How to Wrangle Infinity (an intro to p-adic numbers)
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- čas přidán 30. 01. 2021
- An introduction to p-adics by looking at a strange infinite series.
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I have a colleague who does research with polynomials over p-adic fields. For years our running joke as been me asking, “Yes, but what really is a p-adic number?” All the various definitions and explanations have left me cold in intuition. You have finally broken through this hard head of mine; PLEASE do more videos!
this is kinda similar to programming in a way (in that huge numbers act like -1)
This is actually a pretty interesting topic: can we use any of that p-adic theory in computer programming with two's complement numbers?
@@fyodorx5428 Up to a point you can. If you are looking at -1 and you are using two complement for 8 bit you are at 0xFF 16 bit you are 0xFFFF and 32 bit you are at 0xFFFF_FFFF and 64 bit you are double the F's. Or truncated 2-atic numbers.
"kinda"? What do you mean by "kinda"? This is _exactly_ what goes on behind the math powering 2's component.
In some sense, p-adic integers are the limit as n approaches infinity of the integers mod p^n, so 2-adic integers are just a theoretical ideal binary integer with infinitely many bits.
Addition, subtraction, and multiplication of computer integers is identical to rounded 2-adics. ("Rounded" because the meaning of "most significant" and "least significant" are swapped.) Division can be implemented manually in a surprisingly fast and elegant way using Newton's method and bit shifting. The p-adic valuation is the negative log of the p-adic absolute value, and actually exists as a daughter instruction on many computers called "count trailing zeros".
Sign extension and zero extension, in addition to being used to map the 2-adic machine integers to Real integers, provide a heuristic as to what the rounded off bits are, whether they're all 0s, or copies of the leftmost known bit.
Fun fact: 0x...AAAB = ⅓. Disable overflow checking and multiply it by 3 and see for yourself what you get. This works regardless of how many bits you use. Also raising a number to the power of this will also genuinely give the number's cube root with the only thing extra being making sure overflow checking is disabled.
Freja Holmer in a semi-recent presentation mentioned a schism between mathematicians and programmers, and there aren't many more places where it's more clear than mathematicians talking about p-adics like they're really strange and weird, while programmers have already gotten more than comfortable with the sun of powers of 2 equalling -1.
I'm beginning to like number theory. It's like professional wrestling, where the lines between reality and fantasy become more and more unclear the deeper you look into it. I remember writing off complex numbers a long time ago as being pointless, but then I actually used them for geometrical constructions.
writing off complex numbers as useless is like writing off oxygen as useless xD
@@God0fSnow Well, when you don't see something, it is quite easy to write off. To this day I still don't deal with complex numbers very often.
The electrical system of the country, not to mention radio and phones wouldn’t work without complex numbers.
@@sloaiza81 Yeah, but most people don't know that. Not many are electrical engineers.
Soemthing worthwhile for the p-adics is that this is where we get the notion that 1/infinity is 0. Take an infinite string of 0’s in any p-adic, there are an infinite amount of p’s one can pull out of this. The distance formula for such a number is 1/p^infinity. Now we also know that the distance formula in the p-adics is a simple metric used to determine how far away from a any p^n a number is. As a result, if a number is a p^n then it’s distance formula returns 0 by definition (and you can use the over 1-r to prove this if need be) and so you get 1/p^infinty equals 0.
The iffy bit of magic is to put “+” signs and “…” instead of working with a single big map (which we might call a sum) on a series. The thing that makes it work is to define compatible rules between the big map, operations on the series and usual addition and multiplication.
That gets a bit more technical though.
Well that’s to get the sum to work as a sum.
p-adic numbers can just be defined straight up and then the sum properties are theorems.
The key is to think of series of numbers as maps (functions) from the natural numbers into some target like the for example the real numbers (with usual addition and multiplication)
If we think of the set of possible series (possible maps from N->R), then we can come up with rules that maps series to numbers (sometimes numbers extended with some extra points like infinity).
Some of the ways of assigning a number to series deserve to be called infinite sums. Others may not.
Let s and t be a map(function, think series) from N->R.
Let Z be a map (functional, think infinite sum) from (N->R)->R.
We can add two series to get a new one by adding the corresponding entries at the same place in the series. (s+t)
We can append a number from the left to a series by pushing the series back and placing the number at the front. (x@s)
We can multiply a series by a number by multiplying each entry by that number. (x*s)
Z(x@s) = x+Z(s)
Z(s+t) = Z(s)+Z(t)
Z(x*s) = x*Z(s)
s+o=o+s=s
Z(o)=0
Now suppose that for a,r in R and r
Been searching videos all day to understand p-adic numbers. Your explanation was the most concise and understandable. Thank You!!
Keep it up! This was a really good and informative video. P-adic numbers have come up multiple times on the Numberphile podcast yet Brady hasn’t got a video exclusively about them. You taught me something new!
Make more of these awesome videos, man! The only Numberphile-level video I was able to find on p-adic numbers.
Your explanation on this video was just beautiful and fascinating, I would love to hear you talking about those details you mentioned.
Nice video! Always look forward to your uploads
Thank you! :)
Amazing!
Amazing 😃
Nicely done!
Very clear and interesting video! Loved it!
Nice, very informative, please do more videos on the subject.
Magnificent video!
Amazing explanation, great video
This was awesomeness
Amazing video! 👏 I feel like I actually learned something.
Dude, you are awesome
The "r" in the introductory equation is the inflection point of the decimal. For "r" values less than 1 it represents convergance for addition of terms less than 1. For "r" values greater than 1 it defines "convergange" on the left side of the decimal i.e. values greater than 1. Cool!
Hmmm.
bro keep up the good work. much appreciated
Great video!
Great job introducing p-adics. Love thinking about numbers that are not on the linear plane. And yes, please do some more videos on p-adics and why they are so useful in solving problems where the Real Numbers seem to wear out.
Very clear explanation.
Nice approach to infinity!
well presented video, thank you!
Thank you very much for introducing me to these interesting ideas. It's all very, very exciting.
Very nice explanation
Superb explanation of an idea usually expressed in a hugely complex manner
Very well explained
Definitely want to see the following videos.
Infinite sum of powers of two is the mathematical version of integer overflow.
you REALLY explain well [others LOVE to talk sooo fancy to make sure that nobody understands--- you are VERY different!! keep it UP
Hey I loved this video!! I'm commenting so that the youtube algorithm picks you up:)
No need to invoke the geometric series formula:
Let x = 1 + 1/2 + 1/4 +...
2x = 2*(1+1/2+1/4+...) = 2 + 1 + 1/2 = 2 + x, i.e.,
2x = 2 + x, or
x = 2.
Similarly,
x = 1 + 2 + 4 + ... = 1 + 2( 1 + 2 + 4 +...) = 1 + 2x, or
x = 1 + 2x, so
x = -1.
Very nice ! Your beautiful 'putting the two cases side by side' readily highlights the issue that is arising. In standard arithmetic, only (i) finite numbers and (ii) finite number of + are permitted in an arithmetic expression. So strictly x is undefined in both cases. So both the expressions 2x=2+x and x=1+2x don't mean anything. This is a bit like dividing by zero, so 1/x with x=0 is undefined.
However all is not lost. If the infinite expressions are replaced by "similar looking limit expressions", then x is usually replaced by Lt n -> ∞ [x(n) = b + r(n)], where usually a is fixed finite expression and v(n) is a remainder expression that is never fixed for any n (so it varies as v(n)). In the case of a geometric series x(n) = (1/(1-r) )+ [r^(n+1) / (1-r)], so b = (1/(1-r) ) and v(n) = [r^(n+1) / (1-r)].
Lets take the (1+2+4+...) expression equation x = 1 + 2x. Strictly x(n) ≠ 1 + 2x(n), for any n, but we are hoping that in the limit it will be. So for ease lets assume that x(n) = 1 + 2x(n). If we split x(n) into its fixed and varying parts, x(n) = a + v(n) then the equation becomes (a+v(n)) = 1 + 2(a+v(n)). So this becomes [a +v(n) = (1+2a) + 2v(n)].
If we compare the non varying part then we get a = 1+2a, so a = -1.
By comparing the varying terms, we need Lt n -> ∞ [v(n) = 2v(n)]. In 'very, very loose' terms as v(n) is increasing we can say that the varying finite numbers approach the same infinite value with the property, akin to zero, that ∞ = 2 * ∞ (compare 0 = 2 * 0). We have taken ∞ to represent an 'always increasing finite varying number' with this special property.
So the full solution is (1+2+3+...) = -1 + (an always increasing varying number, given by Lt n -> ∞ [v(n) = 2* v(n)] ) = -1 + ∞
A similar assessment shows (1+1/2 +1/4+1/8+...) = 2 + (an always decreasing varying number, given by Lt n -> ∞ [v(n) = v(n)/2] ) = 2 + 0
Just like two's complement is useful for binary computers, 3-adic numbers seem like the most natural way to represent signed integers in ternary computers.
I've been considering to code a base3 VM, and I was going crazy thinking of all the possible ways to encode ints (there's SO MUCH more than in binary), and trying to decide which should be the default standard. 3-adics seem to have simplified this for me
Great video, please make more (can we have this chillout background music only while culminating the chapter ? because I am trying to relax and do heavy lifting at the same time ☺️)
Excellent, it will be great to know the details you referred in the video, liked the video, have a nice day :D
Subscribed 😀👍
Nice elaboration of the 3b1b video! Grant should see this; he'd probably be pleased.
I liked and enjoyed your presentation in this video. At around the 13:33 mark I got lost and could not follow what you meant by "so in base 2 this infinite sum can be represented as ....1 1 1 an infinite string of ones and we add 'one' to this and we get a string of 0 0 0 which so long as we look at it in base 2 it equals negative 1. So then your conclusion or final step also left me lost in confusion. Can you help with another video to help fill in the gaps in my understanding. One gap might be a lack of fully understanding what a number is in the different number bases and what their purposes might be. Can we better understand "what a number is" by viewing from and across the different base systems? What is a number?
Great explanation, thanks. Clearly 1 + 2 + 4 + ... = -1 does not make sense in base 10, as you explained, only in base 2. So, does this mean now that whenever we write down an infinite series of numbers and give the value it converges to, we need to add to that which base needs to be used, in order for that convergence to make sense? Somehow it now feels like every seemingly nonsensical equation can be made sensical by choosing the right number system...
No, I think changing the number system just makes it intuitively clear that the number converges. There are infinite numbers which are not p-adic. For these there are no convergence.
In essence, yes! It's not just base 2, however, it a 2-adic system (this difference is something I want to discuss more of when I find the time to make a follow-up). For these infinite geometric sums that diverge for rational numbers, the formula only works if we apply to it "analytic continuation". Essentially, this means that if we get a finite answer from the geometric sum formula, it is in reference to some number system in which the sum actually converges.
it's accurate, -1 is the right answer, it's a loop '' number infinity is a loop so you start at positive and find yourself at negative -1
Laughing my soul out when you almost wrote 2-dic numbers hahahahaha
Excelente video and amazed with p-adic numbers
Perfect for the enthusiast amateur maths guy
Beautiful video brother and u got a new subscriber form INDIA !
In my view - this is the BEST explanation.
And THANK you clarifying that it's ONLY VALID to say 1+2+4+8... = -1 when using the 2 system. Is this really 'adic' though?
It's like saying 10 + 10 = 100.
That is also true when you state this is in base-2.
A few other explanations I've seen haven't stated it that way.
Interesting stuff. And gee whiz that 4 at 11:44 is a beaut
Looking forward for more videos. BTW what are you coding ?
I'm actually coding the animations for some of my other videos!
first video that came up when i looked up p-adic numbers....i feel like I'm on the cutting edge of CZcams's math algorithm or somethin
wowowowowowowow,it's so cool!, what a fine design of making video. But i mean what is this about p-adic numbers, this kind of method seems work for all adic number systems.
Loved this! It was super entertaining
3:02 Just a quick question. If we stated the sequence with 1/2 instead of 1. It should converge to 1. But the formula would state a division by 0???
r is still equal to 1/2 in the formula because that is what is multiplied to get the next number. Starting the sequence with 1/2 is the same as subtracting 1 on both sides.
true but according to the formula the first element is now 1/2. So there is a division by 0? Or is this formula only valid for the particular case, starting from 1?
@@suncrafterspielt9479 When I try the formula I don't get a division by zero. The top number is 1/2, the bottom number is 1 - 1/2 which is just 1/2, so I get a half divided by a half. 1 is the answer I was expecting, so it looks okay on my end. Maybe you'll get it to work if you go through the calculation again.
overflow bug
The representation of negative one and negative numbers in general by binary numbers in this way is called twos complement and has always been used in digital computers to represent negative numbers
The one & only reason I will like this video and honestly beg the rest of the mathematical community to do this, is to point out that REAL, NORMAL sum of all numbers is NOT (-1/12). IF you define "SUM" in a different way & literary change the distances between the numbers, then there is a possibility that sums of infinite series can be a number. Anyone who puts 1+2...=-1/12 should have a shit ton of asterisks behind that equation because it is simply NOT true in the reality we live in. What basically you are doing is making a different reality, imposing your rules & then saying that yea this does prove to be correct.
Nothing against P-adic numbers, in fact I am a huge fan & want to see where this field of mathematics takes us, but until we find some REAL life solutions that can be implemented in machines & world systems, we should take most of these sums with the infinite asterisks*** that they come with.
Good intro, But what is the point? And why must the bases be primes?
hmm. Even if a number has an infinite amount of trailing zeroes, it does not mean it is equal to zero. It still may have a non-zero number somewhere infinitely far to the left of the number (if we look initially at the right side of a given number). Otherwise it means this number is equal to infinity and zero at the same time.... very strange. What do I miss here?
It's the difference between recursion and iteration. You can solve recursion. But you can't iterate an infinite number of times. There is no contradiction at all. "..." doesn't tell you how to calculate it. But a recursive definition does. In fact, you can combine the two concepts, where you take a formula with "..." in it, and find the exact value for N iterations, and then continue the "..." with an explicit recursion. If you SUBTRACT the recursive tail, you get the infinite sum; that is "taking the limit". The the recursive sum gives you the bizarre number.
It can be demonstrated by staying clear of any notion of "infinity".
This is ambiguous without recursion, because "..." isn't defined, and the seqence has no identity:
1 + 2 + 4 + 8 + ...
Seq = Head + Tail
S = 1 + 2S
-S = 1
S = -1
S = 1 + 2S
= 1 + 2(1 + 2S)
= (1 + 2) + 4S
= (1 + 2 + 4) + 8S
= 2^N-1 + 2^N S
Limit[S] = S - Tail[S]
// when N is large, we get 2^N-1
S - 2^N S = 2^N-1
// but still...
S(1 - 2^N) = 2^N - 1
-S(2^N - 1) = 2^N - 1
S = -1
10.59. Doesnt it converge to 2 to the power of infinity in the 10 digit system?
This is my first-ever lesson on p-adic numbers...well done, but I have some (admittedly, perhaps naive) questions.........it just seems to me, this infinite series equaling -1 is not an actual mathematical property, but an artifact of this symbolic logic system we use....a glitch, if you will, caused by always getting to kick the can down the road by "carrying the 1" an infinite number of times. It reminds me of the old 0.999999...=1 trick, which would be even more similar if there were 10-adic numbers (but I think you said p must be prime). If this is a true equality, is there any possible operation to derive the infinite series from -1, or the number ......11111 from -1? What is the practical application of p-adic numbers, is this a useful mathematical tool?
I don't think it is a glitch, but it comes down the representation of numbers. We are quite comfortable with the idea that 11 and 3 are representations of the same number in binary and decimal respectively, but so are 1 and 0.999..., both in decimal.
I have found a way by long division to generate negative fractions of the form...333 = -1/3.
What are the practical applications? Well for one, the device you're reading this comment on relies on the sum of powers of two equalling -1 on the transistor level, and if you ask it to print -1 in binary, it will give you a string of all 1s up to whatever precision you asked for. Because it turns out that borrowing forever is easier when subtracting a larger number from a smaller number than having to store a separate symbol for the sign.
This feels like correlation but not causation. The infinite sum formula you used is obviously not supposed to work when common ratio is greater than one, because a common ratio > 1 will give an extra term in the numerator, -a(r)^n. So the formula for a common ratio greater than one should be,
S = {a - a(r)^n} / {1 - r},
as opposed to the one in the video
S = {a} / {1 - r}
What if it’s just coincidence that the omission of that term coincides with whatever the p-adic value is, at least for the numbers you’ve used so far. I guess what I’m trying to say is omission of the one term in the numerator feels like a calculation error and it’s crazy that a calculation error branches into a whole new realm of numbers.
It is misleading here: your result cannot be proved by mathematical induction, because there is a "1" with infinitely many "0" following it and that is not equal to 0.
9:16 triangle orientation feels wrong somehow.
Great vid.
So, it is weird.
I got everytging up to the moment when you turn 1+pow(2,n) into a 2adic number ...1111. This is understandable.
Then, you add one to it and turn everything into ...0000. Still understand it.
I got it that in a sense that ...1111 acts LIKE -1. But I do not understand WHY it should be considered as EQUAL to -1 in this 2adic system. Thus, it is still not obvious why the formula 1+pow(2,n) gives out -1.
Thanks for the video, but I gotta give some constructive feedback. It feels contrived. And I know it's not easy to explain, but a more intuitive example or even an example which makes the usefulness apparent. One thought that comes to mind but which isn't explicitly stated is that we seem to be interested in studying infinities which have regularities in their representation. If you think about the word "infinity" there's almost no information there. But if you think of an infinity that looks like "....2222" or "....3333" there's enough information to study infinite sums that make them up. And we can work backwards from infinite sums to infinities with regularities in representation. And the regularities might be more apparent in different bases. Those are points that I'd highlight personally and seems to capture the essence of the video. The ideas are there but they're more of implied, not as explicit as how I summed it up.
I still learned a lot, even if I think it could be more condensed. Appreciate the video
I would love to know what the 10 adic for -1/12 is now
The proof that ...999.999... = 0 is left as an exercise for the reader.
1 Infinity to get to 2
So does this means all infinite same-base power sums, multipled by 1 less than their base, are the same number, -1?
e.g., in 7-adic numbers: ...6666 = -1
you didn't show how a base-ten number line works in my opinion it would be this 1x10 o plus 2x10 o plus 4x10 o why would the second integer be two times ten to the first when we're only adding two to one not twenty to one. I'm confused. why isn't it the same with the other way? arent the integers solid numbers so shouldn't the 2 bases always be zero as well? what am I missing? also what applications can be applied here in a positive reality, what does this solve? I'm not disagreeing it's interesting I'm just curious. in my opinion we do need new ways of looking at different maths to figure this whole thing out, but if it's true that infinities get us to negative integers, that seems interdimensional to me, is that a possibility? could this be a way to investigate different dimensions with this math?
I thought the sum of all natural numbers up to infinity is actually equal to ( - 1/12 ) isn't it?
These kinds of results aren’t adding up one thing after another and suddenly getting negative numbers.
The game we’re playing is to analytically continuing the notion of a sum.
Another way to say what we’re doing is that we’re coming up with a rule that takes series of numbers and assigns to each of them a number as a result.
There are many ways of doing that.
Some ways of doing that match certain rules that sums obey.
But there are different choices to make in the process and depending on which choices we make, the number we assign to a given series may be different.
There isn’t anything like “the sum” of a divergent series.
It depends on the procedure we apply. Once we pick a procedure it can be perfectly well defined and consistent and so on.
You may say summing the series of all natural numbers is divergent or infinite. That’s the standard notion and is very legit.
You may say that you use a special sum that assigns a value. That can also be legit but you need to say you’re using a non standard sum,
Or you can use another special sum that assigns a value. That’s also fine, as long as you say what you’re doing.
When everyone is initiated we may call at of those a “sum”. On many series they might even give the same results so that it doesn’t really matter too much.
Then we might quibble about whether it makes sense to call these extended notions “the sum” or not in the presence of people who are not initiated in this secret society of mathematicians who know the rules of the game.
It may cause a bunch of confusion.
Very interesting video. Watched it till the end in awe. I still have a problem with it, in particular with " ...1111 + 1 = ...000". This part does not make sense to me, as it seems more undefined than anything else, the same way you could not write 1 + 2 + 4 + ... in the base of 10. I totally agree with 0.999... = 1 because it is well defined from a series point of view, but not with " ...1111 + 1 = 0". And by the way, there is no need to go to base of 2 to see that point. In decimal, it would be like saying ...9999 + 1 = 0, so ...999 = -1. If someone can explain it to me in the decimal base, that would be great. Thanks!
I definitely am not qualified to answer this in any technical sense, but I think just don't overthink it basically. It's just the same as with the repeating decimal. Will you ever see a 1 in the string of digits representing ...1111 + 1? Since you won't, it can't be anything other that 0.
application video?
Some of the best content creators are some of the most obscure as well. I have some questions about the need for complex numbers. What is the principal objection to taking the square root of a negative quantity? Thank you so much for your time and attention!
It forced mathematicians to have a more open mind of what a "number" is.
If all you know is the counting numbers then numbers like 5.3 would not seem "valid". You can't count out 5.3 of something. You have to broaden your view, saying numbers are for counting and measuring. That gives you the real numbers through a stretch of the imagination. The complex numbers were just another stretch. More stretches of the imagination happened later such as quaternions, octonions, sedonians, etc.
@@alexandertownsend3291 Stretch of the imagination certainly applies. So it seems to me, despite the whole apriori approach with the whole "define a function" and "whose domain of definition is thus...", that what we are doing in taking the square root is preforming a binomial factorization for the case of a general rectangle (real number) whose sides happen to be equal. Restricting ourselves to integers momentarily; 21 is a rectangle on the Cartesian plane whose non-trivial sides are 3 by 7. Likewise 25 is so factorized as a 5 x 5 rectangle. We generally only have use of the principle square root. And so too use only the positive factorization of the integer 21. But both of these have stereoscopic alleles in the quadrants of the plane kitty corner. To wit -3 and-7 for 21 and like wise the opposingly signed factors for 25 -> (-5, -5). What I am proposing is that simplicity and consistency would apply to our 'extension field' when we are faced with the same problem in the mirror. That is to say if either -7 × 3, or -3 × 7 have solutions in the second and fourth quadrants of the Cartesian plane then why shouldn't -25 so too? What (really) is the big whoop about talking the square root of a negative quantity? Let me digress briefly to address the issue of all Reals. When we take the square root of a cooperative example - say 36 we get a nice round 6 × 6 square. But take 37 - or Pi for that matter - and what we are essentially doing is taking the geometric mean of the binomial factorization. So the same logic would apply. You would factor negative quantities in quadrants two and four. Am I just terribly confused and ignorant?
He's beautiful!
Very nice teaser to p-adic numbers, and good you finished with the concept of convergence in different bodies. Writing on plain paper ist great and looks so normal, natural. Please fix your cameras focus, so distracting.
4:01 I can't say that that formula worked before. 1+1/2+1/4+.. will get very close to 2, but it will never really reach 2.
Very correct. The actual finite formula for n terms in the geometrical progression is Sn = So/(1-r) + So(r^n+1)/(1-r). The video picks the first term, but doesn't mention what is being done about the second term as n tends to infinity. This means p-adic numbers are "infinitely ignorant" or like colour blindness are "infinitely blind" or even "infinitely incomplete". Infinite carry overs are ignored & like other comments, p-adic numbers are like an "arithmetic overflow tolerant number system", that tries (and surprisingly succeeds) to make sense of what happens when some information is lost during a calculation. However, note that 'ignoring carryovers' is not without precedence in arithmetic. To calculate 1/3 = 0.3 carry 1 repeated, with no final carry completed. Psychologically its easier to accept the 1/3 lack of a final carry since "well, the numbers must be getting smaller so whats wrong with ignoring the infinite sequence of carries, since its likely to lead to a zero effect on the result, right". So conceptually Sn = So/(1-r) + So(r^n+1)/(1-r) has a remainder term. It is conventional to ignore the remainder So(r^n+1)/(1-r) as it tends to zero, and p-adic numbers just say why not also ignore the remainder when the remainder tends to infinity. A similar assumption is made in each case. If you know about infinitesimals then the similarity between the two assumptions becomes more clear. To say the fractional remainder (in Q) tends to zero is to ignore the presence of infinitesimals. A fraction in Q is truely HUGE compared to an infinitesimal.
Geoff
What happens to the carry on the far left, of ...111 + 1? How is this 0? Isn't it 1...000?
What do you mean with "far left"? There is no p-adic number "1...000" .
@@miloszforman6270
My ignorance about how to work with p-adic.
@@fredkerr5720 > What happens to the carry on the far left?
In some sense, it's gone far-far to the left *over* the (countable) infinity. It may be informally represented as *1...:...000.0* with (countably) infinitely many "visible" digits at all positions enumerated by *all* natural numbers to the right of the colon, and then again (even more) infinitely many "invisible" digits at positions *greater* than any natural number (such "big" numbers are called hypernaturals) to the left of the colon. It is similar to the notation developed by Lightstone to represent infinitely small hyperreal numbers (infinitesimals). So there are even more new types of numbers to the story. This description is not exactly accurate, but you can look at it from this point of view.
By the way, since 0.999… = 1 (in the reals) and …999 = −1 (in the 10-adics), then (as Wikipedia puts it) by "blind faith and unabashed juggling of symbols" one may add two equations together and get …999.999… = −1 + 1 = 0, which looks really bizarre.
7:49 This idea has borrowed from programming. Or prorammers took this for their deal. The vivid example of this is so called 'Nuclear Gandhi' of integer overflow en.wikipedia.org/wiki/Nuclear_Gandhi. "However, the aggression level was stored as an 8-bit unsigned integer variable that could only store values in the range from 0 to 255 (or 2^8-1). The negative value would therefore result in an integer overflow, with the value being stored as 255." where 11111111+1 brings zero (1)00000000. Moreover, in many programming languages it is compulsory to point out wether particular variable is unsigned or just signed to represent this value correctly.
en.wikipedia.org/wiki/Signed_number_representations explains 14:15
It is certainly related to signed number representations in binary (and actually, there is a meaningful way to view the 2-adic numbers as infinite-bit binary arithmetic); however, it does not _borrow_ from programming. The p-adic numbers were developed by Kurt Hensel at the end of the 19th century.
How does....2222 + 1 adds to
....all 0s!? To my mind it's 2223.
In base 3. The digit 3 doesn't exist in base 3, so it's instead 10, with the 1 carrying over to the next 2, making it carry over, repeating indefinitely.
@11:58 You should not say "two hundred and one" because that's a number based 3, not based 10. Hundred is only meaningful in base 10.
Instead, you should have said "two zero one".
Did you just prove that we exist in a computer simulation and the sum of an infinite sequence equals -1 because of the matrix numeric system overflow???
...111111 in binary is two's complement for +1
Im very uncomfortable
-1/12 is the same thing as saying i^2/12 ; there are complex numbers implied in that summation.
12:57 starts to write "2-di" instead of "2-adic."
gay
What the f.....