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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.

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Presentation on theme: "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."— Presentation transcript:

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 11

2 Applications to Algebra Applications to Algebra © 2007 Herbert I. Gross next

3 In this lesson we will show how we can extend the rules of arithmetic so that we can paraphrase complicated algebraic expressions and algebraic equations into simpler ones. © 2007 Herbert I. Gross next

4 For example, look at the following two programs… Program 1 Program 2 (1) Start with x (2) Subtract 1 (3) Multiply by 2 (4) Add 3 (5) Multiply by 4 (6) Add 6 (7) Divide by 2 (8) Answer is y (1) Start with x (2) Multiply by 4 (3) Add 5 (4) Answer is y next © 2007 Herbert I. Gross

5 Although Program 1 is more complicated than Program 2, we will show that the two programs are actually equivalent. To illustrate that our claim is at least plausible, let’s pick an input at random, run it through both programs, and show that we get the same output. next © 2007 Herbert I. Gross

6 Suppose, for example, that we pick 7 as our input for each of the two programs. We obtain… next Program 1 Program 2 (1) Start with x (2) Subtract 1 (3) Multiply by 2 (4) Add 3 (5) Multiply by 4 (6) Add 6 (7) Divide by 2 (8) Answer is y (1) Start with x (2) Multiply by 4 (3) Add 5 (4) Answer is y next © 2007 Herbert I. Gross 33

7 Of course the fact that the two programs give the same output when the input is 7 does not prove that the two programs are equivalent. © 2007 Herbert I. Gross next More specifically, recall that for two programs to be equivalent, they must give the same output for each possible input. In other words: if there is even one input for which the two programs give a different output, then the two programs are not equivalent.

8 Shortly, we will use the rules of the game to demonstrate/prove that the two programs are equivalent; but for now, let’s just suppose that the two programs are equivalent. If Program 1 and Program 2 are equivalent, the example in which we replaced x by 7 in both programs clearly shows the advantage of replacing Program 1 by Program 2. Namely, Program 2 used fewer steps to obtain the same answer. Not only is this faster, but experience tells us that the fewer steps we use in a computation, the fewer chances we have of making a computational error. © 2007 Herbert I. Gross

9 Let’s now look at our two programs from the viewpoint of applying the “undoing” process. For example, suppose we know that for a certain input (x) that the output was 14.48, and we want to find the value of x. Let’s first “undo” Program 1. Program 1 (1) Input is x (2) Subtract 1 (3) Multiply by 2 (4) Add 3 (5) Multiply by 4 (6) Add 6 (7) Divide by 2 (8) Output is y x Undoing Program 1 Output is x (8) Add 1 (7) Divide by 2 (6) Subtract 3 (5) Divide by 4 (4) Subtract 6 (3) Multiply by 2 (2) Input is y (1) next next 2.37

10 Now let’s look at the “undoing” of Program 2. Clearly, this is simpler than that of Program 1. © 2007 Herbert I. Gross next Program 2 (1) Input is x (2) Multiply by 4 (3) Add 5 (4) Output is y x Undoing Program 2 (4) Output is x (3) Divide by 4 (2) Subtract 5 (1) Input is y next

11 What we’d like to do in this lesson is to show how the “rules of arithmetic” can be used to paraphrase Program 1 into the equivalent but simpler Program 2. To paraphrase Program 1, we’ll first translate each step of the program into the language of algebra, noting that the output of one step is the input of the next step. © 2007 Herbert I. Gross

12 Program 1 (1) Input is x (2) Subtract 1 (3) Multiply by 2 (4) Add 3 (5) Multiply by 4 (6) Add 6 (7) Divide by 2 (8) Output is y next Program 1 x x x – 1 2(x – 1) [2(x – 1)] + 3 4{[2(x – 1)] + 3} 4{[2(x – 1)] + 3} + 6 ‹ 4{[2(x – 1)] + 3} + 6 › ÷ 2 – 1 To appreciate the power of paraphrasing, we will now use our rules to “simplify” each step as we go along… 2( ) { } + 3 4{ } + 6 › ÷ 2 ‹ next

13 © 2007 Herbert I. Gross In the language of algebra, the result is written as x – 1; and by the “add the opposite” definition of subtraction, this in turn can be written as x Step 1 Start with x Step 2 Subtract 1

14 next © 2007 Herbert I. Gross Step 1 Start with x Step 2 Subtract 1 The result in Step 2 was x Step 3 Multiply the result of step 2 by 2 Remember (by the closure property) that even though x and - 1 are two numbers, their sum, x + - 1, is one number; and to remind us that x is one number, we enclose it within parentheses and then multiply it by 2, giving us 2(x + - 1).

15 next The reason we rewrote x – 1 as x is because the distributive property was expressed in terms of a(b + c), not in terms of a(b – c). © 2007 Herbert I. Gross 2x (x) + 2( - 1) We can then use the distributive property (D1) to rewrite 2(x + - 1) as… Note

16 next © 2007 Herbert I. Gross Step 1 Start with x Step 2 Subtract 1 Step 3 Multiply the result of step 2 by 2 where we’ve enclosed 2x to emphasize that it is one number. Step 4 Add 3 to the result of Step 3 (2x + - 2) + 3 The result of Step 3 was 2x and if we add 3 to it we obtain…

17 next © 2007 Herbert I. Gross Step 1 Start with x Step 2 Subtract 1 Step 3 Multiply the result of step 2 by 2 Step 4 Add 3 to the result of Step 3 2x + 1 By the associative property of addition, (2x + - 2) + 3 can be rewritten as 2x + ( ). And since = 1,we may rewrite the result of Step 4 as…

18 next © 2007 Herbert I. Gross Step 1 Start with x Step 2 Subtract 1 Step 3 Multiply the result of step 2 by 2 Step 4 Add 3 to the result of Step 3 4(2x + 1) Again since 2x + 1 is one number, we enclose it within parentheses; and to indicate that we multiplied it by 4 we write it as… Step 5 Multiply the result of Step 4 by 4

19 next © 2007 Herbert I. Gross 4([2x] + 1) The rules of operations (PEMDAS agreement) allow use of grouping symbols to be minimized. However, to emphasize that 2x represents one number, we rewrite 4(2x + 1) as… 4[2x] + 4(1) Then by using the distributive property, we rewrite 4([2 x] + 1) as…

20 next © 2007 Herbert I. Gross 4 and by the multiplicative identity property, we may rewrite 4(1) as... 8x [(4)(2)]x Then by the associative property of multiplication, we may rewrite 4 [2 x] as… 8x + 4 Therefore, 4[2 x ] + 4(1) may be re-written as… 8x + 4 next

21 © 2007 Herbert I. Gross The result of Step 5 was 8x + 4. Adding 6 to 8x + 4 can be represented as (8x + 4) + 6. By the associative property of addition: this, in turn, can be rewritten as… 8x + (4 + 6), or 8x Step 1 Start with x Step 2 Subtract 1 Step 3 Multiply the result of step 2 by 2 Step 4 Add 3 to the result of Step 3 Step 5 Multiply the result of Step 4 by 4 Step 6 Add 6 to the result of Step 5 next

22 There are usually choices in the way we set up computations. For example, we could have begun Step 6 by writing 6 + (8x + 4) and then, using the commutative property of addition, rewritten it in the form (8x + 4 ) + 6. © 2007 Herbert I. Gross Note The way we actually chose had the advantage of using the associative property to regroup 4 and 6 into 10, in one step. next

23 © 2007 Herbert I. Gross Step 1 Start with x Step 2 Subtract 1 Step 3 Multiply the result of step 2 by 2 Step 4 Add 3 to the result of Step 3 Step 5 Multiply the result of Step 4 by 4 Step 6 Add 6 to the result of Step 5 next Step 7 Divide the result of Step 6 by 2 Dividing by 2 means the same thing as multiplying by 1/2. Hence, we may represent Step 7 in the form 1 / 2 (8x +4).

24 next © 2007 Herbert I. Gross 1 / 2 (8x) + 1 / 2 (10) …and by the distributive and associative properties of multiplication 1 / 2 (8x + 10) becomes… 4x + 5 [ 1 / 2 (8)]x + 1 / 2 (10) which in turn becomes…

25 © 2007 Herbert I. Gross next Step 1 Start with x Step 2 Subtract 1 Step 3 Multiply the result of step 2 by 2 Step 4 Add 3 to the result of Step 3 Step 5 Multiply the result of Step 4 by 4 Step 6 Add 6 to the result of Step 5 Step 7 Divide result of Step 6 by 2 So from Step 7 we see that y = 4x + 5. Step 8 Answer is y

26 © 2007 Herbert I. Gross Program 1 (1) Input is x (2) Subtract 1 (3) Multiply by 2 (4) Add 3 (5) Multiply by 4 (6) Add 6 (7) Divide by 2 (8) Output is y Program 1 x x – 1 2x – 2 2x + 1 8x + 4 8x x + 5 y = 4x + 5 next In summary…

27 Translating y = 4x + 5 into words … © 2007 Herbert I. Gross Program 2 (1) Start with x (2) Multiply by 4 (3) Add 5 (4) Answer is y And, this is exactly what Program 2 also says. we start with x, multiply by 4, and add 5. Hence, we have used the “rules of the game” to show that Programs 1 and 2 are equivalent. next

28 2007 Herbert I. Gross Like the folk-saying, that “It is much easier to scramble an egg than to unscramble one”: it is usually easier to translate a program from “plain English” into the language of algebra than to translate from algebra into “plain English”. However, if we go step by step and use the grouping symbols as directed, the procedure is relatively straightforward. More specifically: starting with… next 1 / 2 ( 4{2(x – 1) + 3} + 6 )

29 2007 Herbert I. Gross Step 1 We first look to locate the input (which in this case is x) next 1 / 2 ( 4{2(x – 1) + 3} + 6 ) In words we write… Start with “x” x

30 2007 Herbert I. Gross Step 2 Since x is inside the parentheses, we next do what the parentheses indicate. That is… next 1 / 2 ( 4{2(x – 1) + 3} + 6 ) In words we write… Subtract 1 x – 1

31 2007 Herbert I. Gross Step 3 & 4 The parentheses are within the braces. Since the order of operations says to multiply before we add, we first multiply what’s inside the parentheses by 2, and then add 3. That is… next 1 / 2 ( 4{2(x – 1) + 3} + 6 ) In words, we write… Multiply by 2 and then add 3 2(x – 1)+ 3

32 2007 Herbert I. Gross Step 5 & 6 The braces are within the large parentheses. They are being multiplied by 4, after which we are adding 6. That is… next 1 / 2 ( 4{2(x – 1) + 3} + 6 ) In words, we write… Multiply by 4 and then add 6 4{2(x – 1) + 3} + 6

33 2007 Herbert I. Gross Step 7 The expression within the parentheses is being multiplied by 1 / 2. Since multiplying by 1 / 2 is the same as dividing by 2... next ( 4{2(x – 1) + 3} + 6 ) ÷ 2 In words, we write… Divide by 2 We may rewrite… 1 / 2 ( 4{2(x – 1) + 3} + 6 ) as…

34 Notice that the instructions we carefully wrote out are exactly the steps that comprise Program 1. © 2007 Herbert I. Gross next Program 1 (1) Start with x (2) Subtract 1 (3) Multiply by 2 (4) Add 3 (5) Multiply by 4 (6) Add 6 (7) Divide by 2 (8) Answer is y

35 In algebraic form, our previous steps could be represented by the following sequence of equalities… © 2007 Herbert I. Gross next Once we have enough experience with the rules of the game of algebra, we can simplify 1/2 ( 4{2(x – 1) + 3} + 6 ) directly from the algebraic expression, without first having to translate the expression into a verbal form.

36 © 2007 Herbert I. Gross next 4x / 2 (8x) + 1 / 2 (10) 1 / 2 ( 8x + 10 ) 1 / 2 ( 8x ) 1 / 2 ( 4{2x } + 6 ) 1 / 2 ( 4{2x + 2[ - 1] + 3} + 6 ) 1 / 2 ( 4{2(x + - 1) + 3} + 6 ) 1 / 2 ( 4{2(x – 1) + 3} + 6 ) ALGEBRAICALGEBRAIC FORMFORM

37 1 / 2 (4{2(x – 1) + 3} + 6) = 77 © 2007 Herbert I. Gross next As an example of how helpful paraphrasing can be, suppose you were called upon to find the value of x for which… Since we have just shown that the expression on the left-hand side of this equation is equivalent to 4x + 5, we may replace the above equation by the equivalent equation… 4x + 5 = 77

38 © 2007 Herbert I. Gross next But to solve this equation, we need only subtract 5 from both sides of the equation to obtain… 4x + 5 = 77 – 5 4x4x72 =

39 © 2007 Herbert I. Gross next …and then divide both sides of the equation by 4 to obtain… 4x + 5 = 77 – 5 4x4x72 = 44 x 18 =

40 © 2007 Herbert I. Gross Program 1 (1) Input is x (2) Subtract 1 (3) Multiply by 2 (4) Add 3 (5) Multiply by 4 (6) Add 6 (7) Divide by 2 (8) Output is y Program next As a “plain English” check: notice that if you start with 18 (that is, replace x by 18), subtract 1, multiply by 2, add 3, multiply by 4, add 6, and then divide by 2; the answer is 77. That is…

41 1 / 2 (154) = 77 © 2007 Herbert I. Gross next Or summarized in the “language of algebra”… 1 / 2 ( ) = 77 1 / 2 (4{37} + 6) = 77 1 / 2 (4{34 + 3} + 6) = 77 1 / 2 (4{2(17) + 3} + 6) = 77 1 / 2 (4{2( x – 1) + 3} + 6) = 77 18

42 next While it is more cumbersome to work with an expression such as 1 / 2 ( 4{2(x – 1) + 3} + 6 ) than with the equivalent expression 4x + 5; some could argue that it might nevertheless be easier because, to paraphrase into the 4x + 5 expression, we have to figure out how best to use the rules of the game Herbert I. Gross (And, it wasn’t all that difficult to use the step-by-step “undoing” of Program 1.) next

43 However, as we shall show in our Keystone example: there are times when a program’s steps cannot be undone in an obvious way. It is in just such cases that replacing an equation by a simpler equivalent equation will be most helpful Herbert I. Gross

44 Equations vs. Expressions © 2007 Herbert I. Gross next The nice thing about working with an equation rather than with an expression is that we are allowed to perform the same operation on both sides of the equation without changing the solution. Thus for example, an alternative method for solving the equation… 1 / 2 (4{2(x – 1) + 3} + 6) = 77 Appendix

45 © 2007 Herbert I. Gross next We can undo multiplying by ½ by multiplying both sides by 2… 1 / 2 (4{2(x – 1) + 3} + 6) = 77 2 × × 2 and obtain… 4{2(x – 1) + 3} + 6 = 154

46 © 2007 Herbert I. Gross next 4{2(x – 1) + 3} + 6 = 154 The last step on the left was, to add 6. Hence, we can subtract 6 from both sides… 4{2(x – 1) + 3} = 148 – 6 and obtain…

47 © 2007 Herbert I. Gross next 4{2(x – 1) + 3} = 148 The last step we did on the left was to multiply by 4. Hence, we can divide both sides by 4… 2(x – 1) + 3 = and obtain… 37

48 © 2007 Herbert I. Gross next 2(x – 1) + 3 = 37 The last step we did on the left was to add 3. Hence, we can subtract 3 from both sides… 2(x – 1) = 34 – 3 and obtain…

49 © 2007 Herbert I. Gross next 2(x – 1) = 34 The last step we did on the left was to multiply by 2. Hence, we can divide both sides by 2… x – 1 = and obtain… 17

50 © 2007 Herbert I. Gross next x – 1 = 17 Finally, the last step we did on the left was to subtract 1. Hence, we can add 1 to both sides … x = 18 and obtain… + 1

51 © 2007 Herbert I. Gross next In Summary x = 18 x – 1 = 17 2(x – 1) = 34 2(x – 1) + 3 = 37 4{2(x – 1) + 3} = 148 4{2(x – 1) + 3} + 6 = / 2 (4{2( x – 1) + 3} + 6) = 77


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