Notes: Chapter 2 Section 2.2: Proof by Induction


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1 Notes: Chapter 2 Section 2.2: Proof by Induction Basic Induction. To prove: n, a W, n a, S n. (1) Prove the base case  S a. (2) Let k a and prove that S k S k+1 Example 1. n N, n i = n(n+1) 2. Example 2. n N, n i2 = n(n+1)(2n+1) 6. Example 3. S n : n! > 3 n, whenever n 7. Proof. Base case, S 7 : LHS = 7! = 5040 and RHS = 3 7 = 2187, so LHS > RHS. (Notice that 6! = 720 < 729 = 3 6.) Assume k 7 and S k is true. Then we have that k! > 3 k. Notice k + 1 > 3. So multiplying inequality, k! > 3 k, on the left by k + 1 and on the right by 3 gives us (k + 1)k! > 3 3 k which implies that (k + 1)! > 3 k+1. In the last example, we used the property of real numbers that x > y and a > b implies that x a > y b. Strong induction. To prove: n, a W, n a, S n. (1) Prove the base case  S a. (2) Let k a and prove that S a, S a+1,..., S k S k+1 Sometimes more that one base case is needed. Special care must be taken when using strong induction to assure that if S t is used to prove S k+1 that a t k. For example, to prove S k+1 if S k 1 is used in the argument, then a second base case is needed since when k + 1 = a + 1 then k 1 = a 1 and S a 1 might not be true. Example 4. 3 and 5 cent stamps are used for postage. Prove that any amount n of postage can be computed using 3 and 5 cent stamps for when n 8. Let S n be the statement that n cents can be computed using 3 and 5 cent stamps. The base cases are: S 8 is true since we can use one 3 and one 5 cent stamp. S 9 is true since we can use three, 3 cent stamps. S 10 is true since we can use two 5 cent stamps. We assume k 10 and S 8,..., S k is true. Then to show S k+1, we see that k = k 2 8. We know that S k 2 is true and so we add a 3 cent stamp to the way that k 2 cents is computed. 1
2 2 Example 5. Find the error in this argument: 3 and 5 cent stamps are used for postage. Prove that any amount n of postage can be computed using 3 and 5 cent stamps for when n 5. Let S n be the statement that n cents can be computed using 3 and 5 cent stamps. The base case is: S 5 is true since we can use one 5 cent stamp. We assume k 5 and S 5,..., S k is true. Then to show S k+1, we use S k 2 so we add a 3 cent stamp to the way that k 2 cents is computed. We now can prove the generalizations of Theorems and Theorem 1. (Generalization of 1.2.1) (The Addition principle of counting.) Suppose k 2 and P = {B 1, B 2,..., B k } is a partition of a finite set S. Then, S = B 1 + B B k. Proof. The base case is for k = 2. We know from Theorem that the base case is true. Now assume that k 2 and S k is true. Then if P = {B 1, B 2,..., B k+1 } is a partition of S, then P = {B 1, B 2,..., B k } is a partition of S and S = B 1 + B B k by the induction assumption. But {S, B k+1 } is a partition of S and again by Theorem 1.2.1, S = S + B k+1, so we have: S = B 1 + B B k + B k+1. Theorem 2. (Generalization of 1.2.1) (The multiplication principle of counting.) Let S 1, S 2,..., S n be nonemty finite sets. Then, S 1 S 2 S n = S 1 S 2 S n. Proof. The base case is for k = 2. We know from Theorem that the base case is true. Now assume that k 2 and the statement is true for k sets, S 1,..., S k. We can view the k + 1tuples in A = S 1 S 2 S k S k+1 as the 2tuples B = (S 1 S 2 S k ) S k+1. The only difference between the sets A and B that in A we have vectors of length k +1 and in B, we have vectors of length 2, but the first coordinate is a vector of length k. Certainly, A = B. By induction we assume that S 1 S 2 S k = S 1 S 2 S k. Thus, B = S 1 S 2 S k S k+1 = ( S 1 S 2 S k ) S k+1. Sequences. First, we have some basic facts about real numbers: Theorem 3. For real numbers, a 1, a 2, b 1 b 2 and c, (1) c(a 1 + a 2 ) = ca 1 + ca 2 (2) If a 1 b 1 and a 2 b 2, then a 1 + a 2 b 1 + b 2.
3 (3) If 0 a 1 < b 1 and 0 a 2 < b 2, then a 1 a 2 b 1 b 2. A sequence is a function whose domain is N. Theorem 4. Let {a n } n=1 and {b n } n=1 be two sequences of real numbers. Then, (1) c n a i = n ca i (2) If a i b i, i [n], then n a i n b i. (3) If 0 a i b i, i [n], then n a i n b i. Example 6. Plus is a binary operation on the real numbers. By associativity we know that any grouping of the sum of 3 numbers gives the same result. Prove that for n 3, any grouping of the n numbers x 1, x 2,..., x n gives the same result. We will prove that every grouping is the same as the natural grouping, 3 ( ( ( (x 1 + x 2 ) + x 3 ) + x 4 ) + x n 1 ) + x n where the plus signs are used in order from left to right. So we may write n x i and group the terms by parenthesis any way we like. Proof. For n = 4 here are all of the possible groupings. (x 1 + x 2 ) + (x 3 + x 4 ) (x 1 + (x 2 + x 3 )) + x 4 ((x 1 + x 2 ) + x 3 ) + x 4 x 1 + (x 2 + (x 3 + x 4 )) x 1 + ((x 2 + x 3 ) + x 4 ) Proof: The base case is for n = 3 which is true by associativity. We use strong induction here. Suppose that for k, where 3 k < n, any grouping of k numbers gives the same result, namely they are all equal to ( ( ( (x 1 + x 2 ) + x 3 ) + x 4 ) + x k 1 ) + x k Thus we can call the sum k x i and group the terms any way we like. Now assume n > 3 We take an arbitrary grouping of x 1, x 2,..., x n. In any grouping, we perform the addition of two numbers at a time. So we can think of the plus signs as being performed in a certain order. Take the one that is to be performed last. On either side of that plus sign, there are two groupings. One is of x 1, x 2,..., x j and the other is of x j+1,..., x n.
4 4 The number of terms in each grouping is less than n. If the number of terms is less than 3, we know that there is no order of operations to worry about. If the number of terms is at least 3, we use the strong induction assumption to conclude that both sides of the the last plus sign can be represented by a sum, with no parenthesis indicated. We have j ( x i ) + ( x i ). i=j+1 We can group together the sums anyway we want. So for the second sum we write: n i=j+1 x i = x j+1 + ( n x i), and for the first sum we assume it is grouped in the natural order. j j ( x i ) + ( x i ) = ( x i ) + (x j+1 + ( x i )) i=j+1 = (( j x i ) + x j+1 ) + ( j+1 = ( x i ) + ( x i ) We continue in this way breaking off the first term of the second sum and grouping it together with the first sum until there is one term left and the natural grouping is assumed on the first sum, that is the order of performing the additions is from left to right. At that point, addition is used to obtain a sum of the terms in the natural order. Here is the last step: x i ) x i ) + ( n 1 ( i=n n 1 x i ) = ( = x i ) + x n Example 7. (The Geometric Sum) x R, x 1, n N, n 1 x i = xn 1. x i
5 Proof. The base case is when n = 1. The LHS of the equality is LHS = 1 1 xi = x 0 = 1. The RHS of the equality is RHS = x1 1 = x 1 = 1. x 1 x 1 Assume that k 1 and that x R, x 1, Then (k+1) 1 k 1 x i = xk 1. x i = k k 1 x i = x i + x k = xk 1 + xk = xk 1 + ()x k = x xk 1 = xk+1 1 5
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