Mathematical induction is a mathematical proof
technique. It is essentially used to prove
that a property P(n) holds for every natural
number n, i.e. for n = 0, 1, 2, 3, and so
on. Metaphors can be informally used to understand
the concept of mathematical induction, such
as the metaphor of falling dominoes or climbing
a ladder:
Mathematical induction proves that we can
climb as high as we like on a ladder, by proving
that we can climb onto the bottom rung (the
basis) and that from each rung we can climb
up to the next one (the step).
The method of induction requires two cases
to be proved. The first case, called the base
case (or, sometimes, the basis), proves that
the property holds for the number 0. The second
case, called the induction step, proves that,
if the property holds for one natural number
n, then it holds for the next natural number
n + 1. These two steps establish the property
P(n) for every natural number n = 0, 1, 2,
3, ... The base step need not begin with zero.
Often it begins with the number one, and it
can begin with any natural number, establishing
the truth of the property for all natural
numbers greater than or equal to the starting
number.
The method can be extended to prove statements
about more general well-founded structures,
such as trees; this generalization, known
as structural induction, is used in mathematical
logic and computer science. Mathematical induction
in this extended sense is closely related
to recursion. Mathematical induction, in some
form, is the foundation of all correctness
proofs for computer programs.Although its
name may suggest otherwise, mathematical induction
should not be misconstrued as a form of inductive
reasoning as used in philosophy (also see
Problem of induction). Mathematical induction
is an inference rule used in formal proofs.
Proofs by mathematical induction are, in fact,
examples of deductive reasoning.
== History ==
In 370 BC, Plato's Parmenides may have contained
an early example of an implicit inductive
proof. The earliest implicit traces of mathematical
induction may be found in Euclid's proof that
the number of primes is infinite and in Bhaskara's
"cyclic method". An opposite iterated technique,
counting down rather than up, is found in
the Sorites paradox, where it was argued that
if 1,000,000 grains of sand formed a heap,
and removing one grain from a heap left it
a heap, then a single grain of sand (or even
no grains) forms a heap.
An implicit proof by mathematical induction
for arithmetic sequences was introduced in
the al-Fakhri written by al-Karaji around
1000 AD, who used it to prove the binomial
theorem and properties of Pascal's triangle.None
of these ancient mathematicians, however,
explicitly stated the induction hypothesis.
Another similar case (contrary to what Vacca
has written, as Freudenthal carefully showed)
was that of Francesco Maurolico in his Arithmeticorum
libri duo (1575), who used the technique to
prove that the sum of the first n odd integers
is n2. The first explicit formulation of the
principle of induction was given by Pascal
in his Traité du triangle arithmétique (1665).
Another Frenchman, Fermat, made ample use
of a related principle, indirect proof by
infinite descent. The induction hypothesis
was also employed by the Swiss Jakob Bernoulli,
and from then on it became more or less well
known. The modern rigorous and systematic
treatment of the principle came only in the
19th century, with George Boole, Augustus
de Morgan, Charles Sanders Peirce, Giuseppe
Peano, and Richard Dedekind.
== Description ==
The simplest and most common form of mathematical
induction infers that a statement involving
a natural number n holds for all values of
n. The proof consists of two steps:
The base case: prove that the statement holds
for the first natural number n0. Usually,
n0 = 0 or n0 = 1; rarely, but sometimes conveniently,
the base value of n0 may be taken as a larger
number, or even as a negative number (the
statement only holds at and above that threshold),
because these extensions do not disturb the
property of being a well-ordered set).
The step case or inductive step: prove that
if the statement holds for any n ≥ n0, it
holds for n+1. In other words, assume the
statement holds for some arbitrary natural
number n ≥ n0, and prove that then the statement
holds for n + 1.The hypothesis in the inductive
step, that the statement holds for some n,
is called the induction hypothesis or inductive
hypothesis. To prove the inductive step, one
assumes the induction hypothesis and then
uses this assumption, involving n, to prove
the statement for n + 1.
Whether n = 0 or n = 1 is taken as the standard
base case depends on the preferred definition
of the natural numbers. In the fields of combinatorics
and mathematical logic it is common to consider
0 as a natural number.
== Example ==
Mathematical induction can be used to prove
that the following statement, P(n), holds
for all natural numbers n.
0
+
1
+
2
+
⋯
+
n
=
n
(
n
+
1
)
2
.
{\displaystyle 0+1+2+\cdots +n={\frac {n(n+1)}{2}}.}
P(n) gives a formula for the sum of the natural
numbers less than or equal to number n. The
proof that P(n) is true for each natural number
n proceeds as follows.
Base case: Show that the statement holds for
n = 0 (taking 0 as a natural).
P(0) is easily seen to be true:
0
=
0
⋅
(
0
+
1
)
2
.
{\displaystyle 0={\frac {0\cdot (0+1)}{2}}\,.}
Inductive step: Show that if P(k) holds, then
also P(k + 1) holds. This can be done as follows.
Assume P(k) holds (for some unspecified value
of k). It must then be shown that P(k + 1)
holds, that is:
(
0
+
1
+
2
+
⋯
+
k
)
+
(
k
+
1
)
=
(
k
+
1
)
(
(
k
+
1
)
+
1
)
2
.
{\displaystyle (0+1+2+\cdots +k)+(k+1)={\frac
{(k+1)((k+1)+1)}{2}}.}
Using the induction hypothesis that P(k) holds,
the left-hand side can be rewritten to:
k
(
k
+
1
)
2
+
(
k
+
1
)
.
{\displaystyle {\frac {k(k+1)}{2}}+(k+1)\,.}
Algebraically:
k
(
k
+
1
)
2
+
(
k
+
1
)
=
k
(
k
+
1
)
+
2
(
k
+
1
)
2
=
(
k
+
1
)
(
k
+
2
)
2
=
(
k
+
1
)
(
(
k
+
1
)
+
1
)
2
{\displaystyle {\begin{aligned}{\frac {k(k+1)}{2}}+(k+1)&={\frac
{k(k+1)+2(k+1)}{2}}\\&={\frac {(k+1)(k+2)}{2}}\\&={\frac
{(k+1)((k+1)+1)}{2}}\end{aligned}}}
thereby showing that indeed P(k + 1) holds.
Since both the base case and the inductive
step have been performed, by mathematical
induction the statement P(n) holds for all
natural numbers n. Q.E.D.
== Variants ==
In practice, proofs by induction are often
structured differently, depending on the exact
nature of the property to be proven.
=== Induction basis other than 0 or 1 ===
If one wishes to prove a statement not for
all natural numbers but only for all numbers
n greater than or equal to a certain number
b, then the proof by induction consists of:
Showing that the statement holds when n = b.
Showing that if the statement holds for some
n ≥ b then the same statement also holds
for n + 1.This can be used, for example, to
show that 2n ≥ n + 5 for n ≥ 3.
In this way, one can prove that some statement
P(n) holds for all n ≥ 1, or even n ≥ −5.
This form of mathematical induction is actually
a special case of the previous form, because
if the statement to be proved is P(n) then
proving it with these two rules is equivalent
with proving P(n + b) for all natural numbers
n with an induction base case 0.
==== Example: forming dollar amounts by coins
====
Assume an infinite supply of 4- and 5-dollar
coins. Induction can be used to prove that
any whole amount of dollars greater than
12
{\displaystyle 12}
can be formed by a combination of such coins.
The amount
n
{\displaystyle n}
is chosen to begin on
12
{\displaystyle 12}
as the statement does not hold true for every
lower number; in particular, it is violated
for
n
=
11
{\displaystyle n=11}
.
In more precise terms, we wish to show that
for any amount
n
≥
12
{\displaystyle n\geq 12}
there exist natural numbers
a
,
b
{\displaystyle a,b}
such that
n
=
4
a
+
5
b
{\displaystyle n=4a+5b}
, where 0 is included as a natural number.
The statement to be shown is thus:
S
(
n
)
:
n
≥
12
⇒
∃
a
,
b
∈
N
.
n
=
4
a
+
5
b
{\displaystyle S(n):\,\,n\geq 12\Rightarrow
\exists \,a,b\in \mathbb {N} .\,\,n=4a+5b}
Base case: Showing that
S
(
k
)
{\displaystyle S(k)}
holds for
k
=
12
{\displaystyle k=12}
is trivial: let
a
=
3
{\displaystyle a=3}
and
b
=
0
{\displaystyle b=0}
. Then,
4
⋅
3
+
5
⋅
0
=
12
{\displaystyle 4\cdot 3+5\cdot 0=12}
.
Step case: Given that
S
(
k
)
{\displaystyle S(k)}
holds for some value of
k
≥
12
{\displaystyle k\geq 12}
(induction hypothesis), prove that
S
(
k
+
1
)
{\displaystyle S(k+1)}
holds, too. That is, given that
k
=
4
a
+
5
b
{\displaystyle k=4a+5b}
for some natural numbers
a
,
b
{\displaystyle a,b}
, prove that there exist natural numbers
a
1
,
b
1
{\displaystyle a_{1},b_{1}}
such that
k
+
1
=
4
a
1
+
5
b
1
{\displaystyle k+1=4a_{1}+5b_{1}}
.
Here we need to consider two cases.
For the first case, assume that
a
≥
1
{\displaystyle a\geq 1}
. By some algebraic manipulation and by assumption,
we see that in that case
k
=
4
a
+
5
b
k
+
1
=
4
a
+
5
b
+
1
=
4
a
+
5
b
−
4
+
5
=
4
(
a
−
1
)
+
5
(
b
+
1
)
=
4
a
1
+
5
b
1
{\displaystyle {\begin{aligned}k&=4a+5b\\k+1&=4a+5b+1\\&=4a+5b-4+5\\&=4(a-1)+5(b+1)\\&=4a_{1}+5b_{1}\end{aligned}}}
where
a
1
=
a
−
1
{\displaystyle a_{1}=a-1}
and
b
1
=
b
+
1
{\displaystyle b_{1}=b+1}
are natural numbers.
This shows that to add
1
{\displaystyle 1}
to the total amount—any amount whatsoever,
so long as it is greater than
12
{\displaystyle 12}
—it is sufficient to remove a single 4-dollar
coin while adding a 5-dollar coin. However,
this construction fails in the case that
a
=
0
{\displaystyle a=0}
, or in words, when there is no 4-dollar coin.
So it remains to prove the case
a
=
0
{\displaystyle a=0}
. Then
k
=
5
b
≥
12
{\displaystyle k=5b\geq 12}
, which implies that
b
≥
3
{\displaystyle b\geq 3}
.
k
=
5
b
k
+
1
=
5
b
+
1
=
5
b
−
15
+
16
=
5
(
b
−
3
)
+
4
⋅
4
=
4
⋅
4
+
5
(
b
−
3
)
=
4
a
1
+
5
b
1
{\displaystyle {\begin{aligned}k&=5b\\k+1&=5b+1\\&=5b-15+16\\&=5(b-3)+4\cdot
4\\&=4\cdot 4+5(b-3)\\&=4a_{1}+5b_{1}\end{aligned}}}
where
a
1
=
4
{\displaystyle a_{1}=4}
and
b
1
=
b
−
3
{\displaystyle b_{1}=b-3}
are again natural numbers.
The above calculation shows that in the case
there are no 4-dollar coins, we can add
1
{\displaystyle 1}
to the amount by removing three 5-dollar coins
while adding four 4-dollar coins.
Thus, with the inductive step, we have shown
that
S
(
k
)
{\displaystyle S(k)}
implies
S
(
k
+
1
)
{\displaystyle S(k+1)}
for all natural numbers
k
≥
12
{\displaystyle k\geq 12}
, and the proof is complete. Q.E.D.
=== Induction on more than one counter ===
It is sometimes desirable to prove a statement
involving two natural numbers, n and m, by
iterating the induction process. That is,
one proves a base case and an inductive step
for n, and in each of those proves a base
case and an inductive step for m. See, for
example, the proof of commutativity accompanying
addition of natural numbers. More complicated
arguments involving three or more counters
are also possible.
=== Infinite descent ===
The method of infinite descent is a variation
of mathematical induction which was used by
Pierre de Fermat. It is used to show that
some statement Q(n) is false for all natural
numbers n. Its traditional form consists of
showing that if Q(n) is true for some natural
number n, it also holds for some strictly
smaller natural number m. Because there are
no infinite decreasing sequences of natural
numbers, this situation would be impossible,
showing by contradiction that Q(n) cannot
be true for any n.
The validity of this method can be verified
from the usual principle of mathematical induction.
Using mathematical induction on the statement
P(n) defined as "Q(m) is false for all natural
numbers m less than or equal to n", it follows
that P(n) holds for all n, which means that
Q(n) is false for every natural number n.
=== Prefix induction ===
The most common form of proof by mathematical
induction requires proving in the inductive
step 
that
∀
k
(
P
(
k
)
→
P
(
k
+
1
)
)
{\displaystyle \forall k(P(k)\to P(k+1))}
whereupon the induction principle "automates"
n applications of this step in getting from
P(0) to P(n). This could be called "predecessor
induction" because each step proves something
about a number from something about that number's
predecessor.
A variant of interest in computational complexity
is "prefix induction", in which one needs
to prove
∀
k
(
P
(
k
)
→
P
(
2
k
)
∧
P
(
2
k
+
1
)
)
{\displaystyle \forall k(P(k)\to P(2k)\land
P(2k+1))}
or equivalently
∀
k
(
P
(
⌊
k
2
⌋
)
→
P
(
k
)
)
{\displaystyle \forall k\left(P\left(\left\lfloor
{\frac {k}{2}}\right\rfloor \right)\to P(k)\right)}
The induction principle then "automates" log
n applications of this inference in getting
from P(0) to P(n). (It is called "prefix induction"
because each step proves something about a
number from something about the "prefix" of
that number formed by truncating the low bit
of its binary representation. It can be viewed
as an application of traditional induction
on the length of that binary representation.)
If traditional predecessor induction is interpreted
computationally as an n-step loop, prefix
induction corresponds to a log n-step loop,
and thus proofs using prefix induction are
"more feasibly constructive" than proofs using
predecessor induction.
Predecessor induction can trivially simulate
prefix induction on the same statement. Prefix
induction can simulate predecessor induction,
but only at the cost of making the statement
more syntactically complex (adding a bounded
universal quantifier), so the interesting
results relating prefix induction to polynomial-time
computation depend on excluding unbounded
quantifiers entirely, and limiting the alternation
of bounded universal and existential quantifiers
allowed in the statement.One can take the
idea a step further: one must prove
∀
k
(
P
(
⌊
k
⌋
)
→
P
(
k
)
)
{\displaystyle \forall k\left(P\left(\left\lfloor
{\sqrt {k}}\right\rfloor \right)\to P(k)\right)}
whereupon the induction principle "automates"
log log n applications of this inference in
getting from P(0) to P(n). This form of induction
has been used, analogously, to study log-time
parallel computation.
=== Complete (strong) induction ===
Another variant, called complete induction,
course of values induction or strong induction
(in contrast to which the basic form of induction
is sometimes known as weak induction) makes
the inductive step easier to prove by using
a stronger hypothesis: one proves the statement
P(m + 1) under the assumption that P(n) holds
for all natural n less than m + 1; by contrast,
the basic form only assumes P(m). The name
"strong induction" does not mean that this
method can prove more than "weak induction",
but merely refers to the stronger hypothesis
used in the inductive step; in fact the two
methods are equivalent, as explained below.
In this form of complete induction one still
has to prove the base case, P(0), and it may
even be necessary to prove extra base cases
such as P(1) before the general argument applies,
as in the example below of the Fibonacci number
Fn.
Although the form just described requires
one to prove the base case, this is unnecessary
if one can prove P(m) (assuming P(n) for all
lower n) for all m ≥ 0. This is a special
case of transfinite induction as described
below. In this form the base case is subsumed
by the case m = 0, where P(0) is proved with
no other P(n) assumed;
this case may need to be handled separately,
but sometimes the same argument applies for
m = 0 and m > 0, making the proof simpler
and more elegant.
In this method it is, however, vital to ensure
that the proof of P(m) does not implicitly
assume that m > 0, e.g. by saying "choose
an arbitrary n < m" or assuming that a set
of m elements has an element.
Complete induction is equivalent to ordinary
mathematical induction as described above,
in the sense that a proof by one method can
be transformed into a proof by the other.
Suppose there is a proof of P(n) by complete
induction. Let Q(n) mean "P(m) holds for all
m such that 0 ≤ m ≤ n". Then Q(n) holds
for all n if and only if P(n) holds for all
n, and our proof of P(n) is easily transformed
into a proof of Q(n) by (ordinary) induction.
If, on the other hand, P(n) had been proven
by ordinary induction, the proof would already
effectively be one by complete induction:
P(0) is proved in the base case, using no
assumptions, and P(n + 1) is proved in the
inductive step, in which one may assume all
earlier cases but need only use the case P(n).
==== Example: Fibonacci numbers ====
Complete induction is most useful when several
instances of the inductive hypothesis are
required for each inductive step. For example,
complete induction can be used to show that
F
n
=
φ
n
−
ψ
n
φ
−
ψ
{\displaystyle F_{n}={\frac {\varphi ^{n}-\psi
^{n}}{\varphi -\psi }}}
where Fn is the nth Fibonacci number, φ = (1
+ √5)/2 (the golden ratio) and ψ = (1 − √5)/2
are the roots of the polynomial x2 − x − 1.
By using the fact that Fn+2 = Fn+1 + Fn for
each n ∈ N, the identity above can be verified
by direct calculation for Fn+2 if one assumes
that it already holds for both Fn+1 and Fn.
To complete the proof, the identity must be
verified in the two base cases n = 0 and n
= 1.
==== Example: prime factorization ====
Another proof by complete induction uses the
hypothesis that the statement holds for all
smaller n more thoroughly. Consider the statement
that "every natural number greater than 1
is a product of (one or more) prime numbers",
which is the "existence" part of the fundamental
theorem of arithmetic. For proving the inductive
step, the induction hypothesis is that for
a given n > 1 the statement holds for all
smaller n > 1. If m is prime then it is certainly
a product of primes, and if not, then by definition
it is a product: m = n1n2, where neither of
the factors is equal to 1; hence neither is
equal to m, and so both are smaller than m.
The induction hypothesis now applies to n1
and n2, so each one is a product of primes.
Thus m is a product of products of primes;
therefore itself a product of primes.
==== Example: dollar amounts revisited ====
We shall look to prove the same example as
above, this time with a variant called strong
induction. The statement remains the same:
S
(
n
)
:
n
≥
12
⟹
∃
a
,
b
∈
N
.
n
=
4
a
+
5
b
{\displaystyle S(n):\,\,n\geq 12\implies \,\exists
\,a,b\in \mathbb {N} .\,\,n=4a+5b}
However, there will be slight differences
with the structure and assumptions of the
proof. Let us begin with the base case.
Base case: Show that
S
(
k
)
{\displaystyle S(k)}
holds for
k
=
12
,
13
,
14
,
15
{\displaystyle k=12,13,14,15}
.
4
⋅
3
+
5
⋅
0
=
12
4
⋅
2
+
5
⋅
1
=
13
4
⋅
1
+
5
⋅
2
=
14
4
⋅
0
+
5
⋅
3
=
15
{\displaystyle {\begin{aligned}4\cdot 3+5\cdot
0=12\\4\cdot 2+5\cdot 1=13\\4\cdot 1+5\cdot
2=14\\4\cdot 0+5\cdot 3=15\end{aligned}}}
The 
base case holds.
Induction hypothesis: Given some
j
>
15
{\displaystyle j>15}
such that
S
(
m
)
{\displaystyle S(m)}
holds for all
m
{\displaystyle m}
with
12
≤
m
<
j
{\displaystyle 12\leq m<j}
.
Inductive step: Prove that
S
(
j
)
{\displaystyle S(j)}
holds.
Choosing
m
=
j
−
4
{\displaystyle m=j-4}
, and observing that
15
<
j
⟹
12
≤
j
−
4
<
j
{\displaystyle 15<j\implies 12\leq j-4<j}
shows that
S
(
j
−
4
)
{\displaystyle S(j-4)}
holds, by inductive hypothesis. That is, the
sum
j
−
4
{\displaystyle j-4}
can be formed by some combination of
4
{\displaystyle 4}
and
5
{\displaystyle 5}
dollar coins. Then, simply adding a
4
{\displaystyle 4}
dollar coin to that combination yields the
sum
j
{\displaystyle j}
. That is,
S
(
j
)
{\displaystyle S(j)}
holds. Q.E.D.
=== Forward-backward induction ===
Sometimes it is more convenient to deduct
backwards, proving the statement for
n
−
1
{\displaystyle n-1}
, given its validity for
n
{\displaystyle n}
. However, proving the validity of the statement
for no single number suffices to establish
the base case; instead, one needs to prove
the statement for an infinite subset of the
natural numbers. For example, Augustin Louis
Cauchy first used forward (regular) induction
to prove the
inequality of arithmetic and geometric means
for all powers of 2, and then used backward
induction to show it for all natural numbers.
== Example of error in the inductive step
==
The inductive step must be proved for all
values of n. To illustrate this, Joel E. Cohen
proposed the following argument, which purports
to prove by mathematical induction that all
horses are of the same color:
Base case: In a set of only one horse, there
is only one color.
Inductive step: Assume as induction hypothesis
that within any set of n horses, there is
only one color. Now look at any set of n +
1 horses. Number them: 1, 2, 3, ..., n, n
+ 1. Consider the sets {1, 2, 3, ..., n} and
{2, 3, 4, ..., n + 1}. Each is a set of only
n horses, therefore within each there is only
one color. But the two sets overlap, so there
must be only one color among all n + 1 horses.The
base case n = 1 is trivial (as any horse is
the same color as itself), and the inductive
step is correct in all cases n > 1. However,
the logic of the inductive step is incorrect
for n = 1, because the statement that "the
two sets overlap" is false (there are only
n + 1 = 2 horses prior to either removal,
and after removal the sets of one horse each
do not overlap).
== Formalization ==
In second-order logic, we can write down the
"axiom of induction" as follows:
∀
P
(
P
(
0
)
∧
∀
k
(
P
(
k
)
⇒
P
(
k
+
1
)
)
⇒
∀
n
(
P
(
n
)
)
)
{\displaystyle \displaystyle \forall P{\Bigl
(}P(0)\land \forall k{\bigl (}P(k)\Rightarrow
P(k+1){\bigr )}\Rightarrow \forall n{\bigl
(}P(n){\bigr )}{\Bigr )}}
,where P(.) is a variable for predicates involving
one natural number and k and n are variables
for natural numbers.
In words, the base case P(0) and the inductive
step (namely, that the induction hypothesis
P(k) implies P(k + 1)) together imply that
P(n) for any natural number n. The axiom of
induction asserts that the validity of inferring
that P(n) holds for any natural number n from
the base case and the inductive step.
Note that the first quantifier in the axiom
ranges over predicates rather than over individual
numbers. This is a second-order quantifier,
which means that this axiom is stated in second-order
logic. Axiomatizing arithmetic induction in
first-order logic requires an axiom schema
containing a separate axiom for each possible
predicate. The article Peano axioms contains
further discussion of this issue.
The axiom of structural induction for the
natural numbers was first formulated by Peano,
who used it to specify the natural numbers
together with four other axioms saying that
(1) 0 is a natural number, (2) the successor
function s of every natural number yields
a natural number (s(x)=x+1), (3) the successor
function is injective, and (4) 0 is not in
the range of s.
In first-order ZFC set theory, quantification
over predicates is not allowed, but we can
still phrase induction by quantification over
sets:
∀
A
(
0
∈
A
∧
∀
k
∈
N
(
k
∈
A
⇒
(
k
+
1
)
∈
A
)
⇒
N
⊆
A
)
{\displaystyle \forall A{\Bigl (}0\in A\land
\forall k\in \mathbb {N} {\bigl (}k\in A\Rightarrow
(k+1)\in A{\bigr )}\Rightarrow \mathbb {N}
\subseteq A{\Bigr )}}
A
{\displaystyle A}
may be read as a set representing a proposition,
and containing natural numbers, for which
the proposition holds. This is not an axiom,
but a theorem, given that natural numbers
are defined in the language of ZFC set theory
by axioms, analogous to Peano's.
== Transfinite induction ==
The principle of complete induction is not
only valid for statements about natural numbers,
but for statements about elements of any well-founded
set, that is, a set with an irreflexive relation
< that contains no infinite descending chains.
Any set of cardinal numbers is well-founded,
which includes the set of natural numbers.
Applied to a well-founded set, it can be formulated
as a single step:
Show that if some statement holds for all
m < n, then the same statement also holds
for n.This form of induction, when applied
to a set of ordinals (which form a well-ordered
and hence well-founded class), is called transfinite
induction. It is an important proof technique
in set theory, topology and other fields.
Proofs by transfinite induction typically
distinguish three cases:
when n is a minimal element, i.e. there is
no element smaller than n;
when n has a direct predecessor, i.e. the
set of elements which are smaller than n has
a largest element;
when n has no direct predecessor, i.e. n is
a so-called limit ordinal.Strictly speaking,
it is not necessary in transfinite induction
to prove a base case, because it is a vacuous
special case of the proposition that if P
is true of all n < m, then P is true of m.
It is vacuously true precisely because there
are no values of n < m that could serve as
counterexamples. So the special cases are
special cases of the general case.
== Equivalence with the well-ordering principle
==
The principle of mathematical induction is
usually stated as an axiom of the natural
numbers; see Peano axioms. However, it can
be proved from the well-ordering principle.
Indeed, suppose the following:
The set of natural numbers is well-ordered.
Every natural number is either 0, or n + 1
for some natural number n.
For any natural number n, n + 1 is greater
than n.To derive simple induction from these
axioms, one must show that if P(n) is some
proposition predicated of n for which:
P(0) holds and
whenever P(m) is true then P(m + 1) is also
true,then P(n) holds for all n.
Proof. Let S be the set of all natural numbers
for which P(m) is false. Let us see what happens
if one asserts that S is nonempty. Well-ordering
tells us that S has a least element, say n.
Moreover, since P(0) is true, n is not 0.
Since every natural number is either 0 or
some m + 1, there is some natural number m
such that m + 1 = n. Now m is less than n,
and n is the least element of S. It follows
that m is not in S, and so P(m) is true. This
means that P(m + 1) is true; in other words,
P(n) is true. This is a contradiction, since
n was in S. Therefore, S is empty.
It can also be proved that induction, given
the other axioms, implies the well-ordering
principle.
Proof. Suppose there exists a non-empty set,
S, of naturals that has no least element.
Let P(n) be the assertion that n is not in
S. Then P(0) is true, for if it were false
then 0 is the least element of S. Furthermore,
suppose P(1), P(2),..., P(n) are all true.
Then if P(n+1) is false n+1 is in S, thus
being a minimal element in S, a contradiction.
Thus P(n+1) is true. Therefore, by the induction
axiom, P(n) holds for all n, so S is empty,
a contradiction.
== See also ==
Combinatorial proof
Recursion
Recursion (computer science)
Structural induction
Proof by exhaustion
== Notes
