Mathematical-Reasoning · Unit 6 · Video 4 · Interactive Practice

Congruent Means the Difference Is a Multiple: A Proof About Sums Mod 8, and Seven Rules for Writing It

IKey Formulas

Statement	Means	Role
$a \equiv b \pmod{n}$	$n \mid (a - b)$	Definition of congruence ( $n \in \mathbb{N}$ ; $a, b \in \mathbb{Z}$ )
$n \mid (a - b)$	$(\exists k \in \mathbb{Z})(a - b = n \cdot k)$	Definition of divides
$a \equiv b \pmod{n}$	$a = b + n \cdot k$ for some $k \in \mathbb{Z}$	Every integer congruent to $b$
$a = 5 + 8k$ , $b = 5 + 8m$	$a + b - 2 = 8(1 + k + m)$	Constructs $q = 1 + k + m$

Key Insight: $34 \equiv 2 \pmod{8}$ even though $34 \neq 2$ . A congruence says only that the difference is a multiple of the modulus.

IICongruence on the Number Line

Two integers are congruent modulo $n$ exactly when their difference is $n$ times an integer.

Modulus n 7 b 2 a 16

IIIThe Know-Show Table

The table works backward from $Q$ and forward from $P$ until both meet at $q = 1 + k + m$ .

Step 1

Step	Know / Show	Reason
$P$	$a \equiv 5 \pmod{8}$ and $b \equiv 5 \pmod{8}$	hypothesis
$P1$	$8 \mid (a - 5)$ and $8 \mid (b - 5)$	def. of congruence
$P2$	$(\exists k, m \in \mathbb{Z})(a - 5 = 8k \text{ and } b - 5 = 8m)$	def. of divides
$P3$	$a = 5 + 8k$ and $b = 5 + 8m$	algebra
$P4$	$a + b - 2 = 8(1 + k + m)$	substitution, algebra
$Q2$	$(\exists q \in \mathbb{Z})(a + b - 2 = 8q)$	?
$Q1$	$8 \mid (a + b - 2)$	?
$Q$	$(a + b) \equiv 2 \pmod{8}$	?

IVConstructing q

For all integers $k$ and $m$ , the number $a + b - 2$ equals $8(1 + k + m)$ , a multiple of $8$ .

k 1 m 2

VQuiz Questions

Problem 1 · Apply the Definition

Given: $a \equiv b \pmod{n}$ means $n \mid (a - b)$ . Which integer is congruent to $3$ modulo $7$ ?

✅ Correct!

(-4) - 3 = -7 = 7 \cdot (-1)

, so

7 \mid (-4 - 3)

❌ Not quite.

(-3) - 3 = -6

, and no integer

k

gives

7k = -6

. Changing the sign of

3

does not give an integer congruent to

3

❌ Not quite.

21 - 3 = 18

, which is not

7

times an integer.

21

is itself a multiple of

7

, so

21 \equiv 0 \pmod{7}

❌ Not quite.

4 - 3 = 1

. Neighbouring integers differ by

1

, not by a multiple of

7

Show solution

Test each option by computing $x - 3$ and asking whether it equals $7k$ for some integer $k$ :

(-3) - 3 = -6, \qquad 21 - 3 = 18, \qquad (-4) - 3 = -7 = 7 \cdot (-1), \qquad 4 - 3 = 1.

Only $-7$ is a multiple of $7$ , so $-4 \equiv 3 \pmod{7}$ . Equivalently, $-4 = 3 + 7 \cdot (-1)$ : start at $3$ and step back one multiple of $7$ .

Problem 2 · From Congruence to an Equation

Given: $a$ is an integer and $a \equiv 5 \pmod{8}$ , where this means $8 \mid (a - 5)$ . Which statement is equivalent to it?

✅ Correct!

8 \mid (a - 5)

means

a - 5 = 8k

for some integer

k

; adding

5

gives

a = 5 + 8k

❌ The roles are swapped. The modulus

8

multiplies

k

, and

5

is the starting point.

a = 5

satisfies the congruence, but

5 = 8 + 5k

would need

k = -\tfrac{3}{5}

❌ Too narrow. That allows only

a = 13

, the case

k = 1

. The definition asks for some integer

k

, so

-3

5

and

21

qualify too.

❌ The divisibility is reversed. The definition says

8

divides

a - 5

, not that

a - 5

divides

8

. For example

a = 6

gives

(6 - 5) \mid 8

, yet

6 - 5 = 1

is not a multiple of

8

Show solution

Unfold the definitions in order. By the definition of congruence, $a \equiv 5 \pmod{8}$ means $8 \mid (a - 5)$ . By the definition of divides, this means there exists an integer $k$ such that $a - 5 = 8k$ . Adding $5$ to both sides gives $a = 5 + 8k$ .

So the integers congruent to $5$ modulo $8$ are $5$ plus multiples of $8$ : $\ldots$ , $-11$ , $-3$ , $5$ , $13$ , $21$ , $\ldots$

Problem 3 · Adapt the Proof

Given: integers $a$ and $b$ with $a \equiv 3 \pmod{7}$ and $b \equiv 5 \pmod{7}$ , so $a = 3 + 7k$ and $b = 5 + 7m$ for some integers $k$ and $m$ . The proof shows $a + b - 1 = 7q$ for an integer $q$ .

What is $q$ ?

Which congruence does this prove?

✅ Correct!

a + b - 1 = 7(1 + k + m)

, so

7 \mid (a + b - 1)

and

(a + b) \equiv 1 \pmod{7}

❌ A term went missing.

3 + 5 - 1 = 7

leaves a

7

that must also be factored, giving

7 \cdot 1

❌ That is $7q$ , not $q$ .

7 + 7k + 7m

is the whole expression; factor out the

7

❌ Check the constant. The constant is

3 + 5 - 1 = 7 = 7 \cdot 1

, so it contributes

1

q

, not

8

❌ Check $q$ . Substitute

a = 3 + 7k

and

b = 5 + 7m

into

a + b - 1

and factor out

7

❌ Not quite. That would need

7 \mid (a + b)

. The proof shows

7 \mid (a + b - 1)

, so the sum is

1

more than a multiple of

7

❌ Check the sign.

(a + b) \equiv 6 \pmod{7}

would need

7 \mid (a + b - 6)

. The proof gives

7 \mid (a + b - 1)

❌ Not quite.

5 - 3 = 2

is a difference, but the proposition is about the sum:

7 \mid (a + b - 1)

❌ Check the congruence. Use the definition:

(a + b) \equiv c \pmod{7}

means

7 \mid (a + b - c)

Show solution

Proof. Let $a$ and $b$ be integers and assume that $a \equiv 3 \pmod{7}$ and $b \equiv 5 \pmod{7}$ . We will prove that $(a + b) \equiv 1 \pmod{7}$ . Since $7 \mid (a - 3)$ and $7 \mid (b - 5)$ , there exist integers $k$ and $m$ such that $a - 3 = 7k$ and $b - 5 = 7m$ . We then see that

a + b - 1 = (3 + 7k) + (5 + 7m) - 1 = 7 + 7k + 7m = 7(1 + k + m).

By the closure properties of the integers, $1 + k + m$ is an integer, so $7 \mid (a + b - 1)$ and hence $(a + b) \equiv 1 \pmod{7}$ . $\blacksquare$

Check with numbers: $a = 10$ ( $k = 1$ ) and $b = 12$ ( $m = 1$ ) give $a + b - 1 = 21 = 7 \cdot 3$ , and $q = 1 + 1 + 1 = 3$ .

Problem 4 · Writing Guidelines

Given: four versions of one sentence from the mod-8 proof. Which version follows all seven writing guidelines?

✅ Correct! It begins with a word, is a complete punctuated sentence, and states the quantifier in English.

❌ Guideline six. The sentence begins with the symbol

a

. Start with a word instead.

❌ Guideline seven. The symbols

\therefore

and

\exists

belong in scratch work such as a know-show table; a written proof uses the words "therefore" and "there exists". It is also not a sentence.

❌ Guideline five (and two). The asterisk is computer notation for multiplication, and the sentence lacks punctuation.

Show solution

Option (b) begins with a mathematical symbol (guideline six). Option (c) uses $\therefore$ and $\exists$ in place of words and is not a sentence (guidelines seven and two). Option (d) writes multiplication with an asterisk and omits the punctuation (guidelines five and two).

Option (a) starts with the word "Since", is a complete sentence with punctuation, and writes "there exists an integer $k$ " in English. It is the sentence the written proof in the video uses.

Solved: 0 / 4