Odd/Even Integer Balance in Predecessor Tree: A Paradox
- In any path, even numbers out-number odd numbers. (e-steps have 2 evens; b-steps
have 2 evens and one odd; s-steps one even and one odd).
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- Every odd number in the predecessor tree has an infinite set of powers-of-2 multiples.
- So, it seems there must be more evens than odd in the predecessor tree.
- But there are equal numbers of evens and odds among the integers. !PARADOX!
- Start with the density of odd numbers in the predecessor tree: 1/2. (From
the infinite sum of the densities of all the elements of all the l.d.a.s)
- That's from using 2/d for the density of every instance of every dn+c formula
- The 2's multiple occurs at 2d for each formula, density 1/d, or 1/2 that of the odd integers.
- The 4's multiple ..., or 1/4 that of the odd integers.
- The infinite sum of 1/2 + 1/4 + 1/8 + .... is 1, or equal to that of the odd integers.
- Use of densities as the measure of the infinite sets succeeds in resolving the paradox.