LOL … make that post #1318.
7777777 is off their rocker, but can I play for fun?
For what it’s worth, I think I’m much more likely to come across the term “…99999” in the context of the 10-adics (where it would indeed denote -1) than the extended reals (where, as you note, it could be taken to converge to infinity).
I was also surprised 7777777 was cool with this. That having been said, in the 10-adic integers (where we may extend infinitely left but not to the right of the decimal point), values do have a unique digit series, where this is understood as not including a sign bit. The unique such digit series of -1 (i.e., the negation of …001) will be …999, as makes sense given the general principle “The last digit of x is the unique digit to which x is equal mod 10”.
In the context of the 10-adics:
…999 x 10 = …9990; i.e., -10.
…999 x 2 = …9998 [write down a large string of 9s, multiply it by 2, and see what happens]; i.e., -2.
…999 x 5 = …9995 [write down a large string of 9s, multiply it by 5, and see what happens]; i.e., -5.
(…999 x5) x 2 = …9990 [write down a large string of 9s followed by a 5, multiply it by 2, and see what happens]; i.e., -10.
n-adics may seem trivial, but there’s actually quite a lot of use in them [e.g., the simplest way I know to understand the structure of the multiplicative group modulo n (that it is cyclic when n is prime, and so on) is through embedding it within the n-adics and understanding the latter]. Also, a lot of fun. For example, here’s something I suspect most people haven’t thought about before:
For every digit, there’s a 10-adic solution to x^5 = x ending in that digit; in fact, for even digits, there will be a unique solution, while for odd digits, there will be two solutions. Two of these (…000 and …001) are old hat, and another follows immediately from our new expression for -1 (…999). But what about the rest?
Well, the full list is:
…00000000
…00000001 and …74218751
…79186432
…33704193 and …7922943
…12890624
…12890625 and …87109375
…87109376
…92077057 and …66295807
…20813568
…25781249 and …99999999
Go ahead, try them out. [Of course, I’ve only given you so many digits to play with, but if you play with them enough, you’ll discover how to keep going. Note that there are other solutions if you only focus on the last 8 digits (such as 25000001); the distinctive thing about these solutions is that they extend infinitely far to the left]. Furthermore, the one ending in 6 and the first one listed ending in 5 even square to themselves, while the ones ending in 2 and 8 cube to their negations (each other). Also, the odd solutions are all even solutions plus or minus the magic number …12890625. Etc., etc.; a plethora of patterns to explore. Isn’t that something?
Do you know how to do long division? Could you get a piece of scratch paper and do 1 divided by 3 for me please?
I don’t think that he has a large enough piece of paper. If what 7777777 is saying makes any sense, he just does not believe in infinitely long decimal fractions. That’s consistent, I suppose, but it does mean that most rational numbers and all irrational numbers cannot be represented by decimal fractions.
We’ve gone from early college to high school to, at last, third grade.
I think he believes in infinitely long decimal fractions, but does not believe that they are equal to rational numbers. So he thinks that 1/3 and 0.333… are two different numbers and that 1/3 - 0.333… > 0 .
Fruitless as it may be I need to check:
7777777, do you believe that the infinite decimal 0.33333… exists?
If so, please tell us what you think 10 x 0.33333… is?
If the criterion is agreeing with your math, then none of us will ever be worthy.
You’re free to join the huge party over in the hell of infinities, though.
Not that anyone cares, but, while the bolded part is true, it’s not actually the insight I meant to refer to; I meant to refer to how the n-adics explain why the multiplicative group modulo n is cyclic when n is an odd prime power. Apologies for the confusion…
Speaking of grade school, it’s worth noting decimal expansions which extend only finitely to the right, but as far as you like to the left, are in many ways better-suited to calculation, and particularly grade-school style calculation, than the more familiar decimal expansions which extend infinitely to the right.
What do I mean by this? Well, does a string of digits represent? It amounts to the coefficients of a polynomial to be evaluated at a particular point. For example, when I say something like “the number 2538”, what this directly amounts to is something like “the number 2x^3 + 5x^2 + 3x + 8, where x = 10”.
And when we add and multiply digit-strings, all we do is diligently follow our polynomial addition and multiplication rules. Easy-peasy…
Oh, with one hitch: just doing the above will often give us coefficients that aren’t single digits. So we follow-up by a normalization routine called “carrying”. This rejiggers the coefficient of each term based on the coefficients of all the lower degree terms.
So far so good. But when we move to infinite digit-streams, there’s a problem with this carrying stage: there may be infinitely many lower degree terms to look at before one can figure out what should become of a particular coefficient.
If we’re only dealing with digit-streams extending to the left, this won’t happen: each position is only affected by finitely many less significant positions. But with digit-streams extending infinitely to the right, we have to wait for infinitely much data before we can say anything.
If you don’t actually care about calculation, then you may not care about this. But if you do care about calculability (perhaps because your grade school arithmetic lessons focused excessively on calculation algorithms, as so many do; perhaps because you enjoy playing with computers; or perhaps just because you like to note what bits of math do and don’t generalize to what contexts), then it’s worth noting that the problem “Alice and Bob will each read to you the digits of a number, and as you listen to them, I want you to tell to me the digits of the sum (or difference, or product, or ratio, or what have you)” is easy to solve if numbers are specified from least significant to most significant digit, possibly never stopping, but impossible to solve if numbers are specified from most significant to least significant digit, possibly never stopping.
An illustration: what’s the first digit of 2.327608… + 3.672391…? Can you figure out any of its digits? Compare that to the problem of figuring out the digits of …8067232 + …1932763.
(This can also be framed topologically; the space of sequences of digits has a natural topology (Cantor space), and in terms of this topology, the above observation is that on leftwards digit-streams, the basic arithmetic operations are continuous, while on rightwards digit-streams they are necessarily discontinuous [at precisely the spots where the .999… = 1 ambiguity manifests].)
I care!
The first calculation Left to Right I can say immediately has a first digit 5, with a possible +1 absolute error.
One step later 5.9 (+0.1)
Two steps 5.99 (+0.01)
Three steps 5.999 (+0.001) (relative error <= 1 part in 6000)
etc.
The second Right to Left something-5 possible +infinity absolute error
One step later something-95 (+infinity)
Two steps something-995 (+infinity)
Three steps something-9995 (+infinity) (relative error <= infinity)
etc.
At first glance it seems to me that practical calculability is best done L to R.
Full disclosure, I am not a grade schooler.
Right. Which means you don’t know what the first digit is. Or what the second or third digits are, for that matter… You have some information about them, but there’s not a single digit you can name. If you start to say “The answer is 5 point…”, you could turn out wrong, and if you start to say “The answer is 6 point…”, you could turn out wrong. So if the protocol is to specify numbers in just the fashion “The answer is d[sub]0[/sub] point d[sub]1[/sub] d[sub]2[/sub] d[sub]3[/sub]…”, you’ll never get started announcing the answer.
Here, you can start announcing “The answer, in R to L order, is 5999…” with complete confidence.
Another way of looking at the successive error bounds is like so: the possible error is “Something ending in 0”, then “Something ending in 00”, then “Something ending in 000”, and so on. In other words, you gain certainty about each digit successively.
The presumption that this counts as infinite uncertainty depends on a particular interpretation of this notation, which needn’t be what we have in mind (almost certainly not, if we’re to take …999 as amounting to -1, and so on).
What is the practical thing to do depends on the purpose to which one is attending. I wouldn’t say 10-adics are useful for modelling most of the things which real numbers are used to model; I’m not actually proposing that. But I’d also say mere left-to-right unending digit-streams are a terrible way of carrying out calculations with reals; arbitrary precision arithmetic is best implemented using some other representation.
One simple such improved representation for fractional components of reals would be to again use rightwards digit-streams, but now allow halfway digits as well (i.e., digits 0.5, 1.5, 2.5, …, 9.5, in addition to the standard ones).
The interpretation of finite sequences of such digits is basically the same (it’s still just a polynomial to be evaluated at a suitable base), and we would interpret infinitary such digit-streams in basically the same way as noted before: the first k digits of such a stream would amount to a claim “The number is >= what these k digits represent on their own, and <= that lower bound + 1/10^k”.
This system has most of the advantages of the traditional representation (e.g., exponentially increasing precision), but is able to handle the aforementioned stream-arithmetic problem because of one difference the recurrent instigators of this thread wouldn’t much care for: failure of uniqueness of representation is much increased. (Now every value has uncountably many representations)
The traditional system has the problem that it is possible to bound a value within a very small interval, yet not be able to express this interval as a subset of any of the finitely-designatable bounding intervals (if the value happens to be on the boundary of two such). This new system gets around that with its network of overlapping intervals.
So if someone were to say to you “What’s 2.327608… + 3.672391…?”, you could now say confidently “It’s (5.5).(4.5)(4.5)(4.5)(4.5)…”! [Less clunkily, if we were to actually discuss this system in detail further, we might want to write “5’.4’4’4’4’…” or some such]
Ack, we don’t want to allow 9.5. The halfway digits should only be those halfway between two genuinely adjacent digits (so there are 9 of them, not 10);
Oh, actually, I misconstrued what you meant by “+1 absolute error” (even though the rest of your post made it clear). I should’ve said:
Wrong. It’s true that you have an at most +1 error in the first digit alone, but your error for the sum, when you propose a lower bound of 5, exceeds that.
In fact, I should let you know: the sum you were calculating was 2.32760830… + 3.67239182… = 6.0000001… . When you said the first digit was 5, you were wrong. Sorry.
A number of posters in this thread have urged others to read the Wikipedia page about 0.999…
In fact, they may have done better to suggest this Wikipedia page instead.
I want to divide this thread by zero.
Exactly.
I think that 1/3 = 1/3
and 0.333333…=0.333333…
just writing that 1/3=0.333333…does not necessarily make them equal
and if they are not equal, there is an infinitesimal difference between them which
never is equal to 0 and therefore cannot be ignored.
This all boils down to how to deal with infinity.
Do you think it is alright to count up to infinity, that you will reach infinity in this way:
1+1+1+1+…= ∞
1+2+3+4+…= ∞
1+2+3+4+…= -1/12
The last one is called Ramanujan summation and is used in physics in the methods of zeta function regularization.
Doesn’t it sound strange that ∞=-1/12 ?
According to your mathematics, and your rules which seem to be settled, you
treat infinity as number, as a negative number, as -1/12.
We all seem to agree that infinity is not a number. I have shown that infinity
cannot be equal to a number, cannot be equal to a negative number, cannot be equal to -1.
Tell me based on this evidence, can infinity be equal to -1/12 if it cannot be equal
to -1?
I again repeat why infinity cannot be equal to -1:
…99999=-1 is true
but according to your way of thinking, you think that …99999=infinity
therefore you will end up at infinity=-1, but this can’t be true. Tell me
is infinity equal to -1? You all said that infinity is not a number,
don’t lie now.
I told you on my post #1247 that you will end up at a problem with infinity,
with the formula ∞=∞ + 1
The number that was sent to you …99999.99999…represents infinity
to you, my witnesses are the users The Hamster King on his post #1289 and Budget Player Cadet on his post #1296. These two witnesses write that
∞=…99999+0.99999…=…99999.99999…=∞+1
I have these two witnesses, so my proof is valid, remember that also the Bible says so, two witnesses is enough, so my prophesy on my post #1247 is fulfilled.
Your problem is that you don’t know what is …99999.99999…because
the truth has not been revealed to you yet, and why it has not happened I have
told you many times.
You think that …99999.99999…is equal to 0 although every digit is different
and there is no context where they could be the same. You think that …99999.99999…is infinity. This number was sent to you but you don’t know it.
You are ignorant. I am not blaming you anything, this is not the last judgement.
You just need to realize that your mathematical system is not settled yet.
It contain errors. You should not teach that it is complete and consistent.
You ridicule me as religious crackpot, but it does not affect me because I have
truth, and it is not my truth. Remember what the Bible says:"“My teaching is not Mine, but His who sent Me. If anyone is willing to do His will, he will know of the teaching, whether it is of God or whether I speak from Myself.…”
Ramanujan summation is not summation in the ordinary sense. 1+2+3+4+… does not have a sum in the ordinary sense, because the partial sums keep on getting bigger without any limit. (And that does not mean that the sum is infinity.)
But it is odd that you are so willing to accept that 1+2+3+4+…= -1/12, but you won’t accept that 3/10+3/100+3/1000+… (i.e., .333…) = 1/3.
I never said that I accept that 1+2+3+4+…=-1/12
I don’t know how can you misunderstand what I wrote.
I said that it is wrong. It is wrong that ∞=-1/12 . That is what I said. I said that it should sound strange, meaning that it is wrong.But
according to your rules, you accept it as true.