Mathematical-Reasoning · Unit 26 · Video 5 · Interactive Practice

Drawing Relations: Shaded Regions and Directed Graphs

IKey Formulas

Relation	Picture
$\{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid y = x^2 \}$	a curve: the point $(x, y)$ is on it exactly when the pair is in the relation
$R = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid y \ge x^2 \}$	the parabola drawn solid, together with the shaded region above it; unbounded, so a drawing shows only a window
A relation on a finite set $A$	a directed graph: one vertex per element of $A$ , one directed edge from $x$ to $y$ per pair $(x, y) \in R$
A pair $(x, x)$	a loop at the vertex $x$ ; the number of directed edges equals the number of pairs

Key Insight: The list of pairs and the directed graph carry the same information: $(3, 2) \in R$ is an arrow from $3$ to $2$ , and an arrow from $2$ to $3$ is a different edge for a different pair.

IIWhich Side of the Parabola?

A pair is in $R$ exactly when its point lies on or above the parabola $y = x^2$ .

Relation

IIIDirected Graphs on {1, 2, 3, 4, 5, 6}

Each ordered pair becomes one directed edge, so counting edges counts the pairs.

Relation

IVQuiz Questions

Problem 1 · Counting Directed Edges

Given: the relation $R = \{ (x, y) \in A \times A \mid x \lt y \}$ on $A = \{1, 2, 3, 4\}$ . How many directed edges does its directed graph have?

✅ Correct!

R = \{ (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4) \}

has

6

pairs, so the digraph has

6

edges and no loops.

❌ Not quite.

12

counts every ordered pair of different elements. But

(2, 1) \notin R

, since

2 \lt 1

is false, so each edge goes one way only.

❌ Not quite.

x \lt x

is false, so there are no loops. Only the

6

pairs with

x \lt y

count.

❌ Not quite.

4

is the number of vertices. Count the ordered pairs of

R

❌ Not quite. List the pairs of

R

; there is one edge per pair.

Show solution

List the pairs one first coordinate at a time: $1 \lt 2, 3, 4$ gives $3$ pairs, $2 \lt 3, 4$ gives $2$ , $3 \lt 4$ gives $1$ , and $4$ is less than no element of $A$ . That is $3 + 2 + 1 = 6$ pairs.

A directed graph has exactly one directed edge per ordered pair, so it has $6$ directed edges. No pair has the form $(x, x)$ , so none of them is a loop, and every arrow points from a smaller number to a larger one.

Problem 2 · Divides on a Larger Set

Given: the relation $x \mid y$ on $A = \{1, 2, 3, 4, 5, 6, 7, 8\}$ . How many directed edges, loops included, does its directed graph have?

✅ Correct! By first coordinate:

8 + 4 + 2 + 2 + 1 + 1 + 1 + 1 = 20

pairs, so

20

directed edges (

8

loops and

12

arrows).

❌ Close, but…

12

is the number of arrows between different vertices. Every

x

divides itself, so add

8

loops.

❌ Not quite. Recount:

2

divides

2, 4, 6, 8

(four pairs) and

4

divides

4

and

8

(two pairs).

❌ Not quite.

64 = 8^2

is the size of

A \times A

; most of those pairs, like

(3, 4)

, are not in the relation.

❌ Not quite. List the pairs one first coordinate at a time.

Show solution

Count the second coordinates $y \in A$ with $x \mid y$ for each first coordinate $x$ :

$x = 1$ : all $8$ elements.
$x = 2$ : $2, 4, 6, 8$ , so $4$ .
$x = 3$ : $3, 6$ , so $2$ .
$x = 4$ : $4, 8$ , so $2$ .
$x = 5, 6, 7, 8$ : only $x$ itself, since $2x \gt 8$ ; so $1$ each.

Total $8 + 4 + 2 + 2 + 4 = 20$ pairs. Each is one directed edge: the $8$ pairs $(x, x)$ are loops and the other $12$ are arrows.

Problem 3 · A Strict Inequality

Given: $U = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid y \gt x^2 \}$ . Which pair is in $U$ ?

✅ Correct!

(-1)^2 = 1

and

2 \gt 1

, so the point lies strictly above the parabola.

❌ Not quite.

2^2 = 4

, so

(2, 4)

is on the parabola. The inequality is strict, so the curve itself is not part of

U

(it would be drawn dashed).

❌ Not quite.

3^2 = 9 \gt 5

, so

(3, 5)

lies below the parabola.

❌ Not quite. The square is

(-2)^2 = 4

, and

-5 \gt 4

is false.

❌ Not quite. Compare the second coordinate with the square of the first.

Show solution

$(-1, 2)$ : $x^2 = 1$ and $2 \gt 1$ , so $(-1, 2) \in U$ .
$(2, 4)$ : $x^2 = 4$ and $4 \gt 4$ is false, so $(2, 4) \notin U$ ; it is on the boundary curve.
$(3, 5)$ : $x^2 = 9$ and $5 \gt 9$ is false.
$(-2, -5)$ : $x^2 = 4$ and $-5 \gt 4$ is false.

The picture of $U$ is the region strictly above $y = x^2$ , with the parabola drawn dashed to show it is excluded.

Problem 4 · Reading a Digraph Back

Given: the directed graph of a relation $R$ on $\{1, 2, 3\}$ has loops at $1$ and $3$ and arrows from $1$ to $2$ , from $2$ to $1$ , and from $2$ to $3$ , and no other edges. Which statement is true?

✅ Correct! The arrow from

2

3

is the pair

(2, 3)

, and there is no arrow from

3

2

, so

(3, 2) \notin R

❌ Not quite. The arrow points from

2

3

, which is the pair

(2, 3)

. Direction matters:

(3, 2)

would need its own arrow from

3

2

❌ Not quite. Count every edge, loops included:

2

loops and

3

arrows give

5

pairs.

❌ Not quite. Vertex

2

has no loop, so

(2, 2) \notin R

❌ Not quite. Turn each edge into its ordered pair.

Show solution

Each directed edge is one ordered pair: the loops give $(1, 1)$ and $(3, 3)$ , and the arrows give $(1, 2)$ , $(2, 1)$ and $(2, 3)$ . So

R = \{ (1, 1), (3, 3), (1, 2), (2, 1), (2, 3) \},

which has $5$ pairs. $(2, 3) \in R$ but $(3, 2) \notin R$ , so $2 \mathrel{R} 3$ and $3 \not\mathrel{R} 2$ . Vertex $2$ has no loop, so $(2, 2) \notin R$ .

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