[defrost]

\1 is a good question that deserves an answer.

\2 is "not always" ..

Consider SumOf 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 ...

an infinite sequence of continuously decreasing numbers, the more you add the smaller the quantity added becomes.

It appears to approach but never reach some finite limit.

Unless, of course, by "Number" you mean "whole integer" | counting number, etc.

It's important to nail down those definitions.

[thaumasiotes]

> \1 is a good question that deserves an answer.

The same argument I mentioned above, that subtracting 0.99999... from 1 will give you a number that is equal to zero, will also tell you that binary ...11111 or decimal ...999999 is equal to negative one. If you add one to the value, you will get a number that is equal to zero.

You might object that there is an infinite carry bit, but in that case you should also object that there is an infinitesimal residual when you subtract 0.9999... from 1.

It works for everything, not just -1. The infinite bit pattern ...[phone] is, according to the geometric series formula, equal to -1/3 [1 + 4 + 16 + 64 + ... = 1 / (1-4)]. What happens if you multiply it by 3?

        ...0101010101
      x            11
    -------------------
        ...0101010101
     + ...01010101010
    -------------------
       ...11111111111

You get -1.

[Dylan16807]

But if you look at limits you get "0" and "diverges".

And decimal "...999999" is an infinity, which should immediately set off red flags and tell you that you need to be extra careful when analyzing it.

In computers your series of 1s is not infinite, there's a modulus that steps in. And this analysis depends on the modulus being an exact power of the base. But you could make a system that's decimal but has a modulus of 999853, for example, and then "-1" would be 999852.

[thaumasiotes]

> In computers your series of 1s is not infinite, there's a modulus that steps in. And this analysis depends on the modulus being an exact power of the base.

That isn't quite correct. The series of 1s really is conceptually infinite. That's why we have sign extension. The analysis (of the sum of all natural powers of 2) will work for any modulus that is an integral power of 2, including a modulus where the integer to which 2 is raised is infinitely large. Such an infinite modulus will still be evenly divided by a finite power of 2 -- as well as by itself -- and so it will disappear whenever you're working in any finite modulus that is a power of 2 -- or when you are working modulo 2^ℕ. The modulus of 2^ℕ will prevent any distinct finite integers from falling into the same equivalence class.

This is what enables you to have an infinite series of leading 1s, or leading patterns, without problems.