Mathematical Investigations IV
Name
Mathematical Induction - Let the Dominoes Fall
Proofs, Old and New
In geometry you proved various theorems and exercises. In developing your proofs, you usually
used what is called a Direct Proof. That is, you started with the given and through a series of
logical steps, you arrived at what you were to prove.
For example, a direct proof would be an appropriate means for you to prove:
Given the figure at the right with
AB
CD and E the midpoint of AD.
A
C
Prove: AEB  DEC
E
B
D
A second form of proof, usually developed in geometry class, is called an Indirect Proof. In this
situation, you started with the given, assumed that the conclusion was false and then derived a
contradiction. For example, an indirect proof would be an appropriate means for proving:
Through a given point, not on a line, there is
exactly one line through the point perpendicular to
the given line.
[The assumption would be that two perpendicular lines exist
and then a contradiction between the given and assumption
would be derived.]
In this unit, we want to look at an additional way to verify relationships. We are going to use the
First Principle of Mathematical Induction.
Mathematical Induction is an important technique for proving certain types of statements. In
particular, it is often used to prove assertions dependent on a positive integer n, or occasionally,
simply for non-negative integers. For example, consider the following propositions Pn.
(1)
Pn: 1 + 3 + 5 + … + (2n – 1) = n2 for n  1
(2)
Pn: 1 + 2 + 22 + … + 2n = 2n+1 – 1 for n  0
(3)
Pn: 4n  4n for n  1
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Mathematical Investigations IV
Name
(4)
Pn:  2 4 n 2  1 is a multiple of 5 for n  1
(5)
Pn: The sum of the interior angles of a convex n-gon is 180(n – 2).
These are different types of examples that all make some statement for an arbitrary value of n.
Proof by Mathematical Induction involves two basic steps. Very often, the method is described
by thinking of a line of dominos. First, we have to be able to make the first domino fall.
Secondly, if any one of the dominos in the line falls, the next one in the line will also fall. If both
of these hold true, then when the first one falls, it will knock over the second one, the second will
knock over the third, etc., so all of the dominos will eventually fall. Now, we need to translate
this concept back into mathematics. If the proposition Pn is true for the starting value of n
(usually n = 0 or 1) and if Pn is true when n is equal to some specific value of k always implies
that it will be true for the next value of k, then this will imply that the statement will be true for
all values of n.
To formalize this:
The First Principle of Mathematical Induction
Let P1, P2, P3, ... be a sequence of statements.
If it can be shown that
(1)
(2)
P1 is true, and that
Pn+1 is true whenever Pn is true,
then all of the statements P1, P2, P3, ... are true.
Note that there are two parts of each proof:
First you must show that a specific statement (usually P1) is true.
Second you must show that the (k+1)th statement is true based on the assumption that the kth
statement is true.
Example 1:
Use Mathematical Induction to show that:
[State that you are using M.Ind.]
Pn: 1 + 3 + 5 + 7 + ... + (2n - 1) = n2
is true for all positive integers.
[Note that P1 is the statement 1 = 12, P2 is the statement 1 + 3 = 22, etc.]
Step 1: Show that P1 is true.
[Usually, this is rather trivial.]
Obviously (21 - 1) = 12.
Step 2: Show that if Pk is true, then Pk+1 is true.
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Mathematical Investigations IV
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Pn is 1 + 3 + 5 + ... + (2k-1) = k2.
(Equation 1)
Assuming that is true, we need to show that Pk 1 ,
1 + 3 + 5 + ... + (2(k) - 1) + (2(k+1) - 1) = (k+1)2,
is true.
[Restate what you wish to prove in terms of Pk+1.]
In order to complete this proof, we are going to start with the left side of the equation for Pn+1
and using substitution and various algebraic properties, derive the right hand side. This is
similar to the technique we used to prove trigonometric identities.
1 + 3 + 5 + ... + (2k-1) + (2(k+1) - 1) = k2 + (2(k+1) - 1)
[Substitution using Equation 1]
= k2 + 2k + 2 - 1
= k2 + 2k + 1
= (k + 1)2
Thus, Pk+1 is true, assuming that Pk is true, which is what we wished to prove.
Therefore, the sum of the first n odd integers is equal to n2.
[Summarize what you have just proven.]
Example 2:
Prove that the sum of the interior angles of a convex n-gon (a polygon with n-sides) is
180(n-2).
Proof by Mathematical Induction
In this case, we will start with n = 3 instead of n = 1.
[It is proper to state the method of proof.]
[I think you know why.]
Step 1:
Let n = 3 and we know that the sum of the measures of the angles of a triangle is
180 = 180(n-2).
Step 2:
Assume that for some k, the sum of the measure of the interior angles of the
convex k-gon is 180(k-2). Show that the sum of the measures of the interior angles of
the convex (k+1)-gon is 180((k+1) - 2).
Proof: Let A1A2A3...AkAk+1 be a convex (k+1)-gon.
From our assumption that Sk is true, we know that
the sum of the angles of the polygon A1A2A3...Ak is
180(k-2). We need to consider how many more
degrees are added when the (k+1) vertex is added.
From the figure at the right, it is clear that the sum of
the angles of A1A2A3...AkAk+1 = (the sum of
A1A2A3...Ak) + (the sum of AkAk+1A1)
= 180(k-2) + 180
= 180(k-2 + 1)
= 180((k+1) - 2)
A3
A2
Ak-1
Ak
A1
Ak+1
which is what we wished to show.
Therefore the sum of the interior angles of any convex n-gon is 180(n-2).
Induction .3
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Mathematical Investigations IV
Name
Exercises
Basic Properties
n
(1) The number   is defined to be the number of unordered pairs of elements you can make
 2
 n  n(n  1)
from a set of n elements. Use induction to prove that   
for every positive
2
 2
integer n  2 .
(2) Use induction to prove that n! is even for all integers n  2 .
(3) Use induction to prove that 52n - 1 is divisible by 24 for every positive integer n.
Series
n
(4) Use induction to prove that  j 
j 1
n(n  1)
.
2
(5) Use induction to prove that 1 + 2 + 22 + … + 2n = 2n+1 – 1 for all integers n  0.
(6) Use induction to prove that 3 + 7 + 11 + . . . + (4n-1) = n(2n + 1) for all integers n  1.
(7) Use induction to prove that, for all positive integers n,
11! + 22! + 33! + . . . + nn! = (n+1)! – 1.
(8) Use induction to prove that, for all positive integers n,
1
1
1
1
n
.


 

1 2 2  3 3  4
n(n  1) n  1
(9) Use induction to prove that, if a  1 , then
n
 ai 
i 0
a n1  1
for all integers n  0 .
a 1
Geometry
(10) Use induction to prove that the number of diagonals in a convex n-gon is
Induction .4
1
n( n  3) .
2
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Mathematical Investigations IV
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Induction .5
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