# Worked-Out Solutions to Exercises: s 1 to 9 in Software Paint QR Code ISO/IEC18004 in Software Worked-Out Solutions to Exercises: s 1 to 9

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infinitely many elements Amazingly enough, you can pair the elements of both sets off one-to-one! The mechanics of this goes a little beyond the scope of this chapter, but you can get an idea of how it works if you divide every element of Weven by 2, one at a time, and then write down the first few elements of the resulting set When you do that, you get {0/2, 2/2, 4/2, 6/2, 8/2, 10/2, } But that s exactly the same as W, because when you perform the divisions, you get {0, 1, 2, 3, 4, 5, } This is one of the strange things infinity can do You can take away every other element of a set that has infinitely many elements and that can be written out as an implied list, and the resulting set is exactly the same size as the original set 7 Let s write out the sets again here as implied lists : A = {1, 1/2, 1/3, 1/4, 1/5, 1/6, } G = {1, 1/2, 1/4, 1/8, 1/16, 1/32, } It s not hard to see that set G contains all those elements, but only those elements, that belong to both sets Therefore A G=G If you start with set A and then toss in all the elements of G, you get the same set A (with certain elements listed twice, but they can count only once) That means set A contains precisely those elements that belong to one set or the other, or both, so A G=A 8 First, isolate all the individual elements of the set {1, 2, 3} They are 1, 2, and 3 Then start the list of subsets by putting down the null set, which is a subset of any set Then assemble all the sets you possibly can, using one or more of the elements 1, 2, 3, and list them You should get {1} {2} {3} {1, 2} {1, 3} {2, 3} {1, 2, 3}
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592 Worked-Out Solutions to Exercises: s 1 to 9
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9 The set {1, {2, 3}} has only two elements: the number 1 and the set {2, 3} You can t break {2, 3} down and have it remain an element of the original set {1, {2, 3}} Therefore, the set of all possible subsets of {1, {2, 3}} is {1} {{2, 3}} {1, {2, 3}} 10 The set {1, {2, {3}}} has two elements: the number 1 and the set {2, {3}} You cannot break {2, {3}} down and have it remain an element of the original set {1, {2, {3}}} Therefore, the set of all possible subsets of {1, {2, {3}}} is {1} {{2, {3}}} {1, {2, {3}}} When writing down complicated sets like these, always be sure that the number of opening braces is the same as the number of closing braces If they re different, you ve made a mistake somewhere
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1 This is the ultimate trivial question In the number-building systems described in this chapter, nothing doesn t represent any number 2 No The number 6 is divisible by 3 without a remainder: 6/3 = 2 Of course, the quotient here, 2, is even There are plenty of other even numbers that can be divided by 3 leaving no remainder 3 When you multiply an odd number by 3, you always get an odd number as the product The reason for this is similar to the reason why any even number times 7 gives you another even number (Proving that was one of the challenge problems in this chapter) For the first few odd numbers, multiplication by 3 always produces an odd number: 3 1=3 3 3=9 3 5 = 15 3 7 = 21 3 9 = 27
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