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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 ...
However, the failure to find a counterexample after extensive search does not constitute a proof that the conjecture is true—because the conjecture might be false but with a very large minimal counterexample. Nevertheless, mathematicians often regard a conjecture as strongly supported by evidence even though not yet proved.
It was first conjectured in 1939 by Ott-Heinrich Keller, [1] and widely publicized by Shreeram Abhyankar, as an example of a difficult question in algebraic geometry that can be understood using little beyond a knowledge of calculus. The Jacobian conjecture is notorious for the large number of attempted proofs that turned out to contain subtle ...
In his fifth and final supplement, published in 1904, he proved this with the counterexample of the Poincaré homology sphere, which is a closed connected three-dimensional manifold which has the homology of the sphere but whose fundamental group has 120 elements. This example made it clear that homology is not powerful enough to characterize ...
For example, a particular statement may be shown to imply the law of the excluded middle. An example of a Brouwerian counterexample of this type is Diaconescu's theorem, which shows that the full axiom of choice is non-constructive in systems of constructive set theory, since the axiom of choice implies the law of excluded middle in such systems.
This is now known to be false. The first counterexample was constructed by Atiyah & Hirzebruch (1961). Using K-theory, they constructed an example of a torsion cohomology class—that is, a cohomology class α such that nα = 0 for some positive integer n —which is not the class of an algebraic cycle. Such a class is necessarily a Hodge class.
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In taxicab geometry, however, only SASAS guarantees triangle congruence. [11] Take, for example, two right isosceles taxicab triangles whose angles measure 45-90-45. The two legs of both triangles have a taxicab length 2, but the hypotenuses are not congruent. This counterexample eliminates AAS, ASA, and SAS.
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