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Topic: Primes Are Fundamentally Additive/Subtractive, Not Multiplicative/Divisive
Replies: 2   Last Post: Mar 8, 2009 8:18 PM

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Osher Doctorow Ph.D.

Posts: 30
From: Southern California
Registered: 3/10/07
Primes Are Fundamentally Additive/Subtractive, Not Multiplicative/Divisive
Posted: Mar 8, 2009 6:39 PM
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>From Osher Doctorow

Wikipedia online has many links and keywords to the various work of Pierre de Fermat as well as to Fibonacci Numbers, Lucas Numbers, and so on, which indicate that Prime numbers are fundamentally additive/subtractive rather than multiplicative/divisive despite what most mathematicians have usually thought as the "overwhelming" evidence of the Fundamental Theorem of Arithmetic.

In reading through Lucas and Fibonacci Numbers, Pell Numbers, and so on, as well as many of Fermat's equations and methods and numbers, you will usually find a preponderance of additive/subtractive expressions rather than multiplicative/divisive. For example, both Fibonacci and Lucas numbers are generated additively and recursively beyond the first 1 or 2 numbers by:

1) Fn = Fn-1 + Fn-2 (n, n-1, n-2 are subscript here), n > 2

and similarly for Lucas (Ln) numbers.

Fermat's method of calculating primes, and also (1) above are essentially additive/subtractive, although (1) doesn't yield all the primes and also yields a considerable number of non-primes. However, (1) yields the first 5 primes in rapid succession, and with similar numbers such as Lucas and Pell and others, we soon get more and more primes.

There is also a strong relationship between additive/subtractive generation and exponential functions and additive/subtractive probability, on the one hand, and between multiplicative/divisive and logarithmic functions and multiplicative/divisible probability on the other hand. The former type of probability, expressed as P(A-->B) = 1 + P(AB) - P(A) and its alternate form P ' (A-->B) = 1 + P(B) - P(A) for P(B) > = P(A), is contrasted with the latter type of probability or "Conditional Probability", P(B|A) = P(AB)/P(A) with P(A) not 0, although I've discovered a variant of the second which is P ' (B|A) = P(B)/P(A) with P(B) < = P(A) if P(A) is not 0. See my thread "Quantum Gravity", now in its over 300th Section on the usenet Forum sci.physics. I will try to discuss some of these relationships and more details in later posts.

Osher Doctorow



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