41 interval notation and review on inequalities

Inequalities

We recall that comparison of two numbers is based on their

locations on the real line.

Inequalities

locations on the real line. Specifically, given two numbers

L and R corresponding to two points on the real line as shown,

Inequalities

+–RL

we define the number to the right to be greater than the

number to the left.

Inequalities

+–RL

number to the left. Hence R is greater than L or that L is

smaller than R .

Inequalities

+–RL

smaller than R .

Inequalities

We write this as L < R or as R > L.

smaller than R .

Inequalities

For example, 2 < 4, –3 < –2, 0 > –1 are true statements

smaller than R .

Inequalities

For example, 2 < 4, –3 < –2, 0 > –1 are true statements and

–2 < –5 , 5 < 3 are false statements.

smaller than R .

Inequalities

We use inequalities to represent segments of the real line, e.g.

“all the numbers x that are greater than 5" as “5 < x".

smaller than R .

Inequalities

We use inequalities to represent segments of the real line, e.g.

“all the numbers x that are greater than 5" as “5 < x".

+–5 x

InequalitiesIn general, we write "a < x" for all In picture,

InequalitiesIn general, we write "a < x" for all the numbers x greater than a

but excluding a. In picture,+

open dot

If we want all the numbers x greater than or equal to a which

includes a, we write it as a < x.

open dot

includes a, we write it as a < x. In picture

solid dot

a ≤ x

open dot

solid dot

a ≤ x

The numbers x such that a < x < b where a < b are all the

numbers x between a and b.

+–a a < x ≤ b b

open dot

solid dot

a ≤ x

numbers x between a and b. A line segment as such is called

an interval.

+–a a < x ≤ b b

open dot

solid dot

a ≤ x

an interval.

+–a a < x ≤ b b

Hence –3 ≤ x < 2 is2–3

open dot

solid dot

a ≤ x

an interval.

+–a a < x ≤ b b

Expressions such as 2 < x > 3 or 2 < x < –3 do not have any

solution.

Hence –3 ≤ x < 2 is2–3

InequalitiesWe note the following algebra about inequalities.

Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.

a. Adding or subtracting the same quantity to both sides

retains the inequality sign,

retains the inequality sign, i.e. if a < b, then a ± c < b ± c.

Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.

b. Multiplying or dividing a positive number c to both sides

retains the inequality sign,

retains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.

Given that 6 < 12, if we multiply –1 to both sides

Given that 6 < 12, if we multiply –1 to both sides then

(–1)6 > (–1)12

– 6 > –12 is true.

(–1)6 > (–1)12

– 6 > –12 is true.

Multiplying by –1 switches the relation of two numbers.

60 +–

(–1)6 > (–1)12

– 6 > –12 is true.

60 +–

12–6 <

(–1)6 > (–1)12

– 6 > –12 is true.

60 +–

12–6–12 <<

(–1)6 > (–1)12

– 6 > –12 is true.

c. Multiplying or dividing by an negative number c reverses

the inequality sign, i.e. if c < 0 and a < b then ca > cb

60 +–

12–6–12 <<

(–1)6 > (–1)12

– 6 > –12 which is true.

Following are the steps for solving linear inequalities.

Inequalities

1. Simplify both sides of the inequalities.

Inequalities

2. Gather the x-terms to one side and the number-terms to the

other sides (use the “change side-change sign” rule).

Inequalities

3. Multiply or divide to get x. If we multiply or divide by

negative numbers to both sides, the inequality sign must be

turned around.

Inequalities

turned around. (This rule can be avoided by keeping the

x-term positive.)

Inequalities

x-term positive.)

Inequalities

Example A. a. Solve –3x + 2 < 11. Draw the solution.

x-term positive.)

Inequalities

–3x + 2 < 11 move the x to the other side to make it positive

x-term positive.)

Inequalities

2 – 11 < 3x

x-term positive.)

Inequalities

2 – 11 < 3x

–9 < 3x

x-term positive.)

Inequalities

2 – 11 < 3x

–9 < 3x divide by 3, no change with the inequality

<3x–9

3 3 –3 < x

x-term positive.)

Inequalities

2 – 11 < 3x

–9 < 3x divide by 3, no change with the inequality

<3x–9

3 3 –3 < x

In picture,

Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.

Draw the solution.

5 < –3x + 1 ≤ 10 subtract 1 from all entries,

Draw the solution.

–6 < –3x ≤ 9

Draw the solution.

–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,

Draw the solution.

–6 –3

–3x –3>

≥9 –3

Draw the solution.

–6 –3

–3x –3>

≥9 –3

–2 > x ≥ –3 or–2–3

Draw the solution.

–6 –3

–3x –3>

≥9 –3

–2 > x ≥ –3 or–2–3

Signs and Inequalities

Draw the solution.

–6 –3

–3x –3>

≥9 –3

–2 > x ≥ –3 or–2–3

One of the most important association to inequalities is

the sign of a given expression.

Draw the solution.

–6 –3

–3x –3>

≥9 –3

–2 > x ≥ –3 or–2–3

the sign of a given expression. Let f be an expression,

following adjectives translate into the following inequalities.

Draw the solution.

–6 –3

–3x –3>

≥9 –3

–2 > x ≥ –3 or–2–3

i. “f is positive” ↔ 0 < f or f > 0

Draw the solution.

–6 –3

–3x –3>

≥9 –3

–2 > x ≥ –3 or–2–3

“f is non–negative” ↔ 0 ≤ f or f ≥ 0

Draw the solution.

–6 –3

–3x –3>

≥9 –3

–2 > x ≥ –3 or–2–3

ii. “f is negative” ↔ f < 0 or 0 > f

Draw the solution.

–6 –3

–3x –3>

≥9 –3

–2 > x ≥ –3 or–2–3

ii. “f is negative” ↔ f < 0 or 0 > f

“f is non–positive” ↔ f ≤ 0 or 0 ≥ f

InequalitiesExample B. Translate the sentence “3x + 12 is positive” into

inequalities and solve for x. Draw the solution.

Being positive translates into the inequality

3x + 12 > 0

3x > –12

x > –4

3x + 12 > 0

3x > –12

x > –4

3x + 12 > 0

3x > –12

x > –4

3x + 12 > 0

b. Let x be the price of a movie ticket. Translate the sentence

“$20 minus the cost of two movie tickets is non–positive”

into an inequality. Solve it and draw the solution.

3x > –12

x > –4

3x + 12 > 0

Two tickets cost $2x,

3x > –12

x > –4

3x + 12 > 0

Two tickets cost $2x, so “$20 minus the cost of two tickets”

is (20 – 2x).

3x > –12

x > –4

3x + 12 > 0

is (20 – 2x). The inequality for “(20 – 2x) is non–positive” is

20 – 2x ≤ 0

3x > –12

x > –4

3x + 12 > 0

20 ≤ 2x

10 ≤ x

20 – 2x ≤ 0

3x > –12

x > –4

3x + 12 > 0

20 ≤ 2x

10 ≤ x

20 – 2x ≤ 0

or 0x10

3x > –12

x > –4

3x + 12 > 0

20 ≤ 2x

10 ≤ x

20 – 2x ≤ 0

or 0x10

So the price of a ticket has to be $10 or more.

InequalitiesIntersection and Union (∩ & U)

Inequalities

Let I be the interval consists of 1 ≤ x ≤ 3 as shown,

Intersection and Union (∩ & U)

Inequalities

and in the set notation we write I = {1 ≤ x ≤ 3 }.

Inequalities

Let J = {2 < x < 4 } be another interval as shown.

Inequalities

The common portion of the two intervals I and J shown here

Inequalities

3I ∩ J:

is called the intersection of I and J and it’s denoted as I ∩ J.

Inequalities

In this case I ∩ J = {2 < x ≤ 3 }.

3I ∩ J:

is called the intersection of I and J and it’s denoted as I ∩ J.

InequalitiesThe merge of the two intervals I and J shown here

2 41 3

I U J:

is called the union of I and J and it’s denoted as I U J.

2 41 3

In this case I U J: = {1 ≤ x < 4 }.

I U J:

2 41 3

I U J:

The Interval Notation

I U J:

We use the infinity symbol “∞” to mean “continues onward and

surpasses all numbers”.

I U J:

–∞ ∞0

surpasses all numbers”. We label the left of the real line with –

∞ and to the right with ∞ as shown below when needed.

I U J:

–∞ ∞0

Infinity “∞” is not a number but we say that –∞ < x < ∞

for all real number x.

I U J:

–∞ ∞0

for all real number x. In fact, the set of all real numbers R is

R = {–∞ < x < ∞ }.

I U J:

–∞ ∞0

for all real number x. In fact, the set of all real numbers R is

R = {–∞ < x < ∞ }. However, we do not write x ≤ ∞ because ∞ is

not a number in particular x can’t be ∞.

InequalitiesUsing the “∞” symbols, we write the line segment

∞aor a ≤ x

∞aor a ≤ x as [a, ∞),

We use ] or [ to include the end points,

∞aor a ≤ x as [a, ∞),

∞aor a < x as (a, ∞),

We use ] or [ to include the end points, ) and ( to exclude the

end points.

∞aor a ≤ x as [a, ∞),

–∞ a

or x ≤ a as (–∞, a],

∞aor a < x as (a, ∞),

–∞ aor x < a as (–∞, a),

end points.

∞aor a ≤ x as [a, ∞),

–∞ a

or x ≤ a as (–∞, a],

∞aor a < x as (a, ∞),

–∞ aor x < a as (–∞, a),

end points.

Let a, b be two numbers such that a < b, we write the intervals

or a ≤ x ≤ b as [a, b],ba

or a < x < b as (a, b),ba

or a ≤ x < b as [a, b),ba

or a < x ≤ b as (a, b],ba

∞aor a ≤ x as [a, ∞),

–∞ a

or x ≤ a as (–∞, a],

∞aor a < x as (a, ∞),

–∞ aor x < a as (–∞, a),

end points. Note the interval notation (a, b) is the same as the

coordinate of a point so its interpretation depends on the context.

Let a, b be two numbers such that a < b, we write the intervals

or a ≤ x ≤ b as [a, b],ba

or a < x < b as (a, b),ba

or a ≤ x < b as [a, b),ba

or a < x ≤ b as (a, b],ba

InequalitiesThe interval [a, a] consists of exactly one point {x = a}.

The empty set which contains nothing is denoted as Φ = { }.

The interval (a, a) or (a, a] or [a, a) = Φ.

Example C. Given intervals I, J , and K, do the set operation.

Draw. Put the answer in the interval and the inequality notation.

I: x > –20–1–4 J: –3 < x ≤ 1K:

a. K U J

b. K ∩ I

I: x > –20–1–4 J: –3 < x ≤ 1K:

a. K U J–3 1

b. K ∩ I

I: x > –20–1–4 J: –3 < x ≤ 1K:

a. K U J–2–3 1

b. K ∩ I

I: x > –20–1–4 J: –3 < x ≤ 1K:

a. K U J

so K U J is –2–3 1

b. K ∩ I

I: x > –20–1–4 J: –3 < x ≤ 1K:

a. K U J

so K U J is –2–3 1

Hence K U J is (–3, ∞) or {–3 < x}.

b. K ∩ I

I: x > –20–1–4 J: –3 < x ≤ 1K:

a. K U J

so K U J is –2–3 1

Hence K U J is (–3, ∞) or {–3 < x}.

b. K ∩ I–3 K

I: x > –20–1–4 J: –3 < x ≤ 1K:

a. K U J

so K U J is –2–3 1

Hence K U J is (–3, ∞) or {–3 < x}.

b. K ∩ I

–1–4 –3 K

I: x > –20–1–4 J: –3 < x ≤ 1K:

a. K U J

so K U J is –2–3 1

Hence K U J is (–3, ∞) or {–3 < x}.

b. K ∩ I

–1–4 –3

The common or overlapping portion is shown.

I: x > –20–1–4 J: –3 < x ≤ 1K:

a. K U J

so K U J is –2–3 1

Hence K U J is (–3, ∞) or {–3 < x}.

b. K ∩ I

–1–4 –3

The common or overlapping portion is shown.

Hence K ∩ I is (–3, –1) or {–3 < x < –1}

Inequalities

c. (K ∩ J) U I

I: x > –20–1–4 J: –3 < x ≤ 1K:

Inequalities

c. (K ∩ J) U I

We do the parenthesis first.

I: x > –20–1–4 J: –3 < x ≤ 1K:

Inequalities

c. (K ∩ J) U I

–2–3 10

I: x > –20–1–4 J: –3 < x ≤ 1K:

K ∩ J :

Inequalities

c. (K ∩ J) U I

so K ∩ J is

–2–3 10

I: x > –20–1–4 J: –3 < x ≤ 1K:

K ∩ J :

Inequalities

c. (K ∩ J) U I

Therefore (K ∩ J) U I is

so K ∩ J is

–2–3 10

1–1–4

I: x > –20–1–4 J: –3 < x ≤ 1K:

K ∩ J :

Inequalities

c. (K ∩ J) U I

so K ∩ J is

–2–3 10

so (K ∩ J) U I = {–4 < x ≤ 1} or (–4, 1]

1–1–4

I: x > –20–1–4 J: –3 < x ≤ 1K:

K ∩ J :

Inequalities

c. (K ∩ J) U I

so K ∩ J is

–2–3 10

so (K ∩ J) U I = {–4 < x ≤ 1} or (–4, 1]

1–1–4

Your turn. Find K ∩ (J U I ). Is this the same as (K ∩ J) U I?

I: x > –20–1–4 J: –3 < x ≤ 1K:

K ∩ J :

41 interval notation and review on inequalities

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