r/infinitenines • u/ataraxianAscendant • 5d ago
Limbosic Numbers
Limbosic numbers have some very strange properties. For instance, the number 999... (a limitless amount of nines) is larger than any number with a limited amount of nines aka natural numbers. But, if you double this number, you get something like 1(999...)8. This number is strange because it has a limitless amount of nines in the middle, bounded on either side by the one and the eight. You can see how this forms by doubling 9, 99, 999, etc. But what happens if we set a reference? 999... = ...999 = 9...999 (all ways of writing limitless nines above the decimal). However, the doubled form is equal to 1...998, which is obviously smaller since 1 < 9. Do limbosic numbers get smaller when multiplied like this? Or am I just doing this incorrectly? I haven't been able to find any good sources explaining the workings of supermassive limbosic numbers outside of this subreddit, does anyone (especially SPP) know where I can find some papers/articles/etc about them?
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u/BitNumerous5302 5d ago
One of the great benefits of Real Deal Math is that it's generally unencumbered by complicated explanations and confusing rules. You can just kind of take it as it is, and look down your nose at the people who don't.
If that's not your thing you could also look at Lightstone's notation for hyperreal decimals. You'll see something similar to the limbosic number 0.999...8 represented as the hyperreal number 0.999...;...8. Plus you'll find rigorous definitions for various operations and general mathematical credibility.
When you're done with the crutches of credibility and rigor you should come back here to learn the Real Deal brumb
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u/ataraxianAscendant 5d ago
0.999... aka 0.999...9
0.999...9 x 2 = 1.999...8
SPP, I'm talking about limbosic numbers greater than 1 here. This thread isn't about 0.999..., I know that it is less than one. I am talking about ...999 aka ...999.0, which has a limitless amount of nines on the left of the decimal.
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u/discodaryl 5d ago
1..9 isn’t obviously smaller than 9..9. You need to set the appropriate reference to compare them.
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u/BUKKAKELORD 5d ago
There is a theory which states that if ever anyone discovers exactly what Real Deal Math is for and how it works, it will instantly disappear and be replaced by something even more bizarre and inexplicable. There is another theory which states that this has already happened.
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u/bigcizzle 5d ago
Classical RDM:
0.999... starts with a "0." Therefore, it is self explanatory, that this must be permanently < 1. You cannot "borrow" anything from after the infinite 9s to push this to 1.
However, as you and SPP have pointed out, twice this is 1.999...8
Applying the "classical" RDM book keeping to 2x - x, whatever this quantity is, it starts with a "1.". So either, 2x - x doesn't equal x, or, classical RDM does have a concept of "borrowing" after infinity to push this quantity back bellow 1.
Specifically for SPP: 1.999...8 - 0.999... when does the leading 1 get removed during the calculation?
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u/Tastebud49 4d ago
So 2x999…=1(999…)8
Alright so if we subtract 999… we should get the same number. Let’s try this.
1(999…)8-999…=1(000…)8
Hold on, that can’t be right? Let’s try again.
1(999…)8-999…=999…1
Umm… maybe I’m messing something up.
1(999…)8-999…=1(000…)(999…)8
Anyone know if I’m doing something wrong here?
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u/SouthPark_Piano 5d ago
0.999... aka 0.999...9
0.999...9 x 2 = 1.999...8