MCS 115 Exam 2 Solutions Apr 26, 2018

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1 MCS 11 Exam Solutions Apr 6, (10 pts) Suppose you have an infinitely large arrel and a pile of infinitely many ping-pong alls, laeled with the positive integers 1,,3,, much like in the ping-pong all prolems we considered in class This time, suppose you dump into the arrel 10 ping-pong alls numered 1-10 as efore and remove numer 1 But next you put in 100 ping-pong alls, numered , and remove numer Then you put in 1000 ping-pong alls, numered , and remove numer 3, and so on How many ping-pong alls remain in the arrel after the stopwatch eeps? Infinitely many? Finitely many? Hint: Can you name one? There will e no ping-pong alls in the arrel when the process ends Suppose ping-pong all numer n remained in the arrel for some n Z + But then what did we remove right after putting all numer n in the arrel? Of course, n can e a really ig numer, ut it is still a positive integer, so the ping-pong all with this numer still got put in the arrel somewhere along the way And that is when all numer n was removed Consider the following infinite collection of real numers: (a) ( pts) Descrie in your own words how these numers are constructed (that is, descrie the procedure for generating this list of numers) The numer in the n-th row is constructed y listing the multiples of n as n,n,3n, after the decimal point and using these as the digits of the numer () (10 pts) Using Cantor s diagonalization argument, find a numer not on the list Justify your answer, that is explain how we know that the numer you found cannot e on the list We can do the same thing we did in class: we will construct a numer m y going down the diagonal

2 and for every digit that is we write down a 4 and for every digit that is not we write down a : m = 04 We know this numer cannot e on the list ecause it differs from every numer on the list in at least one digit Specifically, suppose m were on the list, say in the n-th row In the process of constructing m aove, we would have looked at the n-th digit of the n-th numer If this numer were m, then that digit would have to e or 4 But if it were, we actually would have used 4 as the n-th digit of m And if the digit were 4, then we would have used instead So no matter what, the n-th numer on the list cannot e the same as m 3 (a) (4 pts) Let n e a positive integer and let x and y e integers Explain what x y (mod n) means x y (mod n) means that x y is a multiple of n Another way to say this is that x y is divisile y n Yet another way to say this is that x and y have the same remainders when divided y n () (6 pts) We noted that division modulo n may not e well-defined, ie even if a (mod n) and c d (mod n), it could well e that a/c /d (mod n) Give an example of this: choose a modulus n and find specific integers a,,c,d such that a (mod n) and c d (mod n), ut even though a/c and /d are integers, they are not equivalent modulo n There are many examples Here is one: n = 6, a = 10, = 16, c =, d = 8 Then (mod 6) ecause = 6 8 (mod 6) ecause 8 = 6 10 = = 16 (mod 6) ecause = In class, we defined a golden rectangle as a rectangle with sides a and such that a+ = a (a) (7 pts) Use the aove definition to find the exact value of /a First, let φ = /a Now notice that So a+ = a 1 φ +1 = φ a+ = a + = 1 a +1 = 1 φ +1 multiply oth sides y φ φ+1 = φ sutract φ+1 from oth sides 0 = φ φ 1 By the quadratic formula φ = 1± ( 1) 4( 1) = 1±

3 Since > 1, we can tell 1 is negative But and a were the sides of a rectangle, so they must e positive, and hence /a must e positive too So the correct value is a = 1+ If you are paying attention, you may have noticed this is almost exactly the same argument you were asked to give on the last exam, only F n+1 and F n are replaced y and a () (8 pts) In the diagram elow, suppose that ABCD is a square, ADFE is a rectangle, M is the midpoint of AB, and CE is an arc of a circle centered at M Prove that rectangle ADFE is a golden rectangle Hint: you want to show that if AD = a, then AE = φa, where φ is the golden ratio D C F a A M B E First, note that MB = a/ and BC = a So y the Pythagorean Theorem applied to the right triangle MBC, (a ) a MB = +a a = 4 +a = 4 = a Now, AE = AM +ME = a + a = 1+ a = φa This shows AE/AD = φ, so rectangle ADFE is a golden rectangle Extra credit prolem On your homework, you considered infinite chains of circles like the one elow: and showed that the numer of ways to color such a chain y making each circle either red or lue has greater cardinality than the set of natural numers Z + Now suppose that you are again to color the chain y making each circle red or lue, ut you can only color finitely many of the circles red So for example, the coloring in which the circles alternate etween red and lue is not allowed ecause it would require coloring infinitely many circles red (a) ( pts) Show that there are still infinitely many different ways to color a chain Notice that one such coloring is that we color the first circle red and all the circles after that lue Another, different coloring is that we color the first two circles red and all the circles after those lue A third one that the first three circles are red and all the circles after those are lue And so on Let C n for n = 1,,3, e the chain in which the first n circles are red and the remaining circles are lue Such a chain has

4 finitely many red circles of course But there are infinitely many such chains, one for each positive integer n = 1,,3, Those are not the only ways to color a chain, ut they are already infinitely many ways to color a chain () (10 pts) Does the numer of ways to color a chain have the same cardinality as the set of natural numers? If you think it does, show that it does y finding an appropriate one-to-one correspondence etween all such colorings and the natural numers If you do not think it does, find an argument to show why no such one-to-one correspondence can exist We will show that the numer of ways to color a chain has the same cardinality as the set of natural numers To construct a one-to-one correspondence etween the natural numers and the colorings of the chain, we will list all of the possile colorings in an infinite list As we pointed out in class, such an infinite list is a one-to-one correspondence with the natural numers ecause it assigns to each coloring its position in the list, and that assignment is one-to-one First, notice that since each coloring of the chain is allowed to use only finitely many red circles, there is always a point along the chain after which all of the circles are lue So we first list the chain whose circles are all lue Then we list the chains in which the last red circle is the first circle and all of the circles after that are lue There is actually only one such chain Then we list the chains in which the last red circle is the second one These are the next two chains on the list Then we list the ones in which the last red circle is the third one There are four such chains, the next four on the list elow And so on Here is what the eginning of our list looks like:

5 Every chain with only finitely many red circles has a last red circle, so it will e listed eventually How far down the list it appears determines which natural numer corresponds to it This way, each coloring corresponds to a natural numer Since there are infinitely many colorings, each natural numer also corresponds to some coloring Notice that the way we organized the chains into an infinite list is similar to how one would list the rational numers etween 0 and 1 to show that they are equinumerous with the natural numers, which is what Mindscape 36 asked you to do on your homework

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