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Original post on Math with Bad Drawings
Archivistâs note: Commenter Jefftk suggested using parentheses instead of overlines for representation in text. Thus, for the sake of clarity here, notation such as:
_ 0.91
will be notated as
0.(9)1
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You never know where conversations will turn with Jeff Kaufman. He goes barefoot, hates cucumbers, and donates fantastic chunks of his Google salary to charity. Heâs fun.
One night over ice cream, we found ourselves talking about repeating decimals. After we filled a napkin with the proof that 0.999... = 1, Jeff had a brainstorm, and wrote the following:
0.(9)4
âWhat the heck is that?â I asked.
âItâs a number where the nines go on forever,â he said. âAnd afterwards, thereâs a four.â
âAfter forever?â
âYes,â he said. âExactly.â
âI see. So itâs bigger than this...â
0.(9)3
âBut smaller,â I continued, âthan this.â
0.(9)5
âWell, clearly,â he said. âAll three numbers are identical up until the infinity-th place past the decimal. They differ at the infinity-and-1st place.â
A woman ordering her ice cream eyed us warily.
âWhat about this?â I said, and drew the following:
0.(9)4(8)1
âNaturally,â Jeff said. âThe 9âs go on forever. Then thereâs a 4. Then the 8âs go on forever. Then thereâs a 1.â
âThatâs all logical enough,â I said (perhaps bending the meaning of âlogicalâ). âBut what if thereâs something like this?â
0.(9)(9)
âIs that different,â I continued, âfrom this?â
0.(99)
âThatâs tricky,â Jeff said. âThe first one is nines forever, then nines forever again. But the second one is just double-nines forever â which is the same as nines forever.â
âLet me get this straight,â I said. âYouâre saying this...â
0.(9)(9) > 0.(9)
âAnd also...â
0.(99) = 0.(9)
âExactly,â he said.
By this point weâd filled most of our napkin. I unfolded another so that the treatise could continue to evolve. âIâm not convinced that these are well-ordered,â I said.
âWell-orderedâ is a math term Iâd learned in college. A set of items is âwell-orderedâ if you can put them into a nice, logical order, with no contradictions. For example, tennis playersâ heights are well-ordered. But their skills are not â Anita beats Brianna, and Brianna beats Carla â so youâd expect Anita to beat Carla. But instead (in this real-life example from a friendâs high school tennis team), the reverse happens. Carla beats Anita. Thatâs not well-ordered.
âWell, letâs figure it out,â said Jeff. âIf we can find a case where A > B and also B > A, then youâre right. If not, then everybody everywhere will have to start using these numbers, because theyâre more fun.â
âGood luck with that,â I said.
At this point, my sister joined us for an ice cream. Holding her cone with one hand, her free hand contributed the following to our napkin:
0.((9))1
âOh!â Jeff exclaimed. âCool.â
âI see,â I said. âThe nines go on forever. And then the forevers go on forever. And then, after all that, thereâs a 1.â
This is when things got out of hand (if they hadnât already).
0.((9)1)2
(the nines go on forever, and then thereâs a one, and then that nine-forever-then-a-one pattern goes on forever, and then, at long last, thereâs a two)
0.(((9)1)2)7
(as above, except that repeats forever, and then at the very end, thereâs a 7)
0.(((9)1)27)3(((8(46)))5)
(as above, except that repeats forever, and then thereâs a 3, and then... well, hopefully you get the idea)
âAre these well-ordered?â I asked, as we began to unfold a fourth napkin.
âI donât know,â Jeff said. âTo be honest, Iâm not entirely sure what they are.â
By this point, it was 11pm, closing time at J.P. Licks. The last ones there, we stuffed the napkins into our pockets, grabbed our bags, and made for the door.
âWhat were you guys working on?â one of the employees asked us.
âThe Kaufman decimals,â I told him. âOn napkins today, in textbooks tomorrow.â
He politely rolled his eyes.
Note: As usual, Iâve used imagination to plug the holes in memory â though Jeff is every bit as weird and wonderful as described. Mathematical minds out there: Are the Kaufman decimals well-ordered? Iâm curious mostly about first-order Kaufman decimals. My hunch is that the higher-order ones must run afoul of logic somewhere or other, although Iâd love to see a demonstration of that, too.
0.(((9)1)((27)))8
0.(0)1
Creative Metaphor:
Well... .0(repeating)1 would be what you add to .9(repeating) to get to 1. Right? So wouldnât that make .0(repeating)1 be assigned some random Greek letter (though Iâm pretty sure they are already all assigned so maybe Arabic or Norse letters...)
So letâs say fehu ( |// ) is what you add to .9 (repeating) to get 1.
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Ben Orlin:
Ooh, I like fehu. Gonna keep that one in my back pocket.
I suspect that trying to add these will lead to a whole mess of contradictions. The discussion on my last post seemed to suggest that itâs hard to find a system in which 0.999... doesnât equal 1, but still has a meaning.
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Creative Metaphor:
I also just realized you missed one idea. You have .(9)(9) and .(9) and .(99) and but you didnât have a .(9)9 ... which clearly would be be .(9)(9) > .(9)9 > .(9) = (.99)
And then obviously it follows that if .(9)(9) > .(9) and .(9) = 1 then .(9)(9) > 1.
MWAHAHAHA!
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mrsaturn:
I think that proves that the system is not well ordered.
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Ben Millwood:
I claim that 0.0...1 + 0.9... is not 1 but rather 0.9...1.
Although, come to think of it, I donât know what 0.0...5 + 0.0...5 is. Maybe addition doesnât work very well after all.
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John P:
0.0...1 + 0.9... wouldnât be one in any case. If you can assume âequal cardinality of infinite expansionsâ (Iâm being quite loose with the terminology), then yes, it would be 0.9...1. However, if we donât have âequal infinitiesâ it would be something like 0.9...109... (with just nine repeating) or 0.9...0...1. But that isnât how infinities work, really.
I donât think the Kaufmann decimals would work like that, either. Infinite expansions rarely do.
Also, if you jive with the whole .9... = 1 thing, it still wouldnât be one! Itâd be 1.0...1
I just forgot to cover that case.
Julian Hyde:
Ok, so wait. You can multiply and divide Kaufman decimals, right? So 0.0...1 * 2 = 0.0...2
So therefore 0.0...1 * 10 = 0.0...10. But normally when you have a zero at the end of a decimal you can just remove is (0.500 = 0.50 = 0.5), so 0.0...1 * 10 = 0.0...1. Divide each side by 0.0...1 and you get 10 = 1!
I think this will be a big deal.
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Jeff Kaufman:
I donât think you can multiply and divide these. Iâm not even sure you can add and subtract them. Orlinâs current question is very weak: are they even well ordered?
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Gred:
You can divide by them normally, just not if theyâre no finite distance from 0. You can multiply by them if you remember that infinity minus 1 is infinity, so 4*1.[9]3 = 7.[9]72, because the last 9 times 4 becomes a 6, then the 3 becomes a 12. Then you can at least divide by them some of the time?
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Ben Orlin:
Thatâs clever. I hadnât thought about multiplying â unsurprisingly, it wreaks havoc on place value.
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Abram Demski:
I think this also illustrates why you cannot add these (at least not elegantly).
The value of .(0)1 * 10 is the same as the value of adding up .(0)1 ten times.
More technically:
What weâre doing here is allowing ordinal locations rather than integer locations, in our decimal expansion. This doesnât work well because limit ordinals have no predecessors, but we need predecessors in order to âcarryâ during addition operations!
Ralph Morrison:
It seems that all the Kaufman decimals written thus far are what one might call ârational Kaufman decimalsâ, in that you can represent them in a nice finite way (finitely many numerals, finitely many bars). If you want things to get even CRAZIER, you could look at irrational Kaufman decimals, which have no representation with finitely many numerals and bars. For instance,
0.(1 with a bar)0(1 with a bar)00(1 with a bar)000(1 with a bar)0000...
or, perhaps more trippily, 0.1 with infinitely many bars over the 1. Or you could combine those two.
Whee!
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Ben Orlin:
Yikes! I like 0.(0)1, with infinitely many bars over the zero.
I mean, we could have a notation for infinitely many bars, like a squiggly bar.
But then you could have infinitely many squiggly bars...
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Abram Demski:
So, 0.{0}1.
Joshua Sasmor:
Are these even real numbers? I mean in the Dedekind sense â are they an element of the âstandard analysisâ definition of the reals?
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Ben Orlin:
My hunch is no? Iâm sure youâve got more insight into this than I do, but I canât imagine this notation has any meaningful visual interpretation on the real line. If weâre talking about Dedekind cuts, Iâd suppose 0.(9)1 = 0.(9)2 = ... = 0.(9) = 1.
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Owen Cotton-Barratt:
Yeah, this is correct. You can get lots of these for every rational number (which only see the part of the Kaufman decimal up to the end of the first bar). As produced in the post, you canât get to irrational numbers this way.
Jefftk:
A useful notation when in text: instead of â123 with a barâ write (123). Now I can express â3, then 9 repeated forever, followed by 2, all that after the 3 repeated forever, then 4 repeated foreverâ as: 3((9)2)(4).
Thereâs a list of examples that it sorts properly.
It handles all the examples in the post except for the equivalence between (9) and (99).
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Ben Orlin:
I like your parentheses notation.
Am I right in reading that your sorter puts 0.2(0), 0.2(0)0, and 0.2(0)(0) in that order. Should those be equivalent? Can your sorter do cases of equality?
Also, is your sorter saying 0.5589 < 0.((551)57)4? It seems like the opposite inequality should hold.
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Jefftk:
The sorter doesnât do equality, but Iâd like to figure that out. In the post you have (9) and (99) as equivalent, and I feel like (0) shouldnât matter unless itâs followed by something non-zero (trailing 0s are irrelevant). Right now anything with a distinct string representation will get compared as distinct.
There was a missing comma, but thereâs also a problem. Iâve fixed the comma and moved the test examples into the order that it sorts them, but it does still sort 58 < (51). Uh oh.
Here is a paradox that I think might help illustrate why you donât want to have Kaufman decimals. Do we want to say that the following inequality is true?
0.(9)5 + 0.(0)4 > 0.(9)5 â 0.(0)4
Given that all (9)âs and all (0)âs have the same cardinality (aleph-zero), we are just doing arithmetic with the 5âs and 4âs and the end, so this inequality should be effectively the same as (5 + 4) > (5 â 4). So letâs accept that the inequality is true. Now we add the numbers on the left side.
0.(9)5 + 0.(0)4 = 0.(9)9
That is, an infinite sequence of nines with an extra nine at the end. But how is that different from a plain infinite sequence of nines? You can always accommodate an extra nine into (9), for the same reason that Hilbertâs Hotel can always accommodate an extra guest. So, we have to say that:
0.(9)9 = 0.(9)
Now we do the substraction on the right side.
0.(9)5 â 0.(0)4 = 0.(9)1
And given that
0.(9)1 > 0.(9)
we can expand and we get that
0.(9)5 + 0.(0)4 < 0.(9)5 â 0.0(4)
So, Kaufman decimals allow us to pick a positive number, add another positive number to it, and end up with a positive number smaller than the one we started with. Sort of like saying that 5 + 4 = 0.
Either Kaufmanâs decimals are silly, or they are more sophisticated than I can understand.
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Thomas Fallon:
Well I think the point is that .(9)9 > .(9). It looks like what is going on is instead of assigning a digit to each natural number as we do for decimals representing real numbers, we are now assigning a digit with each ordinal number. If you are unfamiliar with the ordinal numbers, it turns out that they are too numerous to be contained in a set and instead form a proper class. They can be formed by continuing the idea of the set theoretic construction of the natural numbers. We start by identifying the empty set with 0 = â , and then define a successor function S(x) = x âȘ {x}, so 1 = {0}, 2 = {0,1}, 3 = {0,1,2}, etc. This is the successor function used in the axiom of infinity http://en.wikipedia.org/wiki/Axiom_of_infinity. This idea can then be continued by saying Ï is represented by the set of natural numbers, we can then apply the successor function to Ï, so S(Ï) = Ï+1, and we can then form Ï+2, Ï+3, ... , Ï*2. We can then continue applying the successor function to get Ï*2 + 1, Ï*2 + 2, Ï*2 + 3, ... , Ï*3, ..., Ï*4, ... , Ï^2, ..., Ï^2*2, ... , Ï^3 , ... , Ï^4 , ... , Ï^Ï , .... , (Ï^Ï)^2 = Ï^(Ï*2) , ... , Ï^(Ï*3) , ... , Ï^Ï^2 , ..., Ï^Ï^Ï , ...
These all are examples of countable ordinals, but we can then form the set of all countable ordinals which represents the first uncountable ordinal and continue this process even further. Since whenever we form a set of all the ordinals we have constructed so far, we can then take the successor of that to get the next ordinal, this means that no set can contain all of the ordinals and so the ordinals form a proper class. If my explanation is confusing, http://en.wikipedia.org/wiki/Ordinal_number might help, the concept is also explained in Naive Set Theory by Paul Halmos