Mathematical-Reasoning · Unit 4 · Video 1 · Interactive Practice

Change the Set, Change the Truth: Sets, the Roster Method and Variables

IKey Formulas

Idea	Example	What it says
Roster method	$C = \{-3, 3\}$	the integer solutions of $x^2 = 9$ , listed between braces
Dots in the middle	$D = \{1, 3, 5, \ldots, 49\}$	the dots only save writing: $25$ elements, last element $49$
Dots at an end	$F = \{2, 4, 6, \ldots\}$	the pattern continues and there is no last element
Variable with universal set $U$	$y^2 - 2y - 15 = 0$	over $\mathbb{R}$ : $y = 5$ or $y = -3$ ; over $\mathbb{N}$ : only $y = 5$

Key Insight: A set is well-defined when, for any object, we can decide whether it belongs. It need not be listable in full: the integers greater than $4$ have no largest element, yet every integer is either in or out.

IIWhere the Dots Sit

Three dots continue a listed pattern: in the middle the list still ends; at an end it never does.

Set Elements written in 0

IIIOne Sentence, Four Universal Sets

Which values make a sentence true depends on the sentence and on the universal set its variable comes from.

Sentence Universal set

IVTrue for a Whole Range

Some sentences are true on a whole interval, some for every value, some only at scattered natural numbers.

Sentence

VQuiz Questions

Problem 1 · An Infinite Roster

Write the set of integers greater than $-4$ in roster form.

✅ Correct!

-4

is not greater than itself, so the smallest element is

-3

; there is no largest element, so the list ends in dots.

❌ Close, but…

-4 > -4

is false, so

-4

is left out. The smallest element is

-3

❌ Not quite. That roster lists the integers less than

-4

. The integers greater than

-4

lie to its right:

-3, -2, -1, \ldots

❌ Not quite. This list stops at

3

, but

4

5

6

, … are also integers greater than

-4

. A set with no last element needs dots at the end.

❌ Not quite. Find the smallest integer greater than

-4

, then decide on which side the pattern continues.

Show solution

An integer $n$ belongs to the set exactly when $n > -4$ . Since $-4 > -4$ is false, $-4$ is left out, and the smallest element is $-3$ .

Every integer above $-3$ also qualifies and there is no largest one, so the pattern continues to the right without end:

\{-3, -2, -1, 0, 1, \ldots\}

The dots go at the end only: there is a first element ( $-3$ ) but no last element.

Problem 2 · Factor, Don't Divide

Write the set of real numbers that are solutions of $x^2 - 7x = 0$ in roster form.

✅ Correct!

x(x - 7) = 0

gives

x = 0

x = 7

, and both are real numbers.

❌ Close, but… Dividing both sides by

x

assumes

x \ne 0

and throws away the solution

x = 0

: indeed

0^2 - 7 \cdot 0 = 0

. Factor instead.

❌ Not quite.

x - 7 = 0

gives

x = 7

, not

-7

. Check:

(-7)^2 - 7(-7) = 49 + 49 = 98 \ne 0

❌ Not quite. Those solve

x^2 = 7

. Here

7

multiplies

x

x^2 - 7x = x(x - 7)

❌ Not quite. Factor out

x

and use that a product is zero exactly when at least one factor is zero.

Show solution

Factor out $x$ :

x^2 - 7x = x(x - 7) = 0

A product is zero exactly when at least one of its factors is zero, so $x = 0$ or $x - 7 = 0$ , that is, $x = 0$ or $x = 7$ .

Check: $0^2 - 7 \cdot 0 = 0$ ✓ and $7^2 - 7 \cdot 7 = 49 - 49 = 0$ ✓.

\{0, 7\}

Problem 3 · Same Sentence, Two Universal Sets

Consider the sentence $z^2 - z - 12 = 0$ in the variable $z$ .

Universal set ℤ: the integers that make it true

Universal set ℕ: the natural numbers that make it true

✅ Correct! Over

\mathbb{Z}

both

4

and

-3

work; over

\mathbb{N}

only

4

does. The sentence is the same; the universal set changed.

❌ Check the factoring.

(z - 3)(z + 4) = z^2 + z - 12

, which has

+z

, not

-z

. The right factors are

(z - 4)(z + 3)

❌ Close, but…

-3

is an integer, so over

\mathbb{Z}

it stays:

(-3)^2 - (-3) - 12 = 9 + 3 - 12 = 0

❌ Close, but…

-3

is not a natural number, so it is not a value the variable may take over

\mathbb{N}

. Only

4

remains.

❌ Not quite. Factor:

z^2 - z - 12 = (z - 4)(z + 3)

, so

z = 4

z = -3

, and both are integers.

❌ Not quite. Of the two integer solutions

4

and

-3

, keep only those that are natural numbers.

Show solution

Find two numbers with product $-12$ and sum $-1$ : they are $-4$ and $3$ . So

z^2 - z - 12 = (z - 4)(z + 3) = 0,

which is true exactly when $z = 4$ or $z = -3$ .

Check: $4^2 - 4 - 12 = 16 - 16 = 0$ ✓ and $(-3)^2 - (-3) - 12 = 9 + 3 - 12 = 0$ ✓.

Universal set $\mathbb{Z}$ : both $4$ and $-3$ are integers, so the roster is $\{-3, 4\}$ .

Universal set $\mathbb{N}$ : the natural numbers start at $1$ , so $-3$ is excluded and the roster is $\{4\}$ .

Problem 4 · Dots on One Side

An integer multiple of $5$ is a number $5k$ with $k$ an integer. Write the set of integer multiples of $5$ that are less than or equal to $10$ in roster form.

✅ Correct! The largest element is

10

and the pattern continues to the left with no first element, so the dots go on the left only.

❌ Not quite.

k

may be zero or negative:

0 = 5 \cdot 0

-5 = 5 \cdot (-1)

and

-10 = 5 \cdot (-2)

are multiples of

5

that are at most

10

❌ Close, but… Dots on the right would continue the list past

10

15

20

, …, which are greater than

10

. The list stops at

10

❌ Close, but…

-5

-10

-15

, … are also multiples of

5

that are at most

10

. The pattern continues to the left.

❌ Not quite. Decide where the list stops and in which direction the pattern continues.

Show solution

The integer multiples of $5$ are $\ldots, -10, -5, 0, 5, 10, 15, \ldots$ , all of the form $5k$ .

The condition $5k \le 10$ keeps exactly those with $k \le 2$ : the largest element is $5 \cdot 2 = 10$ , and $15, 20, \ldots$ are excluded.

Every $k \le 2$ qualifies, including $k = 0, -1, -2, \ldots$ , so there is no smallest element. The dots go on the left only:

\{\ldots, -10, -5, 0, 5, 10\}

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