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For instance, the first counterexample must be odd because f(2n) = n, smaller than 2n; and it must be 3 mod 4 because f 2 (4n + 1) = 3n + 1, smaller than 4n + 1. For each starting value a which is not a counterexample to the Collatz conjecture, there is a k for which such an inequality holds, so checking the Collatz conjecture for one starting ...
For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.
In general, whenever we multiply both sides of an equation by an expression involving variables, we introduce extraneous solutions wherever that expression is equal to zero. But it is not sufficient to exclude these values, because they may have been legitimate solutions to the original equation.
However, the linear congruence 4x ≡ 6 (mod 10) has two solutions, namely, x = 4 and x = 9. The gcd(4, 10) = 2 and 2 does not divide 5, but does divide 6. Since gcd(3, 10) = 1, the linear congruence 3x ≡ 1 (mod 10) will have solutions, that is, modular multiplicative inverses of 3 modulo 10 will exist. In fact, 7 satisfies this congruence (i ...
It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1. Or x and y can both be treated as unknowns, and then there are many solutions to the equation; a symbolic solution is (x, y) = (a + 1, a), where the variable a may take any value. Instantiating a symbolic solution with specific numbers ...
To date, the only Millennium Prize problem to have been solved is the Poincaré conjecture. The Clay Institute awarded the monetary prize to Russian mathematician Grigori Perelman in 2010. However, he declined the award as it was not also offered to Richard S. Hamilton , upon whose work Perelman built.
[e] This notion of the division a(x)/b(x) results in two polynomials, a quotient q(x) and a remainder r(x), such that a = b q + r and degree(r) < degree(b). The quotient and remainder may be computed by any of several algorithms, including polynomial long division and synthetic division .
Hilbert's tenth problem is the tenth on the list of mathematical problems that the German mathematician David Hilbert posed in 1900. It is the challenge to provide a general algorithm that, for any given Diophantine equation (a polynomial equation with integer coefficients and a finite number of unknowns), can decide whether the equation has a solution with all unknowns taking integer values.