1. The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross By Herbert I. Gross & Richard A. Medeiros next Lesson 13

2. Review of Linearity Review of Linearity © 2007 Herbert I. Gross next

3. Because linearity plays such an important role in the study of mathematics, we want to reinforce what we discussed in Lesson 12 in order to help ensure that you have internalized the full meaning of linear. © 2007 Herbert I. Gross In Lesson 12, we defined a linear relationship to be a relationship between two variables in which the rate of change of one variable with respect to the other variable is constant. next

4. To put this into more tangible terms, suppose we are buying boxes of pens at $4 per box. The relationship between the cost of the pens and the number of boxes we buy is a linear relationship, because the rate of change of the cost with respect to the number of boxes we buy is constant; namely $4 per box. Stated more algebraically, the cost y, in dollars, of x boxes of pens is given by the formula.. © 2007 Herbert I. Gross y = 4x next

5. © 2007 Herbert I. Gross Number of boxes (x) 1$4 2$8$4 10$40$4 100$400$4 Total Cost (4x) Cost/Box (4x ÷ x) next In terms of a table…

6. next When we say that the rate of change of one variable with respect to the other is constant, it doesn't matter which variable is the first one and which is “the other”. © 2007 Herbert I. Gross Rate of Change In terms of the previous table, notice that in the formula y = 4x, we chose to express y in terms of x; in this form it was easy to see that the rate of change of y with respect to x is 4. next In other words… If y is linear in x, then it is also true that x is linear in y.

7. next However, by dividing both sides of y = 4x by 4, we obtain the equivalent formula… © 2007 Herbert I. Gross y = 4x y = x 4 next From the formula above, it is easy to see that x is linear in y and that the rate of change of x with respect to y is 1 / 4. 1 4 4

8. More generally, if the rate of change of y with respect to x is constant, then the rate of change of x with respect to y is also a constant and, in fact, the two constants will be reciprocals of one another. © 2007 Herbert I. Gross next For example, notice that the constants in y = 4x and x = 1 / 4 y, namely 4 and 1 / 4, are reciprocals (the reciprocal of a number is the same thing as the multiplicative inverse of a number). Note

9. next While 4 / 1 is not the same as 1 / 4, it is true that… © 2007 Herbert I. Gross next 4 dollars 1 box Another Point of View 1 box 4 dollars = next

10. © 2007 Herbert I. Gross next In other words… 4 dollars 1 box 4 1 = dollars box () ( ) = 4 dollars per box and… 1 box 4 dollars 1 4 = box dollars () ( ) = 1 / 4 box per dollar are two different ways to say the same thing.

11. If 4/1 and 1/4 were modifying the same noun (unit), then 4/1 ≠ 1/4. © 2007 Herbert I. Gross Adjectives-Nouns (Units) But in our example… 4/1 refers to the noun phrase “dollars per box”, while 1/4 refers to the noun phrase “boxes per dollar”. Clearly, boxes per dollar is not equivalent to dollars per box. But we can use either to figure the cost, however many boxes we buy. next

12. We can use either dollars/box or boxes/dollar to figure out the cost of the pens we bought. So while our formulas y = 4x and x = 1 / 4 y are equivalent, we generally use the formula… y = 4x when we want to talk about “dollars per box”, © 2007 Herbert I. Gross Comment and we use the formula x = 1/4y when we want to talk about “boxes per dollar”. next

13. The formula y = 4x is a special case of the formula… © 2007 Herbert I. Gross y = mx + b where m and b are constants. We will often refer to the formula y = mx + b as the standard form of a linear relationship. The reason for preferring the standard form is that it tells us at once what the rate of change is, and what amount we started with. next

14. Namely, in the standard form, y = mx + b, m is the rate of change of y with respect to x. © 2007 Herbert I. Gross For example, if x denotes the number of boxes of pens, and y denotes the cost of the pens, then m represents the cost of the pens per box, while b represents the initial value of y (that is, the number of pens we started with). Notice that we find the value of b by replacing x by 0 in formula y = mx + b. We then see that y = m(0) + b = 0 + b = b. next

15. In the special case for which b = 0 in the formula y = mx + b, we say that y is directly proportional to x (and equivalently, x is directly proportional to y). © 2007 Herbert I. Gross next If y = mx, the relationship may be rewritten in the form y / x = m. In this form it is easy to see that the ratio of y to x is constant. It is an even stronger statement than saying that the rate of change of y with respect to x is constant. Notes on the Role of b in y = mx + b

16. next Even when b ≠ 0 the relationship y = mx + b is still linear. However, in this case it is no longer true that the ratio of y to x is constant; but it remains true that the rate of change of y with respect to x is constant. © 2007 Herbert I. Gross In the linear relationship y = mx + b the rate of change of y with respect to x depends only on m and is independent of the value of b. This might be easier to see in terms of the following illustration… next

17. You pay $25 to join a swim club. After that you pay $4 each time you go swimming. You can figure how much it costs you altogether (y) to go swimming x times with the formula… © 2007 Herbert I. Gross y = 4x + 25 This has the standard form y = mx + b, where m = 4 and b = 25. Example next

18. So if you only go swimming twice, the formula y = 4x + 25 tells us the cost is $33 (that is, when x = 2, 4x + 25 is 4(2) + 25 = 33). Hence your average cost per swim would be $33 ÷ 2 or $16.50 per swim. However, if you go swimming 4 times (in which case, x = 4 and 4x + 25 = 41), you have to pay a total of $41. Hence, the average cost per swim is $41 ÷ 4 or $10.25 per swim. © 2007 Herbert I. Gross y = mx + b

19. next As the number of times you go swimming increases, the average cost per swim gets closer and closer in value to $4. For example, if you swim 100 times, the total cost is $425, and the average cost is $425 ÷ 100 or $4.25 per swim. And if you went swimming 1,000 times, the average cost per swim would be $4,025 ÷ 1,000 or $4.025 per swim. © 2007 Herbert I. Gross Example

20. next Now suppose that to attract more members, the swim club decides to lower its fee for joining from $25 to $10. The formula for figuring how much it costs you altogether (y) to go swimming x times would now be … © 2007 Herbert I. Gross y = 4x + 10 instead of y = 4x + 25 …but in either case you pay the same $4 charge for each time you swim at the club. Key Point next

21. In the linear relationship y = mx + b, m and/or b can be negative. For example… © 2007 Herbert I. Gross Notes Suppose you are buying pens at a cost of $1.39 per pen, but you have a credit balance of $2.45 at the store. So every pen you buy costs you $1.39, but the $2.45 credit balance is subtracted from the total cost of your purchase. In that case if you buy x pens, the cost in dollars (y) is given by the formula… next y = 1.39x – 2.45

22. The formula y = 1.39x – 2.45 can be rewritten in the standard form… © 2007 Herbert I. Gross y = 1.39x + - 2.45 In this case, m = 1.39 and b = - 2.45. next Notes

23. next In addition to showing us that b can be negative, the formula y = 1.39x + - 2.45 also shows us that m doesn't have to be a whole number. Moreover, m doesn't even have to be positive. © 2007 Herbert I. Gross y = mx + b For example, suppose we have $50 to spend and that we are buying boxes of cookies at $7 per box. In this case, every time we buy 1 more box of cookies, the amount of money we have left decreases by $7. next

24. So if we buy x boxes we spend $7x and since we started with $50, the number of dollars that we have left (y) is given by the formula… © 2007 Herbert I. Gross Note next y = 50 – 7x Notice that if we want to avoid “ deficit spending”, the number of boxes we buy (x) must be less than 8.

25. © 2007 Herbert I. Gross The formula y = 50 – 7x may be rewritten in the form 50 + - 7x, which in standard form becomes… y = - 7x + 50 in which case m = - 7 and b = 50.

26. next And in this form, we see at once that the initial amount is 50 (that is, when x = 0, y = 50, which checks with the fact that we started with $50), and that the constant rate of change is - 7 (which means that every time we buy 1 more box, the amount of money we have left decreases by the constant rate of $7 per box of cookies). © 2007 Herbert I. Gross Note

27. next Even though every linear relationship can be written in the standard form y = mx + b; whether or not a relationship is linear doesn't depend on the form in which it is written. Rather it is usually just a matter of convenience to rewrite any linear relationship in the standard form. In fact, the method we used in paraphrasing the formula y = 50 – 7x into the formula y = - 7x + 50 is a special case of how we use the “rules of the game” to transform linear relationships into the standard form. © 2007 Herbert I. Gross

28. next In particular, by using the properties of arithmetic, any program of any length that consists solely of steps of the form… © 2007 Herbert I. Gross -- “Add a constant", -- “Subtract a constant", -- “Subtract from a constant” -- “Multiply by a constant", -- “Divide by a constant", or -- “ Add or subtract a multiple of the input" next

29. can be paraphrased into the “standard form" program, namely… © 2007 Herbert I. Gross (1) Start with the input.......................... (x) (2) Multiply by a constant.................... (mx) (3) Add a constant........................... (mx + b) (4) The answer is y...................... (y = mx + b) next

30. Let's illustrate our above remark by looking at Program #1 below; a rather complicated program that uses all of the previously mentioned commands. © 2007 Herbert I. Gross Program #1 Step 1Start with the input (x) Step 2Subtract it from 7 Step 3Multiply by 4 Step 4Subtract 6 Step 5Divide by 2 Step 6Add 5 times the input Step 7Subtract 2 Step 8Divide by 3 Step 9 Subtract the result in Step 8 from 11 times the input Step 10Add 5 Step 11Write the output (y) next

31. To transform Program #1 into the standard form, we may begin by translating each step of Program #1 into the language of algebra, and then use the properties of arithmetic to simplify each step before we proceed to the next step… © 2007 Herbert I. Gross next Program #1 Step 1Start with the input (x) Algebraic Format x Step 2Subtract it from 7 7 – x = 7 + - x = - x + 7 = - 1x + 7

32. next © 2007 Herbert I. Gross next continuing Program #1 Step 3Multiply by 4 Algebraic Format 4( - 1x +7) = 4( - 1x) + 4(7) = - 4x + 28 Step 4Subtract 6 = ( - 4x + 28) – 6 = ( - 4x + 28) + - 6 = - 4x + (28 + - 6) = - 4x + 22 Step 5Divide by 2 ( - 4x + 22) ÷ 2 = 1 / 2 ( - 4x + 22) = - 2x + 11 Step 6 Add 5 times the input = ( - 2x + 11) + 5x = ( - 2x + 5x) + 11 = 3x + 11 Step 7 Subtract 2 = (3x + 11) – 2 = 3x + (11 + - 2) = 3x + 9

33. next © 2007 Herbert I. Gross next continuing Program #1 Step 8Divide by 3 Algebraic Format = (3x + 9) ÷3 = 1 / 3 (3x + 9) = x + 3 Step 9 Subtract the result in Step 8 from 11 times the input 11x – (x + 3) = (11x – x) – 3 = 10x + - 3 Step 10Add 5 (10x + - 3) + 5 = 10x + ( - 3 + 5) = 10x + 2 Step 11 Write the output (y)y = 10x + 2

34. next Translated into words, Step 11 of Program #1 told us to start with x, multiply by 10 and then add 2 to obtain our output (y). This may be expressed in the form of Program #2, where… 2007 Herbert I. Gross next Start with the input (x) Multiply by 10 Add 2 Step 1 Step 2 Step 3 Write the output (y) Step 4 Program #2

35. next The fact that Program #2 was derived from Program #1 by using our rules of the game, means that the two programs are equivalent. © 2007 Herbert I. Gross In other words, suppose we want to find the output if the input in Program #1 is 11. As we will show below, this is rather tedious if we use Program #1. However, the process is much less cumbersome if we use the equivalent program, Program #2. next

36. More specifically… © 2007 Herbert I. Gross Program #1Program #2 Step 1Start with the input x 11 Step 1Start with the input x 11 Step 2 Subtract it from 7 (that is, 7 – 11) -4-4 Step 2Multiply by 10 110 Step 3Multiply by 4 - 16 Step 3Add 2 112 Step 4Subtract 6 - 22 Step 4Write the output y 112 Step 5Divide by 2 - 11 Step 6 Add 5 times the input (that is, 5(11) + - 11) 44 Step 7Subtract 2 42 Step 8Divide by 3 14 Step 9 Subtract the result in Step 8 (14) from 11 times the input (121) 107 Step 10Add 5 112 Step 11Write the output (y) 112 next

37. Because subtraction is not commutative (for example 15 – 7 ≠ 7 – 15), it makes a difference whether we say “start with 7 and subtract it from 15” (that is 15 – 7) or whether we say “start with 7 and subtract 15” (that is 7 – 15). Thus, we have to be careful to distinguish between the commands “subtract 7” and “subtract (it) from 7”. Caution next © 2007 Herbert I. Gross

38. On the other hand, because addition is commutative, we do not have to distinguish between such commands as… “start with 7 and add 15” and “start with 7 and add it to 15”. That is; 7 + 15 = 15 + 7. But the Good News… next © 2007 Herbert I. Gross

39. next However, the main point we wish to make in this lesson is that not only is it more convenient to replace Program #1 by Program #2, but that in the form of Program #2 we see immediately that the rate of change of the output in Program #1 with respect to its input is 10. © 2007 Herbert I. Gross Key Point For example, every time the input in Step 1 of Program #1 increases by 1, the output of Program #1 in Step 11 increases by 10. next

40. When the input increases by 1, the output increases by 10. © 2007 Herbert I. Gross Program #1 Step 1Start with the input x 11 Step 2 Subtract it from 7 (that is, 7 – x) -4-4 Step 3Multiply by 4 - 16 Step 4Subtract 6 - 22 Step 5Divide by 2 - 11 Step 6Add 5 times the input 44 Step 7Subtract 2 42 Step 8Divide by 3 14 Step 9 Subtract the result in Step 8 from 11 times the input 107 Step 10Add 5 112 Step 11Write the output (y) 112 next 12 -5-5 - 20 - 26 - 13 47 45 15 117 122 13 -6-6 - 24 - 30 - 15 50 48 16 127 132 14 -7-7 - 28 - 34 - 17 53 51 17 137 142 next

41. The usefulness of knowing that Program #1 is equivalent to Program #2 is shown in the following… Suppose we are asked to find the input of Program #1 if its output is 5,702. It sure helps to paraphrase the question to read instead… What was the input of Program #2 if the output is 5,702 ? © 2007 Herbert I. Gross Note

42. next In this paraphrased form, the answer can be found rather simply as shown in the steps below… © 2007 Herbert I. Gross Program #2 Step 1Start with the input x Step 2Multiply by 10 Step 3Add 2 Step 4Write the output y Undoing Program #2 Step 4Write the input (570) Step 3Divide by 10 (570) Step 2Subtract 2 (5,700) Step 1Start with the output (5,702) next

43. Clearly the paraphrased form of the question was less tedious to work with than the original form. However, it is equally important to observe that the undoing process would not work for Program #1 unless we paraphrased it first. For example, starting with 5,702, the first step we would perform in undoing Program #1 would be to subtract 5 from 5,702. This is not difficult. © 2007 Herbert I. Gross Notes

44. next However, the next step we have to undo involves our having to already know what the input was (otherwise we wouldn't know what 11 times the input was); and therefore this method for undoing Program #1 breaks down. © 2007 Herbert I. Gross Notes But, to check whether we paraphrased the problem correctly, we should replace x, the input, by 570 in Program #1 and verify that the output in this case is indeed 5,702. next

45. © 2007 Herbert I. Gross Program #1 Step 1Start with the input x 570 Step 2Subtract it from 7 - 563 Step 3Multiply by 4 - 2,252 Step 4Subtract 6 - 2,258 Step 5Divide by 2 - 1,129 Step 6Add 5 times the input 1,721 Step 7Subtract 2 1,719 Step 8Divide by 3 573 Step 9 Subtract the result in Step 8 from 11 times the input 5,697 Step 10Add 5 5,702 Step 11Write the output (y) 5,702 Doing this we see that…

46. next In working with the relationship that y is linear in x, we have seen that from a computational point of view, it is convenient to represent the relationship in the form y = mx + b. © 2007 Herbert I. Gross Important Reminder next However, regardless of how y is expressed in terms of x; “y is linear in x” means that the rate of change of y with respect to x is constant.

47. next Many math textbook authors often assume that relationships are linear even when they might not be. For example, if we are told in a problem that John can paint a fence in 4 hours, the authors often say such things as … “Therefore in 1 hour he paints 1/4 of the fence”. © 2007 Herbert I. Gross An interesting Aside next However, it is possible, for example, that he painted half the fence in the first hour and then got tired; and therefore took 3 hours to paint the other half of the fence.

48. next In other words, the authors are assuming that John paints at a constant rate. However, while it is not reasonable to assume that John paints at a constant rate, it is reasonable to assume that water flows through a pipe at a constant rate. © 2007 Herbert I. Gross An interesting Aside next Thus, it would be realistic to assume that if it took 4 hours for the pipe to fill a pool, it would fill 1/4 of the pool in one hour.

49. next The “moral” of this aside is that solving a “real life” mathematics problem involves more than just computational techniques. That is, we often have to know more about the prevailing conditions. © 2007 Herbert I. Gross An interesting Aside next Thus, in wording problems one should be quite explicit; for example, to say such things as… “If John, painting at a constant rate, paints a fence in 4 hours, then in 1 hour he paints 1/4 of the fence”.