Math 2112 Solutions Assignment 5
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1 Math 2112 Solutios Assigmet Idicate which of the followig relatioships are true ad which are false: a. Z Q b. R Q c. Q Z d. Z Z Z e. Q R Q f. Q Z Q g. Z R Z h. Z Q Z a. True. Every positive iteger is a ratioal umber. b. False. For example, 2 is a egative real umber, but is ot ratioal. c. False. For example, 2 3 is ot a iteger. d. False. The iteger is i the set o the right, but ot the oe o the left. e. True. Sice Q R, the Q R Q. f. True. Sice Z Q, the Q Z Q. g. True. Sice Z R, the Z R Z. h. False. For example, 2 3 is i the set o the left, but ot i the set o the right Prove by mathematical iductio that for ay iteger 1 ad all sets A 1, A 2,..., A ad B, (A 1 B (A 2 B (A B (A 1 A 2 A B. Proof: Base Case: Let 1. The LHS A 1 B ad RHS A 1 B. So LHS RHS. Iductive Step: Let k 1. Suppose that (A 1 B (A 2 B (A k B (A 1 A 2 A k B. Cosider (A 1 B (A 2 B (A k1 B. By our iductive hypothesis, we kow that (A 1 B (A 2 B (A k1 B ((A 1 A 2 A k B (A k1 B Thus the result holds by mathematical iductio. ((A 1 A 2 A k B c (A k1 B c (A 1 A 2 A k A k1 B c B c (A 1 A 2 A k A k1 B c (A 1 A 2 A k A k1 B 1
2 For all sets A ad B, (B c (B c A c B. Proof: (B c (B c A c (B c (B c A c c Alterate Represeatio for Set Differece ((B c c (B c A c c De Morga s Laws ((B c c ((B c c (A c c De Morga s Laws (B (B A Double Complemet law B Absorptio Laws Give ay sets A ad B, defie the symmetric differece of A ad B, deoted A B, as follows: A B (A B (B A. Prove each of the followig for all sets A, B, ad C i a uiversal set A. a. A B B A b. A (B C (A B C c. A A d. A A c U e. A A f. If A C B C, the A B. Proof: We start our proof by otig that x A B iff (x A ad x B or (x A ad x B ad x A B iff (x A ad x B or (x A ad x B. a. A B (A B (B A (B A (A B B A. b. x (A B C (x A B ad x C or (x C ad x A B ([(x A ad x B or (x B ad x A] ad x C or (x C ad [(x A ad x B or (x A ad x B] [(x A ad x B ad x C or (x B ad x A ad x C] or [(x C ad x A ad x B or (x C ad x A ad x B] x is i exactly oe of the sets A, B, ad C, or x is i all three of the sets A, B, ad C. Similarly, it ca be show that x A (B C x is i exactly oe of the sets A, B, ad C, or x is i all three of the sets A, B, ad C. Therefore, A (B C (A B C. c. A (A ( A (A c ( A c (A U (A c A A. 2
3 d. A A c (A A c (A c A (A (A c c (A c A c (A A A c A A c U. e. A A (A A (A A A A (A A c. f. Let A, B, ad C be ay sets with A C B C. A B : Suppose x A. Either x C or x C. If x C, the x A C. But A C B C. Thus x B C. Sice x C ad x B C, the x B. O the other had, if x C, the sice x A, x A C. But A C B C. So, sice x C ad x B B, the x B. Hece, i either case, x B. B A : The proof is the same as A B, but with A ad B reversed. Therefore, sice A B ad B A, the A B. 2.7 (CE Explai why the Fiboacci seqeuce appears o the shallow diagoals of Pascal s triagle. Proof: The Fiboacci sequece is defied by F 1, F 1 1 ad F F 1 F 2, 2. Recall also that ( ( k 1 ( k 1 k 1. We cojecture that ( ( ( F ( ( ( F for all. Proof of cojecture: We proceed by strog iductio. Base Cases: F 1 ( ad F1 1 ( 1, so the result is true for, 1. Iductive Step: We will oly prove the eve case. The odd case follows aalgously. Let k >. Assume that the formula holds true for all i such that 3
4 i < 2k. Cosider F 2k F 2k 1 F 2k 2 ( ( ( ( ( 2k 1 2k 2 k 2k 2 2k k 1 1 ( [( ( ] [( ( ] 2k 1 2k 2 2k 2 2k 3 2k [( ( ] ( k k k 1 k 1 k 2 k 1 ( ( ( ( ( 2k 1 2k 1 2k 2 k 1 k k 1 k 1 ( ( ( ( 2k 2k 1 k 1 k... 1 k 1 k As stated above, the odd case follows aagously. Thus the result holds by strog mathematical iductio. ( k 1 k (CE Cosider lightbulbs i a room, umbered 1 to, Determie the umber of ways the lightbulbs ca be tured o or off. By solvig this problem i two differet ways, prove that ( ( 1 ( 2... ( 2 Proof: Light 1 ca be i oe of two positios, amely o or off. Light 2 ca be i oe of two positios, o or off... Thus, there are a total of 2 differet positios the lightbulbs ca be i. O the other had, there are ( ways lights ca be o. There are ( 1 ways 1 light ca be o... Sice every light positio is couted by exactly oe of the biomial coefficiets, we ca deduce that there are ( ( ( 1 ( 2... ( differet positios. Thus, we have ( ( 1 ( , as desired (CE Prove, usig a routes argumet, that ( ( ( ( ( ( ( Proof: Cosider all possible routes o a grid. We label the start positio (, ad the ed positio (,. To get from the start to the ed, we must make rights ad ups. Thus, there are ( 2 differet possible routes. O 4
5 the other had, cosider the checkpoits (2,, (1, 1 ad (, 2. Clearly, every route must pass through exactly oe of these checkpoits. There are ( 2 ways to get to (2, (we choose which of the two steps are rights, ad ( 2 2 ways to get from (2, to (, (agai, we choose which of the remaiig steps are rights. Similarly, there are ( ( routes goig through the checkpoit (1, 1 ad ( ( routes goig through the checkpoit (, 2. Thus there are a total of ( ( ( ( 1( 1 2 ( differet routes. 5
and each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime.
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