Appendices A APPENDIX Algebra A.1 INTRODUCTION TO ALGEBRA: EXPRESSIONS AND EQUATIONS Algebra is rooted in arithmetic. If you are confused about algebra your knowledge of arithmetic can provide guidance. Algebra is more abstract than arithmetic, but a constructive approach to understanding an abstract idea is to work through a concrete example. If you feel confused or amnesic try a very concrete example to get yourself grounded. This is a fundamental problem solving strategy; you will find ample opportunity to apply it—in mathematics as well as in other disciplines. For example, suppose you’re debating whether or not x 2 + y 2 is equal to x + y . Instead of trying to “remember” orsimply guessing, check it out with numbers. For example, if x = 2 and y =3, x 2 + y 2 = √ 2 2 + 3 2 = √ 4 + 9 = √ 13, while x + y = 2 + 3 = 5. Therefore, x 2 + y 2 = x + y. CAUTION From one example that does not “work” we can conclude that the statement does not hold in general. On the other hand, one example that does “work” is inconclusive. For instance, suppose the question of the hour is whether or not x 2 is equal to 2x. You decide to check it out letting x = 2: 2 2 = 4 = 2 · 2. So far so good. You then double-check letting 1051 1052 APPENDIX A Algebra x = 0. Once again all goes smoothly: 0 2 = 0 = 2 · 0. You cannot conclude that squaring a number is the same as doubling it. One counterexample (try any number apart from 0 and 2) puts that idea to rest. In order to keep your bearings algebraically, you must distinguish clearly between expressions and equations. Expressions x 3 −x 2x 2 +2x is an expression, not an equation, just as “an automobile” is an expression, not a sentence. You cannot change the value of an expression without changing its meaning and obtaining a different (nonequivalent) expression. Replacing the expression “a car” by “an automobile” gives an equivalent expression, while replacing it by “a cat” does not. Mathematically, the only thing you can do with an expression is write the same thing in a different way. You can change its form (a stylistic change), but not its content. You can do anything that doesn’t change its value. You can multiply by 1; you can add 0. Generally, we try to simplify expressions, making them look less complicated and easier to manipulate. The Fundamental Principle for working with expressions is preserve the value of the expression. An expression can only be transformed into an equivalent expression. ◆ EXAMPLE A.1 Simplify the following. (a) 2x x (b) x 3 −x 2x 2 +x SOLUTION (a) 2x x = 2 for x = 0 and is undefined for x = 0. 2x x = 2ifx=0; undefined if x = 0. (b) x 3 − x 2x 2 + 2x = x(x 2 − 1) 2x(x + 1) (Factoring changes form, not value.) = x(x − 1)(x + 1) 2x(x + 1) (Factor further.) = x(x + 1) x(x + 1) · x − 1 2 (We know A A = 1 for A = 0.) = x−1 2 for x = 0, x = 1; undefined for x = 0 and x =−1. In other words, x 3 −x 2x 2 +2x is equivalent to the expression x−1 2 except that the former is undefined at x = 0 and x =−1, a fraction being undefined when the denominator is zero. ◆ A.1 Introduction to Algebra: Expressions and Equations 1053 Division by zero is undefined. Why is division by zero undefined? Let’s begin by establishing that a b = c only if a = bc. Suppose we attempt to define a 0 to be some unique number c; then a = 0 · c = 0. But then a must be zero; we cannot define a 0 for a = 0. If a = 0, then the equation 0 = 0 · c holds for any c, not for some unique c,so 0 0 is also undefinable. Equations An equation sets two expressions equal to one another. x 3 − x 2x 2 + 2x = x is an equation. We try to solve equations; in other words, we try to find out what values of x make the equation hold true. The Fundamental Principle for working with equations is to preserve the balance. Therefore you can add or subtract anything from both sides of the equation, you can multiply or divide both sides of the equation by any nonzero quantity. To solve for x means to get x =(an expression with no x’s). Our goal, therefore, is to isolate x. How we proceed depends on what type of equation we have. If x = a is a solution to an equation, then when both sides of the equation are evaluated at x = a, they are equal. Meaning What You Say and Saying What You Mean Math is a language, and a very precise one at that. When we write an expression it has a specified meaning. If you are opening up a new area of inquiry you can “invent” new words, but it is crucial to know the common conventions because that is what allows us to communicate. Sometimes a student will say “when I write ‘3x 2 ’ what I mean is to square 3x.” This is as zany as writing an essay and saying “when I write ‘yellow’ what I mean is ‘magenta.’” Below we review a few conventions worth knowing. Exponents 3x 2 means 3 · x 2 = 3 · x · x. If you mean (3x) · (3x), then you must write (3x) 2 ,or9x 2 .In the expression 3x 2 , 3 is called the coefficient of x 2 . We say that 2 is an exponent;itisthe power to which x is raised. The base is x. An exponent (power) refers to whatever is directly below it. 1054 APPENDIX A Algebra As you see, we can use parentheses to expand the sphere of influence of the exponent. For instance, −5 2 =−5·5=−25, whereas (−5) 2 = (−5) · (−5) =25. This means that if f(x)=3x 2 ,then f(−5)=+75 = f(5);ifg(x) =−3x 2 ,then g(5) =g(−5) =−75. Order of Operations In the absence of parentheses, the convention about order of operations is: First deal with exponents, then multiplication and division, and lastly addition and subtraction. Parentheses can be used to override this order. In the case of nested parentheses, do what is in the innermost set first. Use parentheses whenever they make the meaning more obvious. Note: Multiplication and division are given equal status because they are really the same operation in two different guises; A/B =A · 1 B . Similarly, addition and subtraction are given equal status because A −B = A +(−B). You need to understand how your calculator deals with orders of operation, or make copious use of parentheses to make your intended meaning perfectly clear. For example − 3 ˆ 2 + 2 ∗ 9 − 6 / 3 + 1 must, according to the conventions of order of operations, be interpreted as −(3 2 ) +2 ·9 − 6 3 + 1 =−9+18 − 2 + 1 = 8. Try this on your calculator. (Note: Many calculators have two separate keys, one for subtraction, as in 6 − 3, and another for negative numbers, as in −3. The calculator will often give you an error if you mix up these two keys.) EXERCISE A.1 Evaluate the following functions at the values indicated. (a) f(x)=x−2x 2 (−x +4)−6; find f(−5). (b) g(x) =x −2x 2 −x +4(−6); find g(−2). (c) h(x, y) =xy[xy 3 −y 3 x 2 (x + y)]; find h(2, −1). ANSWERS (a) f(−5)=(−5)−2(25)(9) − 6 =−5−450 − 6 =−461 (b) g(−2) =−2−(2)(4) + 2 − 24 =−2−8+2−24 =−10 − 22 =−32 (c) h(2, −1) = (2)(−1)[(2)(−1) − (−1)(4)(1)] =−2[−2 − (−4)] =−2[−2 + 4] = −2[2] =−4 Square Roots The square root of a nonnegative number A is defined to be the nonnegative number whose square is A. √ 9 = 3 because 3 2 = 9. The solutions to the equation x 2 = 9 are x = 3 and x =−3. They can be expressed as √ 9 and − √ 9. When we write √ 9 we mean only positive 3. Note: − √ 9 = √ −9, the latter A.1 Introduction to Algebra: Expressions and Equations 1055 being undefined in the real number system because there is no real number whose square is −9. Try to take the square root of −9 on your calculator and see what you get. 1 PROBLEMS FOR SECTION A.1 1. For what values of x is each expression undefined? (a) 3 x 2 (b) √ x − 1 (c) 5 x+3 (d) x−2 2x−2 2. Are the expressions below equivalent? If the answer is no, show this with a concrete example or by simplifying the two expressions. Note: Some expressions are not equivalent, but are equal provided that a certain condition holds. For example, the expressions in part (a) are equal provided x = 0. (a) x 3x and 1 3 (b) x+1 x(x+1) and 1 x (c) − A B and −A B and A −B (d) −3 2 and (−3) 2 (e) A B and A 2 B 2 (f) √ x 2 and x (g) xy 2 and y 2 x (h) xy 2 and (xy) 2 3. Answer the questions below. If the answer is no, illustrate this with a concrete example. (a) If we square the numerator (top) and denominator (bottom) of an expression, will we obtain an equivalent expression? Is A B equivalent to A 2 B 2 ? (b) If we square both sides of an equation, will the balance be preserved? If A = B will A 2 = B 2 ?IfA 2 =B 2 does it follow that A =B? (c) If A = B + C, will A 2 = B 2 + C 2 ? (d) Suppose that A = B and A, B = 0. Is 1 A = 1 B ? (e) Suppose that A = B + C and A, B, C = 0. Is 1 A = 1 B + 1 C ? 4. Write each of the following as an algebraic expression. (a) Square x and then multiply by 3. (b) Multiply x by 3 and square the result. (c) Subtract twice x from y and then take twice the square root of the result. (d) The cubed root of 3 times the reciprocal of x. 5. Simplify the following expressions. 1 Some calculators will give an error message. Others give an ordered pair of numbers, indicating a complex number. (0, 3) is shorthand for 0 +3i, where i stands for the complex number √ −1. 1056 APPENDIX A Algebra (a) −2 2 − 3 2 −8 ·4 1/2 + (−3) 2 (b) −x 2 − (−x 2 ) + (−x 2 ) + 3x 2 (c) −x y − x y + 2x −y (d) 3 √ y − 2 3 √ y − √ y 6. (a) If we double an expression, will we have an equivalent expression? (b) If we double the numerator and denominator of an expression, will we have an equivalent expression? (c) If we double both sides of an equation, will the balance be preserved? 7. Evaluate the functions at the values indicated. (a) f(x)=−x 3 −2x 2 +(−x) 2 +x; find f(1)and f(−1). (b) f(x)=− 1 x − 2 x 2 + −3 −x 3 ; find f(2)and f(−2). A.2 WORKING WITH EXPRESSIONS In this section we will work with expressions. Keep in mind that the Fundamental Principle for working with expressions is to preserve the value of the expression. An expression can be transformed only into an entirely equivalent expression. We can change only the form of an expression. Multiplying Out and Factoring: A Brief Introduction One way of changing the form of an expression is to multiply out or factor. The expression x(x +1) is factored; x 2 + x is the equivalent expression multiplied out. The Distributive Law. Multiplication distributes over addition: (a + b) · d =a ·d +b · d. Similarly, (a + b + c) ·d = a · d +b ·d + c ·d. REMARK Multiplication does not distribute over itself. We know that 2 ·3 ·4 =(2 · 3) · 4 = 2 · (3 · 4) = 24 2 ·(3 ·4) =2 · 3 · 2 · 4 = 48. ◆ EXAMPLE A.2 Multiply out the following expression, (2x − 1)(x + 1) SOLUTION (2x − 1) · (x +1) plays the role of d = 2x(x +1) −1(x + 1) = 2x 2 + 2x − x like terms −1 = 2x 2 + x − 1. A.2 Working with Expressions 1057 Generally the first step is not written. We think of multiplying 2x by (x + 1), and simply write out the result, and add to it the product of −1 and (x +1). We write (2x − 1)(x + 1) = 2x 2 + 2x − x − 1. ◆ To factor an expression is to write it as a product. It is the reverse of multiplying out. (You can always check your factoring by multiplying out.) To factor out a common factor is to use the distributive law in reverse. ◆ EXAMPLE A.3 Factor x 3 − x 2 + 4x 7 . SOLUTION x 3 − x 2 + 4x 7 = x 2 (x − 1 + 4x 5 ) ◆ CAUTION Although x 3 − x 2 + 4x 7 + 3 = x 2 (x − 1 + 4x 5 ) − 3, this expression is not fac- tored. Similarly, x 2 − x − 2 = x(x − 1) − 2, but this expression is not factored. x 2 − x − 2 = (x + 1)(x − 2) and this is factored. (See below.) ◆ EXAMPLE A.4 Factor x 3 − x 2 . SOLUTION x 3 −x 2 =x 2 (x − 1) or x 2 (−1)(−x + 1) =−x 2 (1−x),whichever suits us. All are factored forms. ◆ We will take up the topic of factoring more systematically later in Section A.2. Multiplication and Working with Exponents ExponentLaws (1)c m c n =c m+n (2) c m c n = c m−n (3)(c m ) n =c mn From these fundamental laws for working with exponents, it follows that c 0 = 1, because we know that 1 = c n c n = c n−n = c 0 ; c −n = 1 c n , because 1 c n = c 0 c n = c 0−n = c −n ; c 1/p = the pth root of c, because we know that (c 1/p ) p = c 1 = c. Make sure that you can use your calculator to compute quantities along the lines of, say, the fifth root of 32. On many calculators this would be along the lines of 3 2 x y ( 1 / 5 ) = or 3 2 y √ x 5 =. (You ought to get an answer of 2.) 1058 APPENDIX A Algebra Arithmetic Underpinnings 2 ◆ EXAMPLE A.5 Simplify the expressions below. (a) 3 · 1 4 · 1 6 12 3 · 1 4 · 1 6 12 = 3 1 · 1 4 · 3 2 = 3 · 3 4 · 2 = 9 8 (b) 2 3 ·2 4 2 6 2 3 · 2 4 2 6 = 2 7 2 6 = 2 ◆ Extracting the Fundamentals Multiplication of fractions: a b · c d = ac bd . Simplification of fractions: ac cd = a d . ( ac cd = ca cd = c c · a d = 1 · a d = a d ). CAUTION A+B C+B = A C ; e.g., 5 4 = 2+3 1+3 = 2. You can’t “cancel over addition.” Canceling is based on the idea that multiplying by one doesn’t change the value of the expression. You can cancel only when the numerator (top) and denominator (bottom) are each factored (expressed as a product) and have a common factor. Using the fundamentals extracted from arithmetic, we are ready to do some algebra. ◆ EXAMPLE A.6 Simplify x−5 x 2 −1 · x 2 +1 5−x . SOLUTION x − 5 x 2 − 1 · x 2 + 1 5 − x = (x − 5)(x 2 + 1) (x 2 − 1)(5 − x) = x 2 + 1 (x 2 − 1) (x − 5) (5 − x) = x 2 + 1 (x 2 − 1) (x − 5) (−1)(x − 5) = x 2 + 1 (x 2 − 1)(−1) = x 2 + 1 −x 2 + 1 ◆ 2 You might be thinking, “Why arithmetic? I know arithmetic.” We start with arithmetic to emphasize that when you know arithmetic, algebra nicely follows without much ado. A.2 Working with Expressions 1059 Note: −a b = a −b =− a b . Similarly, a−b b−a =−1, because a−b b−a = a−b (−1)(−b+a) = a−b (−1)(a−b) = 1 −1 =−1. ◆ EXAMPLE A.7 Simplify y 4 √ y ( √ xy 2 ) 3 · x 3 y −1 6 √ xy . SOLUTION y 4 √ y ( √ xy 2 ) 3 · x 3 y −1 6 √ xy = y 4+1/2−1 x 3 x 3/2 y 6 · 6x 1/2 y 1/2 = y 7/2 x 3 x 2 y 13/2 = y 7/2−13/2 x 3−2 = y −3 x, or, equivalently, x y 3 . ◆ Addition and Subtraction Arithmetic Underpinnings ◆ EXAMPLE A.8 Add and simplify. 3 · 1 4 · 18 12 + 0.5 − 5 6 = 9 8 + 1 2 − 5 6 . SOLUTION We can only add “like” things, so to add these three terms we must write them with a common denominator. We could use 8 · 2 ·6 as a common denominator but it is less cumbersome to factor the denominators completely and find the least common denominator. 9 2 · 2 · 2 + 1 2 − 5 2 · 3 . The least common denominator is 2 · 2 · 2 · 3 = 24. Finding the Least Common Denominator. Every factor appearing in a denominator must be made a factor of the LCD (least common denominator) and must be raised to the highest power to which it appears in any one denominator. Each term must now be written with the LCD without changing its value. 9 8 · 3 3 = 27 24 ; 1 2 · 12 12 = 12 24 ; 5 6 · 4 4 = 20 24 9 8 + 1 2 − 5 6 = 27 24 + 12 24 − 20 24 = 27 + 12 − 20 24 = 19 24 ◆ Extracting the Fundamentals Addition: We can only add like quantities. Therefore, we can’t add fractions until we have written them with a common denominator. We can’t change the value of a fraction in our eagerness to get a common denominator. a b + c d = ad bd + cb db = ad +cb bd The product of all denominators is always a common denominator. It is generally most efficient to find the least common denominator. Rewrite each fraction with the LCD; then add or subtract. 1060 APPENDIX A Algebra Algebra ◆ EXAMPLE A.9 Simplify x x+1 − 2 x+2 . SOLUTION The terms are already simple. The LCD is (x + 1)(x + 2). Write each fraction with the LCD and then add. x x + 1 = x x + 1 x + 2 x + 2 = (x)(x + 2) (x + 1)(x + 2) ; 2 x + 2 = 2 x + 2 · x + 1 x + 1 = (2)(x + 1) (x + 2)(x + 1) (x)(x + 2) (x + 1)(x + 2) − (2)(x + 1) (x + 2)(x + 1) = (x)(x + 2) − (2)(x + 1) (x + 2)(x + 1) = x 2 + 2x − 2x − 2 (x + 2)(x + 1) = x 2 − 2 (x + 2)(x + 1) . Here we multiplied out the terms of the numerator in hopes that the numerator would simplify (it did), and perhaps even share a common factor with the denominator (it didn’t). We will not benefit by multiplying out the denominator. This last remark deserves to be highlighted. There are very few situations in which it’s helpful to have the denominator of such an expression expanded. Generally, it’s handy to keep denominators in factored form. Then it’s simpler to find an LCD if adding, or to cancel when multiplying. ◆ ◆ EXAMPLE A.10 Simplify x 2 +4x−5 2x 2 −2x − x 4x+6 . SOLUTION See if the terms themselves can be simplified; then, with denominators written in factored form, find the LCD. Write each fraction with the LCD and then add. First term: x 2 + 4x − 5 2x 2 − 2x = (x + 5)(x − 1) 2x(x − 1) = x + 5 2x . Second term: x 4x + 6 = x 2(2x + 3) . x 2 + 4x − 5 2x 2 − 2x − x 4x + 6 = x + 5 2x − x 2(2x + 3) The LCD is 2x(2x + 3). Express each term with the LCD: x + 5 2x = x + 5 2x · 2x + 3 2x + 3 = (x + 5)(2x + 3) (2x)(2x + 3) x (2)(2x + 3) = x (2)(2x + 3) · x x = x 2 (2x)(2x + 3) . . in part (a) are equal provided x = 0. (a) x 3x and 1 3 (b) x+1 x(x+1) and 1 x (c) − A B and −A B and A −B (d) −3 2 and (−3) 2 (e) A B and A 2 B 2 (f) √ x 2 and x (g) xy 2 and y 2 x (h) xy 2 and. is to preserve the value of the expression. An expression can be transformed only into an entirely equivalent expression. We can change only the form of an expression. Multiplying Out and Factoring:. multiplying 2x by (x + 1), and simply write out the result, and add to it the product of −1 and (x +1). We write (2x − 1)(x + 1) = 2x 2 + 2x − x − 1. ◆ To factor an expression is to write it as a product.