Math 283 Spring 2013 Presentation Problems 4 Solutions

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1 Math 83 Spring 013 Presentation Problems 4 Solutions 1. Prove that for all n, n (1 1k ) k n + 1 n. Note that n means (1 1 ) (). n 1 Now, assume that (1 1k ) n n. Consider k n (1 1k ) (1 1n ) n 1 (1 1k ) k k ( ) (n + 1)(n 1) n n (n 1) ( 1 1 n ) n n (n + 1) n. ( n 1 n ) n n. Given x R and n N, use induction to prove that 1 Note that n 1 means x x x 1. n 1 Now, assume that x x(n 1). Consider n x xn. n n 1 x x + x x + x(n 1) xn.

2 n+1 3. Suppose that n N and that x 1,..., x n+1 are odd integers. Prove that x i is odd. Note that n 0 means Assume that (n 1)+1 x i 1 x i x 1. Which is odd since x 1 is odd. n 1 x i is odd when x i is odd for i {1,... n 1} and, since it is odd, let it be j + 1 for some j Z. Consider n+1 x i x n+1 + x n + Notice that x n+1 and x n are odd. Let them be p and p + 1 where p 1, p Z. Thus we have x n+1 + x n + n 1 n 1 x i (p 1 + 1) + (p + 1) + (j + 1) (p 1 + p + j + 1) + 1. x i. Since p 1 + p + j + 1 is an integer, the sum is odd completing the proof. 4. Suppose that n N and that x 1,..., x n+1 are odd integers. Prove that Note that for n 0, we have Now, assume that n 1 1 x i x 1 which is odd since x 1 is odd. n+1 x i is odd. x i is odd when x i, for i {1,..., n 1}, is odd. Consider n+1 x i x n+1 x n n 1 x i. Since x n+1, x n, and This means n 1 x i are odd, let them be a + 1, b + 1 and c + 1, respectively. x n+1 x n n 1 x i (a + 1)(b + 1)(c + 1) (4abc + ab + ac + bc + a + b + c) + 1. Since 4abc + ab + ac + bc + a + b + c is an integer, the product is odd.

3 5. For n N, prove that n ( 1) i i n n(n + 1) ( 1). Notice that when n 1, we have 1 ( 1) i i ( 1) 1 (1) 1 ( 1) 1 1(). n 1 Now, assume that ( 1) i i n 1 (n 1)n ( 1). Consider n n 1 ( 1) i i ( 1) n n + ( 1) i i ( 1) n n n 1 (n 1)n + ( 1) ( 1) n 1 1 n( n 1) n ( n + (n 1)) ( 1)n 1 n n(n + 1) ( 1). 6. Given 0 a i b i for all i N, prove that Notice that if n 1, we have Now, assume that we have n a i n+1 a i a n+1 n a i n a i a 1 b 1 n b i. n b i since a 1 b 1. n b i. Multiplying both sides of the inequality by a n+1 0, n a i a n+1 Where the second inequality holds since a i b i. n b i b n+1 n n+1 b i b i. 7. For n N, prove that for any x, y R where x < y implies x n 1 < y n 1. Notice that for n 1, we have x < y. Now, assume that x n 3 < y n 3. Consider x n 1 x x x n 3 < x x y n 3 < y y y n 3 y n 1.

4 8. Let q be a real number other than 1. Use induction on n to prove that n 1 q i (q n 1)/(q 1). i0 Notice that when n 1, we have i0 0 i0 n 1 Now, assume that q i qn 1 q 1. Consider i0 i0 q i 1 q1 1 q 1. n n 1 q i q n + q i q n + qn 1 q 1 qn+1 q n + qn 1 q 1 q 1 qn+1 1 q Determine for which values of n the inequality n! > n holds, and prove the inequality for all such n by induction. Notice that 8! 4030 and , but 9! and We conjecture that the statements holds for n 9. We have completed the base case. Now assume that (n 1)! > n. Notice that n > 9. Consider n! n (n 1)! > n n > n n. 10. Prove or give a counterexample to the following statement. If P (n) is true for all n N, and P (n) P (n + 1) for all n N, then P (n) is true for all n N. This statement is false since a counterexample is P (1) is false, but P (n) is true for all n N \ {1}. If 0 N (i.e. if we were to define 0 to be a natural number), then the statement would be true since P ( 0) P (0) is true and P (n) P (n + 1). 11. Prove or give a counterexample to the following statement. For n N, This statement is false. Possible counterexamples are n {, 3, 4, 5, 6}. n 18 n 8n + 8 < 1.

5 1. Prove that dn dx n ( e x ) n e x, for n 0. d 0 ( Notice that when n 0, ) e x e x 0 e x. dx 0 Now, assume that dn ( ) e x n e x. Taking the derivative of both sides, we have dx n d n+1 ( ) e x d ( n e x) n e x n+1 e x. dx n+1 dx 13. Prove that dn dx n ( x ) ( 1) n (n 3) n x (n 1)/, for n 1. Notice that d ( ) 1 x dx x 1/ ( 1)1 ((1) 3) x ((1) 1)/. 1 Now, assume that dn ( ) ( 1) n (n 3) x x (n 1)/. Taking the derivative of both sides, we dx n n have d n+1 dx n+1 ( x ) ( 1) n (n 3) n ( 1)n (n 3) n d ( ) x (n 1)/ dx (n 1) x [ (n 1)/)] 1 ( 1)n (n 3)(n 1) n+1 x [ (n 1)/)] / ( 1)n (n 3)(n 1) n+1 x (n+1)/.

6 14. Prove that (n 3) n (n )! n 1 (n 1)! for n 1. This statement is false. Notice that for n 1, we have 3 1 0! and 0! 1. The statement for n > 1 is true. Notice that for n, we have () 3 1! Now, assume (n 3) n (n 3)(n 1) n+1 (n )! n 1 (n 1)! (n 3) n (n 1)! n (n 1)! and consider (n 1) (n 1)! n (n 1)! n n 15. Use induction to show that, for all n 0, <. } {{ } n s Notice that for n 0, we have 1 <. Now, assume that <. Consider } {{ } n s }{{} n+1 s (n )! n 1 (n 1)! (n)! n+1 n!. + 1 } {{ } } n s {{ } n+1 s 3 (1)!. (n 1) < +.

7 16. Prove that for all n 1, F n < F n 1 F n+1 if n is even. Note that F n is the Fibonacci sequence 1, 1,, 3, 5,... given by F 1 1, F 1 and satisfying the recurrence F n F n 1 + F n. Notice that for n, we have F 1 1 < 1 F 1 F 3. Now, assume that F n < F n 1 F n+1 where n is even. Consider the next even situation, F n+ (F n + F n+1 ) F n+1 + F n+1 F n + F n < F n+1 + F n+1 F n + F n 1 F n+1 }{{} ind. hyp. F n+1 (F n+1 + F n + F n + F n 1 ) F n+1 (F n+ + F n+1 ) F n+1 F n Prove that for all n 1, F n > F n 1 F n+1 if n is odd. Note that F n is the Fibonacci sequence 1, 1,, 3, 5,... given by F 1 1, F 1 and satisfying the recurrence F n F n 1 + F n. Notice that for n 1, we have F > 0 1 F 0 F, or F 3 4 > 1 3 F F 4. Now, assume that F n < F n 1 F n+1 where n is even. Consider the next even situation, F n+ (F n + F n+1 ) F n+1 + F n+1 F n + F n > F n+1 + F n+1 F n + F n 1 F n+1 }{{} ind. hyp. F n+1 (F n+1 + F n + F n + F n 1 ) F n+1 (F n+ + F n+1 ) F n+1 F n+3

8 On Diabolical Island, the secret hideout of Professor Evil, there is one and only one road and it goes all the way around the island. There are some number of gas stations along the road, but Professor Evil has supplied them in such a way that the total amount of gas available from all of the stations combined is just enough to go around the island exactly once (in the one and only car on the island). 18. Prove that regardless of how many gas stations there are in total, there is some gas station along the road that will provide enough gas to reach the next gas station. (Hint: Prove this by contradiction.) [By Contradiction] Suppose, by way of contradiction, there is not a gas station with enough gas to reach the next station. This means that every gas station on the island does not contain enough gas to reach the next station. This means that the total amount of gas at all stations would not cover the entire island (since each falls short by some amount). Since the total amount of gas at all stations should cover one trip around the island, this is a contradiction. Therefore, there must be at least one station with enough gas to reach the next station. 19. Prove that regardless of how many gas stations there are in total, there is some gas station along the road where if you begin there with an empty tank of gas, and fill up at every gas station along the way, you will be able to drive all the way around the island. (Hint: Prove this by induction.) We do induction on n, the number of gas stations. Notice that for n 1, there is only one station and it contains enough gas to travel once around the island. This is the station at which we would begin our journey. Assume that if the island has n gas stations, no matter how the gas is distributed among those stations, we can find a gas station where if we began there with an empty tank and filled at each gas station along the way, we would be able to drive all the way around the island. Now consider any gas distribution on an island with n + 1 gas stations. By the previous problem, we know that there is at least one station with enough gas to reach the next station, call this station G and the station it can reach station G + 1. If we consider the simpler island where all of station G + 1 s gas is placed at station G and station G + 1 is eliminated, this would be an island with n gas stations. By the inductive hypothesis, we can find some gas station S where if we begin there with an empty tank of gas, and fill up at every gas station along the way, we can drive all the way around the island. This S is also the starting station for the original island with n + 1 stations since station G can reach station G + 1. The rest of the island can be traveled since it could be traveled in the n gas station island which was the same except for the portion from G to G + 1.

Discrete Math, Spring Solutions to Problems V

Discrete Math, Spring Solutions to Problems V Discrete Math, Spring 202 - Solutions to Problems V Suppose we have statements P, P 2, P 3,, one for each natural number In other words, we have the collection or set of statements {P n n N} a Suppose