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This is an open-book quiz You may (and should) refer to the text as you solve these problems Don t hurry! You ll find worked-out answers in App A The solutions in the appendix may not represent the only way a problem can be figured out If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it! 1 How does the absolute value change if you multiply an integer by 3 2 How does the absolute value change if you start with an integer and multiply by 3 over and over without end 3 Evaluate the following expression: 4 + 32 / 8 ( 2) + 20 / 5 / 2 8 4 Do an S/R proof showing that for any four integers a, b, c, and d abcd = dcba Here s a hint: you solved a problem like this for addition in Chap 4 This proof proceeds in the same way 5 Start with the integer 15, multiply by 45, then divide that result by 25, then multiply that result by 9, then divide that result by 81, and finally multiply that result by 5 What do you end up with 6 Show at least one situation where you can say that a /(b /c) = (a /b)/c where a, b, and c are integers Do not use the trivial case a = 1, b = 1, and c = 1 Find something more interesting!

From eqn (4133), the radius such that P (RCEP ) = 05 is computed as 05 = 1 exp 2 ln(2) RCEP = 11774

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7 Show at least one situation where you can say that (ab)/c = a(b /c) where a, b, and c are integers Do not use the trivial case a = 1, b = 1, and c = 1 Find something more interesting! 8 Suppose you have learned the left-hand distributive law for multiplication over addition, but you have never heard about the right-hand version Show that for any three integers n, m, and p, the right-hand distributive law for multiplication over addition will always work: (m + n)p = mp + np Use the narrative form, not an S/R table 9 Construct an S/R table showing that for any two integers d and g (d + g) = d g 10 Prove that when you want to find the negative of the subtraction of one quantity from another, you can simply switch the order of the subtraction To do this, put together an S/R table showing that for any two integers h and k (h k) = k h

047 05 056 068 1 2 22 33 47 56 1 2

Similarly, the radius such that P (R95 ) = 095 is calculated as 095 R95 R2 1 exp 95 2 = 2 ln(20) 24477 =

You can always divide an integer by another integer, except when the divisor is 0 Then you get a fraction A fraction might not be an integer, but it s still a number

Figure 6-1 shows how you move around on the number line when you divide 3 by 2 First, you take the additive inverse of 3, because you re dividing by a negative quantity That puts you at the point corresponding to 3 Then you reduce your distance from 0 by a factor equal to the absolute value of the divisor That absolute value is 2, so you must move halfway from 3 to 0 That puts you at a point between 1 and 2

For a given value of c, the ellipse in x coordinates is x2 x2 1 2 2 2 + 2 c 1 2 The ellipse in w coordinates is w P 1 w c2 (4135) (4134)

1 2 3 4 5

Remainders In the situation of Fig 6-1, the finishing point is midway between the point for 1 and the point for 2 When you divide 3 by 2, you get 1 with a remainder 1 In this context, the word remainder means portion left over The divisor in this situation is 2, so you have a remainder of 1/( 2), which has the same numerical value as 1/2 The remainder is always added to the whole-number part of the quotient to get the final version of the quotient Here, the arithmetic works out like this:

10 5

3/( 2) = 1 + ( 1/2) = (1 + 1/2) = 1-1/2 The last quantity above is read negative one and a half The whole-number part is separated from the fractional part by a dash This dash is not a minus sign! (The minus sign is much longer) Some writers leave a space between the whole-number part and the fractional part of an expression like this But that can also be confusing, because it could make the above result look like 11/2 That would mean 5-1/2, not 1-1/2

10 5

3 2 1

17 16 15 14 13 12 11 10

-1 Finish here Not at integer point -2 -3

10 5

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