3 comparison statements, inequalities and intervals x

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 0 < x,
0

0
+x is positive

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

0
+– x is negative x is positive

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 that 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 that 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 that “[”, “]” 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) depends 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)

A set is a collection of items and it’s written as {#,#,..}.

For example, S = {A = Abe, B = Bob, C = Cathy}
is a set of 3 people, T = {A = Abe, C = Cathy,
D = Dora, E = Ed} is a set of 4 people.

The union of two sets S and T, written as S U T,
is the combined set of S and T.

Hence S U T = {A, B, C, D, E} consists of all the
people listed in the two sets.

The intersection of two sets S and T, written as S ∩ T,
consists of the common items, i.e. the items listed in
both sets.

both sets. So S ∩ T = {A, C}, the names in both sets.

both. So S ∩ T = {A, C}, the names listed in both sets.
For sets of intervals of numbers, we find their U and ∩
by drawing.

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 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:

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

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:

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

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
Example B. Abe and Bob work at the same shop.
Abe works after 2 pm till no more than 5 pm,
Bob works from exactly 4 pm till before 7 pm,
a. draw each person's schedule on a time line and
write them using the interval notation.

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
31 2 4 5 6 7 8
pm
31 2 4 5 6 7 8
pm
Abe schedule: Bob’s schedule:

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
31 2 4 5 6 7 8
pm
31 2 4 5 6 7 8
pm
A: (2, 5 ]

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
31 2 4 5 6 7 8
pm
31 2 4 5 6 7 8
pm
A: (2, 5 ] B: [4, 7)

b. When will there someone working at the shop?

Stack the schedules as shown.
31 2 4 5 6 7 8 pm
31 2 4 5 6 7 8 pm
A: (2, 5 ]
B: [4, 7)

The answer is the
union of A and B
31 2 4 5 6 7 8 pm
31 2 4 5 6 7 8 pm
A: (2, 5 ]
B: [4, 7)
31 2 4 5 6 7 8 pm
i.e. A U B
A U B:

The answer is the
union of A and B
31 2 4 5 6 7 8 pm
31 2 4 5 6 7 8 pm
A: (2, 5 ]
B: [4, 7)
31 2 4 5 6 7 8 pm
i.e. A U B = (2, 7).
A U B:

The answer is the
union of A and B
31 2 4 5 6 7 8 pm
31 2 4 5 6 7 8 pm
A: (2, 5 ]
B: [4, 7)
So there will be
someone after 2 pm till before 7 pm.
31 2 4 5 6 7 8 pm
i.e. A U B = (2, 7).
A U B:

The answer is the
union of A and B
31 2 4 5 6 7 8 pm
31 2 4 5 6 7 8 pm
A: (2, 5 ]
B: [4, 7)
So there will be
31 2 4 5 6 7 8 pm
i.e. A U B = (2, 7).
A U B:
b. When will both be working at the shop?
31 2 4 5 6 7 8 pm
31 2 4 5 6 7 8 pm
A: (2, 5 ]
B: [4, 7)

The answer is the
union of A and B
31 2 4 5 6 7 8 pm
31 2 4 5 6 7 8 pm
A: (2, 5 ]
B: [4, 7)
So there will be
31 2 4 5 6 7 8 pm
i.e. A U B = (2, 7).
A U B:
31 2 4 5 6 7 8 pm
31 2 4 5 6 7 8 pm
A: (2, 5 ]
B: [4, 7)
31 2 4 5 6 7 8 pm
A ∩ B:
The time both
persons be working
is the intersection of
their schedule,
i.e. A ∩ B

The answer is the
union of A and B
31 2 4 5 6 7 8 pm
31 2 4 5 6 7 8 pm
A: (2, 5 ]
B: [4, 7)
So there will be
31 2 4 5 6 7 8 pm
i.e. A U B = (2, 7).
A U B:
31 2 4 5 6 7 8 pm
31 2 4 5 6 7 8 pm
A: (2, 5 ]
B: [4, 7)
31 2 4 5 6 7 8 pm
A ∩ B:
The time both
persons be working
is the intersection of
their schedule,
i.e. A ∩ B = [4, 5]. So both be there from 4 pm to 5 pm.

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.
For example, {a} ∩ {b} = 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.
9. There were at least 50 people on the bus.
10. There were no more than 50 people on the bus.
11. There were less than 30 people on the bus.
12. There were no less than 28 people on the bus.
Let T be temperature outside.
13. The temperature is no more than –2o.
14. The temperature is at least than 35o.
15. The temperature is positive.

Let M be the amount of money I have.
16. I have at most $25.
17. I have a non–positive amount of money.
18. I have less than $45.
19. 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.
20. We are below the 7th floor.
21. We are above the first floor.
22. We are not below the 3rd floor basement.
23. We are on at least the 45th floor (or higher).
24. We are between the 4th floor basement and the 10th floor.
25. We are in the basement.

3 comparison statements, inequalities and intervals x

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