES:
a. Define modular
arithmetic and perform
addition and
multiplication using the
table for elements.
b. Solve modular
arithmetic problems and
list the properties of real
numbers under in.
3. MODULAR
ARITHMETIC
> is a system of arithmetic for
integers, where numbers “wrap
around” upon reaching a certain
value – the modulus.
>The modern approach to
modular arithmetic was
developed by Carl Friedrich
4. k= Mq +r where 0 < r <
M
When k is positive, simply divide k by M to obtain the
remainder r.
5. k= Mq +r where 0 < r <
M
When k is positive, simply divide k by M to obtain the
remainder r.
25
(mod7)
= 4
6. k= Mq +r where 0 < r <
M
When k is positive, simply divide k by M to obtain the
remainder r.
25
(mod7)
= 4
25 = 7 (3) + 4
7. k= Mq +r where 0 < r <
M
When k is positive, simply divide k by M to obtain the
remainder r.
25
(mod7)
= 4
25
(mod5)
= 0
25 = 7 (3) + 4
8. k= Mq +r where 0 < r <
M
When k is positive, simply divide k by M to obtain the
remainder r.
25
(mod7)
= 4
25
(mod5)
= 0
25 = 7 (3) + 4 25 = 5 (5) + 0
9. k= Mq +r where 0 < r <
M
When k is positive, simply divide k by M to obtain the
remainder r.
25
(mod7)
= 4
25
(mod5)
= 0
25 = 7 (3) + 4 25 = 5 (5) + 0
8
(mod11)
= 8
10. k= Mq +r where 0 < r <
M
When k is positive, simply divide k by M to obtain the
remainder r.
25
(mod7)
= 4
25
(mod5)
= 0
25 = 7 (3) + 4 25 = 5 (5) + 0 8 = 11 (0) + 8
8
(mod11)
= 8
11. k= Mq +r where 0 < r <
M
When k is positive, simply divide k by M to obtain the
remainder r.
25
(mod7)
= 4
25
(mod5)
= 0
25 = 7 (3) + 4 25 = 5 (5) + 0 8 = 11 (0) + 8
8
(mod11)
= 8
0 (mod5)
= 0
12. k= Mq +r where 0 < r <
M
When k is positive, simply divide k by M to obtain the
remainder r.
25
(mod7)
= 4
25
(mod5)
= 0
25 = 7 (3) + 4 25 = 5 (5) + 0 8 = 11 (0) + 8 0 = 5 (0) + 0
8
(mod11)
= 8
0 (mod5)
= 0
13. • Example: Construct the table for the
elements of X and Y in modulo 10.
a. X={2,4,6,8} under multiplication.
b. Y={1,3,5,7,9} under multiplication.
14. d. Determine if each set is closed under
the indicated operation.
e. Find the identity for each set.
f. Find the inverse of every element of
each set.
17. • 2 4 6 8
2 4 8 2 6
4 8 6 4 2
6 2 4 6 8
8 6 2 8 4
d. The set X is closed under
multiplication
e. The identity of X is 6
f. For the inverse of every element of
set X:
The inverse of 2 is 8, since 2•8 = 6
The inverse of 4 is 4, since 4•4 = 6
The inverse of 6 is 6, since 6•6 = 6
The inverse of 8 is 2, since 8•2 = 6
19. • 1 3 5 7 9
1 1 3 5 7 9
3 3 3 5 1 7
5 5 5 5 5 5
7 7 1 5 9 3
9 9 7 5 3 1
d. The set Y is closed under
multiplication
e. The identity of Y is 1
f. For the inverse of every element of
set Y:
The inverse of 1 is 1, since 1•1 = 1
The inverse of 3 is 7, since 3•7 = 1
The inverse of 7 is 3, since 7•3 = 1
The inverse of 9 is 9, since 9•9 = 1
5 has no multiplicative inverse, such
that 5• x = 1
22. Example: If S={(0,0),(1,0),(0,1),
(1,1), where addition + and
multiplication • is defined
under modulo 2 as (a,b)+
(c,d)=(a+c,b+d),and
(a,b)•(c,d)=(ac, bd), then form
the tables and discuss the
closure property.
26. ACTIVITY:
Construct the table for the elements of:
a. S={1,2,4,8} under multiplication modulo 15
b. Z= {0,1,2,3,4,5,6,7} under addition modulo 8
Then, test the properties of real numbers;
Closure
Associative
Identity
Inverses
