Fibonacci Numbers. Justin Stevens. Lecture 5. Justin Stevens Fibonacci Numbers (Lecture 5) 1 / 10

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1 Fibonacci Numbers Lecture 5 Justin Stevens Justin Stevens Fibonacci Numbers (Lecture 5) 1 / 10

2 Outline 1 Fibonacci Numbers Justin Stevens Fibonacci Numbers (Lecture 5) 2 / 10

3 Fibonacci Numbers The Fibonacci numbers show up in many unexpected situations and have many beautiful properties. We present a motivating example: Justin Stevens Fibonacci Numbers (Lecture 5) 3 / 10

4 Fibonacci Numbers The Fibonacci numbers show up in many unexpected situations and have many beautiful properties. We present a motivating example: Example. Consider a board of length n. How many ways are there to tile this board with squares (length 1) and dominos (length 2)? Justin Stevens Fibonacci Numbers (Lecture 5) 3 / 10

5 Fibonacci Numbers The Fibonacci numbers show up in many unexpected situations and have many beautiful properties. We present a motivating example: Example. Consider a board of length n. How many ways are there to tile this board with squares (length 1) and dominos (length 2)? Let the number of tilings of an n-board be f (n). Justin Stevens Fibonacci Numbers (Lecture 5) 3 / 10

6 Fibonacci Numbers The Fibonacci numbers show up in many unexpected situations and have many beautiful properties. We present a motivating example: Example. Consider a board of length n. How many ways are there to tile this board with squares (length 1) and dominos (length 2)? Let the number of tilings of an n-board be f (n). We calculate f (4): Table 1: The five tilings of a 4-board Hence f (4) = 5. For larger values of n, we need a different strategy. Justin Stevens Fibonacci Numbers (Lecture 5) 3 / 10

7 Fibonacci Numbers 2 Observe that if an n-board begins with a square, then we have to tile an n 1 board. However, if the n-board begins with a domino, then we have to tile an n 2 board. Justin Stevens Fibonacci Numbers (Lecture 5) 4 / 10

8 Fibonacci Numbers 2 Observe that if an n-board begins with a square, then we have to tile an n 1 board. However, if the n-board begins with a domino, then we have to tile an n 2 board. Therefore, f (n) = f (n 1) + f (n 2). Justin Stevens Fibonacci Numbers (Lecture 5) 4 / 10

9 Fibonacci Numbers 2 Observe that if an n-board begins with a square, then we have to tile an n 1 board. However, if the n-board begins with a domino, then we have to tile an n 2 board. Therefore, f (n) = f (n 1) + f (n 2). We see that f (1) = 1 and f (2) = 2, so we can compute: f (3) = f (2) + f (1) = = 3 f (4) = f (3) + f (2) = = 5 f (5) = f (4) + f (3) = = 8. Justin Stevens Fibonacci Numbers (Lecture 5) 4 / 10

10 Fibonacci Numbers 2 Observe that if an n-board begins with a square, then we have to tile an n 1 board. However, if the n-board begins with a domino, then we have to tile an n 2 board. Therefore, f (n) = f (n 1) + f (n 2). We see that f (1) = 1 and f (2) = 2, so we can compute: f (3) = f (2) + f (1) = = 3 f (4) = f (3) + f (2) = = 5 f (5) = f (4) + f (3) = = 8. The only difference between this and the Fibonacci numbers are that the latter begins with an extra 1. Therefore, f (n) = F n+1. Using this n-tiling, we can prove several identities regarding Fibonacci numbers. Justin Stevens Fibonacci Numbers (Lecture 5) 4 / 10

11 Property Example. Show that f (m + n) = f (m)f (n) + f (m 1)f (n 1). Justin Stevens Fibonacci Numbers (Lecture 5) 5 / 10

12 Property Example. Show that f (m + n) = f (m)f (n) + f (m 1)f (n 1). The left-hand side is simply the number of tilings of an (m + n)-board. Justin Stevens Fibonacci Numbers (Lecture 5) 5 / 10

13 Property Example. Show that f (m + n) = f (m)f (n) + f (m 1)f (n 1). The left-hand side is simply the number of tilings of an (m + n)-board. For the right-hand side, we have two cases: Justin Stevens Fibonacci Numbers (Lecture 5) 5 / 10

14 Property Example. Show that f (m + n) = f (m)f (n) + f (m 1)f (n 1). The left-hand side is simply the number of tilings of an (m + n)-board. For the right-hand side, we have two cases: If there is no domino at square m, we have f (m)f (n) tilings: 1 2 m 1 m m + 1 m + 2 m + n Justin Stevens Fibonacci Numbers (Lecture 5) 5 / 10

15 Sum Property If there is a domino at square m, we have f (m 1)f (n 1) tilings: 1 2 m 1 m m + 1 m + 2 m + 3 m + n Hence, f (m + n) = f (m)f (n) + f (m 1)f (n 1) as desired. Justin Stevens Fibonacci Numbers (Lecture 5) 6 / 10

16 Sum Property If there is a domino at square m, we have f (m 1)f (n 1) tilings: 1 2 m 1 m m + 1 m + 2 m + 3 m + n Hence, f (m + n) = f (m)f (n) + f (m 1)f (n 1) as desired. Substituting m = a and n = b 1 gives the Fibonacci identity F a+b = F a+1 F b + F a F b 1. Justin Stevens Fibonacci Numbers (Lecture 5) 6 / 10

17 Fibonacci Divisibility Example. Prove that F m F mq for all natural q. Justin Stevens Fibonacci Numbers (Lecture 5) 7 / 10

18 Fibonacci Divisibility Example. Prove that F m F mq for all natural q. We use induction. For q = 1, F m F m. For q = 2, F 2m = F m+1 F m + F m F m 1. Justin Stevens Fibonacci Numbers (Lecture 5) 7 / 10

19 Fibonacci Divisibility Example. Prove that F m F mq for all natural q. We use induction. For q = 1, F m F m. For q = 2, F 2m = F m+1 F m + F m F m 1. Suppose the statement is true for q = k, implying that F m F mk. Justin Stevens Fibonacci Numbers (Lecture 5) 7 / 10

20 Fibonacci Divisibility Example. Prove that F m F mq for all natural q. We use induction. For q = 1, F m F m. For q = 2, F 2m = F m+1 F m + F m F m 1. Suppose the statement is true for q = k, implying that F m F mk. We show it holds for q = k + 1. From the identity with a = mk and b = m: F mk+m = F mk+1 F m + F mk F m 1. Justin Stevens Fibonacci Numbers (Lecture 5) 7 / 10

21 Fibonacci Divisibility Example. Prove that F m F mq for all natural q. We use induction. For q = 1, F m F m. For q = 2, F 2m = F m+1 F m + F m F m 1. Suppose the statement is true for q = k, implying that F m F mk. We show it holds for q = k + 1. From the identity with a = mk and b = m: F mk+m = F mk+1 F m + F mk F m 1. Since F m F mk (hypothesis), F m F mk+m by the linear combination theorem. Hence, the statement is proven for q = k + 1. Justin Stevens Fibonacci Numbers (Lecture 5) 7 / 10

22 Consecutive Fibonacci Numbers Example. Show that consecutive Fibonacci numbers are relatively prime. Justin Stevens Fibonacci Numbers (Lecture 5) 8 / 10

23 Consecutive Fibonacci Numbers Example. Show that consecutive Fibonacci numbers are relatively prime. Proof. gcd(f n+2, F n+1 ) = gcd(f n+1, F n ) = gcd(f n, F n 1 ) = = gcd(f 2, F 1 ) = 1. Justin Stevens Fibonacci Numbers (Lecture 5) 8 / 10

24 Fibonacci GCD Example. Show that gcd(f m, F n ) = F gcd(m,n). Write m = nq + r using the division algorithm. Using the Fibonacci identity, Justin Stevens Fibonacci Numbers (Lecture 5) 9 / 10

25 Fibonacci GCD Example. Show that gcd(f m, F n ) = F gcd(m,n). Write m = nq + r using the division algorithm. Using the Fibonacci identity, F m = F nq+r = F nq+1 F r + F nq F r 1. Justin Stevens Fibonacci Numbers (Lecture 5) 9 / 10

26 Fibonacci GCD Example. Show that gcd(f m, F n ) = F gcd(m,n). Write m = nq + r using the division algorithm. Using the Fibonacci identity, F m = F nq+r = F nq+1 F r + F nq F r 1. Since F n F nq, we can subtract multiples of F n using Euclidean algorithm: gcd(f m, F n ) = gcd(f nq+1 F r + F nq F r 1, F n ) = gcd(f nq+1 F r, F n ). Justin Stevens Fibonacci Numbers (Lecture 5) 9 / 10

27 Fibonacci GCD Example. Show that gcd(f m, F n ) = F gcd(m,n). Write m = nq + r using the division algorithm. Using the Fibonacci identity, F m = F nq+r = F nq+1 F r + F nq F r 1. Since F n F nq, we can subtract multiples of F n using Euclidean algorithm: gcd(f m, F n ) = gcd(f nq+1 F r + F nq F r 1, F n ) = gcd(f nq+1 F r, F n ). Finally, gcd(f nq+1, F n ) = 1 since consecutive Fibonacci numbers are relatively prime: gcd(f m, F n ) = gcd(f r, F n ). Justin Stevens Fibonacci Numbers (Lecture 5) 9 / 10

28 Fibonacci GCD Example. Show that gcd(f m, F n ) = F gcd(m,n). Write m = nq + r using the division algorithm. Using the Fibonacci identity, F m = F nq+r = F nq+1 F r + F nq F r 1. Since F n F nq, we can subtract multiples of F n using Euclidean algorithm: gcd(f m, F n ) = gcd(f nq+1 F r + F nq F r 1, F n ) = gcd(f nq+1 F r, F n ). Finally, gcd(f nq+1, F n ) = 1 since consecutive Fibonacci numbers are relatively prime: gcd(f m, F n ) = gcd(f r, F n ). For example, if m = 182 and n = 65, gcd(182, 65) = 13 and gcd(f 182, F 65 ) = gcd(f 65, F 52 ) = gcd(f 52, F 13 ) = F 13. Justin Stevens Fibonacci Numbers (Lecture 5) 9 / 10

29 Fibonacci GCD Example. Show that gcd(f m, F n ) = F gcd(m,n). Write m = nq + r using the division algorithm. Using the Fibonacci identity, F m = F nq+r = F nq+1 F r + F nq F r 1. Since F n F nq, we can subtract multiples of F n using Euclidean algorithm: gcd(f m, F n ) = gcd(f nq+1 F r + F nq F r 1, F n ) = gcd(f nq+1 F r, F n ). Finally, gcd(f nq+1, F n ) = 1 since consecutive Fibonacci numbers are relatively prime: gcd(f m, F n ) = gcd(f r, F n ). For example, if m = 182 and n = 65, gcd(182, 65) = 13 and gcd(f 182, F 65 ) = gcd(f 65, F 52 ) = gcd(f 52, F 13 ) = F 13. The conclusion is equivalent to gcd(f m, F n ) = F gcd(m,n). Justin Stevens Fibonacci Numbers (Lecture 5) 9 / 10

30 Other Properties The below properties can all be proved using the tiling method: F 1 + F 2 + F F n = F n+2 1. F 1 + F 3 + F F 2n 1 = F 2n. F1 2 + F F F n 2 = F n F n+1. F n+1 = ( n) ( 0 + n 1 ) ( 1 + n 2 ) 2 +. Try them out yourself! Justin Stevens Fibonacci Numbers (Lecture 5) 10 / 10