Mathematical-Reasoning · Unit 26 · Video 1 · Interactive Practice

Conditions on Pairs: Bordering States and an Ellipse

IKey Formulas

Object	Definition	In this video
Cartesian product	$A \times B = \{ (a, b) \mid a \in A \text{ and } b \in B \}$	$A$ = the U.S. states; the pairs of states form $A \times A$
Set picked out by a condition	$R = \{ (x, y) \in A \times A \mid x \text{ and } y \text{ have a land border in common} \}$	$R \subseteq A \times A$
Solution set	$S = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid 4x^2 + y^2 = 16 \}$	the ellipse through $(\pm 2, 0)$ and $(0, \pm 4)$
First and second coordinates of $S$	$\{ x \in \mathbb{R} \mid \exists\, y \in \mathbb{R},\ 4x^2 + y^2 = 16 \}$ and $\{ y \in \mathbb{R} \mid \exists\, x \in \mathbb{R},\ 4x^2 + y^2 = 16 \}$	$[-2, 2]$ and $[-4, 4]$

Key Insight: A universal statement about the pairs, such as "if $(x, y) \in R$ and $(y, z) \in R$ , then $(x, z) \in R$ ", is refuted by one counterexample, but proving one needs an argument that works for every pair.

IINeighbours and Chains in the Border Set

Fixing the first state lists its partners in $R$ ; a chain of two border pairs need not give a third.

Question

IIIWhich Numbers Occur as Coordinates?

A number is a first coordinate of $S$ exactly when its vertical line meets the ellipse $4x^2 + y^2 = 16$ .

Line a 1

IVQuiz Questions

Problem 1 · An Element of the Solution Set

Given: $S = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid 4x^2 + y^2 = 16 \}$ . Which ordered pair is an element of $S$ ?

✅ Correct!

4(\sqrt{3})^2 + (-2)^2 = 12 + 4 = 16

, so

(\sqrt{3}, -2) \in S

❌ Not quite.

4 \cdot 2^2 + (2\sqrt{3})^2 = 16 + 12 = 28

. The point

(1, 2\sqrt{3})

is in

S

, but moving the first coordinate to

2

leaves room only for

y = 0

❌ Close, but…

(1, 4)

satisfies

x^2 + y^2 = 17

, not the given equation:

4 \cdot 1^2 + 4^2 = 20 \neq 16

. The coefficient

4

multiplies

x^2

❌ Not quite.

4(-4)^2 + 0^2 = 64

. The ellipse meets the

y

-axis at

\pm 4

, so the pairs with a

4

in them are

(0, 4)

and

(0, -4)

, with the

4

second.

❌ Not quite. Substitute the first number for

x

and the second for

y

, and check whether

4x^2 + y^2 = 16

Show solution

Test each pair in $4x^2 + y^2$ :

$(2, 2\sqrt{3})$ : $16 + 12 = 28$ , not in $S$ .
$(\sqrt{3}, -2)$ : $4 \cdot 3 + 4 = 16$ , in $S$ .
$(1, 4)$ : $4 + 16 = 20$ , not in $S$ .
$(-4, 0)$ : $64 + 0 = 64$ , not in $S$ .

So the answer is $(\sqrt{3}, -2)$ .

Problem 2 · The Set of First Coordinates

Given: $P = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid x^2 + 9y^2 = 36 \}$ . Find the set of first coordinates, $\{ x \in \mathbb{R} \mid \exists\, y \in \mathbb{R},\ x^2 + 9y^2 = 36 \}$ .

✅ Correct! A real

y

exists exactly when

36 - x^2 \ge 0

, that is, when

-6 \le x \le 6

❌ Close, but…

[-2, 2]

is the set of second coordinates: solving for

x^2 = 36 - 9y^2 \ge 0

gives

y^2 \le 4

❌ Not quite. The condition is

x^2 \le 36

, which bounds

x

6

, not by

36

❌ Not quite. For

x = 7

we would need

9y^2 = 36 - 49 = -13

, and no real number has a negative square.

❌ Not quite. Solve for

y^2

and ask when the right side is not negative.

Show solution

Solve for $y^2$ : $y^2 = \dfrac{36 - x^2}{9}$ . A real $y$ exists exactly when the right side is not negative:

36 - x^2 \ge 0 \iff x^2 \le 36 \iff -6 \le x \le 6.

So the set of first coordinates is the closed interval $[-6, 6]$ . (The same argument with $x^2 = 36 - 9y^2$ gives the set of second coordinates, $[-2, 2]$ .)

Problem 3 · Two Statements About a New Set of Pairs

Given: $Q = \{ (x, y) \in \mathbb{Z} \times \mathbb{Z} \mid |x - y| = 1 \}$ . Consider (I) for all $x, y \in \mathbb{Z}$ , if $(x, y) \in Q$ , then $(y, x) \in Q$ ; and (II) for all $x, y, z \in \mathbb{Z}$ , if $(x, y) \in Q$ and $(y, z) \in Q$ , then $(x, z) \in Q$ . Which is right?

✅ Correct!

|y - x| = |x - y|

, so (I) holds for every pair;

(1, 2) \in Q

and

(2, 3) \in Q

but

|1 - 3| = 2

, so

(1, 3) \notin Q

refutes (II).

❌ Not quite. (II) fails:

(1, 2)

and

(2, 3)

are in

Q

, but

(1, 3)

is not, since

|1 - 3| = 2

❌ Not quite. (I) is true: if

|x - y| = 1

, then

|y - x| = |x - y| = 1

, so

(y, x) \in Q

❌ Not quite. It is the other way round. Reversing a pair keeps the distance

|x - y|

, while two steps of size

1

can add up to a step of size

2

❌ Not quite. Prove (I) for an arbitrary pair, and look for a counterexample to (II).

Show solution

(I) is true. Let $x, y \in \mathbb{Z}$ with $(x, y) \in Q$ . Then $|x - y| = 1$ . Since $|y - x| = |-(x - y)| = |x - y| = 1$ , we have $(y, x) \in Q$ . The argument used nothing about $x$ and $y$ beyond $(x, y) \in Q$ , so it covers every pair.

(II) is false. Take $x = 1$ , $y = 2$ , $z = 3$ . Then $|1 - 2| = 1$ and $|2 - 3| = 1$ , so $(1, 2) \in Q$ and $(2, 3) \in Q$ , but $|1 - 3| = 2$ , so $(1, 3) \notin Q$ . One counterexample refutes the universal statement.

Problem 4 · Second Coordinates of a Parabola

Given: $W = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid y = x^2 - 4 \}$ . Find the set of second coordinates, $\{ y \in \mathbb{R} \mid \exists\, x \in \mathbb{R},\ y = x^2 - 4 \}$ .

✅ Correct!

x^2 = y + 4

has a real solution exactly when

y + 4 \ge 0

, that is,

y \ge -4

❌ Close, but…

\mathbb{R}

is the set of first coordinates: every real

x

gives the pair

(x, x^2 - 4)

. But

y = -5

would need

x^2 = -1

❌ Not quite.

x^2 \ge 0

, but

y = x^2 - 4

4

less:

(0, -4)

and

(1, -3)

are in

W

, so negative second coordinates occur.

❌ Not quite.

[-2, 2]

is where

x^2 - 4 \le 0

, a set of first coordinates. The second coordinate

y = 5

occurs, from

x = 3

❌ Not quite. Fix

y

and ask whether

x^2 = y + 4

has a real solution.

Show solution

Fix $y \in \mathbb{R}$ . A pair $(x, y)$ lies in $W$ exactly when $x^2 = y + 4$ .

If $y \ge -4$ , then $x = \sqrt{y + 4}$ is a real number with $x^2 - 4 = y$ , so $y$ is a second coordinate.
If $y \lt -4$ , then $y + 4 \lt 0$ , and no real $x$ has $x^2 = y + 4$ .

So the set of second coordinates is $[-4, \infty)$ . On the graph, this is the shadow of the parabola $y = x^2 - 4$ on the $y$ -axis, starting at the vertex $(0, -4)$ .

Solved: 0 / 4