Discrete Mathematics: Homework Assignment 1 solutions

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1 Discrete Mathematics: Homework Assignment 1 solutions Due : Thursday, 3 February 2005 Please submit solutions to the following problems: 1. Provide direct proofs of the following statements : (a) If n and m are even then m + n is even. Let n and m be even integers. Then, by definition, there exist integers k and l so that n = 2k and m = 2l. Thus, n + m = 2k + 2l = 2(k + l). k + l is an integer, so n + m is even (by the definition). (b) Show, by using an indirect argument (i.e. by contradiction) that if 100 balls are placed in 9 cups, then one of the cups contains at least 12 balls. Assume that the statement is false. Then, each cup contains at most 11 balls. The total number of balls is then at most 11 times 9 = 99. However, we had 100 balls. This contradiction proves our result. 2. Prove that for all sets A, B and C, A (B C) = (A B) (A C). The proof is in two steps : we will first prove that the LHS is a subset of the RHS, and then we will prove the RHS is a subset of the LHS. Step 1 : Let x LHS. Then, x A (B C). Thus, x A and x B C. Hence, x A and (x B or x C). We consider the two cases (which are not mutually exclusive) : x B Then, x A and x B so x A B. x C Then, x A and x C so x A C. One of these two cases must be true, so x (A B) (A C). We have thus proved LHS RHS. We now prove the RHS LHS. Let y RHS. Then y (A B) or y (A C). We consider the two cases separately : y A B Then, y A and y B. In particular, y A. y A C Then, y A and y C. In particular, y A. Thus, we have that in either case, y A. In the first case, we have y B and in the second case, we have y C so it follows that y B C. Combining these two observations, it follows : y A and y B C so y A (B C). This proves RHS LHS. Hence, RHS = LHS.

2 3. By using mathematical induction, prove that (2n 1) = n 2. Base case : n = 1. The LHS = 1. The RHS = 1 2 = 1. Thus, RHS = LHS for n = 1, proving the base case. Induction step : assume We wish to prove (2k 1) = k (2k 1) + (2k + 1) = (k + 1) 2. By using the induction hypothesis, we observe : (2k 1) + (2k + 1) = k 2 + (2k + 1) Now, we have k 2 + (2k + 1) = (k + 1) 2. Thus, we have established : (2k 1) + (2k + 1) = (k + 1) 2, as required. This completes the induction step. The result follows by mathematical induction. 4. (Page 61, problem 28) We are given n 0 s and n 1 s written around the outside of a circle. (We don t know anything about their arrangement.) Show that we can always find a point so that if we start reading the numbers, going around clockwise, we will always have read out at least as many 0 s as 1 s. We will prove the statement, for any positive integer n, any arrangement of n 0 s and n 1 s allows for the existence of such a starting point. Base case : n = 1. The statement is about a table with a single 0 and a single 1. There is only one possible arrangement (up to rotation of the table). We take our starting point to the left (counterclockwise) of the 0. This constructs a starting point with the desired property. (A picture would also have been acceptable for this base case.) Induction step : we assume that for any table with n 0s and n 1s, we can find a suitable starting point. We wish to prove that if we have a table with n s and n s, then we can again find a suitable starting point. Let us consider a table with n s and n s (arranged arbitrarily). There must exist a 0 followed immediately by a 1 in clockwise order. (Exercise for you : why?) Hide this 0 and this 1. We now have an arrangement of n 0 s and n 1 s on the table. By the induction hypothesis, this admits a suitable starting position. I claim that this starting position works for the table when we unhide our 0 and our 1. We will consider our table in three pieces : the part from our claimed starting point

3 until the hidden 0 and 1, the part consisting of the hidden 0 and 1 and then the part after the hidden 0 and 1. In the first part, the running total of 0s and 1s will be the same whether we consider the 0 and 1 as hidden or not. Hence, at each step, we will have seen as many 0s as 1s. Just before we reach the hidden 0 and 1, the total number of 0s we ve seen will be at least as large as the total number of 1s we ve seen. When we reach the hidden symbols, we first reach the 0. This increases the total number of 0s seen by one, and leaves the total of 1s alone : we still have at least as many zeroes as ones. We now see the 1 : this increases the total number of 1s seen by one. In the third and final part of the table, the running totals of zeroes and ones seen will equal the running totals for the smaller table (i.e. with the symbols hidden) plus 1. From this, we conclude that at each step, the running total of zeroes will be at least as large as the running total of ones. This establishes the claim that this starting point works in the bigger problem. We have thus shown that any arrangement of n+1 0s and of n+1 1s admits a suitable starting place, given the assumption that we can find a starting place with n 0s and 1s. The result now follows by mathematical induction. 5. (Page 69, problems 6 and 7) We define a sequence by : (a) Compute c 2, c 3, c 4 and c 5. (b) Use strong induction to prove that c 1 = 0, c n = c n/2 + n 2 for n > 1 c n < 4n 2 for all n 1 (I will only do part b.) Base case : observe the statement is true for n = 1. Induction step : assume the statement is true for k = 1, 2,..., n. We will prove the statement for n + 1. By the recurrence relation, c n+1 = c (n+1)/2 + (n + 1) 2. Note that n 1 gives 1 (n + 1)/2 n, so, the induction hypothesis applies to c (n+1)/2. We obtain therefore : ( ) 2 ( ) 2 n + 1 n + 1 c (n+1)/2 < 4 4 = (n + 1) Combining this inequality with the value of c n+1 given by the recurrence relation, we obtain : c n+1 < (n + 1) 2 + (n + 1) 2 = 2(n + 1) 2 < 4(n + 1) 2. This proves the induction step. The result now follows by strong induction.

4 6. (Page 62, problem 49) We have n 2 people, each with a pie. They are standing in some arrangement on a football field. When I shout GO!, they each throw their pie at the nearest neighbor. Assume that they all hit their targets. Furthermore, assume that each person has a single nearest neighbor (i.e. there is no ambiguity and so they have no choice as to interpretation of nearest neighbor ). (a) Show that there is an even number n and an arrangement so that everyone is hit by a pie. Consider an arrangement with only two people. (b) Show that if n is odd, there is always at least one person who is NOT hit by a pie. ( Hint: What do you know happens if two people throw their pie at the same target? Use this to simplify the problem somewhat and then use induction.) We observe that if someone is hit by more than one pie, someone must survive. Indeed, there are no longer enough pies to go around to hit everyone. We will now prove the statement by induction. [NOTE : since we are trying to prove the result for odd integers n, we will prove that if we assume the statement for n, we can establish it for n + 2.] Base case : a game with three people. The two closest players must hit each other. The remaining player will hit one of those two and survive. Induction step : assume that for a fixed odd number n, in any game with n players, there will be a survivor. We will then prove that for n + 2 players, there will be a survivor. Consider a game with n + 2 players. The two closest people will hit each other. There are two cases : either no one else hits one of these two, or someone else does hit one of these two. In the first case, we can remove these two players from the game without affecting the remaining pie throws. (Since no one else hits them and they hit no one else.) This leaves us with a game of n players. By the induction hypothesis, there is a survivor. In the second case, one of these two players is hit by more than one pie. By our earlier observation, this means that there must be a survivor. We have thus established that having a survivor in a game of n players implies having a survivor in a game of n + 2 players. Therefore, by mathematical induction, it follows that any game with an odd number of players must have a survivor.

5 Bonus: For extra credit, provide an alternate (direct) proof that (2n 1) = n 2. Observe that we can add up (2n 1) + (2n 1)+ (2n 3)+ (2n 5) = (2n)+ (2n)+ (2n)+... +(2n) = (2n)(n) = 2n 2 It follows that the sum (2n 1) = 1 2 2n2 = n 2.