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To prove the induction step, one assumes the induction hypothesis for n and then uses this assumption to prove that the statement holds for n + 1. Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.
Bernoulli's inequality can be proved for case 2, in which is a non-negative integer and , using mathematical induction in the following form: we prove the inequality for r ∈ { 0 , 1 } {\displaystyle r\in \{0,1\}} ,
In proof by mathematical induction, a single "base case" is proved, and an "induction rule" is proved that establishes that any arbitrary case implies the next case. Since in principle the induction rule can be applied repeatedly (starting from the proved base case), it follows that all (usually infinitely many) cases are provable. [ 15 ]
All horses are the same color is a falsidical paradox that arises from a flawed use of mathematical induction to prove the statement All horses are the same color. [1] There is no actual contradiction, as these arguments have a crucial flaw that makes them incorrect.
The proof uses mathematical induction. The case n = 1 is simply the standard version of Rolle's theorem. For n > 1, take as the induction hypothesis that the generalization is true for n − 1. We want to prove it for n. Assume the function f satisfies the hypotheses of the theorem.
We prove commutativity (a + b = b + a) by applying induction on the natural number b. First we prove the base cases b = 0 and b = S(0) = 1 (i.e. we prove that 0 and 1 commute with everything). The base case b = 0 follows immediately from the identity element property (0 is an additive identity), which has been proved above: a + 0 = a = 0 + a.
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 numbers.
In naive set theory, one would go on to prove by transfinite induction that each ordinal is the order type of the natural order on the ordinals less than , which would imply an contradiction since by definition is the order type of all ordinals, not any proper initial segment of them.