Exploring Z7
The mathematical concept that simplifies each of the calendar questions is basic to
arithmetic modulo seven (mod 7). The following addition table in mod 7 has been
started, replacing Sunday, Monday etc with 0, 1, 2, 3, 4, 5, and 6.
1. Figure out how the table entries were calculated. Then complete the table.
+
0
1
2
3
4
5
6
0
0
1
2
1
1
2
2
3
3
4
5
5
6
6
6
0
1
2
4
3
2. Use the mod 7 addition table to complete each of the following open sentences.
a. 5 + 3 =
b. 3 + 5 =
c. 5 +
=4
d. 3 +
=5
e.
+3=1
f.
+6=0
g. 4 + (3 + 6) =
h. (4 + 3) + 6 =
3. We will use (Z7, +) to reference the mathematical system of the set of numbers
Z7 = {0,1,2,3,4,5, 6} together with the operation of addition. (Z, +) will refer to
the familiar set of Integers and ordinary addition.
The tables for mod 7 addition and ordinary integer addition have many
similarities and some striking differences.
a. List some of the patterns you see in (Z7, +).
b. Compare and contrast (Z7, +) and (Z, +).
Mod 7
1
4. In each of the following, find values (if possible) of a, b and/or c in (Z7, +) that
make the given sentences true statements.
a. a + 5 = 5
b. a + a = 3
c. a + b = a
d. 3 + c = 4
e. 3 + c = 2
f. a + b = a + c
5. For each of the following statements, determine if it is True or False. If you
actually prove that a statement is true by considering all possibilities, write
proven next to it.
a. a + a = a for all a in Z7.
b. The set (Z7, +) has an identity element.
c. For each a in Z7, there is an element x in Z7 such that a + x = 0. [Note: if
a + x = 0 we will write x = −a and read “x is the additive inverse of a.”
d. There are two different elements x and y in Z7 such that 3 + x = 0 and 3 +
y = 0.
e. For every two elements a and b in Z7, a + b is an element of Z7.
f. −(−𝑎) = 𝑎 for all a in Z7.
g. a + b = b + a for all elements a and b in Z7.
h. a + (−𝑏) = b + (−a) for all a, b in Z7.
i. − (a + b) = −a + −b for all a,b in Z7.
j. a + (b + c) = (a + b) + c for all elements a, b, c in Z7.
k. If a = −b then b = −a.
Mod 7
2
6. The same idea used to produce the (Z7, +) table can be extended to
multiplication. Study the entries already given in the mod 7 multiplication table
that follows, then complete the table.
X
0
1
2
3
4
5
6
0
1
0
2
0
3
0
4
5
6
6
2
0
1
5
3
1
6
3
4
6
3
4
7. Use your table to complete the following open sentences.
a. 5  1 =
b.
4=4
c.
 6= 0
d. 4 
=1
e.
5=1
f. 3  (2  5) =
g. (3  2)  5 =
h. (4 
)6=3
i. (−3)  (− 2)
j. 3  2 =
k. 4  (
+ 5) = 4  6 + 4  5
8. As with addition, mod 7 multiplication and ordinary integer multiplication have
many similarities and differences. Compare and contrast (Z7, ) and (Z, ).
Mod 7
3
9. In each of the following, find values of a, b and/or c in Z7 that make the given
sentences true.
a. a  b = a
b. c  c = 1
c. −a  b = a  −b
d. a  b = a  c
e. a  a = a
10. For each of the following statements, decide whether it is True or False. If you
actually prove that a statement is true by considering all possibilities, write
proven.
There is an identity element in (Z7, ).
For any elements a, b in Z7, a  b is an element in Z7.
There is an x in Z7 such that x  0 = 1.
For all a, b, c in Z7, if a  b = a  c and a 0, then b = c.
a  b = a2  b2 for some a, b in Z7 such that a b.
a  b = b  a for all elements a and b in Z7.
a  (b  c) = (a  b)  c for all elements a, b, and c in Z7.
For each a in Z7, there is an element x in Z7 such that a  x = 1. [Note: if
a  x = 1 we will write x = 𝑎−1 and read “x is the multiplicative inverse of
a.”
i. There are two different elements x and y in Z7 such that
6  x = 1 and 6  y = 1.
j. For all non-zero elements a, b in Z7, if a = 𝑏 −1 then b = 𝑎−1 .
k. For all a, b in Z7, if a 0 and b  0, then a  b  0.
l. a  (b  𝑎−1 ) = b for all a, b in Z7 with a 0.
a.
b.
c.
d.
e.
f.
g.
h.
Mod 7
4
11. Use your addition and multiplication tables to solve the following equations for x
in mod 7:
a. 5x = 4
b. 2x = 0
c. 5 + x = 1
d. 2x + 3 = 5
e. 2x = 3
f. x2 = 2
g. x2 = 3
h. −2x = 1
i. Can you always solve an equation of the form ax + b = c? Explain.
12. Refer back to the Calendar Problems, # 7. In mod 7, all the numbers divisible by 7
give you the same day of the week; in numbers, that means they all give you a 0.
In mod 7 arithmetic notation, we write this, for example, as:
49  0 (mod 7) because 49 divided by 7 leaves a remainder
of 0.
We read this as “49 is congruent to 0 mod 7.” Likewise 15  1 (mod 7) because
15 divided by 7 leaves a remainder of 1.
Solve the following equations using this notation.
a.
b.
c.
d.
e.
f.
Mod 7
10  x (mod 7)
7  x (mod 7)
29  x (mod 7)
−3  x (mod 7)
x  4 (mod 7). How many different solutions can you find?
x−3  1 (mod 7)
5