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? _________ 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 ____ is equal to the 1D distance between 7 and ____, and that distance is ____. 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 ____ is equal to the 1D distance between ____ and ____, and that distance is ____. 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 _____________ __________ (or __________) function. © Raelene Dufresne 2014 1 of 6 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 . © Raelene Dufresne 2014 2 of 6 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 © Raelene Dufresne 2014 d x c x 3 of 6 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 © Raelene Dufresne 2014 4 of 6 MATH 150: Absolute Value (One-Dimensional Distance) Exercises: 1. Evaluate (express/simplify without the absolute value function). a) 3 − 15 = b) c) e − π = 3− 2 = 2. Express the following (red) solutions for x as an absolute value equation or inequality. a) -5 2 9 x x b) -14 -4 6 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 ___- dimensional __________ between x and ____ is ____ spaces. a) x + 2 = 5 b) x + 2 < 5 c) x + 2 ≥ 5 d) x + 2 = −5 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 ___ is 12 OR that the distance between ___ and 6 is ____. b) Solve for h graphically in one-dimension (on the h – axis). 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). d) Use your solutions for x to write an equation of the form x − c = d . e) Use the algebraic properties of absolute value to rewrite 6 − 3x = 12 equivalently in your new form in (d). f) Solve 6 − 3x = 12 for x directly and graphically in 1 D (on the 3x – axis). © Raelene Dufresne 2014 5 of 6 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. 7. Prove the following absolute value properties/identities using a proper proof format. a) −a = a b) a − b = b − a 8. Does − x = x ∀x ∈ ? Justify your response using at least two methods from the Rule of Four (algebraically, graphically, numerically, verbally). © Raelene Dufresne 2014 6 of 6
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