View Full Version : Why so many traders needed: Goldbach's Obviousity?

02-21-2002, 02:36 PM
Suppose we are challenged to fill every spot on the whole number, line using as few waves as possible. We could do it with just 1 wave of frequency 1, or 2 staggered waves of frequency 2. If the only frequency we had was 3, it would takes us 3 staggered waves, 4 would take 4 staggered waves, and so on.

If we weren't allowed to stagger, but had to synchronize all our waves to touch down at zero, it would take us a number of wave coefficients equal to one for every prime number up to as high as we wanted to fill the line. So suppose we were given the challenge to fill an entire stretch of number line, with the following restrictions:

1. Where any prime-number frequency wave touches down at zero, and it's first positive whole touchdown, don't count as a fill. We can only fill holes with products, not primes.

2. Zero isn't a frequency, and we aren't allowed to use 1.

3. We are given two sets of separate synchronized waves to work with or overlay, so that one can fill the holes in the other with its own products, counting in the opposite direction from a different zero. One starts at zero and moves right in whole-number increments, the other starts at any even-number increment to the right of zero on the first one, and counts left. Meaning, we can use 2 as a wave coefficient, but we aren't allowed to stagger by 1.

So, a prime number on one line is a hole on that line which must be filled with a product on the other line, for every hole between the two opposite-moving zeros. Can we do it, can we fill every hole? Or will there always be at least one common hole, meaning a spot that is prime whether moving right from our left zero, or left from our right zero, which starts an even-number distance from our left zero?

No we cannot do it, there is no way to fill every hole with these restrictions. No two primes on either line can be an odd number apart, so no prime frequency on one line can fill more than one prime hole on the other, and so we will need a unique prime frequency on the second line to fill every hole on the first line. And since 1 will be a prime on their line, but we can't use 1 on our line - notice we can't use zero either, but our 2's mutually cover their 2's and zeros - we will always end up at least one frequency short!

In other words, if we use a frequency to cover their 1, then we will leave another one of their primes uncovered by some frequency of ours in the form of a product. For every prime on one line to match off as a product on the other, there either has to be, like, a different prime for every prime, or the odd men out have to be at the same spot:)

Put differently, we have to use a frequency for 1, and then 1 as a frequency but, since 1 isn't available, at least two uncovered primes must be somewhere, and the only somewhere spot left is the same spot - duh. Otherwise, they'd each be at a prime frequency in the other guy's number line, meaning another prime couldn't be an odd number away, and would have to be left bare (or else even, and therefore not prime).

Since our 1 cannot match their 1, the total number of OR prime spots must be odd, meaning you need at least one AND to match them off, or something. Like, we each have N prime frequencies to try to fill N+1 primes, so we have to have to at least leave both our +1's unfilled in the same spot, and any uncovered prime must mean another uncovered prime somewhere, and if it's not at the same place, then we have two more, and so on...

It seems like if your stretch is an even number that is twice a prime, those two will midpoints will match off, or if the midpoints are both products, your AND holes will radiate around them, plus a hole at every place on an arbitrarily short stretch of line where your primes don't have as many chances to synch up.

Like, if your even numbers are on a frequency of a prime, like 3, it creates all kinds of weird problems and asymmetries and holes in the crystal, it keeps creating extraneous holes to offset with ANDs. Take 18, like your 11 frequency and your 7 frequency go unused, or something, so they have to be in the same place, not to mention, 5's, 13,s 17's and - of course - 1's! All unused frequencies create AND holes!

But, anyway, the point is you basically need an infinite number of people to fill every hole in a currency chart.


02-21-2002, 03:02 PM
check it out:



I've got more than two weeks left!


02-22-2002, 07:26 AM