MATH 150: Absolute Value (One-Dimensional Distance)
Let’s use the Rule of Four (verbal, numerical, graphical and algebraic representations) to
define and understand Absolute Value (or Modulus) function. First, we will introduce
Absolute Value by way of an example.
If two people are 18 and 25 years old, what is the difference in their ages? 7 years
There could be two answers: 25 − 18 = 7 while 18 − 25 = −7 , since “difference” means the
answer when you subtract. So, “difference” could be a positive, negative or zero result.
In the original question, we probably really wanted to ask, “what is the distance between
the ages”?, because distance is a positive or zero (but NOT negative) result.
It is sometimes practical to have a mathematical term and notation to produce a
nonnegative result from subtraction whenever we wish to determine the distance
between two values on a numberline. This one-dimensional distance is called the absolute
value or modulus function.
The absolute value notation 25 − 18 = 7 = 7 − 0 = 7 says “The absolute value of 25 minus
18 equals the absolute value of 7, which is equivalent to the absolute of 7 minus 0, which
equals 7”. This means that the 1D distance between 25 and 18 is equal to the 1D distance
between 7 and 0 and that distance is 7.
Similarly, the notation 18 − 25 = −7 = 0 − 7 = 7 says “The absolute value of 18 minus 25
equals the absolute value of negative 7, which is equivalent to the absolute value of 0
minus 7, which equals 7”. This means that the 1D distance between 18 and 25 is equal to
the 1D distance between 0 and 7, and that distance is 7.
Any time we refer to a quantity that refers to a one-dimensional distance or a nonnegative (positive or zero) distance, we will need to use the absolute value (or modulus)
function.
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MATH 150: Absolute Value (One-Dimensional Distance)
Absolute Value by the Rule of Four
Verbal: The “absolute value of x” or x measures the non-negative (positive or zero) onedimensional distance from x to zero (0) on the x-axis or numberline. While the value of x
can be either positive, negative or zero, the absolute value of x can ONLY BE POSITIVE
OR ZERO. This is because absolute value is a measurement of distance, which is a
nonnegative measurement (a scalar, not a vector!).
Numerical: Consider −7 .
•
−7 = 0 − 7 = 7 because the distance between 0 and 7 is 7.
•
Equivalently, −7 = −7 − 0 = 7 because the distance between –7 and 0 is 7.
•
Note that −x ≠ x . The reason for this is that x is a variable, whereas -7 is a
constant. This will be clarified graphically.
•
However, it is true that −x = x . We will understand why in Chapter 1.
⎧− x , x < 0
⎪
Algebraic: x = ⎨ x , x = 0 which is a compact symbolic way to say:
⎪ x, x > 0
⎩
•
If x is a negative real number, then x can be substituted with −x .
•
If x is zero, then x can be substituted with x, which is simply 0.
•
If x is a positive real number, then x can be substituted with x.
Graphical: y = x is comprised of 2 linear functions, each defined on half of the x – axis.
The graph of y = x can be made by graphing y = −x when x < 0 and y = x when x ≥ 0 .
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MATH 150: Absolute Value (One-Dimensional Distance)
Properties of Absolute Value: Let x and c be any real number (i.e., on a one-dimensional
numberline like the x - axis) and let d be any non-negative (positive or zero) value.
1. Absolute Value means distance in one dimension (regardless of the direction traveled).
Therefore, the statement x − c = d means that
•
•
•
the distance between x and c is exactly d spaces.
In other words, the number x is a distance of d spaces from c on either side of c.
Therefore, x is either d spaces to the left of c ( x = c − d ) or x is d spaces to the
right of x ( x = c + d ).
Imagine c as the centre of a circle in two dimensions and the value d as the radius
(two dimensional distance between two points) of the circle. The equation
x − c = d refers to all x – coordinates ON the circle on the x - axis.
Absolute Value as one-dimensional distance
d
Radius of a circle as two dimensional distance
d
d
x
c
x
c
2. The statement x − c > d means that
•
•
the distance between x and c is more than d spaces.
In other words, the number x is any number that is more than d spaces to the
right or left of c. Therefore x is either any number to the left of c − d or to the
right of c + d : x < c − d or x > c + d .
•
In the 2D circle representation, the inequality x − c > d
refers to all x –
coordinates OUTSIDE the circle on the x - axis.
Absolute Value as 1D distance
d
Radius of circle as 2D distance
d
c
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d
x
c
x
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MATH 150: Absolute Value (One-Dimensional Distance)
3. The statement x − c < d means that
•
•
the distance between x and c is less than d spaces.
In other words, x is any number that is less than d spaces from the centre c,
therefore x is any number both to the right of c − d and to the left of c + d . This
is equivalent to saying that x is BETWEEN c − d and c + d : c − d < x < c + d .
•
In the 2D circle representation, the inequality x − c < d
refers to all x -
coordinates INSIDE the circle on the x - axis.
Absolute Value as 1D distance
d
Radius of circle as 2D distance
d
c
d
x
c
x
4. Piecewise definition of absolute value of stuff:
a) If “stuff” is negative, then absolute value (or modulus) changes the sign of stuff.
b) If “stuff” is zero, then absolute value (or modulus) does nothing to stuff.
c) If “stuff” is positive, then absolute value (or modulus) does nothing to stuff.
d) The definitions in (b) and (c) can be combined to say: if “stuff” is nonnegative, then
the absolute value (or modulus) does nothing to stuff.
e) This is written in piecewise function notation as:
⎧−stuff, stuff < 0
⎪
⎪⎧−stuff, stuff < 0
stuff = ⎨stuff, stuff = 0 = ⎨
⎪⎩stuff, stuff ≥ 0
⎪stuff, stuff > 0
⎩
(You can read the comma as “if” or “when”.)
5. The following arithmetic rules (properties/axioms) hold for absolute value, if a , b ∈ :
a) a ≥ 0
d) −a = a
b) a ⋅ b = a ⋅ b
e) a + b ≤ a + b (Cauchy-Schwartz Inequality)
c)
a
b
=
a
b
, for b ≠ 0
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MATH 150: Absolute Value (One-Dimensional Distance)
Exercises:
1. Evaluate (express/simplify without the absolute value function).
a) 3 − 15 = −12 = 12
b)
3 − 2 = 3 − 2 Q 3 > 2 , so
(
3 − 2 is positive and the abs value does nothing.
)
c) e − π = − e − π = π − e Q e < π , so e − π is negative and the abs value changes sign
2. Express the following (red) solutions for x as an absolute value equation or inequality.
a)
-5
2
x
9
x
b)
-14
-4
6
−14 ≤ x ≤ 6 ⇒ less than or equal to
x = −5 or x = 9 ⇒ equality
c=
−14 + 6
= −4 and d =
2
∴ x + 4 ≤ 10
c = 2 and d = 7
∴ x −2 = 7
1
2
(6 − (−14)) = 10
Note that the 2 and -4 did not need to be labeled in each numberline; why not?
3. Fill in the blanks and then use a one-dimensional sketch (if possible) to solve the
following equations and inequalities: x + 2 = 5 means that the one - dimensional
distance between x and -2 is 5 spaces.
a) x + 2 = 5
b) x + 2 < 5
c) x + 2 ≥ 5
x = −7 or 3
−7 < x < 3
x ≤ −7 or 3 ≤ x
Use this sketch for parts (a), (b) and (c).
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No solution: x ∈∅ b/c
distance can’t be < 0 .
5
5
-7
d) x + 2 = −5
-2
3
x
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MATH 150: Absolute Value (One-Dimensional Distance)
4. Consider the equation 6 − 3x = 12 .
a) Let h = 3x . Fill in the blanks: 6 − h = 12 means that the distance between 6 and h
is 12 OR that the distance between h and 6 is 12, therefore the absolute value
equation can be written equivalently (by symmetry) as h − 6 = 12 .
b) Solve for h graphically in one-dimension (on the h – axis): h = −6 or 18
c) Use your solutions for h to continue solving for x, and express your solutions for x
in a 1D sketch (on the x – axis).
12
12
-6
h = 3x
18
6
h = −6 or 18 ∴ 3x = −6 or 18 ⇒ x = −2 or 6
(Note: the ⇒ symbol means “leads to” or “implies” or “equivalently produces”. It
can be used like an equal sign for equations or inequalities.)
d) Use your solutions for x to write an equation of the form x − c = d .
x − 2 = 4 … this means that the distance between x and 2 is 4
x = −2 or 6 ⇒
spaces, or equivalently that x is a distance of 4 spaces away from 2.
e) Use the algebraic properties of absolute value to rewrite 6 − 3x = 12 equivalently
in your new form in (d).
−3 ⋅ x − 2 = 12
6 − 3x = 12
3 ⋅ x − 2 = 12
−3x + 6 = 12
(
)
−3 x − 2 = 12
x −2 = 4
f) Solve 6 − 3x = 12 for x directly and graphically in 1 D (on the 3x – axis).
12
12
-6
6
18
3x = −6 or 18
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4
4
3x
-2
2
6
x
x = −2 or 6
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MATH 150: Absolute Value (One-Dimensional Distance)
5. Sketch each function on its grid below:
y = x + 5 for x ≥ −5
(
)
y = − x + 5 for x < −5
y = x +5
6. Write the absolute value expression x + 5 as a piecewise function.
(
(
(
)
)
)
⎧− x + 5 , x + 5 < 0
⎪
⎪
x + 5 = ⎨+ x + 5 , x + 5 = 0
⎪
⎪⎩+ x + 5 , x + 5 > 0
⎧⎪−x − 5, x < −5
which is the graph drawn in #5.
=⎨
⎩⎪x + 5, x ≥ −5
7. Prove the following absolute value properties/identities using a proper proof format.
a) −a = a
b) a − b = b − a
LS = −a
LS = a − b
( )
Alternate proof for (b):
( )(
= −1 ⋅ a
= −1 ⋅ b − a
= −1 ⋅ a
= −1 ⋅ b − a
= 1⋅ a
= 1⋅ b −a
=a
= b −a
= RS
QED
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= RS
QED
)
a − b means the distance
between a and b which is
equivalent to the distance
between b and a which is
expressed as b − a .
∴ a −b = b −a
QED
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MATH 150: Absolute Value (One-Dimensional Distance)
8. Does − x = x ∀x ∈  ? Justify your response using at least two methods from the
Rule of Four (algebraically, graphically, numerically, verbally).
Verbally
Numerically
Graphically
Try a 3 sample
−x = 0 − x = x − 0
Graph of y = −x
values of x.
means the distance
between x and zero
x = 0 : −x = x
which is nonnegative;
meanwhile x is a
−0 = 0 ?
directed
(positive,
0 = 0 true
negative or zero)
position (coordinate)
on a numberline (any x = 4 : −x = x
real number) which
Graph of y = x
−4 = 4 ?
is
not
always
4 = 4 true
nonnegative. The two
values
may
be
sometimes but not x = −5 : −x = x
always equivalent.
− −5 = −5 ?
( )
The two expressions
mean
different
things and those
things can take on
different
values
(distance can’t be
negative), therefore
the answer is NO.
5 = −5 false
We have found (at
least)
ONE
contradictory value
of x, therefore the
answer is NO.
© Raelene Dufresne 2014
Algebraically
( )
−x = −1 ⋅ x
= −1 ⋅ x
= 1⋅ x
= x
≠x
Therefore the
answer is NO.
Which approach
makes the most
sense to you?
The two graphs are
not the same for all
values
of
x,
therefore
the
answer is NO.
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