Re: Surrogate Factoring Theorem

From: Bryan Olson (nameless_at_nowhere.org)
Date: 04/07/05 

Date: Thu, 07 Apr 2005 06:13:11 GMT

Proginoskes wrote:
>>Bryan Olson wrote:
>>>Proginoskes wrote:
>>> > When you're in the field of rational numbers, any number
>>> > is a factor of
>>> > another.
>>>
>>>Well, all but one of them.
>
> Of course.

[...]
> Given: An integer N
> Find: Two rationals p and q (neither of which is +/-1 or +/- N) such
> that p * q = N
>
> is MUCH easier than the following:
>
> Given: An integer N
> Find: Two integers p and q (neither of which is +/-1 or +/-N) such that
> p * q = N.
>
> And THAT'S what all the fuss about factoring is; any 6-year old who
> knows how to manipulate fractions can solve the second quickly (choose
> p = 2, q = N/2, and it'll always work), but NO ONE has been able to
> solve the first quickly.

I'm guessing I can score another "Of course" response, by
pointing out that the second and the first got mixed up
somewhere above. 

-- 
--Bryan

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