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Wyatt's Achievements


Lepton (1/13)



  1. I don’t know, it just feels wrong. Ugh, I’m not sure how to word.
  2. Wouldn’t the answer depend on whether or not 0 and 1 are included in the statement of each number? Say for example they are included, then 0 to 1 will always have two more numbers than 1 to infinity. Those numbers being 0 and 1. Every number that can be made between 0 and 1 (barring the two themselves) can be made in replication essentially speaking by removing {0.} from the front of it: so 0.1 would become one and so on. Admittedly this is perhaps over simplifying the process but I believe the general concept of it still works. However, if 0&1 are included then those would become the only numbers 1-infinity cannot replicate, as 1 would be matching the smallest number starting trailing only zeros until ending in 1. Leaving only zero and one left without a match because one to infinity does not end in a finite number. In the case where they aren’t included in either 0-1 still has more numbers but only by one, that being what I described as one’s normal match up above. [After thought]: for numbers that 0.0 then something i.e. 0.01 it’s probably best to consider the 1- infinity equivalent of natural numbers as putting the zero/zeros at the end of the equation after removing the decimal. So 0.01 would be 10, 0.0054 would be 5400 and so on. Sorry for the poor wording of it. Any idea of how to word it better or issues with the concept? Though admittedly, the layout of what’s included in each part makes to most sense to me set up as follows: [0,1) and [1, infinity) Even still [0,1) would still have one more number than [1, infinity) that being 0. The only way I see them being equal is the set up as follows: (0,1) and [1, infinity)
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