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Mathematical induction can be informally illustrated by reference to the sequential effect of falling dominoes. [1] [2]Mathematical induction is a method for proving that a statement () is true for every natural number, that is, that the infinitely many cases (), (), (), (), … all hold.
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 {,}, from validity for some r we deduce validity for +.
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 ]
The proof is divided into three steps. The idea is to substitute the assumed integral inequality into itself n times. This is done in Claim 1 using mathematical induction. In Claim 2 we rewrite the measure of a simplex in a convenient form, using the permutation invariance of product measures.
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.
Her goal was to use mathematical induction to prove that, for any given p, infinitely many auxiliary primes θ satisfied the non-consecutivity condition and thus divided xyz; since the product xyz can have at most a finite number of prime factors, such a proof would have established Fermat's Last Theorem. Although she developed many techniques ...
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.
Another method of proof is by mathematical induction: We loosen the condition p > q {\displaystyle p>q} to p ≥ q {\displaystyle p\geq q} . Clearly, the theorem is correct when p = q {\displaystyle p=q} , since in this case the first candidate will not be strictly ahead after all the votes have been counted (so the probability is 0).