1. Arithmetic MOD 7 a finite algebra
S = { 0, 1, 2, 3, 4, 5, 6 }
for x and y in S:
x  y = the remainder when x + y is divided by 7
and
x  y = the remainder when x  y is divided by 7

2.  1 2 3 4 5 6  1 2 3 4 5 6

3.  1 2 3 4 5 6  1 2 3 4 5 6
Both tables have CLOSURE

4.  1 2 3 4 5 6  1 2 3 4 5 6
2  5 = 5  2
3  6 = 6  3 4
etc
Both tables have COMMUTATIVITY

5.  1 2 3 4 5 6  1 2 3 4 5 6 Both tables have ASSOCIATIVITY1 2 3 4 5 6  1 2 3 4 5 6
Both tables have ASSOCIATIVITY
(46)  5 = 4(6  5)
3  5 = 4 4
1 =
(36)  4 = 3 (64)
4  4 = 3  3
2 = 2
etc

6.  1 2 3 4 5 6  1 2 3 4 5 6 0 is the IDENTITY for addition1 2 3 4 5 6  1 2 3 4 5 6
0 is the IDENTITY
for addition
x  0 = 0  x = x
1 is the IDENTITY
for multiplication
x  1 = 1  x = x

7.  1 2 3 4 5 6  1 2 3 4 5 6 0 is the IDENTITY for addition1 2 3 4 5 6  1 2 3 4 5 6
0 is the IDENTITY
for addition
1 is the IDENTITY
for multiplication
2 and 5 are
additive INVERSES
3 and 5 are
multiplicative INVERSES

8.  1 2 3 4 5 6  1 2 3 4 5 6 Similar to ordinary real number algebra,1 2 3 4 5 6  1 2 3 4 5 6
Similar to ordinary real number algebra,
there are two operations: “addition” and “multiplication”.
The inverse operations “subtraction” and “division” are defined:
a – b = a + (the additive inverse of b)
eg: 1 – 5 = = 3

9.  1 2 3 4 5 6  1 2 3 4 5 6 Similar to ordinary real number algebra,1 2 3 4 5 6  1 2 3 4 5 6
Similar to ordinary real number algebra,
there are two operations: “addition” and “multiplication”.
The inverse operations “subtraction” and “division” are defined:
a ÷ b = a  (the multiplicative inverse of b)
eg: 2 ÷ 5 = 2  3 = 6

10.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
3x = 2

11.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
3x = 2
3x = 2
+ 2
= 4

12.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
3x = 2
3x = 4
(5)
( 3x) = (4)
= 6 1

13.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
3x = 2
3x = 4
(5)( 3x) = (5)(4)
x = 6

14.  1 2 3 4 5 6  1 2 3 4 5 6

15.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0

16.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0
Use the quadratic formula with: a=2 ; b=4 ; c=5
-b  Vb2 – 4ac 2a

17.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0
Use the quadratic formula with: a=2 ; b=4 ; c=5
-b  Vb2 – 4ac 2a
-4 = additive inverse of 4 = 3 3

18.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0
Use the quadratic formula with: a=2 ; b=4 ; c=5
3  Vb2 – 4ac
42 = 4  4 = 2 2 2a

19.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0
Use the quadratic formula with: a=2 ; b=4 ; c=5
3  V2 – 4ac 2a 5
4ac = 4(2)(5)= 5

20.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0
Use the quadratic formula with: a=2 ; b=4 ; c=5
3  V2 – 5 2a 4
2 – 5 = = 4

21.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0
Use the quadratic formula with: a=2 ; b=4 ; c=5 2
3  √ 4
√ 4 = 2 or 5 2a
3 - 2 = 3 + 5

22.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0
Use the quadratic formula with: a=2 ; b=4 ; c=5
3  2
2a =2(2) = 4 4 2a

23.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0
Use the quadratic formula with: a=2 ; b=4 ; c=5
3  2 4

24.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0
Use the quadratic formula with: a=2 ; b=4 ; c=5 1 5 = = = 4 4 4 4 4

25.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0
Use the quadratic formula with: a=2 ; b=4 ; c=5 1
= 1 2 5
= 5 2 = = = 4 4 4 4 4
To divide by 4 you must multiply by 2 (the mult inv of 4)

26.  1 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0
Use the quadratic formula with: a=2 ; b=4 ; c=5
1 2 =2
5 2 =3

27. x =2 x =3  1 2 3 4 5 6  1 2 3 4 5 6 Solve for x : 2 x2 + 4x + 5 = 01 2 3 4 5 6  1 2 3 4 5 6
Solve for x :
2 x x = 0
the solutions are:
x =2
x =3 or