Mathematical Induction
Consider the statement
“if 8 is even, then %l8# ”
As it stands, this statement is neither true nor false: 8 is a variable and whether the statement is
true or false depends on what value of 8, from what universe, we're talking about. However,
Ða8 − Ñ Ðif 8 is even, then %l8# Ñ
is a (true) proposition. It asserts that a certain statement is true for every 8 in the universe .
The Principle of Mathematical Induction (PMI) is a method for proving statements
of the form Ða8 − Ñ T Ð8Ñ.
Note: Outside of mathematics, the word “induction” is sometimes used differently.
There, it usually refers to the process of making empirical observations and then
generalizing from them to a conclusion: for example, we observe the sun coming up in
the east thousands of times and conclude that “the sun always rises in the east.” This
sort of “induction” is an important part of the scientific method. Of course, it raises a
philosophical problem:
“How can we ever justify drawing a universal conclusion from particular
observations  even if there are thousands of observations?” And if we can't,
then why is the scientific method as effective as it is?
Outside of mathematics, induction is usually viewed as “opposite” to deduction.
Deduction is a style of argument that starts with certain premises (assumptions) and
logically proceeds to a conclusion without reference to any empirical observations. In a
correct deductive argument, the conclusion must be true if the premises are true.
For example, a nonmathematician might observe, correctly, that the statement
“if 8 is a natural number, then 8#  8  %" is prime”
is true for 8 œ "ß #ß $ß ÞÞÞ ß %! and then conclude that the statement is true for all natural
numbers 8. This would be an example of a conclusion drawn by “induction” in the
everyday use of the word. But this conclusion is incorrect: for 8 œ %",
8#  8  %" œ %"#  %"  %" œ %"# is not prime.
“Mathematical induction” is something totally different. It refers to a kind of deductive
argument, a logically rigorous method of proof. It works because of how the natural
numbers are constructed from set theory; as we shall see later, PMI is “built into .”
First we will state PMI and a variation (called PCI, the Principle of Complete Induction) in the
form in which they most often used.
Suppose T Ð8Ñ is a statement about a natural number 8.
Principle of Mathematical Induction (PMI)
 i) T Ð"Ñ is true, and
If  ii) when T Ð5Ñ is true for some particular 5 − ,

then T Ð5  "Ñ must also be true
then a8 −  T Ð8Ñ is true
Principle of Complete Induction (PCI)
 i) T Ð"Ñ is true, and
If  ii) when 5  " and T Ð8Ñ is true for every 8  5 ,

then T Ð5Ñ must also be true
then
a8 −  T Ð8Ñ is true
Both of these principles seem intuitively clear (see the discussion in class). We will see why they
are actually equivalentÞ For the moment, let's just practice using them.
3 œ
8
Example 1 Let T Ð8Ñ be the statement À
3œ"
8Ð8  "Ñ
#
"  ÞÞÞ  8 œ
or more informally
8Ð8 "Ñ
#
Prove that Ða8 − Ñ T Ð8Ñ is true.
Proof i) Step i) is called the base case for the induction  the “starting point.”
T Ð"Ñ is the statement  3 œ
"
3œ"
"Ð""Ñ
# ,
which is true.
ii) Assume T Ð5Ñ is true for some 5 − .
This assumption is called the induction hypothesis or induction assumption. Note: we are not
assuming here that T Ð5Ñ is true for every 5 in   that would be assuming the very thing we're
trying to prove! Rather, we are supposing that 5 is a particular value of 8 for which T Ð8Ñ is
true (such as, for example, 5 œ "Ñ.
Assuming this, we need to show that T Ð5  "Ñ must also be true.
A good idea: write out T Ð5  "Ñ to be clear about what it says. T Ð5  "Ñ reads:
3 œ
5"
3œ"
Ð5  "ÑÐ5  #Ñ
#
or, more informally, "  ÞÞÞ  5  Ð5  "Ñ œ
Ð5  "ÑÐ5  #Ñ
#
To do this:
 3 œ  3  Ð5  "Ñ œ
5"
3œ"
5
3œ"
5Ð5  "Ñ
#
 Ð5  "Ñ œ
5Ð5 "Ñ  #Ð5  "Ñ
#
œ
Ð5  "ÑÐ5  #Ñ
#
Å by the induction hypothesis
which says that T Ð5  "Ñ is true. By PMI, a8 −  T Ð8Ñ is true  that is, T Ð8Ñ is true for all
8 − . ñ
Example 2 (PCI) Prove that every natural number 8  " is either a prime or a product of
primes. If we count a prime number as being a product of primes with “just one factor,” then we
could say that “every natural number bigger than " can be factored into primes.”
This is one part of the Fundamental Theorem of Arithmetic; the other part (which we will not
prove until later) says that the factorization is unique except for the order of the factors.
More formally, we are supposed to prove: a8 −  T Ð8Ñ, where T Ð8Ñ is the (equivalent)
statement:
Ð8 œ "Ñ ” Ð8 is primeÑ ” Ð8 can be factored into primes)
Proof i) T Ð"Ñ is true since " œ ".
ii) Suppose 5  " and assume T Ð8Ñ is true for all 8  5Þ (This is the “induction
hypothesis.”) We must show T Ð5Ñ is true.
If 5 is prime, then T Ð5Ñ is true. Otherwise, since 5  ", we can factor
5 œ :; where :ß ; −  and :ß ;  5Þ The factor : is either prime or (by the induction
hypothesis) : factors into primesà and the same is true for ; . Therefore 5 factors into
primes.
By PCI, we conclude that a8 −  T Ð8Ñ is true.
ñ
Notice the similarities and differences between using PMI and PCI as illustrated in Examples 1
and 2.
With both methods, we need to start by verifying a base case: in Example 1-2, that T Ð"Ñ
is true.
With both methods, we need to prove that T Ð8Ñ is true for a certain value of 8  under a
certain assumption. This is the “induction step.”
a) With MI (Example 1), we need to show, assuming that T Ð5Ñ is true for some
value 5 , that T Ð5  "Ñ is also true À the induction hypothesis only involves a
single natural number 5 , the “immediate predecessor” of 5  "Þ
To illustrate: With PMI, the induction step shows, for example, that if
T Ð$Ñ is true, then T Ð%Ñ must also be true.
b) With PCI (Example 2), we need to show, assuming that
T Ð8Ñ is true for all values of 8 preceding some 5 , that T Ð5Ñ is also true. The
induction hypothesis involves all the natural numbers preceding 5 .
To illustrate: With PCI, then induction step shows, for example, that if
T Ð"Ñ and T Ð#Ñ and T Ð$Ñ are true, then T Ð%Ñ must be true.
The shift from using 5  " in the induction step (in PMI) to using 5 in the induction step (in PCI)
is a just a minor notational shift that is convenient and traditional: it's not really a significant
“difference” in the methods. The main difference is that PCI seems to give us much more to work
with than PMI: with PCI, we assume that all of T Ð"Ñß ÞÞÞß T Ð5  "Ñ are true and try to use this
information prove T Ð5Ñ. PCI seems to give us many more assumptions to use in proving T Ð5Ñ is
true.
In light of this observation, it may be surprising that PMI and PCI turn out to be logically
equivalent: either both are true statements about  or both are false statements about .
If we assume PCI, we can prove PMI is also true, and vice-versa. (We will show this soon.)
However, this does not mean that they are equally useful in a given situation.
In Example 1, all we need to prove T Ð5  "Ñ is T Ð5Ñà any additional assumptions
about T Ð5  "Ñß ÞÞß T Ð"Ñ are completely unnecessary.
In Example 2, it's hard to see how we could prove that 5 factors into primes if the
induction assumption were only about the single number preceding 5  that is, if the
induction assumption were merely that 5  " factors into primes. In the proof in
Example 2, we need to know, somehow, that : and ; are products of primes and that's
what the induction hypothesis using PCI gives us!
Here's a more vivid illustration of PMI and PCI that might fix them in your memory:
PMI states that i) if " is red and ii) if a natural number must be red whenever its
immediate predecessor is red, then all natural numbers must be red.
PCI states that i) if " is red and ii) if a natural number must be red whenever all
its predecessors are red, then all natural numbers must be red.
There is another useful property of , called the well-ordering principle, that can sometimes be
used to prove a statement of the form a8 −  T Ð8Ñ. It is stated in terms of subsets of .
Well-Ordering Principle ÐWOPÑ If E ©  and E Á g, then E contains a smallest element.
Intuitively, this seems clear: if there were a nonempty subset E of  with no smallest
element, we could choose a number +" − E; since E has no smallest element, there
would be an +# − E where +#  +" à since +# is not the smallest element in E, there
would be a stiller smaller number +$ − E. Continuing in this way we could write down
an infinite decreasing sequence of natural numbers: +"  +#  +$  ÞÞÞ  +8  ÞÞÞ and
this is intuitively impossible. Try it!
Suppose, say, +" œ "!$#"(à what could +# be? +$ ? Can you find an “infinite
descending sequence” "!$#"(  +#  +$  ÞÞÞ  +8  ÞÞÞ ?
Notice that in this intuitive argument, we need that E Á g À that assumption is how we
get +" in E to start with. In fact, if E œ g, then E doesn't contain a smallest
element  because E has no elements at all.
Can we justify rigorously why WOP a true statement about ? It turns out (another surprise?)
that all three of WOP, PMI and PCI are logically equivalent. As we will see later, PMI is “builtinto” the system  at a very fundamental level, when  is constructed from set theory and
therefore PCI and WOP are automatically “built-in” too. All three are very fundamental facts
about .
Because PMI, PCI and WOP are logically equivalent, any proof that can be done using one them
could also be done (at least in principle) using the otherÞ Sometimes one is more convenient to
use than the other, and sometimes choosing between them is just a matter of taste.
The following example gives a proof of the result in Example 1 using WOP instead of PMI.
Notice the difference in the approach; but equally important, notice the similarities in the algebra
that comes up. You can decide whether you prefer this argument to the one using PMI in
Example 1.
Ða8 − Ñ T Ð8Ñ, where T Ð8Ñ is the statement:  3 œ
8
Example 3 Use WOP to prove:
3œ"
8Ð8  "Ñ
#
Proof Let E œ Ö8 −  À T Ð8Ñ is false×.
To complete the proof, we want to show that E œ g. We argue by contradiction.
Assume E Á gÞ Then, by WOP, there must be a smallest 5 in E: that is,
3 œ
5
3œ"
5Ð5  "Ñ
#
is false (*) and
T Ð8Ñ is true for natural numbers 8 smaller than 5 .
Since  3 œ
"
3œ"
"Ð"  "Ñ
#
is true, we know 5 Á "ß so 5  "Þ Therefore 8 œ 5  " is still a natural
number and smaller than 5 , so T Ð5  "Ñ must be true:
3 œ
5"
3œ"
Ð5"ÑÐÐ5  " Ñ  "Ñ
#
œ
Ð5"ÑÐ5 Ñ
#
is true.
Then adding 5 to both sides of (**) gives that  3  5 œ
5"
But this equation (do the algebra!) simplifies to  3 œ
3œ"
5
3œ"
(**)
Ð5"ÑÐ5 Ñ
#
5Ð5  "Ñ
#
 5 is true.
 which contradicts ЇÑ.
Since the assumption that E Á g leads to a contradiction, it must be that E œ g; in other words,
3 œ
8
3œ"
8Ð8  "Ñ
#
is true for all 8 − .
ñ
Mathematical induction (in any of the equivalent forms PMI, PCI, WOP) is not just used to prove
equations. Example 2, in fact, uses PCI to prove part of the Fundamental Theorem of Arithmetic.
Examples 4 and 5 illustrate using induction to prove an inequality and to prove a result in
calculus.
Example 4 Prove that /8  "  8 for all natural numbers 8.
More formally, we want to prove Ða8 − Ñ T Ð8Ñ, where T Ð8Ñ is the statement: /8  "  8
Proof i) T Ð"Ñ states that /  #ß which is known to be true. ÐFrom calculus:
/ œ 2.718281828459045ÞÞÞÑ
ii) ÐInduction stepÑ Assume that T Ð5Ñ is true for some particular 5 − : for this 5 , /5  "  5Þ
We need to prove that T Ð5  "Ñ must be true, that is À /5"  "  Ð5  "Ñ œ #  5
/5" œ /5 † /  Ð"  5Ñ † /  Ð"  5Ñ † # œ #  #5  #  5
Å
Å
Å
by the induction hypothesis
because /  #
because 5 − ß #5  5 .
So T Ð5  "Ñ is true.
By PMI, Ða8 − Ñ T Ð8Ñ is true. ñ
Example 5 Prove that if 8 is a natural number, then
product rule for derivatives.Ñ
Proof Let T Ð8Ñ be the statement:
i) T Ð"Ñ is the statement that
.
.B B
. 8
.B B
. 8
.B B
œ 8B8" Þ ÐYou many assume the
œ 8B8"
œ " † B"" œ " † B! œ "
This is true from calculus, but we assume that you could prove it using the definition of
.
derivative: for 0 ÐBÑ œ Bß .B
B œ lim 0 ÐB2Ñ2 0 ÐBÑ œ lim ÐB  22Ñ  B œ lim 22 œ lim " œ "Þ
2Ä!
2Ä!
2Ä!
2Ä!
ii) Assume that
. 5
.B B
. 5"
.B B
. 5
.
† BÑ œ Ð .B
B Ñ † B  B5 Ð .B
BÑ œ Ð5B5" ÑB  B5 Ð"Ñ œ 5B5  B5 œ Ð5  "ÑB5
Å
the product rule
by the induction hypothesis and
because T Ð"Ñ is true
Å
œ
œ 5B5" for some particular 5 −  (the induction hypothesis).
.
5
.B ÐB
œ Ð5  "ÑBÐ5"Ñ" , so T Ð5  "Ñ is true.
By PMI, it follows that Ða8 − Ñ T Ð8Ñ is true. ñ
Rephrasings of PMI and PCI
WOP is stated in terms of sets: every nonempty subset of  contains a smallest element. To
prove that the three statements are equivalent, we will reformulate PMI and PCI in the language
of set theory. Then each of PMI, PCI, WOP will be a statements about certain subsets of 
Suppose T Ð8Ñ is a statement about a natural number 8.
Principle of Mathematical Induction (PMI)
 i) T Ð"Ñ is true, and
If  ii) when T Ð5Ñ is true for some particular 5 − ,

then T Ð5  "Ñ must also be true
then a8 −  T Ð8Ñ is true
PMI rephrased as a statement about subsets of 
Suppose W © Þ
If
i) " − W , and
 ii) a5 Ðif 5 − W , then 5  " − WÑ
then W œ 
When we actually use PMI to prove a statement a8 −  T Ð8Ñ , we are
setting W œ Ö8 −  À T Ð8Ñ is true×. Then
i) checking that “T Ð"Ñ is true” is the same as checking that “" − W ,” and
ii) checking that “if T Ð5Ñ is true for some 5 , then T Ð5  "Ñ must also be true”
is the same as checking that “if 5 − W , then 5  " − W ”
______________________________________________________________________
Principle of Complete Induction (PCI)
 i) T Ð"Ñ is true, and
If  ii) when 5  " and T Ð8Ñ is true for every 8  5 ,

then T Ð5Ñ must also be true
then
a8 −  T Ð8Ñ is true
PCI rephrased as a statement about subsets of 
Suppose W © Þ
If
i) " − W , and
 ii) when 5  " and Ö8 À 8  5× © W , then 5 − W
ß
then W œ 
When we actually use PCI to prove a statement a8 −  T Ð8Ñ , we are
setting W œ Ö8 −  À T Ð8Ñ is true×. Then
i) checking that “T Ð"Ñ is true” is the same as checking that “" − W ,” and
ii) checking that
“when 5  " and T Ð8Ñ is true for every 8  5 , then T Ð5Ñ must also be true”
is the same as checking that
“when 5  " and 8 − W for every 8  5 , then 5 − WÞ
(See comments on the next page)
I stated PCI as I did on the preceding page (with two parts, i) and ii), because it's the set
theoretic version of what we usually do in practice.
However, it's possible to state the set theoretic version of PCI more efficiently (and this is
what the textbook does) without mentioning i) at all. Simply drop “5  "” from ii) and
rewrite PCI as follows:
Suppose W © 
(*)
If a5 −  ÐÖ8 −  À 8  5× © W Ê 5 − W×Ñß
then W œ 
According to (*), to prove that W œ  we should look at all 5 in  (not just 5  ", as in
Part ii) on the preceding page) and argue that Ö8 −  À 8  5× © W Ê 5 − W .
For 5 œ "ß what do we do to show that Ö8 −  À 8  5× © W Ê 5 − W is true? Notice
that Ö8 −  À 8  "× © W is true automatically, because Ö8 −  À 8  5× œ g.
So, the conditional statement Ö8 −  À 8  "× © W Ê " − W is true iff " − Wß that is,
iff T Ð"Ñ is true. And verifying the T Ð"Ñ is true is precisely what Sept i) says on the
previous page.
So, if we are using PCI and check that
(*)
a5 −  ÐÖ8 −  À 8  5× © W Ê 5 − W×Ñ
we have automatically completed Step i)
So why did I break off Step i) as a separate step? It wasn't necessary to do that. But
when 5 œ ", there are no 8's  5 , so when we try to show T Ð"Ñ is true, there aren't any
“previous values of 8” for which T Ð8Ñ is assumed to be true  that is, T Ð"Ñ has to be
established “from scratch” without any “induction assumptions” to work with and
therefore it's usually done in a different way than when 5  "Þ Therefore, in practice, we
might as well state the case 5 œ " separately. It's probably a little clearer that way, even
if not quite so efficient as (*).
As a reminder, we restate WOP here.
III. WOP (Well-Ordering Principle) Suppose E © .
If E Á g, then E contains a smallest element.
The Equivalence of PMI, PCI and WOP
Theorem PMI, PCI, and WOP are equivalent.
Perhaps you feel that WOP is easier to believe than the other two. But (to repeat) PMI,
PCI and WOP are equivalent statements: either all three are true statements in , or
none of them is true in .
Since  is constructed from set theory in a way that makes PMI true (as we shall see),
all three statements are true in .
Proof We will show that the three statements are equivalent by giving three separate arguments
to show that:
i) PMI Ê PCI
ii) PCI Ê WOP
iii) WOP Ê PMI.
i) Prove PMI Ê PCI
Assume that PMI is true. Let W © .
To prove PCI is true:
We assume that a5 −  ÐÖ8 −  À 8  5× © W Ê 5 − W×Ñ
and we need to show that W œ .
(*)
Strategy: We will show (using PMI) that for every 8, Ö"ß #ß ÞÞÞß 8× © W . If that is
true, then 8 − W for every 8, which tells us that  © WÞ Since we already
know W © ß we will conclude that W œ .
a) Since Ö8 À 8  "× œ g © W , (*) gives that " − WÞ So Ö"× © WÞ
b) Assume that for some particular value 5 ,, we have Ö"ß #ß ÞÞÞß 5× © W  that is,
assume Ö8 À 8  5  "× © WÞ By (*), we conclude that 5  " − WÞ Therefore
Ö"ß #ß ÞÞÞß 5ß 5  "× © WÞ
Summarizing: " − W +n. Ð if Ö"ß #ß ÞÞÞß 5× © W , then Ö"ß #ß ÞÞÞß 5  "× © W)
Using PMIß we conclude that for a8 Ö"ß #ß ÞÞÞß 8× © W . Therefore  © W , so (as
outlined in the strategy), W œ . ñ
ii) Prove PCI Ê WOP
Assume PCI is true. Suppose E © .
We want to prove WOP: if E Á g, then E contains a smallest element.
Strategy: We will prove the contrapositive of WOP instead.
Assume E contains no smallest element; we will prove that E œ gÞ
Let W œ   EÞ
i) Since E has no smallest element, " Â E, so " − W . Since " − W is true, the
conditional statement ÐÖ5 À 5  "×Ñ Ê Ð" − WÑ is true.
ii) Suppose 5  " and that Ö8 À 8  5× œ Ö"ß #ß ÞÞÞß 5  "× © W . By definition
of W , this means that "ß #ß ÞÞÞß 5  " are not in E. Therefore 5 Â E (if 5 − E, 5 would be
the smallest element in EÑß so 5 − WÞ
Because of i) and ii), PCI tells us that W œ , and therefore E œ gÞ ñ
iii) Prove WOP Ê PMI
Assume that WOP is true. Suppose W © . We want to prove PMI:
If " − W and a5 Ð5 − W Ê 5  " − WÑ, then W œ Þ
Strategy: PMI, as stated above, has the form T • U Ê Vß
which is equivalent to T • µ V Ê µ U. So PMI is equivalent to
If " − W and W Á , then µ Ða5ÑÐ5 − W Ê 5  " − WÑ
and this is equivalent to
If " − W and W Á , then
Ðb5Ñ Ð 5 − W • 5  " Â WÑ
(#)
Assuming WOP, we will prove (#).
Suppose " − W and W Á Þ Then   W Á g, so by WOP there is a smallest natural
number in   W : call it 8.
Since 8  Wß we know 8  ". Therefore 5 œ 8  " is still natural number, and since 8
is the smallest natural number not in W , we must have 8  " − WÞ Let 5 œ 8  "Þ Then
5 − W but 5  " œ 8  W . This proves (*) (see the strategy). ñ
There are minor generalizations of PMI and PCI that are easy to use. Essentially they say that
the “base case” for the induction can be any integer 5! ; we don't have to start at ".
We put this at the end of these notes because, in fact, we could easily do without them (see the
final two examples). For example, here is the Generalized PMI; there is a similar generalization
for PCI, which you can easily write down.
Notice that here ™ is used in the hypothesis instead of . The only reason for that is to allow the
base case (“starting point”) for the induction to possibly be a negative integer 8! . The induction
could start, say, at 8 œ  $Þ But in that case, the conclusion is NOT W œ ™; just that
W œ Ö8 − ™ À 8  $× œ “all integers the base case.”
Generalized Principle of Mathematical Induction (GPMI)
5! − ™. If
Suppose W © ™ and that
i) 5! − W , and
ii) for any 5 5! ß Ð5 − W Ê 5  " − WÑ
then W œ Ö8 − ™ À 8 5! ×
To use GPMI: If we want to prove that T Ð8Ñ is true for all 8 ( (or all 8  $) we would
i) Show that T Ð(Ñ is true (or that T Ð  $Ñ is trueÑ. Then
ii) Assume that T Ð5Ñ is true for some particular 5 ( (or some 5  $Ñ
and argue that T Ð5  "Ñ must also be true.
Example Prove that #8  8#  "!8  $ for all natural numbers 8 (
We use GPMI with the starting point (base case) 5! œ (Þ
i) "#) œ #(  (#  "!Ð(Ñ  $ œ "##ß so T Ð(Ñ is true
ii) Suppose 5 is some particular natural number with 5 ( and that T Ð5Ñ is true.
We need to show that T Ð5  "Ñ is true: #5"  Ð5  "Ñ#  "!Ð5  "Ñ  $
#5" œ #Ð#5 Ñ  #Ð5 #  "!5  $Ñ œ #5 #  #!5  ' œ Ð5 #  #5  "Ñ  5 #  ")5  &
Å
induction hypothesis
œ Ð5  "Ñ#  5 #  ")5  & œ Ð5  "Ñ#  "!Ð5  "Ñ  5 #  )5  &
œ ÒÐ5  "Ñ#  "!Ð5  "Ñ  $Ó  Ð 5 #  )5  )Ñ
Å
ÐThis part is T Ð5  "Ñà I deliberately rearranged terms, above, to make
T Ð5  "Ñ appearÑ
But 5 #  )5  ) œ Ð5  %Ñ#  #% ! because 5 (Þ Therefore
#5"  Ð5  "Ñ#  "!Ð5  "Ñ  $ß so T Ð5  "Ñ is true.
By GPMI, Ða8 − Ñ Ð 8 ( Ê #8  8#  "!8  $Ñ is true. ñ
Using Generalized GPMI is just a matter of convenience; we could avoid it altogether just by
making a substitution that translates the variable, followed by using PMI. For example, instead
of using GPMI to prove
ЇÑ
Ða8 − Ñ Ð 8 ( Ê #8  8#  "!8  $Ñ
we could “translate”; another way of saying the same thing is
Ða8 − Ñ Ð 8 " Ê #8'  Ð8  'Ñ#  "!Ð8  'Ñ  $Ñ
Since every natural number 8 automatically satisfies 8 "ß the last statement is equivalent to
Ї‡Ñ
Ða8 − Ñ #8'  Ð8  'Ñ#  "!Ð8  'Ñ  $,
which we can prove using PMI.
ÐBe sure you're convinced that Ð‡Ñ and Ї‡Ñ are equivalent.Ñ