Mathematical-Reasoning · Unit 29 · Video 2 · Interactive Practice

Adding Classes in ℤₙ: Why ⊕ and ⊙ Are Well Defined

IKey Formulas

Formula	Meaning
$\mathbb{Z}_n = \{[0], [1], \ldots, [n-1]\}$	the integers modulo $n$ : exactly $n$ congruence classes, each an infinite set
$[a] = [b] \iff a \equiv b \pmod n$	two integers name the same class exactly when they are congruent
$[x] \oplus [z] = [x + z]$	class addition, defined through integer addition
$[x] \odot [z] = [x z]$	class multiplication, defined through integer multiplication

Key Insight: If $[x] = [y]$ and $[z] = [w]$ , then $[x + z] = [y + w]$ and $[xz] = [yw]$ — the corollary on congruence that makes $\oplus$ and $\odot$ well defined.

IIEvery Choice of Representatives

Whichever representatives of $[a]$ and $[c]$ we compute with, the sum and the product land in one class.

n First class Second class Operation

IIIA Rule That Is Not Well Defined

The rule $[x]^{[z]} = [x^z]$ on $\mathbb{Z}_5$ gives one pair of classes several different answers.

Rule

💡 Part three of the corollary is a different statement: a class raised to a fixed natural number $m$ , as in $[2]^2 = [7]^2 = [4]$ in $\mathbb{Z}_5$ , is well defined.

IVQuiz Questions

Problem 1 · Computing in ℤ₇

In $\mathbb{Z}_7$ , write each result as $[r]$ with $0 \le r \le 6$ .

$[4] \oplus [5] = \ ?$

$[4] \odot [5] = \ ?$

✅ Correct!

[4 + 5] = [9] = [2]

and

[4 \cdot 5] = [20] = [6]

, since

9 = 7 + 2

and

20 = 7 \cdot 2 + 6

❌ Not quite. That is the result of the other operation:

\oplus

adds the representatives,

\odot

multiplies them.

❌ Not quite.

9 = 6 + 3

reduces modulo

6

; here the modulus is

7

❌ Check the sum.

[4] \oplus [5] = [9]

; reduce

9

modulo

7

❌ Check the product.

[4] \odot [5] = [20]

; reduce

20

modulo

7

Show solution

By the definitions, $[4] \oplus [5] = [4 + 5] = [9]$ and $[4] \odot [5] = [4 \cdot 5] = [20]$ . Reduce modulo $7$ :

9 = 7 \cdot 1 + 2 \Rightarrow [9] = [2], \qquad 20 = 7 \cdot 2 + 6 \Rightarrow [20] = [6].

Problem 2 · Names of One Class

In $\mathbb{Z}_5$ , which of these is not equal to $[13]$ ?

✅ Correct!

13 - (-3) = 16

is not a multiple of

5

; in fact

-3 \equiv 2 \pmod 5

, so

[-3] = [2]

❌ Not quite.

13 - 3 = 10 = 5 \cdot 2

, so

[13] = [3]

❌ Not quite.

13 - 8 = 5

, so

[13] = [8]

❌ Not quite.

13 - (-2) = 15 = 5 \cdot 3

, so

[13] = [-2]

: negative integers belong to the classes too.

Show solution

$[13] = [b]$ exactly when $5 \mid (13 - b)$ . The differences are

13 - 3 = 10, \quad 13 - 8 = 5, \quad 13 - (-3) = 16, \quad 13 - (-2) = 15.

Only $16$ is not a multiple of $5$ , so $[-3] \ne [13]$ . Since $-3 = 5 \cdot (-1) + 2$ , the class $[-3]$ is $[2]$ .

Problem 3 · Other Representatives in ℤ₆

In $\mathbb{Z}_6$ , $[8] = [2]$ and $[-5] = [1]$ . Compute $[8] \odot [-5]$ and write it as $[r]$ with $0 \le r \le 5$ .

✅ Correct!

[-40] = [2]

, the same as

[2] \odot [1] = [2]

— exactly what well-definedness promises.

❌ Close, but check the sign.

-40 = 6 \cdot (-6) - 4

has

-4

, not a remainder;

-40 = 6 \cdot (-7) + 2

❌ Not quite.

8 + (-5) = 3

is the sum;

\odot

multiplies the representatives.

❌ Not quite. Compute

8 \cdot (-5)

, then find its remainder on division by

6

Show solution

By the definition, $[8] \odot [-5] = [8 \cdot (-5)] = [-40]$ . Since $-40 = 6 \cdot (-7) + 2$ ,

[8] \odot [-5] = [-40] = [2].

With the representatives $2$ and $1$ instead: $[2] \odot [1] = [2]$ . The answers agree, as part two of the corollary guarantees.

Problem 4 · Is the Rule an Operation?

On $\mathbb{Z}_5$ , try to define $[x]^{[z]} = [x^z]$ , using natural numbers as representatives in the exponent. Is this a well-defined operation?

✅ Correct! One pair of classes,

([2], [1])

, gets two different answers, so the rule defines no operation.

❌ Not quite. The power property fixes the exponent

m

and changes only the base; here the exponent's representative changes.

❌ Not quite. Changing the base is harmless; the trouble is changing the exponent from

1

6

❌ Not quite.

\oplus

is well defined on every

\mathbb{Z}_n

; the failure lies in this exponent rule itself.

Show solution

Since $6 \equiv 1 \pmod 5$ , the classes $[1]$ and $[6]$ are equal. Using the representatives $2$ and $1$ , the rule gives $[2^1] = [2]$ . Using $2$ and $6$ , it gives $[2^6] = [64]$ , and $64 = 5 \cdot 12 + 4$ , so $[64] = [4]$ .

[2]^{[1]} = [2] \quad\text{but}\quad [2]^{[6]} = [4], \qquad [1] = [6].

The same input has two outputs, so the rule is not well defined.

Solved: 0 / 4