Problem Set 5 Solutions

Transcription

1 Problem Set 5 Solutions Section 4.. Use mathematical induction to prove each of the following: a) For each natural number n with n, n > + n. Let P n) be the statement n > + n. The base case, P ), is true because 9 > +. We assume that P k) is true, which states that k > + k for some natural number k. We will show P k + ) is true. Now, k+ k > + k ) + k > + k + k+ and hence, k+ > + k+. That is, if P k) is true, then P k + ) is true. Therefore, for all natural numbers n, n > + n. b) For each natural number b with n 6, n > n + ). Let P n) be the statement n > n + ). The base case, P 6), is true because 6 64 > ). Assume P k) is true, which says that k > k + ). We will show that P k + ) is true as well. Consider Now, for k 6, k > k +, so we have k+ k k + ) k + 4k + k + 4k + > k + + 4k + k + 4k + 4 k + ) as desired. So, P k + ) is true wheneverp k) is true. Therefore, by mathematical induction, n > n + ) for all natural numbers n 6. c) For each natural number n with n, + n) n < n. Let P n) be the statement that + n) < n. The base case, P ), is true because + ) 4 ) 64 7 <. If P k) is true, then + ) k k < k. We will show P k + ) is true as well. Consider + ) k+ < k + ) k k k +

2 Notice that k+ < k for k, and so + ) k+ < + ) k+ < k + k + k 4. Thus, if P k) is true, then P k + ) is true. Therefore, for all natural numbers n, + n) n < n. a) Verify that 4) 4 and that ) ) b) Verify that 4) ) ) ) ) 9 ) ) and that 4 ) ) ) ) ) ) ) ) ) ) 9 ) ) ) ) 6 5) ) ) c) For n N with n, make a conjecture for the formula for the product ) ) ) ) n n k ) k n + n d) Based on your work in Parts 4a) and 4b), state a proposition and then use the Extended Principle of Mathematical Induction to prove your proposition. Let P n) be the proposition that for natural numbers n, n i ) i First, for n, we have that P ) is true as i ) i n + n ) 4 4 +

3 Now, suppose for all j k that P k) is true, where k is some natural number greater than. We will show P k + ) is true, which states j k+ j j ) Consider the following: k+ j ) k j ) j k + k + ) ) k + ) ) ) k + k k + ) ) k + k + ) ) k k + ) k + k k + k k + ) k + k + ) So, P k + ) is true whenever P k) is true. Thus, for natural numbers n, n k ) k n + n 9. Can each natural number greater than or equal to 6 be written as the sum of at least two natural numbers, each of which is a or a 5? Justify your conclusion. For example, 6 + +, , and Let P n) be the statement that n can be written as the sum of at least two natural numbers, each of which is a or a 5. For base cases, we have P 6) is true because P 7) is true because P 8) is true because P 9) is true because P 0) is true because Now let k N with k 0. Assume that P 6),..., P k) are true. Since k + k 4) + 5 and 6 k 4 < k, we see that P k 4) is true. So, k 4 is a sum of at least natural numbers, each of which is a or a 5. So, the above equation implies that k + is the sum of at least natural numbers, each of which is a or a 5, and so P k + ) is true. Therefore, for all natural numbers n 6, we can write n as the sum of at least two natural numbers, each of which is a or a 5.

4 0. Use mathematical induction to prove the following proposition: Let x be a real number with x > 0. Then for each natural number n with n, + x) n > + nx. Explain why the assumption x > 0 was used in the proof. Let x be a real number with x > 0. Let P n) be + x) n > + nx. First, P ) is true. For any choice of x > 0, we have + x) > + x x + x + > + x And, since x R +, this last line is always true. x > 0 Let k N with k and assume P k) is true. Then, + x) k > + kx Since x + 0, we see that + x) > 0 as well, so we can multiply both sides of the above inequality by + x). Then, + x) k+ > + kx) + x) + kx + x + kx > + k + )x Hence, if P k) is true, then P k + ) is true. So, for all n N with n, + x) n > + nx when x > 0. In the case where x 0, we have equality in the statement for any choice of n. So by restricting to x > 0, the statement will be true for any choice of x R + and any choice of n N where n.. Prove or disprove each of the following propositions: a) For each n N, nn + ) n n + This is a true statement. We will prove this with mathematical induction. Let n N. Let P n) be the statement P ) is true because +. Assume P k) is true, which says that nn + ) kk + ) n n + k k +

5 for some k N. We will show that P k + ) is true as well. Consider kk + ) + k + )k + ) [ + ] kk + ) k + )k + ) k k + + k + )k + ) kk + ) + k + )k + ) k + k + k + )k + ) k + ) k + )k + ) k + k + as desired. This shows that P k + ) is true whenever P k) is true. So, by mathematical induction, nn + ) n n + for all natural numbers n. b) For each natural number n with n, nn + ) n n + This is a true statement. We will prove this with mathematical induction. Let n N, where n, and let P n) be that Now, P ) is true because that for some k N with k, nn + ) n n + 4 ) kk + ) k k + We will show P k + ) is true as well. Consider. Next, assume P k) is true, which is to say kk + ) + k + )k + ) [ 4 + ] kk + ) k + )k + ) k k + + k + )k + )

6 k )k + ) + k + )k + ) k k + )k + ) k k + 6 as desired. This shows that P k + ) is true whenever P k) is true. So, by mathematical induction, nn + ) n n + for all natural numbers n. c) For each n N, nn + ) nn+)n+). This is a true statement. We will prove this with mathematical induction. Let P n) be that for natural numbers n, nn+) nn+)n+). P ) is true because +)+). Assume P k) is true, which says that kk + ) kk + )k + ) for some natural number n. We will show that P k + ) is true, which says that Now, k + )k + ) k + )k + )k + ) k + )k + ) [ kk + )] + k + )k + ) kk + )k + ) + k + )k + ) kk + )k + ) + k + )k + ) k + )k + )k + ) as desired. This shows that P k + ) is true whenever P k) is true. So, by mathematical induction, nn + )n + ) nn + ) for all n N.

7 Section 4.. Assume that f, f,..., f n,... are the Fibonacci numbers. Prove each of the following: a) For each n N, f 4n is a multiple of. Let P n) be f 4n is a multiple of. Since f 4, P ) is true. Now, if P k) is true, then there exists m N such that f 4k m. Then, f 4k+) f 4k+4 f 4k+ + f 4k+ f 4k+ + f 4k+ + f 4k+ + f 4k f 4k+ + f 4k + f 4k+ + f 4k f 4k+ + f 4k f 4k+ + m) f 4k+ + m) This proves that if P k) is true, then P k + ) is true. So, for all n N, divides f 4n. b) For each n N, f 5n is a multiple of 5. Let P n) be f 5n is divisible by 5 where n is any natural number. Since f 5 5, P ) is true. Now, if P k) is true, then there exists an integer m such that f 5k 5m. Then, f 5k+) 5 5k+5 f 5k+4 + f 5k+ f 5k+ + f 5k+ + f 5k+ + f 5k+ f 5k+ + f 5k+ + f 5k+ f 5k+ + f 5k+ + f 5k+ + f 5k ) + f 5k+ f 5k+ + 4f 5k+ + f 5k 5f 5k+ + f 5k 5f 5k+ + 5m) 5f 5k+ + m) This proves that P k + ) is true whenever P k) is true. So, for all n N, 5 divides f 5n. c) OMIT d) For each n N, f + f f n f n. Let n N and let P n) be the statement f + f f n f n. Since f f, P ) is true. For k N, if P k) is true, then f + f f k f k. Then f + f f k ) + f k+) f + f f k ) + f k+ f k + f k+ f k+ f k+) This proves that if P k) is true, then P k + ) is true. So, for all n N, f + f f n f n.

8 e) For each n N, f + f f n f n+. Let n N and let P n) be f + f f n f n+. P ) is true because f f. For k N, if P k) is true, then f + f f k f k+. Then, f + f f k ) + f k+) f + f f k ) + f k+ f k+ + f k+ f k+ f k+)+ This shows that P k +) is true whenever P k) is true. So, for n N, f +f f n f n+. f) For each n N, f + f f n f n f n+. For n N, let P n) be f + f f n f n f n+. P ) is true because f f f. For k N, if P k) is true, then f + f f k f kf k+. Then, f + f f k ) + f k+ f kf k+ + f k+ f k+ f k + f k+ ) f k+ f k+ This proves that if P k) is true, then P k + ) is true as well. So, for n N, f + f f n f n f n+. g) For each n N such that n 0 mod ), f n is an odd integer. Let P n) be f n+ and f n+ are both odd. Now, P ) is true because f 4 and f 5 5. For k N, if P k) is true, then f k+ and f k+ are both odd. Consider f k+)+ f k+4 f k+ + f k+ Now, by a proposition from the text, Proposition 4.), f k+ is even. And, f k+ is odd by our inductive hypothesis. Therefore, f k+)+ is odd. Also, f k+)+ f k+5 f k+4 + f k+ Again, by proposition 4., f k+ is even. And, we have just proven that f k+4 is odd. Therefore, f k+)+ is odd. This proves that if P k) is true, then P k + ) is true. So, by mathematical induction, if n is a natural number such that n 0 mod ), f n is an odd integer. 4. The quadratic formula can be used to show that α + 5 and β 5 are the two real number solutions to the quadratic equation x x 0. Notice that this implies α α+ and β β +. It may be surprising to find out that these two irrational numbers are closely related to the Fibonacci numbers

9 a) Verify that f α β α β and that f α β α β. f f + 5 ) ) b) Work with the relation f f + f and substitute the expressions for f and f from part a). Rewrite the expression as a single fraction and then in the numerator use α + α αα + ) and a similar equation involving β. Now prove that f α β α β f f + f + α + α) β + β) αα + ) ββ + ) αα ) ββ) α β c) Use induction to prove that for each natural number n, if α + 5 and β 5, then f n αn β n α β. Let P n) be f n αn β n α β. We have previously shown that P ) and P ) are true. Now, f k+ f k + f k αk β k + αk β k

10 αk + α k β k + β k ) ak α + ) β k β + ) ak α ) β k β ) αk+ β k+ This shows that P k + ) is true whenever P k) is true. So, for n N, f n αn β n α β, where α + 5 and β For the sequence a, a,..., a n,..., assume that a and that for each n N, a n+ a n + 5. a) Calculate a through a 6. a a a a + 5 a 4 a a 5 a a 6 a b) Make a conjecture for a formula for a n for each n N. Conjecture: For n N, a n + 5n ). c) Prove your conjecture in Exercise 9b) is correct. Let P n) be a n + 5n ) for n N. P ) is true because a + 5 ). Now, if P k) is true then a k + 5k ). Consider a k+ a k k ) k This shows that P k+) is true whenever P k) is true. So, for all n N, a n +5n ).. For each sequence a, a,..., a n,..., assume that a, a, and that for each n N, a n+ a n+ a n. a) Calculate a through a 6. a a a ) ) 7 a 4 a a 7) ) 5 a 5 a 4 a 5) 7) a 6 a 5 a 4 ) 5) 6 b) Make a conjecture for a formula for a n for each n N. Conjecture: For all natural numbers n, a n n.

11 c) Prove your conjecture in Exercise b) is correct. Let P n) be a n n for n N. P ) is true because a. If P k) is true, then a k k for some natural number k. We want to show that a k+ k+. Consider a k+ a k a k k ) k ) k k + k k k k+ This proves that P k + ) is true whenever P k) is true. So, for all natural numbers n, a n n. 6. For each sequence a, a,..., a n,..., assume that a and that for each natural number n, a n+ a n + n n! a) Compute n! for the first 0 natural numbers.!!! 6 4! 4 5! 0 6! 70 7! ! ! b) Compute a n for the first 0 natural numbers. a a +! a +! 6 a 4 6 +! 4 a ! 0 a ! 70 a ! 5040 a ! 400 a ! 6880 a ! 68800

12 c) Make a conjecture about a formula for a n in terms of n that does not involve a summation or a recursion. Conjecture: for all natural numbers n, a n n!. d) Prove your conjecture in Part c). Let P n) be the statement that a n n! for any natural number n. P is true because a. Now, if P k) is true, then a k k! for some natural number k. Then, a k+ a k + k k! k! + k k! k! + k) k + )! as desired. So, P k + ) is true whenever P k) is true. Therefore, by mathematical induction, a n n! for all natural numbers n.