Mathematical-Reasoning · Unit 6 · Video 3 · Interactive Practice

One Definition, Two Verdicts: Proving a Divisor Divides a Sum and Disproving by Counterexample

IKey Formulas

Formula	Name	What it gives you
$a \mid b \ \text{ means } \ (\exists k \in \mathbb{Z})(b = a \cdot k)$ , with $a \ne 0$	Definition of divides	An equation to work with, and an integer to construct
$b = a s,\ c = a t \ \Rightarrow\ b + c = a(s + t)$	Divisor of a sum	$q = s + t$ , an integer because $\mathbb{Z}$ is closed under addition
$P(n)$ true and $Q(n)$ false	Counterexample to $(\forall n \in \mathbb{Z})(P(n) \to Q(n))$	One such $n$ proves the statement is false
$x = 2n + 1 \qquad (1)$	Numbered equation	A label for an equation a later sentence refers back to

Key Insight: The definition of divides starts both investigations. When it leads to an integer $q$ you can build, you write a proof; when it stalls, try examples with a true hypothesis and look for a false conclusion.

IIVisualization 1 — Why a Divisor of $b$ and $c$ Divides $b + c$

If $b$ is $s$ jumps of $a$ and $c$ is $t$ jumps of $a$ , then $b + c$ is $s + t$ jumps of $a$ .

Example s 3 t −5

IIIVisualization 2 — Which Rows Can Decide a Conditional?

Only a row with a true hypothesis and a false conclusion is a counterexample; every other row is silent or mere evidence.

Claim

💡 The second claim is the converse of the first. A true conclusion in every tested row with a true hypothesis is evidence only, so the next step for it would be to try to write a proof.

IVVisualization 3 — Hunting a Counterexample With an “or” Hypothesis

For all integers $a, b$ : if $5 \mid a$ or $5 \mid b$ , then $5 \mid (5a + b)$ . Which pairs $(a, b)$ break it?

a 5 b 1

VQuiz Questions

Problem 1 · Build q From s and t

Given: $a \mid b$ means $b = a \cdot k$ for some integer $k$ . Let $a = 6$ , $b = 18$ and $c = -42$ , so $b = 6s$ and $c = 6t$ with integers $s$ and $t$ . Following the proof, find the integer $q$ with $b + c = 6q$ .

✅ Correct! With

s = 3

and

t = -7

, the proof's

q = s + t = -4

, and indeed

6 \cdot (-4) = -24 = b + c

❌ Not quite.

10 = 3 - (-7) = s - t

, which is what

b - c = 60

needs. The sum uses

q = s + t

❌ Not quite.

-24

b + c

itself. The integer

q

is the factor that multiplies

a

b + c = 6q

❌ Not quite.

-21 = 3 \cdot (-7) = s \cdot t

. The distributive property gives

6s + 6t = 6(s + t)

, so the integers are added, not multiplied.

❌ Not quite. First find

s = 18/6

and

t = -42/6

, then add them.

Show solution

Since $18 = 6 \cdot 3$ and $-42 = 6 \cdot (-7)$ , we have $s = 3$ and $t = -7$ . Substituting and using the distributive property,

b + c = 6 \cdot 3 + 6 \cdot (-7) = 6\,\big(3 + (-7)\big) = 6 \cdot (-4).

So $q = s + t = -4$ , an integer, and $b + c = -24 = 6 \cdot (-4)$ confirms that $6 \mid (b + c)$ .

Problem 2 · Which Row Is a Counterexample?

Claim: For each integer $n$ , if $5 \mid (n^2 - 1)$ , then $5 \mid (n - 1)$ . Which value of $n$ is a counterexample?

✅ Correct! At

n = 9

n^2 - 1 = 80 = 5 \cdot 16

, so the hypothesis is true, while

n - 1 = 8

and

5 \nmid 8

, so the conclusion is false.

❌ The conclusion is false here, but so is the hypothesis.

n^2 - 1 = 8

and

5 \nmid 8

. A row with a false hypothesis cannot be a counterexample.

❌ Not quite.

n^2 - 1 = 35 = 5 \cdot 7

and

n - 1 = 5 = 5 \cdot 1

: both parts are true, so this row agrees with the claim.

❌ Not quite.

n^2 - 1 = 0

and

n - 1 = 0

, and

5 \mid 0

because

0 = 5 \cdot 0

. Both parts are true.

Show solution

A counterexample needs a true hypothesis and a false conclusion. Check each option:

$n = 3$ : $n^2 - 1 = 8$ , and $5 < 8 < 10$ , so $5 \nmid 8$ . Hypothesis false.
$n = 6$ : $35 = 5 \cdot 7$ and $5 = 5 \cdot 1$ . Both true.
$n = 1$ : $0 = 5 \cdot 0$ for both. Both true.
$n = 9$ : $80 = 5 \cdot 16$ , so the hypothesis is true; $n - 1 = 8$ and $5 < 8 < 10$ , so $5 \nmid 8$ and the conclusion is false.

The integer $9$ is a counterexample that proves the claim is false (as is $4$ , the one from the video).

Problem 3 · An “or” Hypothesis

Claim: For all integers $a$ and $b$ , if $3 \mid a$ or $3 \mid b$ , then $3 \mid (3a + b)$ . Which pair $(a, b)$ is a counterexample?

✅ Correct!

3 \mid 6

makes the hypothesis true, and

3a + b = 22

lies strictly between the consecutive multiples

21

and

24

, so

3 \nmid 22

❌ Not quite.

3 \nmid 1

and

3 \nmid 2

, so neither part of the “or” holds and the hypothesis is false. Such a pair cannot be a counterexample, even though

3a + b = 5

is not a multiple of

3

❌ Not quite. The hypothesis holds through

3 \mid 6

, but

3a + b = 12 = 3 \cdot 4

, so the conclusion is true as well.

❌ Not quite. Both

3 \mid 3

and

3 \mid (-3)

, and

3a + b = 6 = 3 \cdot 2

, so the conclusion is true.

Show solution

Because the hypothesis is an “or”, one true part is enough. We need such a pair with $3 \nmid (3a + b)$ .

Take $(a, b) = (6, 4)$ . The hypothesis is true since $6 = 3 \cdot 2$ . For the conclusion,

3a + b = 3 \cdot 6 + 4 = 22,

and $21 < 22 < 24$ with $21 = 3 \cdot 7$ and $24 = 3 \cdot 8$ consecutive multiples of $3$ , so $3 \nmid 22$ . Hypothesis true, conclusion false: $(6, 4)$ is a counterexample that proves the claim is false.

The other options fail: $(1, 2)$ has a false hypothesis, while $(2, 6)$ gives $12 = 3 \cdot 4$ and $(3, -3)$ gives $6 = 3 \cdot 2$ , both true conclusions.

Problem 4 · Adapt the Proof

Claim: For all integers $a$ , $b$ , $c$ with $a \ne 0$ , if $a \mid b$ and $a \mid c$ , then $a \mid (3b - 2c)$ . A proof starts as before: there are integers $s$ and $t$ with $b = a s$ and $c = a t$ .

Which q satisfies 3b − 2c = a · q?

Which fact shows that this q is an integer?

✅ Correct!

3b - 2c = 3as - 2at = a(3s - 2t)

, and

3s - 2t

is an integer because products and differences of integers are integers.

❌ Check the sign.

3b - 2c = 3as - 2at

; the minus sign carries through the distributive property.

❌ Check the coefficients.

a(s - t) = b - c

, not

3b - 2c

. The

3

and

2

stay in

q

❌ Not quite.

3b - 2c

is the number to be divided. Then

a \cdot q = a(3b - 2c)

, which is not

3b - 2c

in general (for instance,

a = 2

b = c = 2

gives

a \cdot q = 4

while

3b - 2c = 2

❌ Check q. Substitute

b = as

and

c = at

into

3b - 2c

and factor out

a

❌ Not quite.

\mathbb{Z}

is not closed under division:

1 \div 2

is not an integer. Here

3s - 2t

uses only multiplication and subtraction.

❌ Not quite.

a \ne 0

is part of the definition of divides; it says nothing about whether

3s - 2t

is an integer.

❌ Not quite. The associative property rearranges a product; it does not show that

3s - 2t

is an integer.

Show solution

Assume $a \ne 0$ , $a \mid b$ and $a \mid c$ . Then there exist integers $s$ and $t$ with $b = a s$ and $c = a t$ . Substituting,

3b - 2c = 3(as) - 2(at) = a(3s - 2t).

Since $3s$ and $2t$ are integers (closure under multiplication) and their difference is an integer (closure under subtraction), $q = 3s - 2t \in \mathbb{Z}$ . Therefore the previous equation proves that $a \mid (3b - 2c)$ .

No equation needs a number here: the only back-reference is to the previous equation, and a phrase does that job.

Solved: 0 / 4

One Definition, Two Verdicts: Proving a Divisor Divides a Sum and Disproving by Counterexample

IKey Formulas

IIVisualization 1 — Why a Divisor of bbb and ccc Divides b+cb + cb+c

IIIVisualization 2 — Which Rows Can Decide a Conditional?

IVVisualization 3 — Hunting a Counterexample With an “or” Hypothesis

VQuiz Questions

IIVisualization 1 — Why a Divisor of $b$ and $c$ Divides $b + c$