Running head: A GEOMETRIC INTRODUCTION 1


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1 Running head: A GEOMETRIC INTRODUCTION A Geometric Introduction to Mathematical Induction Problems using the Sums of Consecutive Natural Numbers, the Sums of Squares, and the Sums of Cubes of Natural Numbers Nicholls State University Math 58 Fall 0 Galina Landa
2 A GEOMETRIC INTRODUCTION Abstract The Principle of Mathematical Induction (PMI) is a powerful method of proof. The illustrated examples often come from algebra rather than geometry. This paper explores mathematical induction in geometry. The traditional algebraic examples of the PMI involve proving formulas for the sums of consecutive natural numbers, the sums of squares of natural numbers, the sums of cubes, etc. However, it is not always clear how one would arrive at these formulas or in what circumstances they would be useful. This research paper presents several geometry problems to give context to such abstract formulas. It also demonstrates how one can use the PMI to prove important theorems in geometry. The problems considered in this research involve finding the number of lines through n points in a plane; finding the number of diagonals in convex polygons with n sides; and finding the number of squares of all sizes enclosed by a twodimensional checkerboard or the number of cubes of all sizes enclosed by a threedimensional checkerboard with each side length equal to n units.
3 A GEOMETRIC INTRODUCTION 3 Introduction A powerful method of proof, the Principle of Mathematical Induction (PMI) allows us to verify a particular property for all consecutive integers greater than some smallest one. It works like this: we start by checking if our conjecture is true for the first few initial values. Then, after assuming that the conjecture is true for the given element, we check whether we can show that it is also true for the next element. If we manage to complete both steps successfully, then the property is true for all elements. Mathematical induction is used to prove various sum formulas. However, it is not always clear how one would arrive at these formulas, why they would be legitimate candidates for verification, and in what circumstances they would be useful. In this paper, we will discover these formulas in the contexts of several geometry problems and then prove them using mathematical induction. Hopefully, by combining inductive investigation of geometry concepts with inductive proofs of the resulting sums, we will gain a new perspective on the meaning of these formulas and a better understanding of the PMI. The process of mathematical discovery is quite exciting, but a proper mathematical investigation can be time consuming. Thus, in this paper, we will limit our discussion to the following three sums: the sum of consecutive natural numbers the sum of squares of consecutive natural numbers the sum of cubes of consecutive natural numbers These three sums will give names to the three major sections of this paper, which constitute the body of related and original research. This body of research is called The Sums.
4 A GEOMETRIC INTRODUCTION The Sums The Sum of Consecutive Natural Numbers: 3... n We will consider this first sum in the context of the following geometry problem. Problem : Lines and Points Solution: Find the number of lines determined by a given number of points in a plane, no three of which are collinear (Wiscamb, p. 0). We are interested in the finite number of lines, so we will disregard a case of infinitely many lines through a single point. We will start with natural n>. Let us consider the first few cases for the number of points n=, n=3, n=, n=5, and n= (see Table ). Notice that while arbitrary points determine a single line, 3 noncollinear points determine 3 lines, points determine lines, 5 points determine 0 lines, and points determine 5 lines. Table. Lines and Points n, l n, ln 3... ( n ) ln pn pn points lines = ( ) 3 3 3=+ ( 3 3 ) (The table continues on the next page)
5 A GEOMETRIC INTRODUCTION 5 =++3 ( 3) 5 0 0=++3+ (5 ) 0 5 5= (5) 5 Note: Diagrams constructed with The Geometer's Sketchpad software. If we look for patterns in the table, we may notice two interesting details applicable to each case.. The number of lines determined by n points is equal to the sum of natural numbers from to n: ln 3... ( n ). The number of lines at each step is equal to half the product of the number of points at the current step and the number of points at the preceding step: ln ( pn pn )
6 A GEOMETRIC INTRODUCTION Based on these observations, we can make the following conjecture. Conjecture : For natural numbers greater than, 3... ( n ) n( n ) Proof: We can prove this conjecture by the PMI. Let P(n) be 3... ( n ) n( n ), n>. P() is true: ( ). Suppose P(n) is true for n>. 3. We will show that P(n) is true. (We want to show that, nn ( ) 3... n ) By the induction hypothesis, 3... ( n ) n( n ) Let us add n to both sides: 3... ( n ) n n( n ) n Pn ( ) Pn ( )
7 A GEOMETRIC INTRODUCTION 7 n( n ) n nn ( ) nn ( ) Thus, P(n) is true. Then by the PMI, P(n) is true for all natural numbers n>. Therefore, we have proved the formula: 3... ( n ) n( n ), for natural n> From now on, we can use it to find the number of lines determined by a given number of points in a plane, no three of which are collinear. For this particular task, it was convenient to use the formula for the sum of natural numbers from to n. We can also rewrite this formula for the sum of natural numbers from to n (like we did in our proof above for P(n)). Since we have already proven this result, we no longer call it a conjecture, but a theorem. Theorem (The Sum of Consecutive Natural Numbers): For natural numbers greater or equal to, nn ( ) 3... n,n We can now use this result to discover some other mathematical relationships. For example, Avital and Hansen (97) posed the following question.
8 A GEOMETRIC INTRODUCTION 8 Problem a: Diagonals of ngon What is the number of diagonals one can draw in a convex polygon of n vertices? (Avital & Hansen, p. 0) Solution: Clearly, a triangle has no diagonals; we will consider n > 3 (see Table a.) Table a. Diagonals of ngon Polygon n, vertices d, diagonals Triangle 3 0 Quadrilateral = Pentagon 5 5=+3 Hexagon 9=+3+ Heptagon 7 =+3++5 Note: Diagrams constructed with The Geometer's Sketchpad software.
9 A GEOMETRIC INTRODUCTION 9 After considering polygons with, 5,, and 7 sides, we find that the number of diagonals is, 5, 9, and, respectively. The last four numbers can be written as, +3, +3+, and Thus, the number of diagonals in each of these cases is the sum of natural numbers from to n, where n is the number of vertices. Let us denote the number of diagonals d; then for an ngon, d 3... ( n ) This resembles the sum from Theorem, minus, minus (n), and minus n. Then we can use the result from Theorem to find the expression for d: Thus, d 3... ( n ) 3... n ( n ) n Theorem Theorem nn ( ) n n n( n ) n nn ( ) nn ( 3) nn ( 3) 3... ( n ), n 3 From now on, we can use this result to find the number of diagonals in a convex ngon. We have demonstrated that this result follows from Theorem, so we will call it Corollary. Corollary : For natural numbers greater than 3, nn ( 3) 3... ( n ),n
10 A GEOMETRIC INTRODUCTION 0 This result can be verified by the PMI for n>3; however, we will omit the proof here because it is very similar to the proof of Theorem. The Sum of Squares: 3... n Wiscamb (970) suggested introducing the formula for the sum of squares in connection with the following problem. Problem : Checkerboard in D Solution: Given an nxn checkerboard with n small squares each measuring square unit, how many squares of all sizes does it contain? (Wiscomb, p. 03) Let us consider several cases for n=, n=, n=3, and n= (see Table ). Table. Checkerboard in D Squares: x x 3x3 x Total Number of Squares n= n= 5 n=3 9 3 n= Note: Square images above were formatted with MS Word Picture Tools. We can see that that for n=, there is only one square. For n=, there are four x squares and one x square. For n=3, there are nine x squares, four x squares, and one 3x3 square. Finally, for
11 A GEOMETRIC INTRODUCTION n=, there are sixteen x squares, nine x squares, four 3x3 squares, and one x square. The total number of squares in each case is the sum of squares of natural numbers from to n (see Table ). Continuing in this manner, we can conjecture that an nxn checkerboard will contain 3... n squares of all sizes, but what is this sum equal to? How can we guess the closedform expression for the sum of squares of natural numbers? To answer this question, we will try to use an analogy with the formula for the sum of natural numbers from Theorem : nn ( ) 3... n This formula has a polynomial of the second degree on the right side. Thus, we will assume that the sum of squares of natural numbers should add up to a polynomial of the third degree, something like 3 an bn cn d, for real coefficients a, b, c, d. Our guess is that Where, n an bn cn d, (Eq.) a, b, c, d However, what are these real coefficients a, b, c, d? If we rewrite Equation four times for n=, n=, n=3, and n=, respectively, we will get a system of four linear equations in four variables:
12 A GEOMETRIC INTRODUCTION n : n : n 3: n : a b c d 3 a b c d a 3 b 3c d 3 a b c d 30 a b c d 8a b c d 5 7a 9b 3c d a b c d 30 We can solve this system by hand (using a method like Cramer s rule, Gaussian elimination, or several others), or we can use a calculator to save time and paper space. The solutions are: Then, Then we can make the following conjecture. a, b, c, d 0 3 n n n 3 3 n 3n n n(n 3n ) n( n )(n ). 3 an bn cn d 3 Conjecture : For natural numbers greater or equal to, Proof: We can prove this conjecture by the PMI. Let P(n) be, n( n )(n ) 3... n,n n( n )(n ) 3... n,n
13 A GEOMETRIC INTRODUCTION 3. P() is true:. Suppose P(n) is true. ( )( ) 3. We will show that P(n+) is true. (We want to show that, By the induction hypothesis, Let us add ( n ) ( n) (( n) ) 3... n ( n ) ) ( n ) to both sides. Then, n( n )(n ) 3... n 3... n ( n ) n( n )(n ) ( n ) Pn ( ) Pn ( ) n( n )(n ) ( n ) ( n ) n(n ) ( n ) ( n ) n n n ( n ) n 7n ( n ) n (n3) ( n ) ( n) (( n) ) Thus, P(n+) is true. Then by the PMI, P(n) is true for all natural numbers.
14 A GEOMETRIC INTRODUCTION Therefore, we have proved the formula for the sum of squares of natural numbers, n( n )(n ) 3... n,n From now on, we can use it to find the number of squares of all sizes contained in a twodimensional checkerboard. Since we have already proved this result, we will no longer call it a conjecture, but a theorem. Theorem (The Sum of Squares of Consecutive Natural Numbers): For natural numbers greater or equal to, n( n )(n ) 3... n,n We will now consider a similar problem for a threedimensional checkerboard. We will use this problem to introduce the formula for the sum of cubes of natural numbers. The Sum of Cubes: n Problem 3: Checkerboard in 3D Solution: How many cubes of all sizes are there in a nxnxn threedimensional checkerboard? (Wiscomb, p. 0) Let us consider several cases for n=, n=, and n=3 (see Table 3.).
15 A GEOMETRIC INTRODUCTION 5 Table 3. Checkerboard in 3D Cubes: xx xx 3x3x3 Total Number of Cubes n= n= n= Note: Cube images above were formatted with MS Word Picture Tools. We can see that that for n=, there is only one cube. For n=, there are eight x cubes and one x cube. For n=3, there are twentyseven x cubes, eight x cubes, and one 3x3 cube. The total number of cubes in each case is the sum of cubes of natural numbers from to n. Continuing in this manner, we can conjecture that an nxnxn checkerboard will contain n cubes of all sizes, but what is this sum equal to? How can we guess the closedform expression for the sum of cubes of natural numbers? Let us consider the sum of cubes when n=, n=, n=3, and n=. The numbers that we get in each case are perfect squares:, 9, 3, and 00, respectively (see Table 3., left column). They can be written as,3,,0. Consider the sequence, 3,, 0. Notice that we can rewrite the terms of this sequence as, +, ++3, ++3+ (see Table 3., right column).
16 A GEOMETRIC INTRODUCTION Table 3. Checkerboard in 3D Total Number of Cubes k i n= 3 n= ( ) n= n= ( 3) ( 3 ) Note: Total Number of Cubes expressed in two different ways. Notice that in each case, the sum of cubes can be written as the square of the sum of consecutive natural numbers from to n: 3... n ( 3... n) The right side of this expression contains the sum 3... n, for which we already know the formula (from Theorem ): 3... n nn ( ) Then, 3... n n ( n) and therefore, n 3... n ( n) Then we can make the following conjecture. Conjecture 3: For natural numbers greater or equal to, Proof: We can prove this conjecture by the PMI n ( n) 3... n,n
17 A GEOMETRIC INTRODUCTION 7 Let P(n) be. P() is true: n 3... n ( n) 3 ( ). Suppose P(n) is true. 3. We are going to show that P(n+) is true. (We want to show that By the induction hypothesis, n ( n) (( n) ) 3 ( n ) ) n 3... n ( n) Let us add 3 ( n ) to both sides. Then, n ( n) n ( n ) n Pn ( ) Pn ( ) 3 n ( n ) ( n ) ( n ) ( n n ) ( n) ( n) ( n)( n) ( n)(( n) ) ( n) (( n) )
18 A GEOMETRIC INTRODUCTION 8 Thus, P(n+) is true. Then by the PMI, P(n) is true for all natural numbers. Therefore, we have proved the formula for the sum of cubes of natural numbers: n ( n) 3... n,n From now on, we can use it to find the number of cubes of all sizes contained in a threedimensional checkerboard. Since we have already proved this result, we will no longer call it a conjecture, but a theorem. Theorem 3 (The Sum of Cubes of Consecutive Natural Numbers): For natural numbers greater or equal to, n ( n) 3... n,n Recommendations for Further Research There are additional geometry problems that could provide interesting contexts to various formulas used to illustrate the Principle of Mathematical Induction. There are also cases when the PMI proves indispensable to verifying important results in geometry. This mutually beneficial relationship between geometry and the Principle of Mathematical Induction deserves further consideration. Unfortunately, the limited scope of this paper does not allow us to explore these topics in more detail; however, below are some sample questions that could be investigated in future research.
19 A GEOMETRIC INTRODUCTION 9 Questions about n lines in the plane: Given n concurrent lines, how many pairs of supplementary angles are formed? How many pairs of vertical angles are formed? What is the maximum number of regions in the plane determined by n lines? (Wiscamb, 970) Questions about polygons with n sides: What are the sums of interior angles of various convex polygons? Into how many triangles can a convex polygon be divided by nonintersecting diagonals? How many such nonintersecting diagonals are needed to divide a convex polygon into these triangles? What are the answers to these questions for a concave polygon? It takes some time to completely work through questions like these in order to actively explore geometry concepts, to find patterns, to formulate conjectures in terms of natural numbers, and to write induction proofs. However, by not skipping any steps of the process, we can gain not only a better understanding of mathematical induction, but also a feeling for the manner in which mathematics is being created (Avital & Hansen, p. ).
20 A GEOMETRIC INTRODUCTION 0 References Avital, S., & Hansen, R. T. (97). Mathematical induction in the classroom. Educational Studies in Mathematics, 7(), 399. Wiscamb, M. (970). A geometric introduction to mathematical induction. The Mathematics Teacher, 3(5), 00.
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