Exploring the Convergence of 0.999... and the Concept of Infinity

[Hurkyl]

0 < (1/2 x 1/&infin;) < 1/&infin;

same problem. same limit. which is why i was using that as an example.

you guys use limits all the time yet you don't understand one when you see it?

BECAUSE infinity is your limit, you cannot get a number closer to 1 than .999~ but that does not mean they're equal because by imposing that limit you also cancel out the validity of using Achemedian Property to assert that equality. With NO limit on infinity it's very easy to see that Infinity+1 digits of 9 is closer to 1 and infinity+2 digits is even closer... ad infinitum 1 can never be reached and you can archemedes all you want...

How is it the same problem? I can identify some crucial differences that prevent the analogy from holding:

(a) In your toy example, the "limits" 500 and 1/500 are both part of the number system. &infin; and 1/&infin;, however, are not part of the real numbers.

(b) In your toy example, operations like + and * are not defined for every pair of numbers. However, + and * are defined for every pair of real numbers.

(c) In your toy example, the limit was "reachable"; e.g. you could do arithmetic that exceeds the limit (if it were allowed). No such problem exists in the real numbers. (This is merely a conceptual restatement of (b). )

Furthermore

(d) You are confusing the word "limit" (as in a limit is something that cannot be surpassed) with the word "limit" (as in a limit is something to which a sequence converges)

(e) Did you not notice that the proof I supplied does not use the Archmedian property? The proof works for all ordered fields, including those that are non-Archmedian.

(f) Things are only "easy to see" because:

(f1) You insist on treating "infinity" as if it behaves identically to finite things.
(f2) You haven't supplied a definition to which you intend to adhere, thus you aren't restricted by petty things like logical consistency.

 

[matt grime]

WHen i see 0.999... i know what that means. i add 9/10 to 9/100 to 9/1000 etc, now when you have your thing that's 'closer' to 1, which we'll notate as 0.9...|9 for the sake of a better argument, i think i need to add 9/10,9/100 and so on but what does the 9 after the bar mean i should add? 9 over 10^what?

 

[ram2048]

Matt. if .999~ = 1, BUT you said you believe in larger than infinite, then infinity+1 number of 9's = what? 1.000~9? infinity+2 9's = 1.000~99?

no matter how mant 9's you add you never get to 1. it doesn't even enter into it what 9 over 10^n the digit location is in. 9 is NOT 10.

and Hurkyl the differences you state have no relevance, what matters is there is a stated upwards limit on the number. And about not being able to GET to infinity but you can GET to 500, you should be able to draw a logical conclusion that if you COULD get to infinity, the result of 1/infinity would NOT be 0 because 1/500 or 1/100billion or 1/100bazomajillion with the same upwards limit would not be 0.

saying .999~ = 1 and 1/infinity=0 is a convenience. nothing more

 

[matt grime]

ram, you miss the point entirely. i know what decimal notation means, and i know how to use it.

now you want in your system to have 0.999999... an infinite number of nines, and then to add one more on to the right. so what does that mean? what are you adding in terms of numbers? the 9 at the n'th spot in the proper expansion means add 9/10^n, so what is it you think you are adding on in your notation when you add one more 9 to the right of all the others. you must be able to tell us.

 

[matt grime]

you also don't appear to understand the idea of cardinality. when you say the number of nines is infinity plus one, you are using cardinals, and implictly we are saying there are aleph-0 of them, and in the arithmetic of cardinals aleph-0 + 1 = aleph-0, it is the cardinality of the set of naturals union a singleton set which is the same as the card of the set of naturals.

saying 0.9999...=1 follows from the definition of the real numbers and decimal expansion, it is not a convenience.


note also that 3 is not 6, yet mod 3 they are equivalent...

 

[matt grime]

if you COULD get to infinity

as we can't do that in the real numbers your argument is vacuous

 

[hello3719]

Ram2048, k then how would you represent

1/3 + (1/3)^2 + (1/3)^3 + ... + (1/3)^n as n goes to infinity

do you agree that it converges to one specific number ? if yes, then try to write that number.

 

[pallidin]

Ram, how can it be made more clear?? Infinity has no specific numerical value and it never can have one. Infinity is a mathematical concept of relationship that is dynamic; it changes and changes and changes... always increasing in numerical value for positive infinity, or always decreasing for negative infinity.
There is no end to this sequence, because it is exactly that: A concept based entirely on having no end.

 

[pallidin]

To the paradox of always being able to divide a distance in half, thereby never mathematically reaching it's destination(for a collapse event):
In standard mathematical division this paradox is correct, and no supercomputer on Earth could EVER collapse a halving division sequence into zero, or "contact"
But where is the flaw?
Halving to zero is inherently impossible because zero is, itself, an infinity. So, "contact" cannot occur at zero, because at zero THERE IS NOTHING TO CONTACT WITH!
So, contact must occur at a defined numerical value specific to the event, and zero is NOT a numerical value, it is an infinity. As such, we can arbitrarily define the "contact" value as "greater than zero", thus, collapse is possible.

 

[pallidin]

We are go for launch: 3...2...1...0... lift-off.

Yeah, mathematically incorrect, as it would suggest: 3...2..1... wait for infinity... lift-off.

So, instead they say. 3...2...1... lift-off.

Get it?

 

[ram2048]

To the paradox of always being able to divide a distance in half, thereby never mathematically reaching it's destination(for a collapse event):
In standard mathematical division this paradox is correct, and no supercomputer on Earth could EVER collapse a halving division sequence into zero, or "contact"
But where is the flaw?
Halving to zero is inherently impossible because zero is, itself, an infinity. So, "contact" cannot occur at zero, because at zero THERE IS NOTHING TO CONTACT WITH!
So, contact must occur at a defined numerical value specific to the event, and zero is NOT a numerical value, it is an infinity. As such, we can arbitrarily define the "contact" value as "greater than zero", thus, collapse is possible.

Palladin: but that would work the same way for convergence to ANY value, not necessarily with zero, so that rationalization holds no significance.

9/10 +9/100 +9/1000 converges to 1 but can NEVER reach 1. these guys will continue to try and confound you with mathematical notations all which have no basis in logic whatsoever, but the truth of the matter is simply that 9 is NOT 10 and infinite 9's is not 1.

but that issue is concluded, working on root 2 now :D

forgot what the heck I'm supposed to do with the thing, however...

 

[ram2048]

Matt: cardinality doesn't enter into it. by defining a default infinity I'm creating a new numberline to extrapolate logical numerical transforms on.

i can say infinity(d) is my 0 and everything i do to it is taken from that point.

certain things will logically transform back to the realm of known reals, like infinity(d) - infinity(d) = 0 or infinity(d)/infinity(d) = 1.

 

[pallidin]

Palladin: but that would work the same way for convergence to ANY value, not necessarily with zero, so that rationalization holds no significance.

9/10 +9/100 +9/1000 converges to 1 but can NEVER reach 1. these guys will continue to try and confound you with mathematical notations all which have no basis in logic whatsoever, but the truth of the matter is simply that 9 is NOT 10 and infinite 9's is not 1.

but that issue is concluded, working on root 2 now :D

forgot what the heck I'm supposed to do with the thing, however...

I am fully aware that it works for both convergence and divergence. I was simply stating an example on one aspect to keep it short.

Ram, perhaps you are not understanding certain things correctly, and I do mean that gently.
Basically, you are suggesting an infinite mathematical series within a finite mathematical framework. This is not possible.
You are certainly correct that .999 forever cannot equal 1. OK. So what? 1.999~ will never equal 2. Agreed, and no one has a problem with that.
But, do you see what you are doing?
Your are forcefully introducing an infinity(.999 ~) in a direct relationship with a finite value(1) on your same number sequence scheme.
This cannot be done, because you are eventually saying that infinity and finity co-exist. Even if it could, under co-existance, what happens?
Think about it.
Infinity takes over by virtue of property and finity becomes non-existant.

 

[Hurkyl]

... There is no end to this sequence, because it is exactly that: A concept based entirely on having no end.

Mathematicians are somewhat less vague about how they define things. For the record, the symbols +&infin; and -&infin; are defined precisely so that they are the endpoints of the extended real line. Topologically, extending the real numbers to include +&infin; and -&infin; is exactly analogous to extending the interval (0, 1) to include 0 and 1.

but the truth of the matter is simply that 9 is NOT 10 and infinite 9's is not 1.

2 is not 1, and 4 is not 2, and 1/2 is not 2/4, yet these manage to be equal...

 

[Hurkyl]

Ok, a reminder for both ram and pallidin:


Mathematically, a decimal is a function from the integers into a set of digits. If f is a decimal, we interpret f(n), intuitively, to mean "the digit in the n-th place of f".


In particular, 0.999~ is the function f where:
f(n) = 0 (for n >= 0)
f(n) = 9 (for n < 0)


In this way, we've managed to represent an "infinite mathematical series within a finite mathematical framework".


Arithmetic can be defined on these decimals representations directly, rather than defining it by translating decimals into real numbers, doing the arithmetic, then translating back to decimals. For example, addition would then be defined as (and this is just arithmetic you learned in elementary school, except it computes from left to right):

Suppose f and g are decimal numbers. We define a new decimal, h, to be the sum of f and g by:

For each position m:

(Computing the carry)
If it is never true that f(n) + g(n) >= 10 for n < m, then carry = 0.
Otherwise, choose n to be the highest integer smaller than m
so that f(n) + g(n) >= 10. If it is true that f(p) + g(p) = 9 for all
p between m and n, then carry = 1. Otherwise carry = 0.

(Computing the digit)
If f(m) + g(m) + carry < 10, then we set h(m) = f(m) + g(m) + carry.
Otherwise, we set h(m) = f(m) + g(m) + carry - 10.


For example, let's add .451 and 1.573
First, let's recall that these decimals are really functions from the integers into digits, so let's write them down. Let's call the first f and the second g:
f(n) = 0 (n >= 0)
f(-1) = 4
f(-2) = 5
f(-3) = 1
f(n) = 0 (n <= -4)

g(n) = 0 (n >= 1)
g(0) = 1
g(-1) = 5
g(-2) = 7
g(-3) = 3
g(n) = 0 (n <= -4)

So now, let's compute h := f + g.

Notice that n = -2 is the only place where f(n) + g(n) >= 10, and
n = -1 is the only place where f(n) + g(n) = 9.

Let's first compute h(m) for m >= 1.
The carry is zero because 0 is between m and -2, and f(0) + g(0) < 9.
Furthermore, f(m) + g(m) = 0, therefore h(m) := 0.

When computing h(0), the carry is 1 because f(-2) + g(-2) >= 10, and
f(n) + g(n) = 9 for all n between 0 and -2. Therefore, h(0) = f(0) + g(0) + 1 = 0 + 1 + 1 = 2.

Similarly, the carry is 1 when computing h(-1). (There is no n between -1 and -2, thus it is vacuously true that f(n) + g(n) = 9 for all n between -1 and -2)
Furthermore, f(-1) + g(-1) + carry >= 10, so:
h(-1) = f(-1) + g(-1) + carry - 10 = 4 + 5 + 1 - 10 = 0

And then we get h(-2) = 2, h(-3) = 4, and h(n) = 0 for n <= -4. Thus,
we've proven that

.451 + 1.573 = 2.024


Now, let's do another example; 0.999~ + 0.111~.
Again, let's write these as functions:
f(n) = 0 (n >= 0)
f(n) = 9 (n < 0)
g(n) = 0 (n >= 0)
g(n) = 1 (n < 0)

When we apply the addition algorithm to compute the sum, h, we get:
h(n) = 0 (n >= 1)
h(n) = 1 (n <= 0)

Note, however, that when we apply the addition algorithm to 1 + 0.111~ we get:

h(n) = 0 (n >= 1)
h(n) = 1 (n <= 0)

Since we get the same function in both cases, we get the same decimal number. 0.999~ + 0.111~ = 1 + 0.111~


This is the "use" of defining that 0.999~ = 1. If we did not define it to be so, then addition as I've defined it above doesn't behave like addition is "supposed to behave". However, if we make this identification, one can then prove that all of the arithmetic algorithms we learned in elementary school behave how they're supposed to behave (e.g. that dividing by 9 then multiplying by 9 is a no-op).


One thing to note is that the elementary school division algorithm involves computing a remainder once you stop; having a remainder turns out to be unnecessary with the decimals in the standard mathematical definition. That they are unnecessary relies entirely on putting the pseudoparadoxes of nonfininte cardinals and ordinals to practical use.

 

[ram2048]

Now, let's do another example; 0.999~ + 0.111~.
Again, let's write these as functions:
f(n) = 0 (n >= 0)
f(n) = 9 (n < 0)
g(n) = 0 (n >= 0)
g(n) = 1 (n < 0)

When we apply the addition algorithm to compute the sum, h, we get:
h(n) = 0 (n >= 1)
h(n) = 1 (n <= 0)

Note, however, that when we apply the addition algorithm to 1 + 0.111~ we get:

h(n) = 0 (n >= 1)
h(n) = 1 (n <= 0)

Since we get the same function in both cases, we get the same decimal number. 0.999~ + 0.111~ = 1 + 0.111~

assuming equal infinity number of digits <for simplicity "infinity(d)"> of each non-terminating decimal. (which would be the logical and sane way to compute that sum).

.999~ + .111~ =

h(n) = 0 (n >= 1)
h(n) = 1 (n <= 0 BUT > h(n)=-Infinity(d) )

that computation yields 10, not 11.

granted it's off the chart of any computer, but there's still no denying that 1 + 9 = 10.

 

[Hurkyl]

assuming equal infinity number of digits

I presume you are referring to the nonzero digits.

The set of positions where 0.999~ has a nonzero digit is the same as the set of positions where 0.111~ has a nonzero digit (in particular, that set is exactly the set of negative integers). What do you think needs to be "assumed"?

<for simplicity "infinity(d)">

How is that simplicity? You've:

yet to fully define that term (either via a list of axioms or by fully specifying just what it is)

you've given no reason to think that it has anything to do with "numbers of things"

if it does have to do with "numbers of things", you've given no reason to think that it should correspond to the number of negative integers

(which would be the logical and sane way to compute that sum).

The logical way to compute the sum would be either directly from the definition of addition, or some alternative algorithm that is equivalent to the definition. I happen to think the definition is a sane method as well, because it corresponds directly to the way we learned how to add decimals in elementary school.

(n <= 0 BUT > h(n)=-Infinity(d) )

First, I presume you meant "n <= 0 but n > -Infinity(d)"

Secondly, I'm not sure about the point you are making by bringing this up. All integers are greater than -Infinity(d), and since n is restricted to the integers, it is fairly pointless to specify that n > -Infinity(d), since it is true for all n.

that computation yields 10, not 11.

What computation? Are you forgetting about carry?

 

[ram2048]

the very first one to the far right. 9 + 1 with no carry

in all arithmetic answers stem from the right, that's how we do things.

111
+999
1110 <-- note the 0

your assumption that a 1 would carry for that digit has no basis in logic because there will always be a digit for which there is no carry.

this is assuming exactly infinite(d) number of decimal digits for both numbers added. (logical)

and yes i have "defined" my default infinity. twice in this thread i believe. and have proven that it DOES work and yields perfect numbers as well.

i totally expect you to immediately come back and say "where" as you normally do so i will wait for that expectantly... :|

 

[matt grime]

you may feel you have defined it but you haven't done so in any rigorous sense (listen to a mathematician for once). in particular you've not explained what number you get when you have infinity+1 9s in the expansion .9999...|9.


so to repeat myself again: i know what 0.999... is, but what on Earth does it mean for you to add one more 9 to the right of all the others?

as you say the number of 9s is infinite you are talking about cardinality. go and read your own post.


also please answer the other issue raised: clear 1 is not 2, 2 not 4, but 1/2 is 2/4. if you understood equivalence relations (of cauchy sequences or rationals! or in this case of localizations) you might begind to understand wherein your ignorance lies.

moreoever, just because you only know how to add from the right doesn't mean that we can't add from the left, it is a very easy algorithm to define, you might care to try it at some point

 

[lvlastermind]

Matt. if .999~ = 1, BUT you said you believe in larger than infinite, then infinity+1 number of 9's = what? 1.000~9? infinity+2 9's = 1.000~99?

no matter how mant 9's you add you never get to 1. it doesn't even enter into it what 9 over 10^n the digit location is in. 9 is NOT 10.

and Hurkyl the differences you state have no relevance, what matters is there is a stated upwards limit on the number. And about not being able to GET to infinity but you can GET to 500, you should be able to draw a logical conclusion that if you COULD get to infinity, the result of 1/infinity would NOT be 0 because 1/500 or 1/100billion or 1/100bazomajillion with the same upwards limit would not be 0.

saying .999~ = 1 and 1/infinity=0 is a convenience. nothing more

---Proof that 2=1 if division by zero was allowed---

a = x [true for some a's and x's]
a+a = a+x [add a to both sides]
2a = a+x [a+a = 2a]
2a-2x = a+x-2x [subtract 2x from both sides]
2(a-x) = a+x-2x [2a-2x = 2(a-x)]
2(a-x) = a-x [x-2x = -x]
2 = 1 [divide both sides by a-x]

All of the steps are perfectly legal except for the last one, dividing both sides by a-x. What is a-x? Well, a=x (step 1), so a-x=0. In the last step, we divided by zero. That's not allowed. And this puzzle is a good example of why
it is not allowed.

---1/infinity not allowed---

The reason why 1/infinity is not allowed is because infinity doest have a value. Infinity can be .33333... or it can be .5555555... or it can be .888888... The point is that 1/.333333... and 1/.55555555... and 1/.8888888... would yield different answers thus making it undefined.

---Infinity not real number---

Infinity is not a number, anyway (not in arithmetic or algebra, see Transfinite Numbers). We can add infinities, and multiply them (sort of). But, we don't get bigger infinities, we get the same infinity. That's interesting. But, if we multiply infinity to both sides of an equation, we are in big trouble. It is the same as dividing by zero. In our little puzzle, when we divided both sides by a-x, that was the same as multiplying both sides by infinity. It is meaningless. It is not allowed in mathematics.

Arithmetic with infinity is not allowed, because infinity is not a number. And, just like our little puzzle, we get answers that make no sense. Calculus is essentially the field in which we deal with infinity and division by zero. And, we never deal directly with infinity or division by zero. We always see what happens when a number gets large without bound or gets closer and closer to zero.

http://www.jimloy.com/algebra/two.htm

---Why 1/0 equals infinity---

What's 1/0? Infinity, right? We said above that we can't divide by zero. But, can't we divide by zero, if we're careful? Let's look at 1/0, more closely. In Calculus, we deal with problems like this by using limits. In other words, we don't look at 1/0, we look at 1/x (the graph of y = 1/x is shown) when x gets close to zero. Well, when x gets close to zero(asymptote), 1/x gets very large without bounds it is infinity. Not so fast, x also gets close to zero on the negative side. Then 1/x becomes a very large negative number, without bounds, it is negative infinity. So, the answer to the question, "What is 1/0?" is "plus-or-minus infinity." Kind of a wild answer, isn't it? It is not exactly simple.

---Why Division by infinity is not allowed...---

If division by infiniy were allowed then 2 would equal 1.

(1)infinity=(2)infinity
(divide both sides by infinity(if infinity were a real #))

1=2

---IN SIMPLE---

1/0 and 1/infinity are not allowed because you cannot divide an 1 apple into zero peices just the same as you cannot divide an apple into infinite peices.

 

[ram2048]

matt you wouldn't even understand my cardinals if i told you. you'd be all "infinity(d+1)" isn't a number blah blah or "infinity(e)" isn't defined.

so what are you pushing for?

and |v|astermind: in my system (1)Infinity(d) = (2)Infinity(d) is false so stuff like that wouldn't happen.

default infinity is very functional like that.

 

[matt grime]

for the fourth time: you say that you can get closer to 1 by adding another 9 after all the nines in .99999...

call this number 0.9...|9

but which real number does that represent?

i am perfectly capable of knowing that w and w+1 are distinct ordinals. you have not defined how to add two (infinite) cardinals, or what that addition represents. for instance in the usual arithmetic of the cardinals the sum of A and B is the cardinality of the union of sets of individual cardinality A and B. Thus aleph-0 +1 = aleph-0, yet w the first infinite ordinal is not equal to w+1.

 

[Hurkyl]

i totally expect you to immediately come back and say "where" as you normally do so i will wait for that expectantly... :|

Yes, I am going to say "where", because that is your #1 problem. I have this wildly optimistic hope that one of these times I'll say it and you'll understand why it's a problem.

An integral part of why the decimal system works is because the positions are indexed by integers, and that there is no largest nor smallest integer.

And you are talking about digits at positions like infinity(d), but there can't be digits at those positions because they're not integers.

 

[Zurtex]

O.K, just answer me these two questions please, if I multiply $\pi$ by 10 will it create a "0"?

If so does that not mean there is a last digit to $\pi$?

 

[ram2048]

it absolutely WILL make a "0"

we may never be able to compute to that digit EVER, but it does exist

 

[ram2048]

And you are talking about digits at positions like infinity(d), but there can't be digits at those positions because they're not integers.

that's why it's a NEW SYSTEM. as i have stated MANY a time.

"your system won't work because it's not like ours"

well if it was exactly like yours what would be the point of claiming it was MY system... OR new?

:|

it works, and one day your math will be obselete.

 

[Hurkyl]

You seem to have difficulty making clear when you are talking about the standard system and when you are talking about your system.

 

[matt grime]

as you claim to have a 'better' system for the real numbers, it ought to exactly concur with the real numbers (wherein 0.999...=1, so you're screwed from the start).

you've still to explain what the 9 you add to the right of all the other 9s in 0.9999... signifies, and to state why it must be true that since x is not y, a statement P(x) cannot be the same as statement Q(y), such as Hurkyl's 1,2 and 2,4 thing with 1/2=2/4.

 

[Zurtex]

it absolutely WILL make a "0"

we may never be able to compute to that digit EVER, but it does exist

Then you are saying there is a last digit to $\pi$

This is what like the 60th time your argument has now been disproved?

 

[ram2048]

as you claim to have a 'better' system for the real numbers, it ought to exactly concur with the real numbers (wherein 0.999...=1, so you're screwed from the start).

you've still to explain what the 9 you add to the right of all the other 9s in 0.9999... signifies, and to state why it must be true that since x is not y, a statement P(x) cannot be the same as statement Q(y), such as Hurkyl's 1,2 and 2,4 thing with 1/2=2/4.

why should my system agree with something that is wrong?

fractions are a completely different notation system, do that in decimals :| and none of that 9xInfinity=10xInfinity "omg 9 =10" crap. that's just stupid.

and Zurtex you have disproved NOTHING. Good work again.

 

[matt grime]

let's stay away from opinion shall we.

in your claim to get closer to 1 than 0.99999... by 'adding a nine' after all the nines already there, what does that actually mean numerically?

I mean i can say that there is an integer between 1 and two, called derek, but it doesn't mean there actually is one there does it?

so what does it mean to add a nine after all the nines in 0.99...?

and you've still not answered any of the 1/2 2/4 thing. and who on Earth says 9=10? all that that proves is infinity is not in the multipliactive group of real numbers, which is not surprising because, erm, it isn't there. after all 0*9=0*10, but that doesn't imply 9=10.

so what does the notation mean of adding that nine after all the infinite number of nines there? this is the sixth time i believe I've asked you and you're yet to provide an answer.

 

[Zurtex]

and Zurtex you have disproved NOTHING. Good work again.

Once again you show your non understanding of mathematics, stop bugging people who give their help free and willingly and actually go and learn some maths.

 

[Hurkyl]

why should my system agree with something that is wrong?

Are you going to tell me that you're not talking about the standard system here either?

 

[arildno]

why should my system agree with something that is wrong?

fractions are a completely different notation system

Oh how interesting!
I never knew that!
(Here I've gone about believing that the decimal notation is simply a shorthand for sums of certain fractions, but now I know differently!)

 

[ram2048]

and you've still not answered any of the 1/2 2/4 thing. and who on Earth says 9=10? all that that proves is infinity is not in the multipliactive group of real numbers, which is not surprising because, erm, it isn't there. after all 0*9=0*10, but that doesn't imply 9=10.

YOU PEOPLE are saying 9 = 10. in order for .999~ to be = to 1, there MUST exist a digit that is 10. since we're using base 10, a DIGIT of 10 cannot exist so that means you're wrong from the start.

so what does the notation mean of adding that nine after all the infinite number of nines there? this is the sixth time i believe I've asked you and you're yet to provide an answer.

Set infinity(d) as cardinal f(n) n=0 for the system would be a way of describing what I'm doing :|

and you had to ask 6 times because i told you already even if i described it to you, you'd just say "that's stupid, our system doesn't work like that, you're wrong"