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Fermat's little theorem and some proofs; Gödel's completeness theorem and its original proof; Mathematical induction and a proof; Proof that 0.999... equals 1; Proof that 22/7 exceeds π; Proof that e is irrational; Proof that π is irrational; Proof that the sum of the reciprocals of the primes diverges
For example, direct proof can be used to prove that the sum of two even integers is always even: Consider two even integers x and y. Since they are even, they can be written as x = 2a and y = 2b, respectively, for some integers a and b. Then the sum is x + y = 2a + 2b = 2(a+b). Therefore x+y has 2 as a factor and, by definition, is even. Hence ...
In 1954 Zarankiewicz claimed to have solved Turán's brick factory problem about the crossing number of complete bipartite graphs, but Kainen and Ringel later noticed a gap in his proof. Complex structures on the 6-sphere. In 1969 Alfred Adler published a paper in the American Journal of Mathematics claiming that the 6-sphere has no complex ...
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.
In mathematics, certain kinds of mistaken proof are often exhibited, and sometimes collected, as illustrations of a concept called mathematical fallacy.There is a distinction between a simple mistake and a mathematical fallacy in a proof, in that a mistake in a proof leads to an invalid proof while in the best-known examples of mathematical fallacies there is some element of concealment or ...
List of mathematical proofs; List of misnamed theorems; List of scientific laws; List of theories; Most of the results below come from pure mathematics, but some are from theoretical physics, economics, and other applied fields.
In which case, if P 1 (S) is the set of one-element subsets of S and f is a proposed bijection from P 1 (S) to P(S), one is able to use proof by contradiction to prove that |P 1 (S)| < |P(S)|. The proof follows by the fact that if f were indeed a map onto P ( S ), then we could find r in S , such that f ({ r }) coincides with the modified ...
The following is a rough outline of a proof. [1] First, some preliminary definitions: Write the polynomial f ( x ) {\displaystyle f(x)} as ∑ i = 0 n a i x b i {\displaystyle \sum _{i=0}^{n}a_{i}x^{b_{i}}} where we have integer powers 0 ≤ b 0 < b 1 < ⋯ < b n {\displaystyle 0\leq b_{0}<b_{1}<\cdots <b_{n}} , and nonzero coefficients a i ≠ ...