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Rational numbers have irrationality exponent 1, while (as a consequence of Dirichlet's approximation theorem) every irrational number has irrationality exponent at least 2. On the other hand, an application of Borel-Cantelli lemma shows that almost all numbers, including all algebraic irrational numbers , have an irrationality exponent exactly ...
This shows that any irrational number has irrationality measure at least 2. The Thue–Siegel–Roth theorem says that, for algebraic irrational numbers, the exponent of 2 in the corollary to Dirichlet’s approximation theorem is the best we can do: such numbers cannot be approximated by any exponent greater than 2.
Again this new bound is best possible in the new setting, but this time the number √ 2 is the problem. If we don't allow √ 2 then we can increase the number on the right hand side of the inequality from 2 √ 2 to √ 221 /5. Repeating this process we get an infinite sequence of numbers √ 5, 2 √ 2, √ 221 /5, ... which converge to 3. [1]
Dov Jarden gave a simple non-constructive proof that there exist two irrational numbers a and b, such that a b is rational: [28] [29] Consider √ 2 √ 2; if this is rational, then take a = b = √ 2. Otherwise, take a to be the irrational number √ 2 √ 2 and b = √ 2. Then a b = (√ 2 √ 2) √ 2 = √ 2 √ 2 · √ 2 = √ 2 2 = 2 ...
A method analogous to piece-wise linear approximation but using only arithmetic instead of algebraic equations, uses the multiplication tables in reverse: the square root of a number between 1 and 100 is between 1 and 10, so if we know 25 is a perfect square (5 × 5), and 36 is a perfect square (6 × 6), then the square root of a number greater than or equal to 25 but less than 36, begins with ...
In fact, 1 ⁄ 2 is such an ε. Because the irrational numbers are dense in the reals, no matter what δ we choose we can always find an irrational z within δ of y, and f(z) = 0 is at least 1 ⁄ 2 away from 1. If y is irrational, then f(y) = 0.
If the approximate ratio of two factors (/) is known, then a rational number / can be picked near that value. N u v = c v ⋅ d u {\displaystyle Nuv=cv\cdot du} , and Fermat's method, applied to Nuv , will find the factors c v {\displaystyle cv} and d u {\displaystyle du} quickly.
If n is even, a complex number's nth roots, of which there are an even number, come in additive inverse pairs, so that if a number r 1 is one of the nth roots then r 2 = –r 1 is another. This is because raising the latter's coefficient –1 to the n th power for even n yields 1: that is, (– r 1 ) n = (–1) n × r 1 n = r 1 n .