Mathematical-Reasoning · Unit 29 · Video 3 · Interactive Practice

Addition and Multiplication Tables for ℤ₂, ℤ₃ and ℤ₅

IKey Formulas

Formula	Meaning
$[x] \oplus [y] = [r]$ , $\ x + y \equiv r \pmod n$ , $\ 0 \le r \le n - 1$	an addition-table entry: add the representatives, keep the remainder
$[x] \odot [y] = [r]$ , $\ x y \equiv r \pmod n$ , $\ 0 \le r \le n - 1$	a multiplication-table entry: multiply, keep the remainder
$[x] \oplus [y] = [y] \oplus [x]$ , $\ [x] \odot [y] = [y] \odot [x]$	both tables are symmetric across the main diagonal
$[n-1] = [-1]$ , so $[n-1] \odot [y] = [-y]$	the last row of a multiplication table is quick to write

Key Insight: In an addition table each class appears exactly once in every row and every column; a multiplication table has no such pattern, since row $[0]$ holds only $[0]$ .

IIReading a Cell of ℤₙ

Each cell combines its row label with its column label and reduces to a remainder from $0$ to $n - 1$ .

n Table

IIIThe Last Row: [n − 1] = [−1]

Because $[n-1] = [-1]$ , row $[n-1]$ of the multiplication table is $[0], [-1], [-2], \ldots$ reduced.

n 5

IVQuiz Questions

Problem 1 · One Cell, Two Tables

In $\mathbb{Z}_5$ , write the entry in row $[2]$ , column $[4]$ of each table the way the table shows it: $[r]$ with $0 \le r \le 4$ .

Addition table: $[2] \oplus [4] = \ ?$

Multiplication table: $[2] \odot [4] = \ ?$

✅ Correct!

2 + 4 = 6 = 5 + 1

and

2 \cdot 4 = 8 = 5 + 3

❌ Not quite.

[6] = [1]

is true, but a table entry is written as

[r]

with

0 \le r \le 4

❌ Not quite.

[8] = [3]

is true, but a table entry is written as

[r]

with

0 \le r \le 4

❌ Check the sum. Add

2 + 4

and keep the remainder on division by

5

❌ Check the product. Multiply

2 \cdot 4

and keep the remainder on division by

5

Show solution

[2] \oplus [4] = [6], \quad 6 = 5 \cdot 1 + 1 \ \Rightarrow\ [1]; \qquad [2] \odot [4] = [8], \quad 8 = 5 \cdot 1 + 3 \ \Rightarrow\ [3].

Problem 2 · A Whole Row of ℤ₅

In the multiplication table for $\mathbb{Z}_5$ , the columns are $[0], [1], [2], [3], [4]$ in that order. Which list is row $[2]$ ?

✅ Correct! The products

0, 2, 4, 6, 8

reduce to

0, 2, 4, 1, 3

❌ Not quite. The last two products,

6

and

8

, still need reducing modulo

5

❌ Not quite. That is row

[2]

of the addition table.

❌ Not quite. The last two are swapped:

2 \cdot 3 = 6 \equiv 1

and

2 \cdot 4 = 8 \equiv 3

Show solution

Row $[2]$ holds $[2] \odot [y] = [2y]$ for $y = 0, 1, 2, 3, 4$ :

[0], [2], [4], [6] = [1], [8] = [3].

Each nonzero class appears once in this row — in $\mathbb{Z}_5$ that happens for every nonzero row.

Problem 3 · Where [0] Sits in ℤ₇

In the addition table for $\mathbb{Z}_7$ , in which column(s) of row $[3]$ does $[0]$ appear?

✅ Correct!

3 + 4 = 7 \equiv 0

, and each class appears exactly once in a row of an addition table.

❌ Not quite.

3 + 3 = 6

, so that cell is

[6]

❌ Not quite. Column

[0]

of the addition table copies the row label:

[3] \oplus [0] = [3]

❌ Not quite. Every class appears once in every row of an addition table,

[0]

included.

Show solution

Row $[3]$ is $[3 + y]$ for $y = 0, \ldots, 6$ : $[3], [4], [5], [6], [0], [1], [2]$ . The entry $[0]$ is in column $[4]$ , because $3 + 4 = 7 = 7 \cdot 1 + 0$ , and nowhere else.

Problem 4 · Counting Zeros in ℤ₄

In the multiplication table for $\mathbb{Z}_4$ , consider only the rows and columns for $[1], [2], [3]$ . How many of those nine entries are $[0]$ ?

✅ Correct! Only

[2] \odot [2] = [4] = [0]

❌ Not quite. Check

[2] \odot [2] = [4]

, and

4 \equiv 0 \pmod 4

❌ Not quite.

[2] \odot [3] = [6] = [2]

, not

[0]

6 = 4 + 2

❌ Not quite. Write out the nine products and reduce each modulo

4

Show solution

The nonzero block of the table for $\mathbb{Z}_4$ is

$\odot$	$[1]$	$[2]$	$[3]$
$[1]$	$[1]$	$[2]$	$[3]$
$[2]$	$[2]$	$[0]$	$[2]$
$[3]$	$[3]$	$[2]$	$[1]$

using $4 \equiv 0$ , $6 \equiv 2$ and $9 \equiv 1 \pmod 4$ . Exactly one entry is $[0]$ : two nonzero classes with product $[0]$ .

Solved: 0 / 4