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In logic a counterexample disproves the generalization, and does so rigorously in the fields of mathematics and philosophy. [1] For example, the fact that "student John Smith is not lazy" is a counterexample to the generalization "students are lazy", and both a counterexample to, and disproof of, the universal quantification "all students are ...
Formal mathematics is based on provable truth. In mathematics, any number of cases supporting a universally quantified conjecture, no matter how large, is insufficient for establishing the conjecture's veracity, since a single counterexample could immediately bring down the conjecture.
The assumption that if there is a counterexample, there is a minimal counterexample, is based on a well-ordering of some kind. The usual ordering on the natural numbers is clearly possible, by the most usual formulation of mathematical induction; but the scope of the method can include well-ordered induction of any kind.
Such counterexamples do not disprove a statement, however; they only show that, at present, no constructive proof of the statement is known. One weak counterexample begins by taking some unsolved problem of mathematics, such as Goldbach's conjecture , which asks whether every even natural number larger than 4 is the sum of two primes.
The conjecture was disproved in 1966, with a counterexample involving a count of only four different 5th powers summing to another fifth power: 27 5 + 84 5 + 110 5 + 133 5 = 144 5. Proof by counterexample is a form of constructive proof, in that an object disproving the claim is exhibited.
Models And Counter-Examples (Mace) is a model finder. [1] Most automated theorem provers try to perform a proof by refutation on the clause normal form of the proof problem, by showing that the combination of axioms and negated conjecture can never be simultaneously true, i.e. does not have a model.
To form a counterexample, take the smallest non-abelian group B ≅ S 3, the symmetric group on three letters. Let A denote the alternating subgroup , and let C = B / A ≅ {±1 }. Let q and r denote the inclusion map and the sign map respectively, so that
In this tiling of the plane by congruent squares, the green and violet squares meet edge-to-edge as do the blue and orange squares. In geometry, Keller's conjecture is the conjecture that in any tiling of n-dimensional Euclidean space by identical hypercubes, there are two hypercubes that share an entire (n − 1)-dimensional face with each other.
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