1 = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + ...
1/2 = 1/3 + 1/9 + 1/27 + ...
and each term has a similar construction where it can be represented by the sum of smaller numbers and so on.
thus 1 = lim n -> inf (1/n + 1/n + ...) = 0 + 0 + 0 + 0 +...
0 + 0 = 0
this 1 = 0.
However we're just making this assumption, it's a pretty standard assumption since we often use assumptions likes these with infinite series, we made it earlier claiming that the series of 1/2^n for n \in N >0 is 1.
However we can use hitomi density reasoning to make it reasonable.
As we should all know if you've ever browsed on /sci/ before Hitomi's number claims to be the most dense number such that induction or any standard arithmetic is arbitary when applied to it. The algebra isn't exactly closed, but its existence is all we need here.
the series 1/2^n as n increases 2^n becomes infinitely dense where it is now stuck in the Hitomi set and can now be considered a Hitomi number h. 1/h = 0 closing the series.
Thus we can claim the series is completely one.
so one can be rewritten as 1 = 1/h + 1/h + 1/h +...
well if the number that defines it's length is also h then h*1/h = 1.
And now we can all admit 0 * infinity is 1.
How can we pretend the delta function makes sense if this isn't true?
So I offer an amendment:
a completion of some sort.
That h*1/h only describes one and no other number.
One of the things we understand about Hitomi numbers is that by its existence regular arithmetic doesn't work very well on it.
h + 1 = h.
h plus infinitely dense amount of ones is h.
but this concept applies for any number infinitely dense number of 2s, 3s, ns.
so h + h = h thus 2h = h.
h*1/h only describes one.
If you want to merge the two algebras you need to accept the order.
2*{h*1/h} = 2
2 cannot merge with h*1/h because you're applying two different algebras.
One of infinities and one of definite numbers.
1/2 = 1/3 + 1/9 + 1/27 + ...
and each term has a similar construction where it can be represented by the sum of smaller numbers and so on.
thus 1 = lim n -> inf (1/n + 1/n + ...) = 0 + 0 + 0 + 0 +...
0 + 0 = 0
this 1 = 0.
However we're just making this assumption, it's a pretty standard assumption since we often use assumptions likes these with infinite series, we made it earlier claiming that the series of 1/2^n for n \in N >0 is 1.
However we can use hitomi density reasoning to make it reasonable.
As we should all know if you've ever browsed on /sci/ before Hitomi's number claims to be the most dense number such that induction or any standard arithmetic is arbitary when applied to it. The algebra isn't exactly closed, but its existence is all we need here.
the series 1/2^n as n increases 2^n becomes infinitely dense where it is now stuck in the Hitomi set and can now be considered a Hitomi number h. 1/h = 0 closing the series.
Thus we can claim the series is completely one.
so one can be rewritten as 1 = 1/h + 1/h + 1/h +...
well if the number that defines it's length is also h then h*1/h = 1.
And now we can all admit 0 * infinity is 1.
How can we pretend the delta function makes sense if this isn't true?
So I offer an amendment:
a completion of some sort.
That h*1/h only describes one and no other number.
One of the things we understand about Hitomi numbers is that by its existence regular arithmetic doesn't work very well on it.
h + 1 = h.
h plus infinitely dense amount of ones is h.
but this concept applies for any number infinitely dense number of 2s, 3s, ns.
so h + h = h thus 2h = h.
h*1/h only describes one.
If you want to merge the two algebras you need to accept the order.
2*{h*1/h} = 2
2 cannot merge with h*1/h because you're applying two different algebras.
One of infinities and one of definite numbers.
