Mathematical-Reasoning · Unit 26 · Video 3 · Interactive Practice

x R y Notation: Testing the Divides Relation Three Ways

IKey Formulas

Notation or relation	Meaning
$x \mathrel{R} y$ and $x \not\mathrel{R} y$	$(x, y) \in R$ and $(x, y) \notin R$ ; a generic relation is often written $\sim$
$R_{\lt} = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid x \lt y \}$	$2 \lt 5$ says $(2, 5) \in R_{\lt}$ , while $(5, 2) \notin R_{\lt}$
$D = \{ (m, n) \in \mathbb{Z} \times \mathbb{Z} \mid m \mid n \}$	$m \mid n$ means $m \neq 0$ and $n = mq$ for some integer $q$
$\operatorname{dom}(D)$ and $\operatorname{range}(D)$	the nonzero integers, and all of $\mathbb{Z}$

Key Insight: For nonzero integers, $x \mid x$ always, $x \mid y$ does not force $y \mid x$ ( $2 \mid 4$ but $4 \nmid 2$ ), and $x \mid y$ with $y \mid z$ always gives $x \mid z$ .

IIThe Divides Relation as a Set of Pairs

Each filled cell $(m, n)$ is a pair in $D$ ; row $0$ stays empty and column $0$ fills up.

IIITwo Links Make a Direct Link

If $y = xs$ and $z = yt$ , then $z = x(st)$ : the chain $x \mid y$ , $y \mid z$ always yields $x \mid z$ .

x 2 s 3 t 4

IVQuiz Questions

Problem 1 · Which Pair Is in D?

Given: the divides relation $D = \{ (m, n) \in \mathbb{Z} \times \mathbb{Z} \mid m \mid n \}$ . Which pair is an element of $D$ ?

✅ Correct!

12 = (-3)(-4)

and

-4

is an integer, so

-3 \mid 12

and

(-3, 12) \in D

❌ Not quite. A divisor must be nonzero, so no pair with first coordinate

0

is in

D

❌ Not quite. Order matters:

(12, 3) \in D

would mean

12 \mid 3

, and no integer

q

has

3 = 12q

❌ Not quite.

6 = 4q

needs

q = \tfrac{3}{2}

, which is not an integer, so

4 \nmid 6

❌ Not quite.

(m, n) \in D

means

m \neq 0

and

n = mq

for an integer

q

Show solution

$(0, 5)$ : $0$ is never a divisor, so $(0, 5) \notin D$ .
$(12, 3)$ : $3 = 12q$ forces $q = \tfrac{1}{4}$ , so $12 \nmid 3$ .
$(4, 6)$ : $6 = 4q$ forces $q = \tfrac{3}{2}$ , so $4 \nmid 6$ .
$(-3, 12)$ : $12 = (-3) \cdot (-4)$ with $-4 \in \mathbb{Z}$ , so $-3 \mid 12$ .

So $(-3, 12) \in D$ , which in the $x \mathrel{R} y$ style is written $-3 \mid 12$ .

Problem 2 · Zero in D

Given: the divides relation $D$ on $\mathbb{Z}$ . Decide whether each pair is in $D$ .

Is $(-4, 0) \in D$ ?

Is $(0, 0) \in D$ ?

✅ Correct!

0 = (-4) \cdot 0

, so

-4 \mid 0

; but

0

is never a divisor, so

(0, 0) \notin D

even though

0 = 0 \cdot q

for every

q

❌ Not quite.

m \mid n

is a statement, not the fraction

\tfrac{m}{n}

. Here

0 = (-4) \cdot 0

with

0 \in \mathbb{Z}

, so

-4 \mid 0

❌ Not quite. The definition of

m \mid n

requires

m \neq 0

, so the first coordinate

0

rules the pair out, whatever the second coordinate is.

Show solution

$(-4, 0)$ . The divisor $-4$ is nonzero, and $0 = (-4) \cdot 0$ with $0$ an integer, so $-4 \mid 0$ and $(-4, 0) \in D$ . In the same way every nonzero integer divides $0$ , which is why $\operatorname{dom}(D)$ is all the nonzero integers.

$(0, 0)$ . The definition of divides applies only to a nonzero divisor, so $0 \nmid 0$ and $(0, 0) \notin D$ . No pair with first coordinate $0$ lies in $D$ .

Problem 3 · Three Questions for Less Than

Given: $R_{\lt} = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid x \lt y \}$ . Consider (1) for every $x \in \mathbb{R}$ , $x \lt x$ ; (2) for all $x, y \in \mathbb{R}$ , if $x \lt y$ , then $y \lt x$ ; (3) for all $x, y, z \in \mathbb{R}$ , if $x \lt y$ and $y \lt z$ , then $x \lt z$ . Which of them are true?

✅ Correct!

0 \lt 0

is false,

1 \lt 2

holds while

2 \lt 1

fails, and

x \lt y \lt z

gives

x \lt z

❌ Not quite. A single counterexample refutes (1):

0 \lt 0

is false. In fact no real number is related to itself by

\lt

❌ Close, but… That is the pattern for divides on the nonzero integers. For

\lt

, statement (1) fails:

x \lt x

is false for every real

x

❌ Not quite. (3) is true: if

x \lt y

and

y \lt z

, then

z - x = (z - y) + (y - x)

is a sum of two positive numbers, so

x \lt z

❌ Not quite. Test each statement: look for a counterexample, or give an argument for every case.

Show solution

(1) is false. For $x = 0$ , the statement $0 \lt 0$ is false, so $(0, 0) \notin R_{\lt}$ .

(2) is false. $1 \lt 2$ , so $(1, 2) \in R_{\lt}$ , but $2 \lt 1$ is false, so $(2, 1) \notin R_{\lt}$ .

(3) is true. Let $x, y, z \in \mathbb{R}$ with $x \lt y$ and $y \lt z$ . Then $y - x \gt 0$ and $z - y \gt 0$ , so $z - x = (z - y) + (y - x) \gt 0$ . Hence $x \lt z$ .

Problem 4 · Products That Are Not Negative

Given: the relation $\sim$ on $\mathbb{R}$ with $x \sim y$ if and only if $xy \ge 0$ . Which of the three questions (every $x$ related to itself; $x \sim y$ implies $y \sim x$ ; $x \sim y$ and $y \sim z$ imply $x \sim z$ ) gets the answer no?

✅ Correct!

x \cdot x = x^2 \ge 0

and

yx = xy

settle the first two, while

0

is related to everything and so links

1

-1

, whose product is

-1 \lt 0

❌ Not quite.

x \cdot x = x^2 \ge 0

for every real

x

, so every

x

is related to itself.

❌ Not quite.

0 \cdot 2 = 0 \ge 0

, so

0 \sim 2

. In general

yx = xy

, so the order never matters.

❌ Not quite. The number

0

is related to every real number, so it can link a positive number to a negative one:

1 \sim 0

and

0 \sim -1

, but

1 \cdot (-1) = -1

❌ Not quite. Try a chain that passes through

0

Show solution

First question: yes. For every $x \in \mathbb{R}$ , $x \cdot x = x^2 \ge 0$ , so $x \sim x$ .

Second question: yes. If $x \sim y$ , then $yx = xy \ge 0$ , so $y \sim x$ .

Third question: no. $1 \cdot 0 = 0 \ge 0$ and $0 \cdot (-1) = 0 \ge 0$ , so $1 \sim 0$ and $0 \sim -1$ ; but $1 \cdot (-1) = -1 \lt 0$ , so $1 \not\sim -1$ . This one counterexample settles it.

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