Induction and Summation
Let P(n) be the property that
We use induction to prove that P(n) is true for all integers .
Basis step. If n=1, then
Inductive step. Assume as induction hypothesis that P(n) is true for some arbitrary, but fixed integer . i.e.,
We have to show that P(n+1) is also true, or equivalently
We have
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Induction and Inequality
Let P(n) be the property
We prove that P(n) is true for all .
Basis step. If n=3, then
Inductive step. Assume as induction hypothesis that P(n),
is true for some arbitrary, but fixed integer . We have to show that P(n+1),
is also true.
We have
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Another Induction Proof
Let P(n) be the property
where a and b are some nonnegative real numbers. We prove that P(n) is true for all .
Basis step. If n=0 we have to prove
which is true because .
Inductive step. Assume as induction hypothesis that P(n),
is true for some arbitrary, but fixed integer . We have to show that P(n+1),
is also true. To prove the latter implication, let us assume (1) . We then need to show .
From the induction hypothesis and (1) we may infer (2) by modus ponens.
Since a is nonnegative, we get (3) from (2).
Since is nonnegative, we get (4) from (1).
Putting (3) and (4) together, we obtain
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Strong Mathematical Induction
The inductive step of mathematical induction requires one to show that P(k+1) is true whenever P(k) is true. This can be generalized as follows.
Principle of Strong Mathematical Induction
Let P(n) be a predicate defined on the natural numbers and let a and b be natural numbers with . Suppose the following two statements are true:Again the first part is called the basis step, and the second part, the inductive step.Then P(n) is true for all natural numbers n with .
- are true.
- For all natural numbers , if P(i) is true for all natural numbers i with , then P(k+1) is true.
This principle looks stronger than plain mathematical induction, because the inductive hypothesis
P(i) is true for all natural numbers i with ,is more general.
But one can show that strong mathematical induction can be derived from the standard mathematical induction principle.
Divisibility by a Prime
Let P(n) be the property
n is divisible by a prime number.We prove that P(n) is true for all integers n with n> 1.
Basis step. If n=2, then P(n) is true because 2 is a prime and every number divides itself.
Inductive step. Let k be some integer with and assume, as induction hypothesis, that P(i) is true for all integers i with .
We have to show that P(k+1) is true, i.e., that k+1 is divisible by a prime.
We distinguish two cases.
Case (i). If k+1 is a prime, it is divisible by a prime (namely by itself).
Case (ii). If k+1 is not a prime, then it is a product of two integers s and t, such that 1 < s < k+1 and 1 < t < k+1.
By the induction hypothesis, t is divisible by a prime. By the transitivity of divisibility, is also divisible by a prime.
We have shown that in either case k+1 is divisible by a prime, which completes the proof. height6pt width4pt
The strong induction principle was applied here for values a=b=2.
Inductively Defined Structures
Many data structures are formally defined by mathematical induction. We have seen several examples, for instance, lists of elements of a certain type t.
Basis.
The empty list, denoted by [] or nil, is a t-list.
Induction.
If L is a t-list and a is an object of type t, then a::L is a t-list.
The key observation is that such a definition uniquely specifies a set of data objects and that the inherent inductive structure can be used to prove properties about the data type or functions defined on it.
Another example is the set of all strings that can be composed from characters of a given set . This set can be defined inductively as follows:
Basis.
The empty string, denoted by , is an element of .
Induction.
If a is a symbol in and s is a string in , then is also a string in .