Appendix A(1).pdf

MATENA1
Numbers, Inequalities and Absolute Values
Appendix A

A number is a mathematical object used to
count, measure and label.

count, measure and label. Numbers are
classified into sets, called number systems,
such as the natural numbers and the real
numbers.

numbers.
I We start with the natural numbers:
N = {0, 1, 2, . . .}.

numbers.
I We start with the natural numbers:
N = {0, 1, 2, . . .}.
I To this set we add the negative whole
numbers and get the integers:
Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . .}.

I The rational numbers, denoted Q,
are ratios of integers. Thus, any rational
number r can be expressed as r = m
n
where m and n are integers and n 6= 0.

n
I Examples of rational numbers are
1
2
−
3
7
46 =
46
1
0.17 =
17
100

n
I Examples of rational numbers are
1
2
−
3
7
46 =
46
1
0.17 =
17
100
I Division by 0 is not allowed, so
expressions like 3
0 and 0
0 are undefined.

I Some numbers, such as
√
2, π, log10 2,
can’t be expressed as a ratio of integers
and are therefore called irrational
numbers.

I Some numbers, such as
√
2, π, log10 2,
can’t be expressed as a ratio of integers
and are therefore called irrational
numbers.
I The set of real numbers, denoted R,
consists of the rational numbers
together with the irrational numbers.

Complex numbers
Real numbers Imaginary numbers
Rational numbers Irrational numbers
Integers
Natural numbers
Note: The complex numbers will be discussed in
Appendix H.

I If the number is rational, the correspon-
ding decimal is repeating. For example,
1
2
= 0.50000... = 0.50

1
2
= 0.50000... = 0.50
2
3
= 0.666666... = 0.6

1
2
= 0.50000... = 0.50
2
3
= 0.666666... = 0.6
157
495
= 0.317171717... = 0.317

1
2
= 0.50000... = 0.50
2
3
= 0.666666... = 0.6
157
495
= 0.317171717... = 0.317
9
7
= 1.285714285714... = 1.285714

1
2
= 0.50000... = 0.50
2
3
= 0.666666... = 0.6
157
495
= 0.317171717... = 0.317
9
7
= 1.285714285714... = 1.285714
I The bar indicates that the sequence of
digits repeats forever.

I On the other hand, if the number is
irrational, the decimal is non-repeating:
√
2 = 1.414213562373095 . . .

√
2 = 1.414213562373095 . . .
π = 3.141592653589793 . . .

√
2 = 1.414213562373095 . . .
π = 3.141592653589793 . . .
I If we stop the decimal expansion of any
number at a certain place, we get an
approximation to the number.

√
2 = 1.414213562373095 . . .
π = 3.141592653589793 . . .
I If we stop the decimal expansion of any
number at a certain place, we get an
approximation to the number. For
instance,
π ≈ 3.14159265

I We say a is less than b and write
a < b if b − a is positive.

I Equivalently, we say b is greater than
a and write b > a.

a and write b > a.
I When we write a ≤ b (or b ≥ a) we
mean that either a < b or a = b and we
read it as “a is less than or equal to b”.

a and write b > a.
I When we write a ≤ b (or b ≥ a) we
mean that either a < b or a = b and we
read it as “a is less than or equal to b”.
I For instance,
7 < 7.4 < 7.5 − 3 < π 3 < π
√
2 < 2
√
2 ≤ 2 2 ≤ 2.

We use this order property of R to represent
real numbers as points on a line, which is
called a real number line, or simply a real
line.

line.
−3 −2 −1 0 1 2 3 4
−2.63 −3
7
1
2
√
2 π

line.
−3 −2 −1 0 1 2 3 4
−2.63 −3
7
1
2
√
2 π
The positive direction (to the right) is
indicated by an arrow.

We will make use of sets of numbers.

I A set is a collection of objects, and
these objects are called the elements
of the set.

of the set.
I If S is a set, then the notation a ∈ S
means that a is an element of S.

of the set.
I On the other hand, a /
∈ S means that a
is not an element of S.

of the set.
I On the other hand, a /
∈ S means that a
is not an element of S.
I For example, −3 ∈ Z but π /
∈ Z.

I Some sets can be described by listing
their elements between braces.

I For instance, the set A consisting of all
positive integers less than 7 can be
written as
A = {1, 2, 3, 4, 5, 6}.

I For instance, the set A consisting of all
positive integers less than 7 can be
written as
A = {1, 2, 3, 4, 5, 6}.
I We can also write A in set-builder
notation as
A = {x ∈ Z | 0 < x < 7}.
Note: A chain of inequalities like 0 < x < 7 means
that x greater than 0 and x is less than 7.

Sets can be combined to form new sets.

I If A and B are sets, then their union
A ∪ B is the set consisting of all
elements that are in A or B (or in both
A and B).

A and B).
I The intersection of A and B is the set
A ∩ B consisting of all elements that
are both in A and B.

A and B).
I The intersection of A and B is the set
A ∩ B consisting of all elements that
are both in A and B. In other words,
A ∩ B is the common part of A and B.

Example: Find A ∪ B and A ∩ B if A =
{1, 2, 3, 4} and B = {3, 4, 5, 6}.

{1, 2, 3, 4} and B = {3, 4, 5, 6}.
Solution:
A ∪ B = {1, 2, 3, 4, 5, 6}
and

{1, 2, 3, 4} and B = {3, 4, 5, 6}.
Solution:
A ∪ B = {1, 2, 3, 4, 5, 6}
and
A ∩ B = {3, 4}

Suppose A = {1, 2} and B = {1, 2, 3, 4}.

Suppose A = {1, 2} and B = {1, 2, 3, 4}.
I Here every element of A is also an
element of B.

Suppose A = {1, 2} and B = {1, 2, 3, 4}.
element of B.
I In this case we say that A is a subset
of B and write A ⊆ B.

Suppose A = {1, 2} and B = {1, 2, 3, 4}.
element of B.
I For example, N ⊆ Q and Q ⊆ R.

Suppose A = {1, 2} and B = {1, 2, 3, 4}.
element of B.
I The empty set, denoted by ∅, is the set
that contains no element.

Suppose A = {1, 2} and B = {1, 2, 3, 4}.
element of B.
I The empty set, denoted by ∅, is the set
that contains no element. The empty
set is a subset of every other set.

Exercise: Let A = {a, b, c}. State whether
each of the following is true or false:
1. a ∈ A
2. {a} ∈ A
3. b ⊆ A
4. {a} ⊆ A
5. {a, b, c} ⊆ A
6. ∅ ⊆ A
Try this exercise on your own before looking
at the solution on the next slide.

Solution:
1. True.
2. False, {a} ⊆ A, but {a} /
∈ A.
3. False, b ∈ A, but b 6⊆ A.
4. True.
5. True.
6. True. (The empty set is a subset of
every set.)

Exercise: List all the elements of the
following sets:
1. {x ∈ R | x2
− 3x = 4}
2. {x ∈ Z | −3 ≤ x ≤ 3 and x2
−3x 6= 4}
3. {x ∈ N | x is an odd and x < 10}
4. {x ∈ Z | −3 < x < 3 and x2
= 9}

Solution:
1. {−1, 4}
2. {−3, −2, 0, 1, 2, 3}
3. {1, 3, 5, 7, 9}
4. ∅

Exercise: Describe the following sets using
set-builder notation:
1. {2, 4, 6, 8}
2. {0, 1, 8, 27, 64, 125, . . .}
3. {−3, 3}

Solution:
1. {x ∈ Z | x = 2k, k ∈ N & 1 ≤ k ≤ 4}
2. {x ∈ Z | x = k3
, k ∈ N}
3. {x ∈ Z | x2
= 9}

I An interval is a subset of R that
contains all real numbers between two
endpoints.

endpoints.
I If a < b, the open interval from a to b
is the set
(a, b) = {x ∈ R | a < x < b}.

endpoints.
is the set
(a, b) = {x ∈ R | a < x < b}.
Notice that the endpoints of the interval
a and b are excluded.

endpoints.
is the set
(a, b) = {x ∈ R | a < x < b}.
Notice that the endpoints of the interval
a and b are excluded. This is indicated
by round brackets ( ) and by open dots
on the real line.

I The closed interval from a to b is the
set
[a, b] = {x ∈ R | a ≤ x ≤ b}.

set
[a, b] = {x ∈ R | a ≤ x ≤ b}.
Here the endpoints of the interval are
included.

set
[a, b] = {x ∈ R | a ≤ x ≤ b}.
Here the endpoints of the interval are
included. This is indicated by square
brackets [ ] and by solid circles on the
real line.

I We also need to consider infinite
intervals such as
(a, ∞) = {x ∈ R | x > a}.

I We also need to consider infinite
intervals such as
(a, ∞) = {x ∈ R | x > a}.
This does not mean ∞ is a number.
The notation (a, ∞) denotes the set of
all real numbers greater than a, so ∞
simply indicates that the interval extends
indefinitely far in the positive direction.

Interval Set description Picture
(a, b) {x ∈ R | a < x < b}
a b
[a, b] {x ∈ R | a ≤ x ≤ b}
a b
[a, b) {x ∈ R | a ≤ x < b}
a b
(a, b] {x ∈ R | a < x ≤ b}
a b
(a, ∞) {x ∈ R | x > a}
a
[a, ∞) {x ∈ R | x ≥ a}
a
(−∞, b) {x ∈ R | x < b}
b
(−∞, b] {x ∈ R | x ≤ b}
b
(−∞, ∞) R

Exercise:
Consider the following intervals:
A = (1, 4] B = [3, 5)
C = (2, ∞) D = (−∞, 6)
Determine the following intersections and
unions and write the answers in interval
notation:
1. A ∩ C
2. C ∩ D
3. A ∪ B
4. C ∪ D

Solution:
1. A ∩ C = (2, 4]
2. C ∩ D = (2, 6)
3. A ∪ B = (1, 5)
4. C ∪ D = (−∞, ∞) = R

Rules for inequalities:
1. If a < b, then a + c < b + c.
2. If a < b and c < d, then
a + c < b + d.
3. If a < b and c > 0, then ac < bc.
4. If a < b and c < 0, then ac > bc.
5. If 0 < a < b, then
1
a
>
1
b
.

Example: Solve the inequality 1 + x < 7x + 5.

Solution:
1 + x < 7x + 5

Solution:
1 + x < 7x + 5
1 + x − 1 < 7x + 5 − 1 (Rule 1)

Solution:
1 + x < 7x + 5
1 + x − 1 < 7x + 5 − 1 (Rule 1)
x < 7x + 4

Solution:
1 + x < 7x + 5
1 + x − 1 < 7x + 5 − 1 (Rule 1)
x < 7x + 4
x − 7x < 7x + 4 − 7x (Rule 1)

Solution:
1 + x < 7x + 5
1 + x − 1 < 7x + 5 − 1 (Rule 1)
x < 7x + 4
x − 7x < 7x + 4 − 7x (Rule 1)
− 6x < 4

Solution:
1 + x < 7x + 5
1 + x − 1 < 7x + 5 − 1 (Rule 1)
x < 7x + 4
x − 7x < 7x + 4 − 7x (Rule 1)
− 6x < 4
−6
−6
x >
4
−6
(Rule 4)

Solution:
1 + x < 7x + 5
1 + x − 1 < 7x + 5 − 1 (Rule 1)
x < 7x + 4
x − 7x < 7x + 4 − 7x (Rule 1)
− 6x < 4
−6
−6
x >
4
−6
(Rule 4)
x > −
2
3

Solution continued. . .
We can represent the solution graphically as
−2 0 2 4 6
−2
3

−2 0 2 4 6
−2
3
and write it in set-builder notation as
S =

x ∈ R | x −
2
3

.

−2 0 2 4 6
−2
3
S =

x ∈ R | x −
2
3

.
In interval notation the solution set is
S =

−
2
3
, ∞

.

Example: Solve the inequalities
4 ≤ 3x − 2 13.

4 ≤ 3x − 2 13.
Solution:
4 ≤ 3x − 2 13

4 ≤ 3x − 2 13.
Solution:
4 ≤ 3x − 2 13
6 ≤ 3x 15 (Rule 1)

4 ≤ 3x − 2 13.
Solution:
4 ≤ 3x − 2 13
6 ≤ 3x 15 (Rule 1)
6
3
≤ x
15
3
(Rule 3)

4 ≤ 3x − 2 13.
Solution:
4 ≤ 3x − 2 13
6 ≤ 3x 15 (Rule 1)
6
3
≤ x
15
3
(Rule 3)
2 ≤ x 5

0 1 2 3 4 5 6

0 1 2 3 4 5 6
S = {x ∈ R | 2 ≤ x 5} .

0 1 2 3 4 5 6
S = {x ∈ R | 2 ≤ x 5} .
In interval notation the solution set is
S = [2, 5) .

Example: Solve the inequality x2
− 5x + 6 ≤ 0.

− 5x + 6 ≤ 0.
Solution: First we factorize the expression to get
(x − 2) (x − 3) ≤ 0.

− 5x + 6 ≤ 0.
(x − 2) (x − 3) ≤ 0.
If x = 2 or x = 3, then (x − 2) (x − 3) = 0.

− 5x + 6 ≤ 0.
(x − 2) (x − 3) ≤ 0.
If x = 2 or x = 3, then (x − 2) (x − 3) = 0. We
call the number x = 2 and x = 3 critical points.

− 5x + 6 ≤ 0.
(x − 2) (x − 3) ≤ 0.
If x = 2 or x = 3, then (x − 2) (x − 3) = 0. We
call the number x = 2 and x = 3 critical points.
The numbers 2 and 3 divide the real line into three
intervals:
−3 −2 −1 0 1 2 3 4
(−∞, 2) (2, 3)(3, ∞)

We want to know if (x − 2) (x − 3) 0 or
(x − 2) (x − 3) 0 on these intervals.

(x − 2) (x − 3) 0 on these intervals. To find out
we use the following table:
Interval x − 2 x − 3 (x − 2)(x − 3)
(−∞, 2) − − +
(2, 3) + − −
(3, ∞) + + +

(x − 2) (x − 3) 0 on these intervals. To find out
we use the following table:
Interval x − 2 x − 3 (x − 2)(x − 3)
(−∞, 2) − − +
(2, 3) + − −
(3, ∞) + + +
Hence, the solution set is
S = {x ∈ R | 2 ≤ x ≤ 3} = [2, 3] .

Example: Solve x3
+ 3x2
4x.
Solution: We first factorize the expression:
x3
+ 3x2
4x

Example: Solve x3
+ 3x2
4x.
x3
+ 3x2
4x
x3
+ 3x2
− 4x 0

Example: Solve x3
+ 3x2
4x.
x3
+ 3x2
4x
x3
+ 3x2
− 4x 0
x x2
+ 3x − 4

0

Example: Solve x3
+ 3x2
4x.
x3
+ 3x2
4x
x3
+ 3x2
− 4x 0
x x2
+ 3x − 4

0
x (x − 1) (x + 4) 0

Example: Solve x3
+ 3x2
4x.
x3
+ 3x2
4x
x3
+ 3x2
− 4x 0
x x2
+ 3x − 4

0
x (x − 1) (x + 4) 0
Note that x (x − 1) (x + 4) = 0 iff x = 0 or
x = 1 or x = −4.

Example: Solve x3
+ 3x2
4x.
x3
+ 3x2
4x
x3
+ 3x2
− 4x 0
x x2
+ 3x − 4

0
x (x − 1) (x + 4) 0
Note that x (x − 1) (x + 4) = 0 iff x = 0 or
x = 1 or x = −4. These are our critical
points.

The solutions x = −4, x = 0, and x = 1 divide the
real line into four intervals (−∞, −4), (−4, 0), (0, 1)
and (1, ∞).

and (1, ∞).
Interval x x − 1 x + 4 x(x − 1)(x + 4)
(−∞, −4) − − − −
(−4, 0) − − + +
(0, 1) + − + −
(1, ∞) + + + +

and (1, ∞).
Interval x x − 1 x + 4 x(x − 1)(x + 4)
(−∞, −4) − − − −
(−4, 0) − − + +
(0, 1) + − + −
(1, ∞) + + + +
S = {x ∈ R | −4 x 0 or x 1}
= (−4, 0) ∪ (1, ∞) .

Example: Solve for x if
x − 1
x − 4
1.

x − 1
x − 4
1.
Solution:
x − 1
x − 4
1

x − 1
x − 4
1.
Solution:
x − 1
x − 4
1
x − 1
x − 4
− 1 0

x − 1
x − 4
1.
Solution:
x − 1
x − 4
1
x − 1
x − 4
− 1 0
x − 1 − (x − 4)
x − 4
0

x − 1
x − 4
1.
Solution:
x − 1
x − 4
1
x − 1
x − 4
− 1 0
x − 1 − (x − 4)
x − 4
0
x − 1 − x + 4
x − 4
0

x − 1
x − 4
1.
Solution:
x − 1
x − 4
1
x − 1
x − 4
− 1 0
x − 1 − (x − 4)
x − 4
0
x − 1 − x + 4
x − 4
0
3
x − 4
0

Now, 3 is always positive, so
3
x − 4
0

3
x − 4
0
⇐⇒ x − 4 0

3
x − 4
0
⇐⇒ x − 4 0
⇐⇒ x 4.

3
x − 4
0
⇐⇒ x − 4 0
⇐⇒ x 4.
S = {x ∈ R | x 4}
= (4, ∞) .

x2
− x − 6
(x + 1)2
0.

x2
− x − 6
(x + 1)2
0.
Solution:
x2
− x − 6
(x + 1)2
0

x2
− x − 6
(x + 1)2
0.
Solution:
x2
− x − 6
(x + 1)2
0
(x − 3) (x + 2)
(x + 1)2 0

x2
− x − 6
(x + 1)2
0.
Solution:
x2
− x − 6
(x + 1)2
0
(x − 3) (x + 2)
(x + 1)2 0
Hence, the critical points are x = −2, x = −1 and
x = 3.

x2
− x − 6
(x + 1)2
0.
Solution:
x2
− x − 6
(x + 1)2
0
(x − 3) (x + 2)
(x + 1)2 0
Hence, the critical points are x = −2, x = −1 and
x = 3. These numbers divide the real line into the
intervals (−∞, −2) , (−2, −1) , (−1, 3) and (3, ∞).

Interval x − 3 x + 2 (x + 1)2 x2
− x − 6
(x + 1)2
(−∞, −2) − − + +
(−2, −1) − + + −
(−1, 3) − + + −
(3, ∞) + + + +
Therefore, the solution set is
S = (−2, −1) ∪ (−1, 3) .

Example: Solve for x if x2
+ x 1.

+ x 1.
Solution: We first rewrite the inequality as
x2
+ x − 1 0.

+ x 1.
x2
+ x − 1 0.
Using the quadratic formula, we see that
x2
+ x − 1 = 0 ⇐⇒ x =
−1 ±
p
1 − 4(−1)
2

+ x 1.
x2
+ x − 1 0.
x2
+ x − 1 = 0 ⇐⇒ x =
−1 ±
p
1 − 4(−1)
2
⇐⇒ x =
−1 ±
√
5
2

+ x 1.
x2
+ x − 1 0.
x2
+ x − 1 = 0 ⇐⇒ x =
−1 ±
p
1 − 4(−1)
2
⇐⇒ x =
−1 ±
√
5
2
Hence, our critical points are x = −1+
√
5
2 and
x = −1−
√
5
2 .

These numbers divide the real line into the intervals

−∞, −1−
√
5
2

,

−1−
√
5
2 , −1+
√
5
2

and

−1+
√
5
2 , ∞

.


−∞, −1−
√
5
2

,

−1−
√
5
2 , −1+
√
5
2

and

−1+
√
5
2 , ∞

.
Interval x + 1+
√
5
2 x + 1−
√
5
2 x2
+ x − 1

−∞, −1−
√
5
2

− − +

−1−
√
5
2 , −1+
√
5
2

− + −

−1+
√
5
2 , ∞

+ + +


−∞, −1−
√
5
2

,

−1−
√
5
2 , −1+
√
5
2

and

−1+
√
5
2 , ∞

.
Interval x + 1+
√
5
2 x + 1−
√
5
2 x2
+ x − 1

−∞, −1−
√
5
2

− − +

−1−
√
5
2 , −1+
√
5
2

− + −

−1+
√
5
2 , ∞

+ + +
Thus, the solution set is
−∞,
−1 −
√
5
2
!
∪
−1 +
√
5
2
, ∞
!

I The absolute value of a number a,
denoted by |a|, is the distance from a to
0 on the real number line.

I Distances are always positive or 0, so we
have
|a| ≥ 0 for every number a.

I Distances are always positive or 0, so we
have
|a| ≥ 0 for every number a.
I For example,
|3| = 3 | − 3| = 3 |0| = 0
|
√
2 − 1| =
√
2 − 1 |3 − π| = π − 3

In general,
|a| =

a if a ≥ 0
−a if a 0.

In general,
|a| =

a if a ≥ 0
−a if a 0.
Note: Recall that the symbol
√
means “the
positive square root of”.

In general,
|a| =

a if a ≥ 0
−a if a 0.
√
means “the
positive square root of”. Thus
√
r = s means
s2
= r and s ≥ 0.

In general,
|a| =

a if a ≥ 0
−a if a 0.
√
means “the
√
r = s means
s2
= r and s ≥ 0. Therefore, the equation
√
a2 = a is not always true. It is true only
when a ≥ 0.

In general,
|a| =

a if a ≥ 0
−a if a 0.
√
means “the
√
r = s means
s2
√
when a ≥ 0. If a 0, then −a 0, so we have
√
a2 = −a.

In general,
|a| =

a if a ≥ 0
−a if a 0.
√
means “the
√
r = s means
s2
√
when a ≥ 0. If a 0, then −a 0, so we have
√
a2 = −a. We thus have the equation
√
a2 = |a|
which is true for all values of a.

Example: Express |3x − 2| without using
the absolute-value symbol.

Solution:
|3x − 2| =

3x − 2 if 3x − 2 ≥ 0
−(3x − 2) if 3x − 2 0.

Solution:
|3x − 2| =

3x − 2 if 3x − 2 ≥ 0
−(3x − 2) if 3x − 2 0.
=





3x − 2 if x ≥
2
3
2 − 3x if x
2
3
.

The graph of the absolute value function y = |x|
looks as follows:
−3 −2 −1 1 2 3
1
2
3
y = |x|
x
y
Note that the y-values are always non-negative
(y ≥ 0).

Properties of absolute values:
Suppose a and b are any real numbers and
n is an integer. Then:
1. |ab| = |a||b|
2.
a
b
=
|a|
|b|
(b 6= 0)
3. |an
| = |a|n
.

Properties of absolute values:
Suppose a 0. Then:
4. |x| = a if and only if x = ±a
5. |x| a if and only if −a x a
6. |x| a if and only if x a or
x −a.

Example: Solve |3x + 5| = 1 .
Solution: Using Property 4, we have
3x + 5 = 1 or 3x + 5 = −1

3x + 5 = 1 or 3x + 5 = −1
Hence, 3x = −4 or 3x = −6.

3x + 5 = 1 or 3x + 5 = −1
Hence, 3x = −4 or 3x = −6. Thus, x = −4
3
or x = −2.

Example: Solve |x − 4| 1.
Solution: By Property 5, |x − 4| 1 is
equivalent to
−1 x − 4 1

equivalent to
−1 x − 4 1
Therefore, adding 4 to each side, we have
3 x 5

equivalent to
−1 x − 4 1
Therefore, adding 4 to each side, we have
3 x 5
Hence, the solution set is the open interval
(3, 5).

Example: Solve |2x + 3| ≥ 5.

Solution: By Property 4 and 6, |2x + 3| ≥ 5
is equivalent to
2x + 3 ≤ −5 or 2x + 3 ≥ 5

is equivalent to
2x + 3 ≤ −5 or 2x + 3 ≥ 5
In the first case 2x ≤ −8, which gives
x ≤ −4.

is equivalent to
2x + 3 ≤ −5 or 2x + 3 ≥ 5
x ≤ −4. In the second case 2x ≥ 2, which
gives x ≥ 1.

is equivalent to
2x + 3 ≤ −5 or 2x + 3 ≥ 5
x ≤ −4. In the second case 2x ≥ 2, which
gives x ≥ 1. Hence, the solution set is
S = {x ∈ R | x ≤ −4 or x ≥ 1}
= (−∞, −4] ∪ [1, ∞) .

The Triangle Inequality: If a and b are
any real numbers, then
|a + b| ≤ |a| + |b|.

Example: If |x − 4| 0.1 and |y − 7|
0.2, use the Triangle Inequality to estimate
|(x + y) − 11|.

|(x + y) − 11|.
Solution: We use the Triangle Inequality with
a = x − 4 and b = y − 7:

|(x + y) − 11|.
a = x − 4 and b = y − 7:
|(x + y) − 11| = |(x − 4) + (y − 7)|

|(x + y) − 11|.
a = x − 4 and b = y − 7:
|(x + y) − 11| = |(x − 4) + (y − 7)|
≤ |x − 4| + |y − 7|

|(x + y) − 11|.
a = x − 4 and b = y − 7:
|(x + y) − 11| = |(x − 4) + (y − 7)|
≤ |x − 4| + |y − 7|
0.1 + 0.2 = 0.3

|(x + y) − 11|.
a = x − 4 and b = y − 7:
|(x + y) − 11| = |(x − 4) + (y − 7)|
≤ |x − 4| + |y − 7|
0.1 + 0.2 = 0.3
Thus, |(x + y) − 11| 0.3.

Prescribed tut problems
I Appendix A:
4, 5, 10, 16, 24, 27, 37, 42, 45, 50, 51,
55, 61, 63, 68

Appendix A(1).pdf

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Appendix A(1).pdf