Mathematical-Reasoning · Unit 6 · Video 1 · Interactive Practice

Divisibility: Definition, Notation, and Non-Divisors

IKey Formulas

Statement	Meaning	How to show it
$m \mid n$ (with $m \neq 0$ )	$(\exists q \in \mathbb{Z})(n = m \cdot q)$	Produce one integer $q$ and check $n = m \cdot q$
$m \nmid n$	$(\forall q \in \mathbb{Z})(n \neq m \cdot q)$	Rule out every integer $q$
$m \mid 0$ for every $m \neq 0$	$0 = m \cdot 0$	The integer $q = 0$ works
$m \mid n$ versus $\frac{m}{n}$	A statement versus a number	$4 \mid 32$ is true, while $\frac{4}{32} = \frac{1}{8}$

Key Insight: The definition of divides is a there-exists statement, so a proof that $m \mid n$ must produce the integer $q$ . Zero is a multiple of every nonzero integer and a divisor of none.

IIDivides Means Landing on a Multiple

$m \mid n$ exactly when $n$ is one of the multiples $m \cdot q$ with $q \in \mathbb{Z}$ .

Video example m 4 n 32

IIIChaining Divisibility

In each example with $a \mid b$ and $b \mid c$ , the integer $c$ also lands on a multiple of $a$ .

Video example a 2 s (b = a·s) 3 t (c = b·t) 4

💡 Three examples are evidence for the conjecture "if $a \mid b$ and $b \mid c$ , then $a \mid c$ ", not a proof; the next video proves it with a know-show table.

IVWeekdays and Clock Times

Counts that land on the same weekday differ by a multiple of $7$ ; on a clock, by a multiple of $12$ .

Cycle Count A 3 Count B 10

VQuiz Questions

Problem 1 · Apply the Definition

Definition: a nonzero integer $m$ divides an integer $n$ provided that there is an integer $q$ with $n = m \cdot q$ . Which of these is a true divides statement?

✅ Correct!

-54 = 6 \cdot (-9)

and

-9 \in \mathbb{Z}

; a negative

q

counts.

❌ Not quite.

32 = 5 \cdot q

forces

q = \frac{32}{5}

, which is not an integer, so

5 \nmid 32

❌ Check the order.

9 \mid 3

asks for

3 = 9 \cdot q

, which forces

q = \frac{1}{3} \notin \mathbb{Z}

. It is

3 \mid 9

that is true.

❌ Not a valid statement. The definition requires the divisor to be nonzero, so

0 \mid 7

is not allowed at all.

Show solution

Test each statement by looking for an integer $q$ .

$5 \mid 32$ : $32 = 5 \cdot q$ forces $q = \frac{32}{5} \notin \mathbb{Z}$ , so no integer works and $5 \nmid 32$ .
$9 \mid 3$ : $3 = 9 \cdot q$ forces $q = \frac{1}{3} \notin \mathbb{Z}$ , so $9 \nmid 3$ .
$6 \mid -54$ : $q = -9$ works, since $6 \cdot (-9) = -54$ . True.
$0 \mid 7$ : the divisor must be nonzero, so this is not a valid divides statement.

The answer is $6 \mid -54$ .

Problem 2 · Relationship, Not Fraction

Which description of $8 \mid 2$ is correct?

✅ Correct! The only candidate is

q = \frac{1}{4}

, which is not an integer, so

8 \nmid 2

❌ That is the fraction.

\frac{2}{8} = \frac{1}{4}

is a number, but

8 \mid 2

is a statement about two integers, true or false.

❌ Check the order.

8 = 2 \cdot 4

shows

2 \mid 8

. The statement

8 \mid 2

asks for

2 = 8 \cdot q

❌ Not quite. The definition requires

q

to be an integer, and

\frac{1}{4}

is not one.

Show solution

By the definition, $8 \mid 2$ means there is an integer $q$ with $2 = 8 \cdot q$ . The only number solving this equation is $q = \frac{2}{8} = \frac{1}{4}$ , which is not an integer. So no integer $q$ works, and $8 \mid 2$ is a false statement.

Reversing the order gives the true statement $2 \mid 8$ , since $8 = 2 \cdot 4$ .

Problem 3 · Negating the Definition

Given: $m$ is a nonzero integer, $n$ is an integer, and $m \mid n$ means $(\exists q \in \mathbb{Z})(n = m \cdot q)$ . Which statement means $m \nmid n$ ?

✅ Correct! The negation of a there-exists statement is a for-all statement with the open sentence negated.

❌ The quantifier must change. This says only that one

q

fails, for example

10 \neq 3 \cdot 2

. One failing

q

is not enough: every integer

q

must fail.

❌ Half the negation. The quantifier changed, but the open sentence

n = m \cdot q

must also be negated.

❌ The roles are swapped. This statement is about

m

being a multiple of

n

, not about

n

being a multiple of

m

Show solution

Start from $m \mid n \iff (\exists q \in \mathbb{Z})(n = m \cdot q)$ . By the theorem on negations of quantified statements, $\neg(\exists x)P(x) \equiv (\forall x)\neg P(x)$ , so

m \nmid n \iff \neg(\exists q \in \mathbb{Z})(n = m \cdot q) \iff (\forall q \in \mathbb{Z})(n \neq m \cdot q).

For $3$ and $10$ this reads $(\forall q \in \mathbb{Z})(10 \neq 3 \cdot q)$ : every integer $q$ must be ruled out.

Problem 4 · Clock Arithmetic

Given: a $12$ -hour clock (no morning or evening) now reads $5$ o'clock, and in $4$ hours it will read $9$ o'clock. Numbers of hours that reach the same time differ by a multiple of $12$ . Which number of hours from now also gives $9$ o'clock? (Negative numbers count hours ago.)

✅ Correct!

40 - 4 = 36 = 12 \cdot 3

, so

12 \mid (40 - 4)

and

40

hours from now is

9

o'clock.

❌ Not quite.

48

is a multiple of

12

, so

48

hours brings the clock back to

5

o'clock. Compare with

4

48 - 4 = 44 = 12 \cdot q

forces

q = \frac{11}{3} \notin \mathbb{Z}

❌ That is the reading, not the count. Nine hours after

5

o'clock is

2

o'clock, and

9 - 4 = 5

is not a multiple of

12

❌ Check the sign. Four hours ago it was

1

o'clock, and

-4 - 4 = -8 = 12 \cdot q

forces

q = -\frac{2}{3} \notin \mathbb{Z}

Show solution

A count $h$ gives $9$ o'clock when $12 \mid (h - 4)$ . Test each option.

$h = 40$ : $40 - 4 = 36 = 12 \cdot 3$ , and $3 \in \mathbb{Z}$ . ✓
$h = 48$ : $48 - 4 = 44$ , and $44 = 12 \cdot q$ forces $q = \frac{11}{3} \notin \mathbb{Z}$ .
$h = 9$ : $9 - 4 = 5$ , and $5 = 12 \cdot q$ forces $q = \frac{5}{12} \notin \mathbb{Z}$ .
$h = -4$ : $-4 - 4 = -8$ , and $-8 = 12 \cdot q$ forces $q = -\frac{2}{3} \notin \mathbb{Z}$ .

Directly: $40 = 36 + 4$ is three full turns of the clock plus $4$ hours, so the clock reads $9$ o'clock. The answer is $40$ .

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