600 Worked-Out Solutions to Exercises: s 1 to 9

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Table A-4 Solution to Prob 3 in Chap 5 As you read down the left-hand column, each statement is equal to every statement above it

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Statements 4 + 32 / 8 ( 2) + 20 / 5 / 2 8 4 + 32 / [8 ( 2)] + 20 / 5 / 2 8 4 + 32 / ( 16) + 20 / 5 / 2 8 4 + [32/( 16)] + [(20/5)/2)] 8 4 + ( 2) + [(20/5)/2] 8 4 + ( 2) + (4/2) 8 4 + ( 2) + 2 8 4 + ( 2) + 2 + ( 8) 4 Reasons Begin here Group the multiplication Do the multiplication Group the divisions Do the division 32/( 16) = 2 Do the division 20/5 = 4 Do the division 4/2 = 2 Convert the subtraction to negative addition Do the additions from left to right

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Table A-5 Solution to Prob 4 in Chap 5 As you read down the lefthand column, each statement is equal to every statement above it

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Statements abcd a(bcd ) a(dcb) Reasons Begin here Group the last three integers Result of the challenge where it was proved that you can reverse the order of a product of three integers Commutative law for the product of a and (dcb) Ungroup the first three integers Mission accomplished

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(dcb)a dcba QED

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4 See Table A-5 Follow each statement and reason closely so you re sure how it follows from the statements before This proves that for any four integers a, b, c, and d, abcd = dcba 5 Here are the steps in the calculation process, using parentheses where needed: 15 ( 45) = 675 675 / ( 25) = 27 27 ( 9) = 243 243 / ( 81) = 3 3 ( 5) = 15 6 This always works if c = 1 Then a can be any integer, and b can be any integer except 0 Here s the original equation: (a /b)/c = a /(b /c)

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Plug in 1 for c Then you get (a /b) /1 = a /(b/1) which simplifies to a /b = a /b Here is an opportunity get some extra credit Does the original rule work when a = 1, but you let b and c be any integers except 0 Does it work when b = 1, but you let a be any integer and c be any integer except 0 7 This, again, always works if c = 1 Then a and b can be any integers Here s the original equation: (ab)/c = a(b /c) Plug in 1 for c Then you get (ab)/1 = a(b /1) which simplifies to ab = ab Now, can you guess what s coming Another extra-credit workout! Does the original rule hold true when a = 1, but you let b be any integer and c be any integer except 0 Does it work when b = 1, but you let a be any integer and c be any integer except 0 8 Let s start with the distributive law of multiplication over addition in its left-hand form You ll notice that we re using different letters of the alphabet That will help keep you from sinking into an abc rut with the naming of variables Otherwise, the law is stated in exactly the same way We can write the original form of the law like this: p(m + n) = pm + pn where n, m, and p are integers Now, the solution is only a matter of applying the commutative law for multiplication three times, once for each of the three products above: (m + n)p = mp + np QED That s all there is to it! 9 The variables have unfamiliar names for the same reason as in Prob 8 See Table A-6 This proves that for any two integers d and g, (d + g) = d g 10 Again, unfamiliar variable names can keep your attention on the way things work, without getting stuck in the abc routine of rote memorization See Table A-7 This proves that for any two integers h and k, (h k) = k h

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