Download A Geometric Introduction to Mathematical Induction

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 584
Fall 2014
Galina Landa
1
A GEOMETRIC INTRODUCTION
2
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 two-dimensional
checkerboard or the number of cubes of all sizes enclosed by a three-dimensional checkerboard
with each side length equal to n units.
A GEOMETRIC INTRODUCTION
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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.
A GEOMETRIC INTRODUCTION
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The Sums
The Sum of Consecutive Natural Numbers:
1  2  3  ...  n
We will consider this first sum in the context of the following geometry problem.
Problem 1: Lines and Points
Find “the number of lines determined by a given number of points in a plane, no three of
which are collinear” (Wiscamb, p. 402).
Solution:
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>1. Let us consider the first few cases for the
number of points n=2, n=3, n=4, n=5, and n=6 (see Table 1). Notice that while 2 arbitrary points
determine a single line, 3 non-collinear points determine 3 lines, 4 points determine 6 lines, 5
points determine 10 lines, and 6 points determine 15 lines.
Table 1. Lines and Points
n,
ln ,
ln  1 2  3  ...  (n 1)
points
2
lines
1
1=1
3
3
3=1+2
(The table continues on the next page)
ln 
1
( pn  pn 1 )
2
1
(3  2)  3
2
A GEOMETRIC INTRODUCTION
5
4
6
6=1+2+3
1
(4  3)  6
2
5
10
10=1+2+3+4
1
(5  4)  10
2
6
15
15=1+2+3+4+5
1
(6  5)  15
2
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.
1. The number of lines determined by n points is equal to the sum of natural numbers from 1
to n-1:
ln  1 2  3  ...  (n 1)
2. 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:
1
ln  ( pn pn 1 )
2
A GEOMETRIC INTRODUCTION
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Based on these observations, we can make the following conjecture.
Conjecture 1: For natural numbers greater than 1,
1
1  2  3  ...  (n  1)  n(n  1)
2
Proof:
We can prove this conjecture by the PMI.
Let P(n-1) be
1
1  2  3  ...  (n  1)  n(n  1) , n>1
2
1. P(2) is true:
1
2  1   2  (2  1)  1
2
2. Suppose P(n-1) is true for n>1.
3. We will show that P(n) is true. (We want to show that,
1  2  3  ...  n 
n(n  1)
)
2
By the induction hypothesis,
1
1  2  3  ...  (n  1)  n(n  1)
2
Let us add n to both sides:
1
1  2  3  ...  (n  1)  n  n(n  1)  n
2
P ( n 1)
P ( n 1)
A GEOMETRIC INTRODUCTION
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
n(n  1)  2n
2

n(n  1  2)
2

n(n  1)
2
Thus, P(n) is true. Then by the PMI, P(n-1) is true for all natural numbers n>1.
Therefore, we have proved the formula:
1
1  2  3  ...  (n  1)  n(n  1) , for natural n>1
2
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
1 to n-1. We can also rewrite this formula for the sum of natural numbers from 1 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 1 (The Sum of Consecutive Natural Numbers):
For natural numbers greater or equal to 1,
1  2  3  ...  n 
n(n  1)
,n
2
We can now use this result to discover some other mathematical relationships. For example,
Avital and Hansen (1976) posed the following question.
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Problem 1a: Diagonals of n-gon
“What is the number of diagonals one can draw in a convex polygon of n vertices?” (Avital
& Hansen, p. 404)
Solution:
Clearly, a triangle has no diagonals; we will consider n > 3 (see Table 1a.)
Table 1a. Diagonals of n-gon
Polygon
n, vertices
d, diagonals
Triangle
3
0
Quadrilateral
4
2=2
Pentagon
5
5=2+3
Hexagon
6
9=2+3+4
Heptagon
7
14=2+3+4+5
Note: Diagrams constructed with The Geometer's Sketchpad® software.
A GEOMETRIC INTRODUCTION
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After considering polygons with 4, 5, 6, and 7 sides, we find that the number of diagonals is 2, 5,
9, and 14, respectively. The last four numbers can be written as 2, 2+3, 2+3+4, and 2+3+4+5.
Thus, the number of diagonals in each of these cases is the sum of natural numbers from 2 to n-2,
where n is the number of vertices. Let us denote the number of diagonals d; then for an n-gon,
d  2  3  4  ...  (n  2)
This resembles the sum from Theorem 1, minus 1, minus (n-1), and minus n. Then we can use the
result from Theorem 1 to find the expression for d:
d  2  3  4  ...  (n  2)
 1  2  3  ...  n   1  (n  1)  n
Theorem1

n(n  1)
1  n  1 n
2
Theorem1

n(n  1)  4n
2

n(n  1  4)
2

n(n  3)
2
Thus,
2  3  4  ...  (n  2) 
n(n  3)
,n  3
2
From now on, we can use this result to find the number of diagonals in a convex n-gon. We have
demonstrated that this result follows from Theorem 1, so we will call it Corollary 1.
Corollary 1: For natural numbers greater than 3,
2  3  4  ...  (n  2) 
n(n  3)
,n
2
A GEOMETRIC INTRODUCTION
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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 1.
The Sum of Squares: 12  22  32  ...  n2
Wiscamb (1970) suggested introducing the formula for the sum of squares in connection with the
following problem.
Problem 2: Checkerboard in 2D
“Given an n-x-n checkerboard with n2 small squares each measuring 1 square unit, how
many squares of all sizes does it contain?” (Wiscomb, p. 403)
Solution:
Let us consider several cases for n=1, n=2, n=3, and n=4 (see Table 2).
Table 2. Checkerboard in 2D
Squares:
2x2
n=1
1x1
1
3x3
n=2
4
1
n=3
9
4
1
n=4
16
9
4
4x4
Total Number of Squares
12  1
12  22  5
12  22  32  14
1
12  22  32  42  30
Note: Square images above were formatted with MS Word Picture Tools.
We can see that that for n=1, there is only one square. For n=2, there are four 1x1 squares and one
2x2 square. For n=3, there are nine 1x1 squares, four 2x2 squares, and one 3x3 square. Finally, for
A GEOMETRIC INTRODUCTION
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n=4, there are sixteen 1x1 squares, nine 2x2 squares, four 3x3 squares, and one 4x4 square. The
total number of squares in each case is the sum of squares of natural numbers from 1 to n (see
Table 2).
Continuing in this manner, we can conjecture that an nxn checkerboard will contain
12  22  32  ...  n 2
squares of all sizes, but what is this sum equal to? How can we guess the closed-form 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 1:
1  2  3  ...  n 
n(n  1)
2
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 an3  bn 2  cn  d , for real coefficients a, b, c, d . Our guess is that
12  22  32  ...  n2  an3  bn2  cn  d ,
(Eq.1)
Where,
a, b, c, d 
However, what are these real coefficients a, b, c, d ?
If we rewrite Equation 1 four times for n=1, n=2, n=3, and n=4, respectively, we will get a
system of four linear equations in four variables:
A GEOMETRIC INTRODUCTION
n  1:
n  2:
n  3:
n  4:
12
a  b  c  d  1
 3
2
2 a  2 b  2c  d  5

 3
2
3 a  3 b  3c  d  14
43 a  42 b  4c  d  30

a  b  c  d  1
8a  4b  2c  d  5


27a  9b  3c  d  14
64a  16b  4c  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:
1
1
1
a  ,b  ,c  , d  0
3
2
6
Then,
an3  bn 2  cn  d 
1
1
1
 n3  n 2  n
3
2
6
3
2
2n  3n  n

6
2
n(2n  3n  1)

6

n(n  1)(2n  1)
.
6
Then we can make the following conjecture.
Conjecture 2: For natural numbers greater or equal to 1,
12  22  32  ...  n 2 
n(n  1)(2n  1)
6
,n
Proof:
We can prove this conjecture by the PMI.
Let P(n) be,
12  22  32  ...  n 2 
n(n  1)(2n  1)
6
,n
A GEOMETRIC INTRODUCTION
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1. P(1) is true:
12 
1 (1  1)(2 1  1)
1
6
2. Suppose P(n) is true.
3. We will show that P(n+1) is true. (We want to show that,
12  22  32  ...  n2  (n  1)2 
(n  1)  (n  1)  1 (2(n  1)  1)
6
)
By the induction hypothesis,
12  22  32  ...  n 2 
n(n  1)(2n  1)
6
Let us add (n  1)2 to both sides. Then,
12  22  32  ...  n2  (n  1)2
P(n)

n(n  1)(2n  1)
 (n  1)2
6
P(n)






n(n  1)(2n  1)  6(n  1)2
6
(n  1)  n(2n  1)  6(n  1) 
6

2

2
(n  1) 2n  n  6n  6
6
(n  1) 2n  7n  6


6
(n  1)  n  2 (2n  3)
6
(n  1)  (n  1)  1 (2(n  1)  1)
6
Thus, P(n+1) is true. Then by the PMI, P(n) is true for all natural numbers.
A GEOMETRIC INTRODUCTION
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Therefore, we have proved the formula for the sum of squares of natural numbers,
12  22  32  ...  n 2 
n(n  1)(2n  1)
6
,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 2 (The Sum of Squares of Consecutive Natural Numbers):
For natural numbers greater or equal to 1,
12  22  32  ...  n2 
n(n  1)(2n  1)
,n
6
We will now consider a similar problem for a three-dimensional checkerboard. We will use this
problem to introduce the formula for the sum of cubes of natural numbers.
The Sum of Cubes: 13  23  33  ...  n3
Problem 3: Checkerboard in 3D
“How many cubes of all sizes are there in a n-x-n-x-n three-dimensional checkerboard?”
(Wiscomb, p. 404)
Solution:
Let us consider several cases for n=1, n=2, and n=3 (see Table 3.1).
A GEOMETRIC INTRODUCTION
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Table 3.1 Checkerboard in 3D
1x1x1
1
2x2x2
n=1
Cubes:
n=2
8
1
n=3
27
8
3x3x3
Total Number of Cubes
13  1
13  23  9
1
13  23  33  36
Note: Cube images above were formatted with MS Word Picture Tools.
We can see that that for n=1, there is only one cube. For n=2, there are eight 1x1 cubes and one
2x2 cube. For n=3, there are twenty-seven 1x1 cubes, eight 2x2 cubes, and one 3x3 cube. The total
number of cubes in each case is the sum of cubes of natural numbers from 1 to n.
Continuing in this manner, we can conjecture that an nxnxn checkerboard will contain
13  23  33  ...  n3
cubes of all sizes, but what is this sum equal to? How can we guess the closed-form expression for
the sum of cubes of natural numbers?
Let us consider the sum of cubes when n=1, n=2, n=3, and n=4. The numbers that we get in each
case are perfect squares: 1, 9, 36, and 100, respectively (see Table 3.2, left column).
They can be written as 12 ,32 ,62 ,102 . Consider the sequence 1, 3, 6, 10. Notice that we can rewrite
the terms of this sequence as 1, 1+2, 1+2+3, 1+2+3+4 (see Table 3.2, right column).
A GEOMETRIC INTRODUCTION
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Table 3.2 Checkerboard in 3D
n=1
n=2
n=3
n=4
Total Number of Cubes
ki2
13  1
1  23  9
13  23  33  36
13  23  33  44  100
9  3  (1  2)2
36  62  (1  2  3)2
100  102  (1  2  3  4)2
1  12
3
2
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 1 to n:
13  23  33  ...  n3  (1  2  3  ...  n)2
The right side of this expression contains the sum 1  2  3  ...  n , for which we already know the
formula (from Theorem 1):
1  2  3  ...  n 
n( n  1)
2
n (n  1)
2
Then,
1  2  3  ...  n
and therefore,
1  2  3  ...  n 
2

4
n (n  1)
2
3
3
2
3
3
2
4
Then we can make the following conjecture.
Conjecture 3: For natural numbers greater or equal to 1,
n (n  1)
2
1  2  3  ...  n 
3
3
3
3
4
Proof:
We can prove this conjecture by the PMI.
2
,n
A GEOMETRIC INTRODUCTION
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Let P(n) be
n (n  1)
2
1  2  3  ...  n 
3
3
3
3
2
4
1. P(1) is true:

3
1
1 (1  1)
2
2
4
1
2. Suppose P(n) is true.
3. We are going to show that P(n+1) is true. (We want to show that
1  2  3  ...  n
3
3
3
3
 (n  1)3
(n  1)2 ((n  1)  1)2
)
4
By the induction hypothesis,
n (n  1)
2
1  2  3  ...  n 
3
3
3
3
2
4
Let us add (n  1)3 to both sides. Then,
13  23  33  ...  n3  (n  1)3 
P(n)
n2 (n  1)2
3
  n  1
4
P(n)
n2 (n  1)2  4(n  1)3
4
2
(n  1) (n2  4n  4)

4
2
(n  1) (n  2)2

4
2
 (n  1)(n  2) 


2



 (n  1)((n  1)  1) 


2


(n  1)2 ((n  1)  1)2

4
2
A GEOMETRIC INTRODUCTION
18
Thus, P(n+1) 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  1)
2
1  2  3  ...  n 
3
3
3
3
2
,n
4
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 1,
13  23  33  ...  n3 
n2 (n 1)2
,n
4
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.
A GEOMETRIC INTRODUCTION
19
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, 1970)
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 non-intersecting 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. 411).
A GEOMETRIC INTRODUCTION
20
References
Avital, S., & Hansen, R. T. (1976). Mathematical induction in the classroom. Educational
Studies in Mathematics, 7(4), 399411.
Wiscamb, M. (1970). A geometric introduction to mathematical induction. The Mathematics
Teacher, 63(5), 402404.