induction

induction

Induction

is the name given to a certain kind of proof, and also to
a (related) way of defining a

function

.
For a proof, the statement to
be proved has a suitably

ordered

set of cases.
Some cases (usually one, but possibly zero or more than one), are proved
separately, and the other cases are deduced from those.
The

deduction

goes by

contradiction

, as we shall see.
For a function, its

domain

is suitably ordered.
The function is first defined on some (usually nonempty)
subset of its domain, and is then defined at other points

x

in terms of its values at points

y

such that

y

<

x

.

1

Elementary proof by induction

Proof by induction is a

variety

of

proof by contradiction

, relying,
in the

elementary

cases, on the fact that every non-empty set of
natural numbers has a least

element

.
Suppose we want to prove a statement

F

⁢

(

n

)

which involves a

natural
number

n

.
It is enough to prove:

1) If

n

∈

ℕ

, and

F

⁢

(

m

)

is true for

all

m

∈

ℕ

such that

m

<

n

, then

F

⁢

(

n

)

is true.

or, what is the same thing,

2) If

F

⁢

(

n

)

is false, then

F

⁢

(

m

)

is false for

some

m

<

n

.

To see why, assume that

F

⁢

(

n

)

is false for some

n

.
Then there is a

smallest

k

∈

ℕ

such that

F

⁢

(

k

)

is false.
Then, by

hypothesis

,

F

⁢

(

n

)

is true for all

n

<

k

.
By (1),

F

⁢

(

k

)

is true, which is a contradiction.

(If we don’t regard induction as a kind of proof by contradiction, then we
have to think of it as supplying some kind of

sequence

of proofs, of
unlimited length.
That’s not very satisfactory, particularly for

transfinite
inductions

, which we will get to below.)

Usually the initial case of

n

=

0

, and sometimes a few cases, need
to be proved separately, as in the following example.
Write

B

n

=

∑

k

=

0

n

k

2

.
We claim

B

n

=

n

3

3

+

n

2

2

+

n

6

⁢

for all

⁢

n

∈

ℕ

Let us try to apply (1). We have the

inductive hypothesis

(as it is called)

B

m

=

m

3

3

+

m

2

2

+

m

6

⁢

for all

⁢

m

<

n

which tells us something

if

n

>

0

. In particular, setting

m

=

n

-

1

,

B

n

-

1

=

(

n

-

1

)

3

3

+

(

n

-

1

)

2

2

+

n

-

1

6

Now we just add

n

2

to each side, and verify that the right side becomes

n

3

3

+

n

2

2

+

n

6

.
This proves (1) for nonzero

n

.
But if

n

=

0

, the inductive hypothesis is

vacuously true

, but of no use.
So we need to prove

F

⁢

(

0

)

separately, which in this case is

trivial

.

Textbooks sometimes distinguish between

weak

and

strong

(or

complete

) inductive proofs.
A proof that relies on the inductive hypothesis (1) is said to go
by strong induction. But in the sum-of-squares

formula

above,
we needed only the hypothesis

F

⁢

(

n

-

1

)

, not

F

⁢

(

m

)

for all

m

<

n

.
For another example, a proof about the

Fibonacci sequence

might
use just

F

⁢

(

n

-

2

)

and

F

⁢

(

n

-

1

)

.
An

argument

using only

F

⁢

(

n

-

1

)

is referred to as weak induction.

2

Definition of a function by induction

Let’s begin with an example, the function

ℕ

→

ℕ

,

n

↦

a

n

, where

a

is some

integer

>

0

.
The inductive definition reads

a

0

=

1

a

n

=

a

⁢

(

a

n

-

1

)

⁢

for all

⁢

n

>

0

Formally, such a

definition

requires some justification, which runs roughly
as follows.
Let

T

be the set of

m

∈

ℕ

for which the following
definition “has no problem”.

a

0

=

1

a

n

=

a

⁢

(

a

n

-

1

)

⁢

for

⁢

0

<

n

≤

m

We now have a

finite sequence

f

m

on the interval

[

0

,

m

]

, for
each

m

∈

T

.
We verify that any

f

l

and

f

m

have the same values
throughout the

intersection

of their two domains.
Thus we can define a single function on the

union

of the
various domains.
Now suppose

T

≠

ℕ

, and let

k

be the least element of

ℕ

-

T

.
That means that the definition has a problem when

m

=

k

but not when

m

<

k

.
We soon get a contradiction, so we deduce

T

=

ℕ

.
That means that the union of those domains is all of

ℕ

, i.e.
the function

a

n

is defined, unambiguously, throughout

ℕ

.

Another inductively defined function is the Fibonacci sequence, q.v.

We have been speaking of the inductive definition of a

function

,
rather than just a sequence (a function on

ℕ

), because the
notions extend with little change to transfinite inductions.
An illustration

par excellence

of inductive proofs and
definitions is Conway’s theory of

surreal numbers

.
The numbers and their

algebraic

laws of composition are defined
entirely by inductions which have no special starting cases.

3

Minor variations of the method

The reader can figure out what is meant by “induction starting at

k

”,
where

k

is not necessarily zero.
Likewise, the term “downward induction” is self-explanatory.

A common variation of the method is proof by induction on a

function

of the index

n

.
Rather than spell it out formally, let me just give an example.
Let

n

be a

positive integer

having no

prime factors

of the form

4

⁢

m

+

3

.
Then

n

=

a

2

+

b

2

for some integers

a

and

b

.
The usual textbook proof uses induction on a function of

n

, namely the number of prime factors of

n

.
The induction starts at

1

(i.e. either

n

=

2

or prime

n

=

4

⁢

m

+

1

), which in this

instance

is the only part of the proof that is not quite easy.

4

Well-ordered sets

An

ordered set

(

S

,

≤

)

is said to be

well-ordered

if any nonempty subset
of

S

has a least element.
The criterion (1), and its proof, hold without change for any well-ordered
set

S

in place of

ℕ

(which is a well-ordered set).
But notice that it won’t be enough to prove that

F

⁢

(

n

)

implies

F

⁢

(

n

+

1

)

(where

n

+

1

denotes the least element

>

n

, if it exists).
The reason is, given an element

m

, there may exist
elements

<

m

but no element

k

such that

m

=

k

+

1

.
Then the induction from

n

to

n

+

1

will fail to “reach”

m

.
For more on this topic, look for “

limit ordinals

”.

Informally, any variety of induction which
works for ordered sets

S

in which a segment

S

x

=

{

y

∈

S

|

y

<

x

}

may be

infinite

, is called “transfinite induction”.

5

Noetherian induction

An ordered set

S

, or its order, is called

Noetherian

if any non-empty
subset of

S

has a

maximal element

.
Several

equivalent

definitions are
possible, such as the “

ascending chain condition

”:
any strictly increasing sequence of elements of

S

is finite.
The following result is easily proved by contradiction.

Principle of Noetherian induction:

Let

(

S

,

≤

)

be
a set with a Noetherian order, and let

T

be a subset of

S

having
this

property

: if

x

∈

S

is such that the condition

y

>

x

implies

y

∈

T

, then

x

∈

T

. Then

T

=

S

.

So, to prove something “

F

⁢

(

x

)

” about every element

x

of a Noetherian set,
it is enough to prove that “

F

⁢

(

z

)

for all

z

>

y

” implies “

F

⁢

(

y

)

”.
This time the induction is going downward, but of course that is only a
matter of notation.
The opposite of a Noetherian order, i.e. an order in which any strictly
decreasing sequence is finite, is also in use; it is called a partial
well-order, or an ordered set having no infinite

antichain

.

The standard example of a Noetherian ordered set is the set of ideals
in a

Noetherian ring

.
But the notion has various other uses, in

topology

as well as

algebra

.
For a nontrivial example of a proof by Noetherian
induction, look up the

Hilbert basis theorem

.

6

Inductive ordered sets

An ordered set

(

S

,

≤

)

is said to be inductive if any

totally ordered

subset
of

S

has an

upper bound

in

S

.
Since the

empty set

is totally ordered,
any inductive ordered set is non-empty.
We have this important result:

Zorn’s lemma:

Any inductive ordered set has a maximal element.

Zorn’s lemma is widely used in

existence proofs

, rather than in proofs
of a property

F

⁢

(

x

)

of an arbitrary element

x

of an ordered set.
Let me sketch one typical application.
We claim that every

vector space

has a basis.
First, we prove that if a free subset

F

, of a vector space

V

,
is a

maximal

free subset (with respect to the order relation

⊂

), then it is a basis.
Next, to see that the set of free subsets is inductive, it is enough
to verify that the union of any

totally ordered set

of free subsets
is free, because that union is an upper bound on the totally ordered set.
Last, we apply Zorn’s lemma to conclude that

V

has a maximal free
subset.

Title

induction

Canonical

name

Induction

Date of creation

2013-03-22 12:24:27

Last modified on

2013-03-22 12:24:27

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

14

Author

Daume (40)

Entry type

Topic

Classification

msc 03B48

Classification

msc 03F07

Related topic

TransfiniteInduction

Related topic

PrincipleOfFiniteInduction

Defines

inductive order

Defines

partial well-order

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