Discrete Math, Spring 2012 - Solutions to Problems V

1. Suppose we have statements P1,P2,P3,..., one for each natural number. In other words, we have the collection or set of statements {Pn | n ∈ N}. a. Suppose further that if Pn is false for some n then also Pm is false for some m less than n. Does it follow that Pn is false, for all n?

b. Suppose instead that, for any n, if Pm is true for all natural numbers m less than n, then Pn is true. Does it follow that Pn is true, for all n?

a. Consider S = {n ∈ N | Pn is false}. By hypothesis, S has no least element. By the well-ordering principle, it follows that S is empty. Hence Pn is true, for all n. So it certainly does not follow that Pn is false, for all n.

b. We use strong induction to show that Pn is true, for all n. Thus we have to show: i. P1 is true; ii. for n > 1, if Pm is true for all m < n, then Pn is true. Condition ii holds by hypothesis, so we only have to prove i. The case n = 1 of our hypothesis says that if Pm is true for all m < 1 then P1 is true. Well there are no natural numbers less than one, and so Pm is true for all m < 1. (To put it another way, there is no m < 1 for which Pm is false as there are no natural numbers m < 1 at all – the condition is vacuously satisfied.) Thus P1 is true. Hence, by strong induction, Pn is true, for all n.

2. Show that the sum of the first n odd natural numbers is n2, for any natural number n. Pn 2 We need to show that i=1(2i − 1) = n , for any natural number n. Writing P (n) for this identity, we’ll prove by induction that P (n) holds for all n. It’s clear that P (1) holds: 2 Pk 2 2 · 1 − 1 = 1 . Suppose P (k) holds for a given natural number k, i.e., i=1(2i − 1) = k . Then k+1 k X X (2i − 1) = (2i − 1) + (2k + 1) i=1 i=1 = k2 + 2k + 1 (using P (k)) = (k + 1)2. Pn 2 Thus P (k) implies P (k + 1) for any natural number k. By induction, i=1(2i − 1) = n , for all natural numbers n.

Pn i+1 2 n+1 Pn 3. Show that i=1(−1) i = (−1) i=1 i, for all n ∈ N. We prove this by induction using n X i = n(n + 1)/2 (∗) i=1 which we proved in class. The base case (−1)1+112 = (−1)1+11 is clear. Assume then that the identity holds for a natural number k. We need to show it also holds for k + 1. We have k+1 k X X (−1)i+1i2 = (−1)i+1i2 + (−1)k+2(k + 1)2. i=1 i=1 2

By our inductive hypothesis, this is k X (−1)k+1 i + (−1)k+2(k + 1)2 = (−1)k+1k(k + 1)/2 + (−1)k+2(k2 + 2k + 1), i=1 using (∗). We rewrite the right side as (−1)k+2(k2 + 2k + 1 − k2/2 − k/2) = (−1)k+2(k2/2 + 3k/2 + 1) = (−1)k+2(k + 1)(k + 2)/2 k+1 X = (−1)k+2 i, i=1 using (∗) once more. Thus the identity holds for k + 1. By induction, it therefore holds for all natural numbers n.

4. Consider the Fibonacci sequence 1, 1, 2, 3, 5, 8,..., given by

f1 = 1,

f2 = 1,

fn = fn−1 + fn−2, for n > 2.

a. Show that f3n is even, for all natural numbers n.

b. Show that f3n−1 and f3n−2 are both odd, for all natural numbers n.

a. We show by induction that f3n is even, for all n. Clearly, f3 = 2 is even. Assume that f3k is even for a natural number k. Then

f3(k+1) = f3k+3

= f3k+2 + f3k+1

= (f3k+1 + f3k) + f3k+1

= 2f3k+1 + f3k,

so that f3(k+1) is also even. By induction, we conclude that f3n is even, for all n.

b. We show by induction that f3n−1 is odd, for all n. (The proof that f3n−2 is odd, for all n, is effectively identical.) First, it’s obvious that f3·1−1 = f2 is odd. Next, suppose f3k−1 is odd, for a natural number k. Then

f3(k+1)−1 = f3k+2

= f3k+1 + f3k

= (f3k + f3k−1) + f3k

= 2f3k + f3k−1,

so that f3(k+1)−1 is also odd. Hence, by induction, f3n−1 is odd, for all n.

5. Consider 1 2 n + + ··· + . 2! 3! (n + 1)! For n = 1, 2, 3, this is 1/2, 5/6, 23/24 respectively. Find the general statement suggested by these special cases and prove it. 3

The general statement is 1 2 n (n + 1)! − 1 + + ··· + = . 2! 3! (n + 1)! (n + 1)! (Note (n + 1)! = 2, 6, 24 for n = 1, 2, 3 respectively.) We’ll prove the general statement using induction. The base case n = 1 is clear: 1 2! − 1 = . 2! 2! Now suppose the statement holds for a natural number k. Then 1 2 k k + 1 (k + 1)! − 1 k + 1 + + ··· + + = + 2! 3! (k + 1)! (k + 2)! (k + 1)! (k + 2)! (k + 2)((k + 1)! − 1) + (k + 1) = (k + 2)! (k + 2)! − 1 = , (k + 2)! and so the statement also holds for k + 1. Thus, by induction, the statement holds for all natural numbers n.

6. Consider 1! 1 + 2! 2 + ··· + n! n. For n = 1, 2, 3, 4, this equals 1, 5, 23, 119 respectively. Find the pattern suggested by these special cases and show it holds in all cases. The general statement is 1! 1 + 2! 2 + ··· + n! n = (n + 1)! − 1 which we now prove by induction. The base case 1!1 = 2! − 1 is clearly true. For the inductive step, we assume 1! 1 + 2! 2 + ··· + k! k = (k + 1)! − 1, for some k. Then 1! 1 + 2! 2 + ··· + k! k + (k + 1)! (k + 1) = (k + 1)! − 1 + (k + 1)! (k + 1) = (k + 2) (k + 1)! = 1 = (k + 2)! − 1, and so the statement also holds for k + 1. By induction, the statement holds for all natural numbers n.

7. Prove the inequality 1 1 1 √ √ + √ + ··· + √ > n, 1 2 n for any n > 1. √ √ Using n = n/ n, we rewrite the inequality as 1 1 1 n √ + √ + ··· + √ > √ , for n > 1, 1 2 n n or equivalently √ √ √ n n n √ + √ + ··· + √ > n, for n > 1 1 2 n 4

Now each term on the left hand side is ≥ 1 and the first is surely > 1 (since n > 1) and thus the sum of the n terms must be > n. We can also prove the inequality by induction. The base case n = 2 is the statement 1 √ 1 + √ > 2. 2 √ Multiplying both sides by 2, we see that this holds if and only if √ 2 + 1 > 2 which is true and so we’ve established the base case. Suppose now that 1 1 1 √ √ + √ + ··· + √ > k, 1 2 k for some natural number k ≥ 2. Then 1 1 1 1 √ 1 √ + √ + ··· + √ + √ > k + √ . 1 2 k k + 1 k + 1 It suffices therefore to show that √ 1 √ k + √ > k + 1, for k ≥ 2. k + 1 √ Multiplying both sides by k + 1, we see that this is equivalent to √ √ k + 1 k + 1 > k + 1, for k ≥ 2. √ √ But k + 1 > k, for k > 0, so certainly √ √ √ √ k + 1 k + 1 > k k + 1 = k + 1, for k ≥ 2. This establishes the inductive step and hence the inequality holds for all n ≥ 2.

8. Suppose we have a collection of real numbers {am,n | m, n ∈ N} (that is, we’re given a real number for each ordered pair of natural numbers) that satisfies a. a1,1 = 0, b. if am,n = 0, for some m and n, then am,n+1 = 0, c. if am,n = 0, for some m and all n, then am+1,1 = 0. Show that am,n = 0, for all m and n.

Here’s a way of thinking about the given conditions. Imagine the numbers am,n listed in an array with infinitely many rows and columns:

a1,1 a1,2 ...... a1,n ...... am,1 am,2 ...... am,n ... am+1,1 am+1,2 ...... am+1,n ...... Condition a says the (1, 1) entry is zero; b says that if the n-th entry in row m is zero then the n + 1-st entry in row m is also zero; c says that if all the entries in a row are zero then the first entry in the next row is also zero. 5

Let P (m) be the statement

am,n = 0, for all n ∈ N. In other words, P (m) says that each entry in the m-th row is zero. We’ll show by induction that P (m) holds for all m, so that am,n = 0 for all m and n. We first prove P (1) using induction (a proof by induction inside a proof by induction). We have a1,1 = 0. If a1,l = 0, then also a1,l+1 = 0 (by b with m = 1, n = l). Hence (by induction) a1,n = 0, for all n.

Now suppose P (k) holds, i.e., ak,n = 0, for all n. Then c gives ak+1,1 = 0. Further, by b, if ak+1,l = 0, then ak+1,l+1 = 0. Hence (by induction) ak+1,n = 0, for all n. That is, P (k + 1) holds. By induction, P (n) holds, for all n.