1.5 comparison statements, inequalities and intervals y

Comparison Statements, Inequalities and Intervals

The following adjectives or comparison phrases are
translated into inequalities in mathematics:

“positive” vs. “negative”,
“non–positive” vs. ”non–negative”,
“more/greater than” vs. “less/smaller than”,
“no more/greater than” vs. “no less/smaller than”,
“at least” vs. ”at most”,

“Positive” vs. “Negative”
0

A quantity x is positive means that x > 0,
0
+x is positive

and x is negative means that x < 0.
0
+– x is negative x is positive

and x is negative means that x < 0.
0
+– x is negative x is positive
The phrase “the temperature T is positive” is “0 < T”.

“Non–Positive” vs. “Non–Negative”

A quantity x is non–positive means x is not positive,
or “x ≤ 0”,

or “x ≤ 0”,
0
+–
x is non–positive

or “x ≤ 0”, and x is non–negative means x is not
negative, or “0 ≤ x”.
0
+–
x is non–positive
x is non–negative

0
+–
x is non–positive
x is non–negative
The phrase
“the account balance A is non–negative” is “0 ≤ A”.

0
+–
x is non–positive
x is non–negative
The phrase
“More/greater than” vs “Less/smaller than”

0
+–
x is non–positive
x is non–negative
The phrase
Let C be a number, x is greater than C means
“C < x”,
C
x is more than C

0
+–
x is non–positive
x is non–negative
The phrase
Let C be a number, x is greater than C means
“C < x”, and x is less than C means “x < C”.
Cx is less than C
x is more than C

“No more/greater than” vs “No less/smaller than”
and “At most” vs “At least”

A quantity x is “no more/greater than C”
is the same as “x is at most C” and means “x ≤ C”.

+–
x is no more than C C
x is at most C

A quantity x is “no–less than C” is the same as
“x is at least C” and means “C ≤ x”.
+–
x is no more than C x is no less than CC
x is at most C x is at least C

+–
x is no more than C x is no less than C
“The account balance A is no–less than 500”
is the same as “A is at least 500” or that “500 ≤ A”.
C

+–
x is no more than C x is no less than C
“The temperature T is no–more than 250o”
is the same as “T is at most 250o” or that “T ≤ 250o”.
“The account balance A is no–less than 500”
is the same as “A is at least 500” or that “500 ≤ A”.
C

We also have the compound statements such as
“x is more than a, but no more than b”.

In inequality notation, this is “a < x ≤ b”.

+–
a a < x ≤ b b

+–
a a < x ≤ b b
and it’s denoted as: (a, b]
where “(”, “)” means the end points are excluded
and “[”, “]” means the end points are included.

+–
a a < x ≤ b b
A line segment as such is called an interval.

Therefore the statement “the length L of
the stick must be more than 5 feet but no
more than 7 feet” is “5 < L ≤ 7”
+–
a a < x ≤ b b

or that L must be in the interval (5, 7].
+–
a a < x ≤ b b
75
5 < L ≤ 7
or (5, 7]
L

or that L must be in the interval (5, 7].
+–
a a < x ≤ b b
75
5 < L ≤ 7
or (5, 7]Following is a list of interval notation.
L

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

or a ≤ x ≤ b as [a, b],ba

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

Note: The notation “(2, 3)”
is to be viewed as an interval or as
a point (x, y) depending on the context.

or a ≤ x < b as [a, b),ba
or a < x ≤ b as (a, b],ba

Using the “∞” symbol which means to “surpass all
finite numbers”, we may write the rays
∞
a
or a ≤ x, as [a, ∞),

∞
a
or a ≤ x, as [a, ∞),
∞a
or a < x, as (a, ∞),

∞
a
or a ≤ x, as [a, ∞),
–∞ a
or x ≤ a, as (–∞, a],
∞a
or a < x, as (a, ∞),
–∞ a
or x < a, as (–∞, a),

Intersection and Union (∩ & U) of Intervals

Let I = [1, 3] as shown,
31
I:
and let J = (2, 4) be another interval as shown,
42
J:

31
I:
42
J:
The common portion of the two intervals I and J
shown here 31
I:
42
J:

31
I:
42
J:
shown here 31
I:
42
J:
2
3
32

31
I:
42
J:
shown here 31
I:
42
J:
2
3
3
I ∩ J:
is called the intersection of I and J.
2

31
I:
42
J:
shown here 31
I:
42
J:
2
3
3
I ∩ J:
is called the intersection of I and J.
It’s denoted as I ∩ J and in this case I ∩ J = (2, 3].
2

The merge of the two intervals I and J shown here
31
I:
42
J:

31
I:
42
J:
2
3
2 41 3

31
I:
42
J:
2
3
I U J:
is called the union of I and J and it’s denoted as I U J.
2 41 3

31
I:
42
J:
2
3
I U J:
2 41 3
In this case I U J = [1, 4).

31
I:
42
J:
2
3
I U J:
2 41 3
Example A. Given intervals I, J, and K, perform the
following set operation. Draw the solution and write
the answer in the interval notation.
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
a. K U J

31
I:
42
J:
2
3
I U J:
2 41 3
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
a. K U J
–3 1
0
KWe have

31
I:
42
J:
2
3
I U J:
2 41 3
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
a. K U J
–2
–3 1
0
K
J
We have

31
I:
42
J:
2
3
I U J:
2 41 3
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
a. K U J
–2
–3 1
0
K
J
and K U J is
–3
0
We have
the union:

31
I:
42
J:
2
3
I U J:
2 41 3
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
a. K U J
–2
–3 1
0
so K U J = (–3, ∞).
K
J
and K U J is
–3
0
We have
the union:

b. K ∩ I
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
c. (K ∩ J) U I

We have 10
b. K ∩ I
–1–4 –3 K
I
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
c. (K ∩ J) U I

We have 10
b. K ∩ I
–1–4 –3
The intersection is the overlapping portion as shown
K
I
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
c. (K ∩ J) U I

We have 10
b. K ∩ I
–1–4 –3
so K ∩ I is or (–3, –1).
K
I
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
0
–1–3
c. (K ∩ J) U I

We have 10
b. K ∩ I
–1–4 –3
so K ∩ I is or (–3, –1).
K
I
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
0
–1–3
c. (K ∩ J) U I
Do the parenthesis first.

We have 10
b. K ∩ I
–1–4 –3
so K ∩ I is or (–3, –1).
K
I
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
0
–1–3
c. (K ∩ J) U I
therefore K ∩ J is
–2–3 1
0
–2
0
1
K ∩ J is the overlap of

We have 10
b. K ∩ I
–1–4 –3
so K ∩ I is or (–3, –1).
K
I
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
0
–1–3
c. (K ∩ J) U I
Hence (K ∩ J) U I is
–2–3 1
0
–2
0
1
–2
0
1
–1–4

We have 10
b. K ∩ I
–1–4 –3
so K ∩ I is or (–3, –1).
K
I
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
0
–1–3
c. (K ∩ J) U I
–2–3 1
0
–2
0
1
or (K ∩ J) U I = [–4, 1]
–2
0
1
–1–4

We have 10
b. K ∩ I
–1–4 –3
so K ∩ I is or (–3, –1).
K
I
0
–1–4
J: x > –2 K: –3 < x ≤ 1I:
0
–1–3
c. (K ∩ J) U I
–2–3 1
0
–2
0
1
or (K ∩ J) U I = [–4, 1]
–2
0
1
–1–4
Your turn: Is (K ∩ J) U I = K ∩ (J U I )?

Absolute Value Inequalities
The geometric meaning of the absolute value of x, denoted
as |x|, is “the distance between x and 0”, e.g. |–2| = |2| = 2.

Hence “|x| < c” means “the distance between x and 0 is < c”.
(provided that c is not negative in which case no such x exists).

Example B. Translate the meaning of |x| < 7 and draw x.

0
We are to draw all numbers which are within 7 units from the
number 0.

-7 < x < 7-7-7 7
0
x
number 0.
x

-7 < x < 7-7-7 7
0
x
number 0.
x
The open circle means the end points are not included in the
solution.

-7 < x < 7-7-7 7
0
x
number 0.
x
solution.
Intervals may be described using distance inequalities,
e.g. [1, 3] = {x’s where x is “no more than 1 unit away from 2”}.
which are absolute value inequalities.

-7 < x < 7-7-7 7
0
x
number 0.
x
solution.
Intervals may be described using distance inequalities,
e.g. [1, 3] = {x’s where x is “no more than 1 unit away from 2”}.
which are absolute value inequalities. Hence, all simple linear
inequalities corresponds to intervals, i.e. as their solutions.
Following are the two basic cases.

I. (One Piece | |–Inequalities)
If |x| < c then –c < x < c.

Because |x – y| means “the distance between x and y”,
the expression |x – y| < c means the “the distance between x
and y is less than c”.

and y is less than c”. We use this geometric meaning to solve
these problems by simple drawings.

Example C. Translate the meaning of |x – 2| < 3 and solve.

|x – 2| < 3
Translate the symbols to a geometric description.

|x – 2| < 3
the distance between x and 2

|x – 2| < 3
the distance between x and 2 less than 3

|x – 2| < 3
2
Draw

|x – 2| < 3
2
x x
right 3left 3
Draw

|x – 2| < 3
–1 5
2
x x
right 3left 3
Draw

|x – 2| < 3
Hence –1 < x < 5.
–1 5
2
x x
right 3left 3
Draw

I. (Two Piece | |–Inequalities)
If |x| > c then x < –c or c < x.

Example D. Translate the meaning of |x| ≥ 7 and draw.

The geometric meaning of the inequality is that “the distance
between x and 0 is 7 or more”.

between x and 0 is 7 or more”. In picture
-7-7 7
0
x x
end point
included

x < –7 or 7 < x-7-7 7
0
x x
end point
included

Note that the answer uses the word “or” for expressing the
choices of two sections–not the word “and”.
x < –7 or 7 < x-7-7 7
0
x x
end point
included

choices of two sections–not the word “and”. The word “and”
is used when there are multiple conditions to be met.
x < –7 or 7 < x-7-7 7
0
x x
end point
included

x < –7 or 7 < x-7-7 7
0
x x
the expression |x – y| > c means the “the distance
between x and y is more than c”.
end point
included

x < –7 or 7 < x-7-7 7
0
x x
the expression |x – y| > c means the “the distance between x
and y is more than c”.
We recall that we may break up | | for multiplication, i.e.
|x * y| = |x| * |y|.
end point
included

The interval [a, a] consists of one point {x = a}.
The empty set which contains nothing is denoted as
Φ = { }, and interval (a, a) = (a, a] = [a, a) = Φ.

Exercise. A. Draw the following Inequalities. Translate each
inequality into an English phrase. (There might be more than
one way to do it)
1. x < 3 2. –5 ≤ x 3. x < –8 4. x ≤ 12
5. x ≥ 3 6. –5 > x 7. x ≥ –8 8. x > 12
Exercise. B. Translate each English phrase into an inequality.
Draw the Inequalities.
Let P be the number of people on a bus.
1. There were at least 50 people on the bus.
2. There were no more than 50 people on the bus.
3. There were less than 30 people on the bus.
4. There were no less than 28 people on the bus.
Let T be temperature outside.
5. The temperature is no more than –2o.
6. The temperature is at least 35o.
7. The temperature is positive.
Inequalities

Inequalities
Let M be the amount of money I have.
8. I have at most $25.
9. I have a non–positive amount of money.
10. I have less than $45.
11. I have at least $250.
Let the basement floor number be given as a negative number
and let F be the floor number that we are on.
12. We are below the 7th floor.
12. We are above the first floor.
13. We are not below the 3rd floor basement.
14. Our floor is at least the 45th floor.
15. We are between the 4th floor basement and the 10th floor.
16. We are in the basement.

C. Translate and solve the expressions geometrically.
Draw the solution.
1. |x| < 2 2. |x| < 5 3. |–x| < 2 4. |–x| ≤ 5
5. |x| ≥ –2 6. |–2x| < 6 7. |–3x| ≥ 6 8. |–x| ≥ –5
9. |3 – x| ≥ –5 10. |3 + x| ≤ 7 11. |x – 9| < 5
12. |5 – x| < 5 13. |4 + x| ≥ 9 14. |2x + 1| ≥ 3
15. |x – 2| < 1 16. |3 – x| ≤ 5 17. |x – 5| < 5
18. |7 – x| < 3 19. |8 + x| < 9 20. |x + 1| < 3

D. Let I, J, and K be the following intervals:
1. K U I
0
1–5 J: x < –1 K: –3 < x ≤ 3I:
9. (K ∩ J) U I
11. Is (K U J) U I = K U (J U I)?
Is (K ∩ J) ∩ I = K ∩ (J ∩ I)?
Is (K ∩ J) U I = K ∩ (J U I)?
Draw the following intervals and write the answers
in the interval notation.
2. K U J 3. J U I 4. (K U J) U I
5. K ∩ I 6. K ∩ J 7. J ∩ I 8. (K ∩ J) ∩ I
10. K ∩ (J U I)
12. Apu works from 2 pm to before 10 pm,
Bobo works from after 4 pm to exactly midnight.
a. When are they both working?
b. When is that at least one of them is working?

E. Express the following intervals as absolute value inequalities
in x.
1. [–5, 5]
2. (–5, 5) 3. (–5, 2) 4. [–2, 5]
5. [7, 17] 6. (–49, 84) 7. (–11.8, –1.6) 8. [–1.2, 5.6]
–2 4
–15 –2
8 38
0 a
–a –a/2
a – b a + b
10.
13.
11.
12.
9.
14.

(Answers to odd problems) Exercise A.
1. My nephew is less than 3 years old
3. Today the temperature is less than –8o
-8x
3x
3 x
5. Your child must be at least 3 years old to enter the kinder
garden
7. The temperature of the fridge must be at least -8o or else
your food won’t last
-8 x
Inequalities

Exercise B.
1. 𝑃 ≥ 50
3. 𝑃 < 30
50 P
30P
5. 𝑇 ≤ – 2
-2T
7. 𝑇 > 0
0 T
Inequalities

9. 𝑀 ≤ 0
0M
11. 𝑀 ≥ 250
250 M
13. 𝐹 > 1
1 F
15. −4 < 𝐹 < 10
-4 F 10
Inequalities

Exercise C.
1. [−5, 3] 3. (−∞, 1)
3–5 1
5. (−3, 1)
–3 -1
7. [−5, 1)
–5 -1
9. [−5, 1)
–5 -1
11. Is (𝐾 ∪ 𝐽) ∪ 𝐼 = 𝐾 ∪ (𝐽 ∪ 𝐼)? Yes
Is (𝐾 ∩ 𝐽) ∩ 𝐼 = 𝐾 ∩ (𝐽 ∩ 𝐼)? Yes
Is (𝐾 ∩ 𝐽) ∪ 𝐼 = 𝐾 ∩ (𝐽 ∪ 𝐼)? No
𝐾 ∩ 𝐽 ∪ 𝐼 = [−5,1)
and 𝐾 ∩ 𝐽 ∪ 𝐼 = (−3,1)

Exercise D.
1. |𝑥| < 2
-2 2
x x
0
3. |– 𝑥| < 2
-2 2
x x
0
5. |𝑥| ≥ – 2
x x
0
7. |– 3𝑥| ≥ 6
-2 2
x x
0

9. |3 – 𝑥| ≥ – 5
x x
0
11. |𝑥 − 9| < 5
4 14
x x
9
13. |4 + 𝑥| ≥ 9
-13 5
x x
-4
15. |𝑥 – 2| < 1
1 3
x x
2

17. |𝑥 − 5| < 5
0 10
x x
5
19. |8 + 𝑥| < 9
-17 -1
x x
-8
Exercise E.
1. |𝑥| ≤ 5 3. 𝑥 < 5 5. 𝑥 − 12 ≤ 5
7. |𝑥 + 6.7| < 5.1 9. |𝑥 − 1| ≥ 3 11. 𝑥 + 8.5 > 6.5
13.|𝑥 +
3𝑎
4
| ≥
𝑎
4

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