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60 = 2 × 2 × 3 × 5, the multiplicity of the prime factor 2 is 2, while the multiplicity of each of the prime factors 3 and 5 is 1. Thus, 60 has four prime factors allowing for multiplicities, but only three distinct prime factors.
One advantage of this proof over the others is that it shows not only that a polynomial must have a zero but the number of its zeros is equal to its degree (counting, as usual, multiplicity). Another use of Rouché's theorem is to prove the open mapping theorem for analytic functions.
An example where it does not is given by the isolated singularity of x 2 + y 3 z + z 3 = 0 at the origin. Blowing it up gives the singularity x 2 + y 2 z + yz 3 = 0. It is not immediately obvious that this new singularity is better, as both singularities have multiplicity 2 and are given by the sum of monomials of degrees 2, 3, and 4.
In the polynomial + the only possible rational roots would have a numerator that divides 6 and a denominator that divides 1, limiting the possibilities to ±1, ±2, ±3, and ±6. Of these, 1, 2, and –3 equate the polynomial to zero, and hence are its rational roots (in fact these are its only roots since a cubic polynomial has only three roots).
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 ...
A perfect power has a common divisor m > 1 for all multiplicities (it is of the form a m for some a > 1 and m > 1). The first: 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100 (sequence A001597 in the OEIS ). 1 is sometimes included.
However, these counterexamples rely on −1 having a square root. If we take a field where −1 has no square root, and every polynomial of degree n ∈ I has a root, where I is any fixed infinite set of odd numbers, then every polynomial f(x) of odd degree has a root (since (x 2 + 1) k f(x) has a root, where k is chosen so that deg(f) + 2k ∈ I).
It is also not a multiple of 5 because its last digit is 7. The next odd divisor to be tested is 7. One has 77 = 7 · 11, and thus n = 2 · 3 2 · 7 · 11. This shows that 7 is prime (easy to test directly). Continue with 11, and 7 as a first divisor candidate. As 7 2 > 11, one has finished. Thus 11 is prime, and the prime factorization is ...