Encyclopedia
Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers. 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. Indeed, the validity of mathematical induction is logically equivalent to the well-ordering principle.
Mathematical induction should not be misconstrued as a form of inductive reasoning, which is considered non-rigorous in mathematics. In fact, mathematical induction is a form of deductive reasoning and is fully rigorous.
The first known proof by mathematical induction appears in
Francesco Maurolico's Arithmeticorum libri fuo . Maurolico used the technique to prove that the sum of the first
n odd integers is
n2.
Informal description
Mathematical induction is used to prove that every statement in an infinite sequence of statements is true. It is done by
- proving that the first statement in the infinite sequence of statements is true, and then
- proving that if any statement in the infinite sequence of statements is true, then so is the next one.
Formal description
The simplest and most common form of mathematical induction proves that a statement holds for all natural numbers
n and consists of two steps:
- The basis: showing that the statement holds when n = 0.
- The inductive step: showing that if the statement holds for n = m, then the same statement also holds for n = m + 1.
The proposition following the word "if" in the inductive step is called the
induction hypothesis . To perform the inductive step, one assumes the induction hypothesis and then uses this assumption to prove the statement for
n =
m + 1.
This method works by first proving the statement is true for a starting value, and then proving that the
process used to go from one value to the next is valid. If these are both proven, then any value can be obtained by performing the process repeatedly. It may be helpful to think of the domino effect; if you have a long row of dominoes standing on end, and you can be sure that:
- The first domino will fall
- Whenever a domino falls, its next neighbor will also fall,
then you can conclude that
all of the dominoes will fall, and this fact is inevitable.
Another analogy can be to consider an infinite set of identical lily pads, all equally spaced on a pond. If a frog wishes to traverse the pond, he must:
- Determine if the first lily pad will hold his weight.
- Prove that he can jump from one lily pad to another.
Thus, he can conclude that he can jump to all of the lily pads.
The basic assumption or
axiom of induction is, in logical symbols,
-
where
P is the proposition in question and
n is a natural number.
Step 1. prove
P - the formula holds for integer
0.
Step 2. prove that for all natural number
k,
P implies
P. To do this one assumes
P and shows that it implies
P. This does
not mean substituting into
P - this is a very common mistake, which consists in assuming what is to be proved.
Together 1 & 2 imply that
P holds for all
n greater than or equal to
0. In general if P is proved, where s can be a negative integer , then P holds for all n greater than or equal to s.
Example
Suppose we wish to prove the statement:
for all natural numbers
n; call this statement P. This is a simple formula for the sum of the positive natural numbers less than or equal to number
n. The proof that the statement is true for all natural numbers
n proceeds as follows.
Check if it is true for
n = 0. Notice that the empty sum is 0. And 0 / 2 = 0. So the statement is true for
n = 0. Thus we have that P holds.
Now we have to show that if the statement holds when
n =
m, then it also holds when
n =
m + 1. This can be done as follows.
Assume the statement is true for
n =
m, i.e.,
Adding
m + 1 to both sides gives
By algebraic manipulation we have for the right-hand side
Thus we have
This is just P. This proof is conditional: we made the assumption that P is true, and from that assumption we derived P. Symbolically, we have shown that
However, by induction:
- We have P, i.e. the equation holds if n has the value 0.
- Since P implies P, we get P.
- Similarly since P implies P, we get P.
- With P, P follows.
- From P, we get P.
- Et cetera.
- We may conclude that P holds for any natural number n. Q.E.D.
Variants
In practice, proofs by induction are often structured differently, depending on the exact nature of the property to be proved.
Starting at some other number
If we want to prove a statement not for all natural numbers but only for all numbers greater than or equal to a certain number
b then:
- Showing that the statement holds when n = b.
- Showing that if the statement holds for n = m = b then the same statement also holds for n = m + 1.
This can be used, for example, to show that
n2 > 2
n for
n = 3. A more substantial example is a proof that
In this way we can prove also claims
P that hold 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 that we intend to prove is
P then proving it with these two rules is equivalent with proving
P for all natural numbers
n with the first two steps.
Building on n = 2
In mathematics, many standard functions, including operations such as "+" and relations such as "=", are binary, meaning that they take two arguments. Often these functions possess properties that implicitly extend them to more than two arguments. For example, once addition
a +
b is defined and is known to satisfy the
associativity property +
c =
a + , then the trinary addition
a +
b +
c makes sense, either as +
c or as
a + . Similarly, many axioms and theorems in mathematics are stated only for the binary versions of mathematical operations and relations, and implicitly extend to higher-arity versions.
The identity relation "=" has an inherently binary character that can be expressed thus: if any
two horses are of the same color, then
all horses are of the same color.
Suppose that we wish to prove a statement about an
n-ary operation implicitly defined from a binary operation, using mathematical induction on
n. Then it should come as no surprise that the
n = 2 case carries special weight. Here are some examples.
Example: product rule for the derivative
In this example, the binary operation in question is multiplication . The usual product rule for the
derivative taught in
calculus states:
This can be generalized to a product of
n functions. One has
In each of the
n terms, just one of the factors is a derivative; the others are not.
When this general fact is proved by mathematical induction, the
n = 0 case is trivial, . The
n = 1 case is also trivial, And for each
n ≥ 3, the case is easy to prove from the preceding
n − 1 case. The real difficulty lies in the
n = 2 case, which is why that is the one stated in the standard product rule.
Example: Pólya's proof that there is no "horse of a different color"
In this example, the binary relation in question is equality, =, applied to color of horses. The argument is essentially identical to the one above, but the crucial
n = 2 case fails, causing the entire argument to be invalid.
In the middle of the 20th century, a commonplace colloquial locution to express the idea that something is unexpectedly different from the usual was "
That's a horse of a different color!". George Pólya posed the following exercise: Find the error in the following argument, which purports to prove by mathematical induction that all horses are of the same color:
- If there's only one horse, there's only one color.
- Suppose 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 and . Each is a set of only n horses, therefore with each there is only one color. But the two sets overlap, so there must be only one color among all n + 1 horses.
Beginning the induction at 1, the
n = 1 case is trivial , and the inductive step is correct in all cases
n = 3. However, the logic of the inductive step is incorrect when
n = 2, because the statement that "the two sets overlap" is false. Indeed, the
n = 2 case is clearly the crux of the matter; if one could prove the
n = 2 case, then all higher cases would follow from the transitive property of equality.
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 performs a basis step and an inductive step for
n, and in each of those performs a basis step and an inductive step for
m. See, for example, the proof of commutativity at
addition of natural numbers. More complicated arguments involving three or more counters are also possible.
Complete induction
Another generalization, called complete induction , allows that in the second step we assume not only that the statement holds for
n =
m but also that it is true for
n smaller than or equal to
m.
Perhaps surprisingly, in this form of induction, it is not necessary to prove that the proposition is true in the first case! That is because it is vacuously true that the proposition holds in all cases before the first case, simply because there are no cases before the first case. Note that the proof then of the step needs to be able to work with an empty antecedent; the first proof above is not of this kind .
This can be used, for example, to show that
where fib is the
nth Fibonacci number and F = /2 . Given that fib = fib + fib, it can be proven that that the statement holds for
m + 1 if we can assume that it already holds for both
m and
m − 1. .
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 prime numbers", and assume that for a given
m > 1 it 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 =
n1 n2, 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. Then
m is a product of products of primes; i.e. a product of primes. Note both that the base case was never explicitly considered, and that the hypothesis that
all smaller numbers than
m are products of primes was used, since the factors of
m are
a priori unknown.
This generalization, complete induction, can be derived from the ordinary mathematical induction described above. Suppose P is the statement that we intend to prove by complete induction. Let Q mean P holds for all
m such that 0 ≤
m =
n. Apply mathematical induction to Q. Since Q is just P, we have the base case. Now suppose Q is given and we wish to show Q. Notice that Q is the same as P and P and ... P. The hypothesis of complete induction tell us that this implies P. If we add that to Q, we get P and P and ... P and P which is just Q. So using mathematical induction, we get that Q holds for all natural numbers
n. But Q implies P, so we have the conclusion of strong induction, namely that P holds for all natural numbers
n.
Transfinite induction
The last two steps can be reformulated as one step:
- Showing that if the statement holds for all n < m then the same statement also holds for n = m.
This is in fact the most general form of mathematical induction and it can be shown that it is not only valid for statements about natural numbers, but for statements about elements of any well-founded set, that is, a set with a
partial order that contains no infinite descending chains .
This form of induction, when applied to
ordinals , 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 m is a minimal element, i.e. there is no element smaller than m
- when m has a direct predecessor, i.e. the set of elements which are smaller than m has a largest element
- when m has no direct predecessor, i.e. m is a so-called limit-ordinal
Strictly speaking, it is not necessary in transfinite induction to prove the basis, 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.
Proof or reformulation of mathematical induction
The principle of mathematical induction is usually stated as an axiom of the natural numbers; see Peano axioms. However, it can be proved in some logical systems. For instance, it can be proved if one assumes:
- The set of natural numbers is well-ordered.
- Every natural number is either zero, 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, we must show that if P is some proposition predicated of
n, and if:
- P holds and
- whenever P is true then P is also true
then P holds for all
n.
We first show that if P is true for all
k <
m, then P is also true. If
m is zero, then P is true. If
m =
k + 1, then P is true because
k <
m and so P is true which means that P is true. The rest follows from applying the principle transfinite induction .
Generalization
Transfinite induction Given a well-ordered set, W, and a proposition, P, such that whenever
m is such that P is true for all
k <
m, then P is also true. We can show that P is true for all
n. Let S be the subset of W for which P is false. We first show that S has no minimal element. Consider any element
m. Either P is true for every
k <
m or there is a
k <
m such that P is false. In the first case, P must be true and
m does not belong to S and so
m is not the least element of S. In the second case, P is false for some
k <
m and so
k is in S and smaller than
m. Thus ,
m cannot be the least element of S. Now if S has no minimal element and is a subset of a well-ordered set, then S must be empty and so P must be true for all
n.
See also
External links
References
- Donald Knuth. The Art of Computer Programming, Volume 1: Fundamental Algorithms, Third Edition. Addison-Wesley, 1997. ISBN 0-201-89683-4. Section 1.2.1: Mathematical Induction, pp.11–21.
- Kolmogorov, A. N. and Fomin, S. V.; Introductory Real Analysis, Section 1.3.8: Transfinite induction, pp. 28-29. Silverman, R. A. , Dover, New York, 1975. ISBN 0-486-61226-0.
