4 ms·

Infinity is an infinitely high number. What is Infinity plus one? It must still be Infinity, since it's already infinitely high and can't go any higher. But

by naet 4y ago

Infinity is an infinitely high number. What is Infinity plus one? It must still be Infinity, since it's already infinitely high and can't go any higher. But if you have an odd number and add one, you get an even number (and vice versa). For Infinity = Infinity + 1 to hold true, Infinity must be both odd and even. Or something like that. It's probably more accurate to say Infinity is neither odd nor even. If you add 2+2+2+2... up to infinity, it might appear even. But that could also be equivalent to 1+3+1+3+1... to infinity, which would be alternating even and odd. Or 1+2+2+2+2... which would always be odd. Infinite quantities bend or break a lot of typical rules that other rational numbers follow.

Swizec 4y ago
If I remember my college set theory correctly, this is because there are infinite infinities. 2+2+2+… is a different infinity than 2+1+2+1+… You can construct infinite sets of numbers that do not overlap. Odd vs. Even infinity is the example our professor used in class to show the concept. I don’t remember enough to replicate here.
- greggertt 4y ago
  Wow way to kill the enjoyment of a nonsense verse
  - Swizec 4y ago
    Nah the verse was wonderful, I thoroughly enjoyed it.
- thaumasiotes 4y ago
  In the ordinal-numbers model that your notation leans toward, 2+2+2+... and 2+1+2+1+... are definitely the same infinity, both ω. On the other hand, you reminded me that conditionally convergent series may converge to different sums depending on how you choose to group their terms, which is fun. (That is, [(s_1 + s_2) + s_3 + (s_4 + s_5) + s_6 + ...] may have a different sum than [(s_1 + s_2) + (s_3 + s_4) + (s_5 + s_6) + ...] does.)† † I hope this is true; I can't find a cite for it. The more famous theorem is that you can get any sum you want by reordering the terms, but I think regrouping is good enough to get you different sums, if not arbitrary sums. ----- Thinking about it more, the idea would be that the sum of a series is of course the limit of the sequence of partial sums. By artfully parenthesizing the terms of an original series, we can reduce it to a new series with a different sequence of partial sums. This sequence must be a subsequence of the original sequence -- parenthesizing the series essentially just means ignoring certain partial sums. So then the question, if we're willing to start waving our hands, is "when (infinite) sequence S has limit L, and (infinite) sequence S' is a subsequence of S, must S' also have limit L?". The infinitude of S' is going to require the answer to be yes, so regrouping will end up not being good enough to change the sum of a conditionally convergent series. Sorry. :( However, if a series does not converge or diverge, such that its sequence of partial sums has multiple limit points, artfully parenthesizing the terms of that series should be able to get you a new series that converges on any limit point of the old one.
thaumasiotes 4y ago
> What is Infinity plus one? It must still be Infinity, since it's already infinitely high and can't go any higher. This is a mistake. That is how the extended reals behave. There are other systems. The diagonalization argument will already tell you that if you're at infinity, you can still go higher. This is also true for ordinal numbers and nonstandard analysis. It is the general case; the behavior in the extended reals is a special case. > But if you have an odd number and add one, you get an even number (and vice versa). For Infinity = Infinity + 1 to hold true, Infinity must be both odd and even. This is much worse; your conclusion is unrelated to your premise. You would need an additional premise, that infinity is either even or odd, which you didn't bother to supply. The same argument will tell you that either (a) 1.3 is even and 2.3 is odd, or (b) 1.3 is odd and 2.3 is even. (a) and (b) are both false.

by naet 4y ago

Swizec 4y ago

If I remember my college set theory correctly, this is because there are infinite infinities. 2+2+2+… is a different infinity than 2+1+2+1+… You can construct infinite sets of numbers that do not overlap. Odd vs. Even infinity is the example our professor used in class to show the concept. I don’t remember enough to replicate here.

greggertt 4y ago
Wow way to kill the enjoyment of a nonsense verse
- Swizec 4y ago
  Nah the verse was wonderful, I thoroughly enjoyed it.
thaumasiotes 4y ago
In the ordinal-numbers model that your notation leans toward, 2+2+2+... and 2+1+2+1+... are definitely the same infinity, both ω. On the other hand, you reminded me that conditionally convergent series may converge to different sums depending on how you choose to group their terms, which is fun. (That is, [(s_1 + s_2) + s_3 + (s_4 + s_5) + s_6 + ...] may have a different sum than [(s_1 + s_2) + (s_3 + s_4) + (s_5 + s_6) + ...] does.)† † I hope this is true; I can't find a cite for it. The more famous theorem is that you can get any sum you want by reordering the terms, but I think regrouping is good enough to get you different sums, if not arbitrary sums. ----- Thinking about it more, the idea would be that the sum of a series is of course the limit of the sequence of partial sums. By artfully parenthesizing the terms of an original series, we can reduce it to a new series with a different sequence of partial sums. This sequence must be a subsequence of the original sequence -- parenthesizing the series essentially just means ignoring certain partial sums. So then the question, if we're willing to start waving our hands, is "when (infinite) sequence S has limit L, and (infinite) sequence S' is a subsequence of S, must S' also have limit L?". The infinitude of S' is going to require the answer to be yes, so regrouping will end up not being good enough to change the sum of a conditionally convergent series. Sorry. :( However, if a series does not converge or diverge, such that its sequence of partial sums has multiple limit points, artfully parenthesizing the terms of that series should be able to get you a new series that converges on any limit point of the old one.

thaumasiotes 4y ago

> What is Infinity plus one? It must still be Infinity, since it's already infinitely high and can't go any higher. This is a mistake. That is how the extended reals behave. There are other systems. The diagonalization argument will already tell you that if you're at infinity, you can still go higher. This is also true for ordinal numbers and nonstandard analysis. It is the general case; the behavior in the extended reals is a special case. > But if you have an odd number and add one, you get an even number (and vice versa). For Infinity = Infinity + 1 to hold true, Infinity must be both odd and even. This is much worse; your conclusion is unrelated to your premise. You would need an additional premise, that infinity is either even or odd, which you didn't bother to supply. The same argument will tell you that either (a) 1.3 is even and 2.3 is odd, or (b) 1.3 is odd and 2.3 is even. (a) and (b) are both false.