4 A Catalogue of Functions and Inequalities 8. SECTION E Modulus Function

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1 4 A Catalogue of Functions and Inequalities 8 SECTION E Modulus Function By the end of this section you will be able to visualize the geometric interpretation of the modulus function derive some inequalities involving the modulus function The first part of this section is straightforward. However for later parts you will need to know your work on inequalities to derive properties of the modulus function. E1 Introduction to the Modulus Function Definition (4.11). Let be a real number then the modulus of is denoted by and is defined as What does this mean? If is positive or zero then the value of 5? if 0 if 0 and if is negative then. What is By definition (4.11) we have 5 5. What is the value of 5? By definition (4.11) we have What is the value of Similarly. What is the value of 1 1. What is the value of? 3 3. What do you notice about your results? They are all positive. What is 0 equal to? 0 0 What does the graph of look like? Note that the modulus of negative values of, positive and zero we have 1 3? 1?, is positive and for. Therefore the graph of y is:

2 4 A Catalogue of Functions and Inequalities 9 Figure 4 What do you notice about the graph of y? Symmetrical about the y ais, that is. We will prove this result later in this section. Also the graph of y is above the ais and only touches it at 0. Hence the modulus of a real number is always positive or zero and we can write this as an inequality (4.1) 0 What are the values of which satisfies? By the above definition if 0 (4.11) if 0 we have if or. Therefore has the solutions or. The modulus function is also called the distance function or the absolute function. The geometrical interpretation of the modulus function is the distance on the real number line. For eample is the distance from to zero. Consider the above eample. We can visualize represented by: as the distance of from zero is and this is = = Figure 5

3 4 A Catalogue of Functions and Inequalities 10 The equation has the solution or. E Inequalities of the Modulus Function Visualizing the modulus function as the distance function is particularly useful for inequalities. What does represent? The distance of from zero is less than or equal to and can be illustrated as: What is the difference between and? Figur represents all the real numbers less than or equal to. The notation e 6 represents all the real numbers between to. This is the set of all real numbers between which is the interval,. What values of satisfy? This can be illustrated by: to and is denoted by < Figure 7 The values of which satisfy is the set which is the open interval,. This is all the real numbers between to but ecluding and. Eample 1 Determine the set of real numbers,, such that 0.5. Solution. What does 0.5 mean in terms of the distance function? Means that the distance of from zero is less than 0.5 and is illustated by: Figure The set of real numbers which satisfy 0.5 is

4 4 A Catalogue of Functions and Inequalities 11 We can solve general inequalities such as means. What does a mean? a where a 0 by algebraic The distance from zero is less than a and can be illustrated on the real line as: Figure 9 a means that lies between a to a and is denoted by a a which we can write in symbolic form as (4.13) a a a We will prove a version of this result later in this section. Eample Determine the set of real numbers,, such that 3 5. Solution. a How can we interpret 3 5 as distance on the real number line? This inequality, 3 5, means that the distance from 3 is less than 5 and we can illustrate this by: a Figure < 5 a a Hence the set of real numbers which satisfy the given inequality 3 5 is 8 which is the open interval, 8 We can also solve this algebraically as follows: By (4.13) Adding 3 Hence all the real numbers between Simplifying to 8 satisfies the given inequality Eample 3 Solve the inequality 3 5.

5 4 A Catalogue of Functions and Inequalities 1 Solution. How can we interpret 3 5 as distance on the real number line? We can rewrite 3 5 as 3 5. That is This 3 5 means that the distance from 3 illustrate this by: is less than 5 and we can Figure 10 We can show this algebraically: By (4.13) Subtracting 3 8 Simplifying Eample 4 Solve the inequality 3 5. Solution. We have By (4.13) Subtracting 3 8 Simplifying 4 1 Dividing by E3 Properties of the Modulus Function This subsection is more demanding because the proofs of the results require you to thoroughly understand the definition of the modulus function and inequalities of real numbers as discussed in section D. Proposition (4.14). For all we have (a) (b) Proof.

6 4 A Catalogue of Functions and Inequalities 13 (a). Since (b) If 0 If 0 then 0 then therefore by definition (4.11) we have our required result,. Because 0 therefore Proposition (4.15). Let a and a 0, then a a a. Comment. How do we prove this proposition? Since we have the implication going both ways,, in our proposition so we have to prove and its converse. Hence the proof below is split into two halves, first we prove a a a and then we prove a a a. Remember our work on mathematical logic from chapter 1, for a P Q we assume P and deduce Q. Similarly for Q P proof we assume Q and deduce P. Proof. ( ). Assume a. Required to prove a a. If 0 then so we have a. Why? Because by assumption we have If 0 then so we have a a a. Multiplying by 1 Remember multiplying by a negative number changes the inequality. We have a and a which means that lies between a normally written as a a. Hence we have proven. proof to a and is ( ). Now lets go the other way. Assume a a. Required to prove a. We can use proof by contradiction. Suppose a. We first consider 0 (positive or zero) and then 0 (negative). If 0 then a. Contradiction! Why?

7 4 A Catalogue of Functions and Inequalities 14 Because by assumption we have a ( is less than or equal to a ) but now a ( is greater than a ). Our supposition a a. must be false so therefore Similarly if 0 then a. Multiplying this, a, by 1 the inequality and we have Contradiction! Why? a which is equivalent to a Because by assumption we have a ( a changes is less than equal to ) and now by deduction a ( a is greater than ). Our supposition a false, therefore a. Note that this proposition earlier result: (4.13) a a a but includes equality. a a a is similar to the stated must be Proposition (4.16). For all we have. Proof. Using the above proposition (4.15) with a that is implies Proposition (4.17). For all and y we have y Proof. We consider the 4 possible cases: y (1) If both 0 and y 0 then y 0 y y y and we have () If both 0 and y 0 then y 0 and we have y y y y Remember by the definition of the modulus function if 0 and y 0 and y y respectively. (3) If 0 and y 0 then y 0 and we have then

8 4 A Catalogue of Functions and Inequalities 15 Because 0 so y y y y y y y Because y 0 so y y (4) If 0 and y 0 then y 0 and we have Because 0 so y y y y Because y 0 so y y Proposition (4.18). For all we have Proof. We use the above proposition (4.17) to prove this result: 1 1 By (4.17) y y 1 Because 1 1 Triangle Inequality (4.19). Let both and y be real numbers. Then we have the important inequality Proof. We consider y y y and prove this is less than or equal to take the square root of both sides. Hence we have y y Because y 0 y y Epanding y y y Because by (4.16) y y We have y Factorizing y y Because by (4.17) y y and then y y. Taking the square root keeps the same inequality, therefore we have our required result, y y.

9 4 A Catalogue of Functions and Inequalities 16 You are asked to provide another proof of this triangular inequality in the Eercises. Net we look at a where 0. Let a be a real number and 0. In general the inequality a means the distance from a is less than and can be illustrated on the real number line as: a a a Figure 11 Like the last section the Greek letter is used to represent a small positive real number. Do not let this Greek letter put you off. The symbol is generally used in mathematics to represent small positive real numbers. One of the greatest mathematician of 0 th Century Paul Erdos used to call young children epsilons. Paul Erdos was born to a Hungarian Jewish family in Budapest in 1913 and died in Warsaw at a conference in Figure 1 Figure 1 We can solve the given inequality, a, by algebraic means: a a By (4.13) a a Adding a The set of real numbers,, which lie between a to a is called the - neighbourhood of a. Proposition (4.0). Let a be a real number and 0. If for every 0 we have then a. a Proof. By (4.1) we know the modulus function is greater than or equal to zero, a 0. Because for every 0 we have 0 a

10 4 A Catalogue of Functions and Inequalities 17 therefore by proposition (4.9) we have 0 a implies a 0 a 0 implies a 0. Hence a which is our result. What does proposition (4.0) mean? Proposition (4.0) means that if can be made as close as we please to a then a.