Math 3000 Section 003 Intro to Abstract Math Homework 6

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1 Math 000 Section 00 Intro to Abstract Math Homework 6 Department of Mathematical and Statistical Sciences University of Colorado Denver, Spring 01 Solutions April, 01 Please note that these solutions are only suggestions; different answers or proofs are always possible. Section 6.1: The Principle of Mathematical Induction 1. Exercise 6.10: Let r 1 be a real number. Use induction to prove that a + ar + ar + + ar n 1 an Solution: We will prove by weak mathematical induction. 1. Base Step: For n 1, the formula a1 a is satisfied.. Induction Hypothesis: Now assume that a+ar+ +ar n 1 an for n 1,,..., k.. Inductive Step: Then consider a+ar+ +ar k 1 +ar k where a+ar+ +ar k 1 ak by our induction hypothesis. Using some basic algebra, we find that a + ar + ar + + ar k 1 + ar k ak ak + ark + ar k ak + r k r k+1 ak+1. Section 6.: A More General Principle of Mathematical Induction. Exercise 6.: Prove that if A 1, A,..., A n are any n sets, then A 1 A A n A 1 A A n. Solution: We will proof by strong mathematical induction. 1. Base Step: For n, the identity A 1 A A 1 A is true by De Morgan s law.. Induction Hypothesis: Now assume that A 1 A n A 1 A n for n,,..., k.. Inductive Step: Then consider A 1 A k A k+1 and let B A 1 A k so that A 1 A k A k+1 B A k+1 B A k+1 * by our base step or induction hypothesis for n, and B A 1 A k A 1 A k ** by our induction hypothesis for n k. Combining equations * and **, we obtain A 1 A k A k+1 B A k+1 A 1 A k A 1 A k. Remark: Note how the inductive step from k to k + 1 in the proof of Exercise 6.10 only depends on the induction hypothesis for n k weak induction, whereas the proof of Exercise 6. also uses the induction hypothesis for n which forms the base step strong induction.

2 Math Intro to Abstract Math Homework 6, UC Denver, Spring 01 Solutions Section 6.: Proof by Minimum Counterexample. Exercise 6.0: Use proof by minimum counterexample to prove that for every nonnegative integer n. n + Solution: Note that n + n 1 + n is clearly divisible by. The simplicity of this statement makes it a good example to practice a proof by minimum counterexample. Proof by Minimum Counterexample. By contradiction, let us assume that there exists a nonnegative integer n such that n +, and let m be the smallest such integer so we assume that m is the minimum counterexample. Because is divisible by, it follows that m 1 and k + k+1 for all k 0, 1,..., m 1. In particular, then m 1 + m 0 mod and by the laws of modular arithmetic we find that m 1 + m m + m+1 0 mod in contradiction to our initial assumption that m + m+1. Hence our assumption must have been wrong, and n + for every nonnegative integer n. Section 6.4: The Strong Principle of Mathematical Induction 4. Exercise 6.4: A sequence {a n } is defined recursively by a 1 1, a 4, a 9, and a n a n 1 a n + a n + n for n 4. Conjecture a formula for a n and prove that your conjecture is correct. Solution: We conjecture that a n n for every positive integer n clearly true for n 1,,. Proof. Because our conjecture is true for n 1,,, it suffices to verify the recursive relationship defined by the identity a n 1 a n + a n + n a n by algebraic substitution: a n 1 a n + a n + n n 1 n + n + n n n + 1 n 4n n 6n n 6 n a n. 5. Exercise 6.6 : Consider the following sequence of equalities: a What is the next equality in this sequence? b Now develop a general conjecture and prove that your conjecture is correct by induction. Solution: For a, it is not difficult to see that the next equality in this sequence is both of which equal 189. For b, we conjecture that a general identity is given by n n n + n + 1 kn +1 k n + n + 1 for every nonnegative integer n which by the above is true for n 0, 1,,, 4.

3 Math Intro to Abstract Math Homework 6, UC Denver, Spring 01 Solutions Proof. We can use the known relationship that n n k n to write n n n + n + 1 kn +1 k k k * n + 1 n n n + 1 n n + 1 n 4 + n n 4 + 4n + 6n + 4n n + n + 1 n 4 n n + n + n + 1 n + n + 1. Remark: Note that the above proof is a direct proof and does not use the principle of mathematical induction. The only induction proof that I found myself is more complicated and also uses the formula for the sum of cubes: n n k n 4. Proof by Induction. The base step is established by the initial observations, so let us now assume that the identity is correct for all nonnegative integers up to n 1 and show that it is also true for n. Like in the direct proof, we start with * but now split up the second sum: n 1 k k + +1 k + k +1 k + + n 1 kn +1 k + n n kn 1 +1 Because each sum on the right-hand side satisfies our induction hypothesis, it follows that k n + n 1 + n 1 + n n 0 + n 1 n 1 k + n n + n n 1 n + n 4 n4 + n clearly n k n n +1 n4 +n is much faster so that the proof continues as in **. Additional Exercises for Chapter 6 6. Exercise 6.8 : In class or by Result 6.5 in the book, we have shown that n a Use this formula to determine a new formula for nn + 1n n b Use that formula and a to determine a new formula for n 1 ** k.

4 Math Intro to Abstract Math Homework 6, UC Denver, Spring 01 Solutions 4 c Use a and b to determine a new formula for n d Use mathematical induction to verify the formulas in b and c. Solution: For a, we can write n k 4 n k n. For b, then we find n k 1 n k n k n4 6 n nn 1 after some algebra. For c, we initially distinguish the two cases in which n is even or odd: n 1 k+1 k m k 1 m m+1 k 1 m k mm 1m+1 mm+1m+1 mm + 1 if n m for m N; k m+1m+1m+ mm+1m+1 m + 1m + 1 if n m + 1 for m N. If we now substitute m n/ and m + 1 n + 1 for the former, and m + 1 n + 1/ and m+1 n for the latter, we obtain that n 1k+1 k 1 n in both cases. For d, we first successfully verify the above formulas for small values of n and then compute n k k + 4n + 1 nn + 1n + 1 1n n + 1 n + n + 6n + 6 n + 1n + 7n + 6 n + 1n + n + n + 1 n n n k 1 k 1 + n + 1 nn 1n + 1 n n + 1 n n + 6n + n + 1n + 5n + n + 1n + 1n + n + 1 n n n 1 k+1 k 1 k+1 k 1 n+ n + n + 1 n + n + 1 n+ n + 1 nn + 1 n+ n n+ n + 1n + n n+ n + n + n Exercise 6.40: Prove that 4 n > n Solution: There are many possible ways to prove this result; here is a short induction proof.

5 Math Intro to Abstract Math Homework 6, UC Denver, Spring 01 Solutions 5 1. Base Step: The inequality 4 n > n holds for n 1,, because 4 > 1, 16 > 8, 64 > 7.. Induction Hypothesis: Now assume that 4 n > n for n 1,,..., k, where k.. Inductive Step: We need to show that 4 k+1 > k + 1 and use the observation that we can equivalently write 4 k k 4 k +4 k +4 k +4 k > k+1 k +k +k+1 now we should have an immediate idea on what to do. Namely, by our induction hypothesis and because k, we know that 4 k > k kk k kk > k > 1 and thus 4 k k 4 k + 4 k + 4 k > k + k + k + 1 k Exercise 6.50: Evaluate the proposed proof for the following result. Result. For every positive integer n, n 1 n. Proof. We proceed by induction. Since 1 1 1, the formula holds for n 1. Assume that k 1 k for a positive integer k. We prove that k + 1 k + 1. Observe that k + 1 k k 1 + k + 1 k + 1 k + k + 1 k + 1 k + 1 k + 1. Solution: The first four sentences of this proof are perfect: the reader is informed about the type of proof induction, the induction is properly anchored at n 1, the weak induction hypothesis is stated correctly, and the reader is alerted about the subsequent inductive step. However, although the correct idea of the following computation is apparent, its presentation is not acceptable. In particular, the fifth sentence is logically false: from the stated assumptions, it cannot be observed that k + 1 k + 1 ; otherwise the proof would be trivial. To correct the proof, the fifth sentence including the computation should be changed as follows: Observe that k k 1 + k + 1 k + k + 1 k + 1. Please let me know if you have any questions, comments, corrections, or remarks. Important Reminder Regarding Spring 01 Drop Dates For non-clas students, the last day to drop or withdraw from this course without a petition and special approval from the academic dean is April, 01 at 5 PM. After this date, a dean s signature is required. For CLAS students, the last day to drop or withdraw from this course with signatures from the faculty and dean but without a full petition is April 16, 01 at 5 PM. After this date, all schedule changes require a full petition. Please talk to me in due time if you are not sure whether you should drop the course.