Search results
Results from the WOW.Com Content Network
Description. The simplest and most common form of mathematical induction infers that a statement involving a natural number n (that is, an integer n ≥ 0 or 1) holds for all values of n. The proof consists of two steps: The base case (or initial case): prove that the statement holds for 0, or 1. The induction step (or inductive step, or step ...
Proof by exhaustion, also known as proof by cases, proof by case analysis, complete induction or the brute force method, is a method of mathematical proof in which the statement to be proved is split into a finite number of cases or sets of equivalent cases, and where each type of case is checked to see if the proposition in question holds. [1]
In proof by exhaustion, the conclusion is established by dividing it into a finite number of cases and proving each one separately. The number of cases sometimes can become very large. For example, the first proof of the four color theorem was a proof by exhaustion with 1,936 cases. This proof was controversial because the majority of the cases ...
The argument above makes the implicit assumption that the set of + horses has the size at least 3, [3] so that the two proper subsets of horses to which the induction assumption is applied would necessarily share a common element. This is not true at the first step of induction, i.e., when + =.
Structural induction is used to prove that some proposition P(x) holds for all x of some sort of recursively defined structure, such as formulas, lists, or trees. A well-founded partial order is defined on the structures ("subformula" for formulas, "sublist" for lists, and "subtree" for trees). The structural induction proof is a proof that the ...
Transfinite induction requires proving a base case (used for 0), a successor case (used for those ordinals which have a predecessor), and a limit case (used for ordinals which don't have a predecessor). Transfinite induction is an extension of mathematical induction to well-ordered sets, for example to sets of ordinal numbers or cardinal ...
Clearly the theorem is true if p > 0 and q = 0 when the probability is 1, given that the first candidate receives all the votes; it is also true when p = q > 0 as we have just seen. Assume it is true both when p = a − 1 and q = b, and when p = a and q = b − 1, with a > b > 0. (We don't need to consider the case. a = b {\displaystyle a=b}
Dilworth's theorem. In mathematics, in the areas of order theory and combinatorics, Dilworth's theorem states that, in any finite partially ordered set, the maximum size of an antichain of incomparable elements equals the minimum number of chains needed to cover all elements. This number is called the width of the partially order.