# how to use barcode in c#.net Worked-Out Solutions to Exercises: s 1 to 9 in Software Maker QR Code ISO/IEC18004 in Software Worked-Out Solutions to Exercises: s 1 to 9

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1 When a negative number is raised to an even positive integer power, the result is always a positive number When a negative number is raised to an odd positive integer power, the result is always a negative number 2 The answers, along with explanations, are as follows (a) If we raise a base of 2 to increasing integer powers starting with 1, we get this sequence: ( 2)1, ( 2)2, ( 2)3, ( 2)4, ( 2)5, When we multiply these out, we get 2, 4, 8, 16, 32, The numbers alternate between negative and positive, and their absolute values double with each repetition This sequence runs away toward both positive infinity and negative infinity ! (b) If we do the same thing with a base of 1, we get ( 1)1, ( 1)2, ( 1)3, ( 1)4, ( 1)5, Multiplying these out gives us 1, 1, 1, 1, 1, The numbers simply alternate between 1 and 1 (c) If we carry out the same process with a base of 1/2, we get ( 1/2)1, ( 1/2)2, ( 1/2)3, ( 1/2)4, ( 1/2)5, Multiplying these out produces 1/2, 1/4, 1/8, 1/16, 1/32, The numbers again alternate between negative and positive, and their absolute values get half as large with each repetition This sequence converges toward 0 from both sides 3 Here are the answers Note how they mirror the results of Prob 2 (a) If we raise a base of 2 to smaller and smaller negative integer powers starting with 1, we get this sequence: ( 2) 1, ( 2) 2, ( 2) 3, ( 2) 4, ( 2) 5, This is the same as 1/( 2)1, 1/( 2)2, 1/( 2)3, 1/( 2)4, 1/( 2)5,
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612 Worked-Out Solutions to Exercises: s 1 to 9
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When we multiply these out, we get 1/( 2), 1/4, 1/( 8), 1/16, 1/( 32), which is the same as 1/2, 1/4, 1/8, 1/16, 1/32, The numbers alternate between negative and positive, and their absolute values get half as large with each repetition This sequence is identical to the solution for Problem 2(c) (b) If we do the same thing with a base of 1, we get ( 1) 1, ( 1) 2, ( 1) 3, ( 1) 4, ( 1) 5, This is the same as 1/( 1)1, 1/( 1)2, 1/( 1)3, 1/( 1)4, 1/( 1)5, Multiplying these out gives us 1/( 1), 1/1, 1/( 1), 1/1, 1/( 1), which is the same as 1, 1, 1, 1, 1, The numbers simply alternate between 1 and 1 This result is identical with the solution for Prob 2(b) (c) Finally, let s do the process with a base of 1/2 We get the sequence ( 1/2) 1, ( 1/2) 2, ( 1/2) 3, ( 1/2) 4, ( 1/2) 5, This is the same as 1/( 1/2)1, 1/( 1/2)2, 1/( 1/2)3, 1/( 1/2)4, 1/( 1/2)5, which is the same as 1/( 1/2), 1/(1/4), 1/( 1/8), 1/(1/16), 1/( 1/32), which can be simplified to 2, 4, 8, 16, 32, That s the same thing we got when we solved Problem 2(a)
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4 The expression can be simplified to a sum of individual terms when we remember that squaring any quantity (that is, taking it to the second power) is the same thing as multiplying it by itself Then we can use the results of the final Challenge in Chap 5 Step-by-step, we get: ( y + 1)2 = ( y + 1)( y + 1) = yy + y1 + 1y + (1 1) = y2 + y + y + 1 = y 2 + 2y + 1 5 This problem can be solved just like Prob 4, but we have to pay careful attention because of the minus sign: ( y 1)2 = ( y 1)( y 1) = yy + y( 1) + ( 1y) + [( 1) ( 1)] = y 2 + ( y) + ( y) + 1 = y 2 2y + 1 6 Let s start with the generalized addition-of-exponents (GAOE) rule as it is stated in the chapter text Here it is again, with the exponent names changed for variety! For any number a except 0, and for any rational numbers p and q, a pa q = a (p + q) Suppose r is another rational number Let s multiply both sides of the above equation by a r This gives us (a pa q)a r = a( p + q)a r According to the rule for the grouping of factors in a product, we can take the parentheses out of the left-hand side of the above equation and get a pa qa r = a( p + q)a r The left-hand side of the equation now contains the expression we want to evaluate Let s consider (p + q) to be a single quantity We can then use the GAOE rule on the righthand side of this equation, getting a pa qa r = a[( p + q) +r] Again, we invoke the privilege of ungrouping, this time to the addends in the exponent on the right-hand side This gives us a pa qa r = a( p + q + r) QED Mission accomplished!
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