Mathematical-Reasoning · Unit 28 · Video 2 · Interactive Practice

Three Remainders, Three Sets: The Integers Modulo 3

IKey Formulas

Formula	Meaning
$C[k] = \{ x \in \mathbb{Z} \mid x \equiv k \pmod{3} \}$	the integers congruent to $k$ modulo $3$
$x = 3q + r$ , $\ r \in \{0, 1, 2\}$	the Division Algorithm: $x \in C[r]$
$C[0] \cup C[1] \cup C[2] = \mathbb{Z}$	every integer has a remainder $0$ , $1$ or $2$
$C[0] \cap C[1] = C[0] \cap C[2] = C[1] \cap C[2] = \emptyset$	no integer has two remainders

Key Insight: A new label is not a new set: $C[3] = C[0]$ and $C[4] = C[1]$ , because $3 \equiv 0$ and $4 \equiv 1 \pmod{3}$ .

IIPlacing an Integer

Write $x = 3q + r$ with $0 \le r \lt 3$ ; the remainder $r$ names the one set that holds $x$ .

Integer x −14

IIIAny Label, Only Three Sets

$C[k]$ is one of $C[0]$ , $C[1]$ , $C[2]$ : the one named by the remainder of $k$ .

Label k 4

IVQuiz Questions

Problem 1 · Place a Negative Integer

Which set contains $-14$ ?

✅ Correct!

-14 = 3(-5) + 1

, so the remainder is

1

❌ That copies the remainder of $14$ .

14 = 3 \cdot 4 + 2

, but negating gives

-14 = 3(-4) - 2

, and

-2

is not a remainder.

❌ A remainder must be $0$ , $1$ or $2$ . Go one multiple of

3

lower:

-14 = 3(-5) + 1

❌ Not quite.

3

does not divide

-14

Show solution

We need $-14 = 3q + r$ with $0 \le r \lt 3$ . The multiple of $3$ just below $-14$ is $-15 = 3(-5)$ , so

-14 = 3(-5) + 1.

The remainder is $1$ , hence $-14 \in C[1]$ . Check: $-14 - 1 = -15 = 3(-5)$ , so $-14 \equiv 1 \pmod{3}$ .

Problem 2 · Another Name for $C[2]$

Which of these sets is equal to $C[2]$ ?

✅ Correct!

11 = 3 \cdot 3 + 2

, so

11 \equiv 2 \pmod{3}

and

C[11] = C[2]

❌ Not quite.

-2 = 3(-1) + 1

, so

C[-2] = C[1]

, not

C[2]

❌ Not quite. Find the remainder of each label on division by

3

Show solution

$C[k] = C[r]$ where $r$ is the remainder of $k$ , because $k \equiv r \pmod 3$ and congruence is symmetric and transitive. The remainders: $-2 = 3(-1) + 1$ , $6 = 3 \cdot 2 + 0$ , $4 = 3 \cdot 1 + 1$ , $11 = 3 \cdot 3 + 2$ .

So $C[-2] = C[1]$ , $C[6] = C[0]$ , $C[4] = C[1]$ and $C[11] = C[2]$ .

Problem 3 · A Number and Its Negative

Which sets contain $1000$ and $-1000$ ?

✅ Correct! Like

79

and

-79

, they land in different sets, so

1000 \not\equiv -1000 \pmod{3}

❌ Negating does not keep the remainder.

-1000 = 3(-333) - 1

, and

-1

is not a remainder; go one multiple lower.

❌ Not quite. Compute each remainder separately.

Show solution

1000 = 3 \cdot 333 + 1, \qquad -1000 = 3(-334) + 2.

So $1000 \in C[1]$ and $-1000 \in C[2]$ . They lie in different sets, so they are not congruent modulo $3$ ; indeed $1000 - (-1000) = 2000$ and $2000 = 3 \cdot 666 + 2$ is not a multiple of $3$ .

Problem 4 · Modulo 5

For each integer $k$ let $D[k] = \{ x \in \mathbb{Z} \mid x \equiv k \pmod{5} \}$ . Which statement is true?

✅ Correct! Each integer has exactly one remainder

0, 1, 2, 3, 4

on division by

5

❌ Not quite.

5 \equiv 0 \pmod 5

, so

D[5] = D[0]

❌ Not quite.

-1 = 5(-1) + 4

, so

D[-1] = D[4]

❌ Not quite.

7 \equiv 2 \pmod 5

, so

D[7] = D[2]

, and a nonempty set is not disjoint from itself.

Show solution

The argument of the video works with $5$ in place of $3$ . Every integer $x$ is $5q + r$ with exactly one $r \in \{0, 1, 2, 3, 4\}$ , so $D[0], \ldots, D[4]$ are pairwise disjoint with union $\mathbb{Z}$ , and every $D[k]$ equals $D[r]$ for the remainder $r$ of $k$ .

So (a) is true; $D[5] = D[0]$ , $D[-1] = D[4]$ , and $D[2] \cap D[7] = D[2] \ne \emptyset$ .

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