Is 0.999... Truly Equal to 1 in the Realm of Infinity?

[Zurtex]

Not trying to waste your time at all, just showing you that your system does not work at all ever. But let's go back to basics, you agreed that:

0.3\overline{3} = \sum_{n=1}^{\infty} \frac{3}{10^n}

Well sorry to say but if you work that out using simple sums to infinity of geometric seris you get:

\sum_{n=1}^{\infty} \frac{3}{10^n} = \frac{1}{3}

 

[ram2048]

The point is to show relative consistency; if you believe the standard system to be flawed, then your system must be flawed as well because your system can be used to create the standard system.

not if the flaw is inherent in the creation of the standard. eg: rounding processes.

Here's a puzzler. If you are only introducing remainders in the case of dividing things, then what is e^(-ln 3)?

i have no idea what that is. please explain -ln

has a finite non zero length, thus there must be a hole in the real line at this point, are you saying that .999... is the hole? are there other holes? If there is a hole at .333... why is it that .1(base 3) exists is it not right in the middle of the hole that you have placed at .333...?

i'm saying that because these numbers have infinity included in their nature the gaps between them are beyond human comprehension and can only be measured or described by "infinitessimal" notations.

Choose any of the 9's in 0.999~.
Consider the 9 in the next position to the right.
This 9 gets shifted underneath the 9 you chose when you multiply by 10.
Thus, 9.999~ has a 9 in the same position as the 9 you chose in 0.999~.

If this argument holds for any 9 you choose, then we have proved that 0.999~ cannot have a 9 in a position that 9.999~ does not.

yes but consider a static infinity.

let's say you have two numbers of .999. Each with \infty_d
number of 9's as decimal digits.

you multiply one of them by 10.

compare the two. if they both have \infty_d number of 9's still you have created a gap at the very end such that when you subtract the two there is not an integer result.

if you "create" another 9 (wackiness) then one has \infty_d+1(9's) and the other has \infty_d. hence they're using different infinities and the calculation is incorrect.

Well sorry to say but if you work that out using simple sums to infinity of geometric seris you get: \sum_{n=1}^{\infty} \frac{3}{10^n} = \frac{1}{3}

sums to infinity are wrong which is what I've been trying to say forever now :D

Zeno's paradox is not obliterated by "sums to infinity" within the confines of the problem set forth it creates a true statement that the destination cannot be reached

 

[russ_watters]

you should care because the traditional way of viewing it leads to inaccuracies. such that something cut infinitely results in nothing.

it's fine by me if you want to embrace something that produces the wrong results due to fallacious logic.

go ahead

Math is a tool with which we can model the real world. Its of course necessary that the capabilities of the tool exceed the requirements of any concievable application.

What you see as a flaw in math is actually just part of its power and flexibility. But the problem of course, is that in order to get the correct answers from math, you have to use it correctly:

I think it is your system that is full of holes, it simply does not work. Your time would be better spent actually attempting to learn how the real number system works rather then trying to reinvent a square wheel. It's your time.

Not only do you see holes that aren't there, you're trying to patch them with a wire mesh. Yeah - it really would be better for you if you learned not only how math works, but how to use and apply it correctly.

 

[Zurtex]

sums to infinity are wrong which is what I've been trying to say forever now :D

Then you think 0.\overline{3} is inherently wrong anyway and thus this discussion is pointless.

If you think sums to infinity are wrong you must then believe that we have got values such as e, \pi and \sqrt{2} wrong. Please tell use how you define such values.

 

[Hurkyl]

not if the flaw is inherent in the creation of the standard. eg: rounding processes.

It's straightforward to show that this methodology is consistent; the "danger" is that it might wind up just saying that everything is equal.

i have no idea what that is. please explain -ln

ln is the natural logarithm (LN, but lowercase)

i'm saying that because these numbers have infinity included in their nature the gaps between them are beyond human comprehension and can only be measured or described by "infinitessimal" notations.

If you can measure or describe them by infinitessimal notations, doesn't that place them in the realm of human comprehension?

Something needs clarified here; are you saying that the real numbers has holes, or just the decimal numbers (whatever you think the decimal numbers are)?

yes but consider a static infinity.

I can't; I don't know what a "static infinity" is... but from

you have created a gap at the very end

I'm assuming you mean an order type that has an end. As I stated, if you are using an order type that has an end, then this argumentat is correct. However, this argument fails when the order type has no end. Since there is no last integer, the decimals as they are mathematically defined has no end, so your argument fails.

sums to infinity are wrong which is what I've been trying to say forever now :D

Are you disagreeing with the claim that

<br /> \lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{3}{10^n} = \frac{1}{3}<br />

? Or are you disagreeing with

<br /> \sum_{n=1}^{\infty} \frac{3}{10^n} = \lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{3}{10^n}<br />

?

 

[ram2048]

disagreeing with

\sum_{n=1}^{\infty} \frac{3}{10^n} = \lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{3}{10^n}

that.

the sum will never equal its limit because infinity is infinite

when you say .333 that's a sum when you say 1/3 that's its limit

i'm sure I'm not saying it correct but it should be correct enough to understand.

.333 approaches 1/3 but never reaches it.

.999 approaches 9/9 but never reaches it.

ps> i have no knowledge of how to perform logarithmic calculations and don't have a sci calc handy (not that it would help since i wouldn't know what to put into begin with..)

 

[Zurtex]

Ahh well that clears it up, you simply don't understand what infinity, a limit, a sum to infinity or a recurring decimal is.

 

[Hurkyl]

the sum will never equal its limit because infinity is infinite

Ok, then this is one precise point where we differ.


Mathematically, \sum_{n=1}^{\infty} a_n means \lim_{m \rightarrow \infty} \sum_{n=1}^m a_n. (i.e. it's neither a theorem nor an assumption; this is how an "infinite sum" is defined)

 

[ram2048]

ah so i have to work at getting that abolished first then eh?

because it makes no logical sense that the sum of a convergent series equals it's limit when the very definition of convergent means it will never reach its limit.

so there we go.

 

[arildno]

the very definition of convergent means it will never reach its limit.

?

 

[russ_watters]

the very definition of convergent means it will never reach its limit.

so there we go.

This is another definition on which you are wrong (or 'don't agree with' - which in this case is pretty much the same thing). "Convergent" means it does reach its limit. "Divergent" means it doesn't.

 

[ram2048]

Convergent: - Mathematics. The property or manner of approaching a limit, such as a point, line, function, or value.

hmm Wolfram Mathworld studiously avoids defining convergent...

possibly another conspiracy... ;D

 

[Hurkyl]

A sequence is convergent iff it has a limit.


BTW, have you considered the sequence: 0, 0, 0, 0, ...?


I think what you meant to say is that, generally, each term of the sequence will be inequal to the limit. (But, of course, that is not always true)

because it makes no logical sense that the sum of a convergent series equals it's limit when the very definition of convergent means it will never reach its limit.

I think what you meant to say is that it violates your "common sense".

Even with your definitions this doesn't follow; even if the partial sums don't "reach" the limit, why should that suggest anything about the infinite sum?

 

[ram2048]

for fun's sake let's go back to Zeno's Paradox

1/2 + 1/4 + 1/8 + 1/16...

every step only completes half of the remaining value converging towards 1.

BUT let us consider the last digit

1/\infty

here's where we get the problem i think

1/\infty would complete only half distance theoretically as well, BUT current math doesn't believe beyond infinity.

AND

they believe that anything divided by infinity is the same number

hence you COULD have 2/\infty and complete the whole remaining distance and it would be the same as adding 1/\infty

pure conjecture of course. i think that in order to continue our understanding of math we have to find ways to define stuff better. having 20 different definitions for infinity is a pain in the patoot

 

[Integral]

having 20 different definitions for infinity is a pain in the patoot

I sure understand what you are saying, that is why I cannot figure out why you insist on having so many definitions of infinity.

If you would simply make an effort to understand the SINGLE definition used by Mathematicians your world would be much simpler.

In the extended real line infinity is simply defined as being larger then all real numbers. A single simple definition. Thus any real number times infinity is still infinity because it is still larger then any real number. Likewise 1 over infinity must be smaller (in absolute value) then all real numbers, the only thing that meets this condition is 0, therefore \frac 1 \infty = 0 simple and consistent. You only add unnecessary complexity with your infinite number of infinites.

You are right that a set of infinitely shrinking intervals does indeed contain something (I think you phrased it that no matter how much you divide it up there is still something there). That something is a single point. Will you agree that the "length" of a single point is 0? That is 1-1=0? So does it not make sense to say that the length of the interval resulting in infinite divisions (which results in a single point) has length zero?

Will wait for your reply to make sure we are on the same page.

 

[matt grime]

Ram seems to be having the standard issue that sequences can tend to limits and not actually reach them after a finite number of steps. That aand the facthe thinks there's a last term in an infinite sequence.

Would you like to list the 20 different definitions of infinity? There's only one meaning for infinite, and no need to use the word infinity if it confuses you.

 

[ram2048]

In the extended real line infinity is simply defined as being larger then all real numbers. A single simple definition. Thus any real number times infinity is still infinity because it is still larger then any real number.

so you consider infinity to be a number governed by its own rules such that infinity is greater than itself (greater than all real numbers)

or is it not a number, such that it doesn't break that definition being greater than itself should infinity + 1 > infinity

and as far as the interval being a single point i don't believe it so. I still believe everything is further infinitely divisible beyond infinity.

 

[arildno]

ram_1024:
(I won't be part of this ego-doubling process from your side; however, I'm not a churl, so I won't start halving it either)

I'd like to point out a few implications of your own ideas, rather than making clear to you why standard maths is consistent, and hence, why your attacks on it is quixotic at best.

(Others on this forum are by far more competent than myself in doing such an explanation, and if you had bothered to read, and tried to digest what they have patiently written to you, you would have stopped these ridiculous attacks long ago)

Implications of your ideas:
a)
Now, you say that a sequence: 0.1, 0.01, 0.001 and so on does not converge to 0,
but to some number 0...1 (with an infinite number of zeroes in between)
Clearly then, a sequence: 0.2, 0.02, 0.002 goes to some number:
0...2 (Right?)

In particular, this is perfectly in accordance that each term in the second sequence is the double value of the same term in the first sequence:
For example, 0.02=2*0.01, so it should make sense that the end value,
0...2 is the double of 0...1 (Correct?)

b) Look now at the sequence:
1, 0.1, 0.01 and so on.
At every single instance, the term in this sequence is 10 times bigger than the same term in the sequence 0.1, 0.01 and so on.

Hence, by your logic, 1, 0.1, 0.01 must go to the number 0...10
(You can't escape this conclusion, sorry about that!)

c)
However, the first sequence is simply a subsequence of the last one; i.e. subsequences in your system doesn't go to/converge to the same values as the sequence itself.
In short, the whole convergence concept is blown to smithereens, and it is meaningless in the first instance to say that something "produces" or goes to
something at all; i.e. every single utterance you have made is mumbo-jumbo and nothing else.

Have a good day!

 

[matt grime]

there is nothing contradictory about this infinity+1 not being greater than infinity, since the definition is that it is greater than all real numbers; infinity is not a real number, so infinity+1 is still infinity, unless you're talking ordinals, when w and w+1 are distinct ordinals, but that is a different system from the extended real line again. amazingly it always seems that these crackpot attempts at showing inconsistency are inconsistent, but also that someone non-crackpot has thought it through and offered something that is consistent and does it properly: surreal numbers, hyperreal numbers, the extended complex plane, ordinals, cardinals...

 

[Hurkyl]

Suppose I do a thorough analysis of red M&Ms. Does that mean there are no green M&Ms?

The destination, indeed, does not appear in Zeno's analysis, but there's no reason in particular to think that Zeno's analysis covers the entirety of the motion in question.

However, there is not a "remaining distance" either. Zeno's analysis covers every position up to (but not including) the destination. If the destination is 1 meter away, then Zeno's analysis covers all positions x where 0 <= x < 1. The proof of this requires the Archmedian property posessed by the real numbers: for every real number r there is an integer n such that n > r.

It goes roughly as follows: let x be any real number in 0 <= x < 1. Let y = 1/(1-x). By the archmedian property, there is an integer n > y. By induction, we can find an integer m with 2^m > n. Thus, there is an integer m with 2^m > y, so 1/(2^m) < 1-x and 1 - 1/(2^m) > x. However, 1 - 1/(2^m) is the "current position" of the runner after the m-th step of Zeno's analysis. Thus, the position x has been considered by Zeno.

1/2 + 1/4 + 1/8 + 1/16...

every step only completes half of the remaining value converging towards 1.

BUT let us consider the last digit

Did you mean last term?

And why would there be a last term? Each term is of the form 1/2^m where m is an integer and there is no last integer...

so you consider infinity to be a number governed by its own rules such that infinity is greater than itself (greater than all real numbers)

The system Integral describes is called the "extended real numbers". In this system, infinity and -infinity are extended real numbers that are not real numbers.

 

[russ_watters]

so you consider infinity to be a number governed by its own rules such that infinity is greater than itself (greater than all real numbers)

Looks like we also need to define "real numbers" for you. The word "real" isn't arbitrary as you are using it, it has a specific definition in math as well. http://en.wikipedia.org/wiki/Real_numbers

In mathematics, the real numbers are intuitively defined as numbers that are in one-to-one correspondence with the points on an infinite line—the number line. The term "real number" is a retronym coined in response to "imaginary number".

Notice, while the real number line is infinite, "infinity" is not a point on the line, hence "infinity" is not a real number. Your above objection is based on a misunderstanding of the definition of "real numbers." (you also characterized "infinity" as a "digit" a few posts ago - also incorrect).

Seriously, ram, you have a lot to learn about math. What we're talking about here is largely high school stuff. If you would only accept that you have a lot to learn and decide to learn it, you'd be much better off.

 

[ram2048]

i can goto 5 sites and pull a different definition of real numbers, irrational numbers, and infinity.

and you talk to ME about inconsistency. I can't use your words because your definitions are "Mumbo jumbo"

i say something then you correct me with a different definition that means the same thing.

the point of the matter is, you say the number line is infinitely divisible but then you say when you divide it by infinity "it can't be determined" or "is close enough to zero we'll just call it zero"

i'm trying to establish a notational system that allows for infinite division AND allows for division beyond that.

you guys say the current system works, why mess with it, but totally balk at the idea of replacing it with something that works BETTER.

sentimental value i guess... :|

 

[ram2048]

Implications of your ideas:
a)
Now, you say that a sequence: 0.1, 0.01, 0.001 and so on does not converge to 0,
but to some number 0...1 (with an infinite number of zeroes in between)
Clearly then, a sequence: 0.2, 0.02, 0.002 goes to some number:
0...2 (Right?)

these are not even converging sequences under my notation. you would have .00...02, .00...002, .00...0002, . . .

Have a good day!

i didn't! you cursed me didn't you!

 

[ram2048]

Did you mean last term?

And why would there be a last term? Each term is of the form 1/2^m where m is an integer and there is no last integer...

of course i meant last term. Integral has me talking about digits and then you jump on me when i get a word wrong. okay okay spanish inquisition over here..

and yes there is no last integer by MY notation, but there IS by the current one. such that 1/Infinity = 0 = limit.

it's your system though, you KNOW it better than me why do i have to tell you?

 

[Hurkyl]

i can goto 5 sites and pull a different definition of real numbers, irrational numbers, and infinity.

I wouldn't doubt it. There are often many equivalent ways of defining something.

For instance, (IIRC) the first rigorous definition of the real numbers was that a real number is an equivalence class of cauchy sequences of rational numbers. (In laymen's terms, any real number is identified with the sequences of fractions that approach it)

The usual modern definition preferred would be to define the real numbers as a complete ordered field. (In laymen's terms, +, -, *, /, and < all work "properly", and there are no "holes")

And it can be proved that the first definition satisfies the requirements of the second definition. Conversely, anything that satisfies the second definition is isomorphic (in laymen's terms, the same) to the first definition.


(Incidentally, 5? Really? I can only think of 3 definitions of real numbers you could be reasonably expected to find online, and only one definition of irrational. Different definitions of infinity wouldn't surprise me, because there are a lot of different concepts that are (sloppily) called "infinity")

the point of the matter is, you say the number line is infinitely divisible but then you say when you divide it by infinity "it can't be determined" or "is close enough to zero we'll just call it zero"

Actually, we say "Please be more specific".

"Infinitely divisible", taken entirely literally, means that it can be divided into an infinite number of terms. (Notice I did not say "an infinity of terms") In particular, the real line can be divided into c (= |R|) terms.

c is a thing called a cardinal number. It is not a finite cardinal, so it must be an infinite cardinal. (Some call it a transfinite cardinal, simply because so many laymen get confused when things are called infinite)

Division is not well-defined on cardinals, because multiplication is not very nice. For example, 1 * c = 2 * c. If you could divide by c, you would get 1 = 2, which is bad.

i'm trying to establish a notational system that allows for infinite division AND allows for division beyond that.

Learning math might be helpful to see how to do this.

Here's a simple approach to defining such a system:

Consider all (real) rational functions in x. E.G. things like 6, 7 + x^2, and (1 + 3x + 4x^5)/(4x + 3x^7)

+, -, *, and / can all be defined and function "properly" in this field. You can order this field by decreeing that x is bigger than any real number, and then extending the definition of < to accommodate this decree. So, for example, 7 + x^2 is infinite, because it is bigger than any real number. Proof:

Let r be any real number.
r < x
1 < x
r = 1 * r < 1 * x < x * x < x * x + 7
Thus r < x^2 + 7

Similarly, (1 + 3x + 4x^5)/(4x + 3x^7) is an infinitessimal.

If you don't like x, maybe you could use w (omega), a common symbol for transfinite numbers.


There's another number system (whose technical details are very difficult to follow) called the hyperreals which have transfinites and infinitessimals, but, for the most part, behave exactly like the real numbers. You might find information on this by searching for "non-standard analysis". Last time I went looking, there was actually an undergraduate calculus text in PDF format somewhere on the web that develops calculus using nonstandard analysis (i.e. with infinitessimals and transfinite integers, etc) instead of the usual way.

you guys say the current system works, why mess with it, but totally balk at the idea of replacing it with something that works BETTER.

We're "balking" because you are claiming the current system doesn't work.

of course i meant last term. Integral has me talking about digits and then you jump on me when i get a word wrong. okay okay spanish inquisition over here..

Or, maybe I'm just making sure I knew what you meant.

 

[ram2048]

i never claimed it didn't work, i claimed it was inaccurate ;D

i walk on bridges and fly by airplane. i wouldn't do so if i didn't trust our current set of math at all

in any case, there's little point in arguing as it seems apparent that such things are not going to change overnight.

i thank you for the envigorating discussion, and humbly apologize for getting bent out of shape at you in previous threads, Hurk

 

[matt grime]

yes there is no last integer by MY notation, but there IS by the current one. such that 1/Infinity = 0 = limit.

it's your system though, you KNOW it better than me why do i have to tell you?

There is no last integer in 'our' system, that you think we think there is is your error.

 

[ram2048]

ooooooooooooooooookay

define a number less than 1/infinity but greater than 0 in your system

 

[matt grime]

what do you mean by 1/infinity? infinity isn't a real number, so why should i be able to do that? have you not learned anything from this thread? you aren'ty dealing with the real numbers when you write that kind of thing.


in robinsonian analysis i believe the object you are talking about is labelled epsilon.

 

[Zurtex]

Well I'd like to thank all the people who tried to help ram out on this thread, whether or not he still spouting rubbish you have really made me feel a lot better about numbers . I still can't believe your trying to get the point across and have not just banned him from posting on the maths forums and ignored him here, well done your good people.

 

[Hurkyl]

define a number less than 1/infinity but greater than 0 in your system

This question, in general, doesn't make sense.

It cannot make sense in the context of the real numbers.

In the context of the extended real numbers, this cannot be done, since 1/infinity is 0.

In the context of the hyperreal numbers, if we rephrase your challenge (so that it makes sense) as:
"If w is a positive, infinite number, then define a number less than 1/w but greater than 0"
then an answer to your challenge would be 1/(2w).

 

[ram2048]

i somehow wrote that all wrong :O

what i meant to say is the "last" integer in your system would be a function of infinity such that nint(infinity) = that integer (theoretically if you could USE that function)

but that's still getting off the point.

if you COULD then 1/2^nint(∞) would be the closest step for your consideration.

forgot where i was going with this O_O

meh...

 

[Zurtex]

ram2048 what is a "last integer"?

That makes less sense to me than anything else you have written.

 

[ram2048]

hell if i know, Hurkyl was talking about integers and blah blah no last integer..

but there IS a last integer because you have an upwards limit of infinity.

you can't define that number without it being an expression "of infinity" itself, so it kinda defies itself.

but there was a point i was trying to make such that a sum to infinity, even AT infinity does not equal its limit in such a convergent sieries as Zeno's paradox.

as with everything "Infinity related" it's all theory and you have to apply logic. if every "step" in the process or "term" computed is half the remaining, there will never be a process that is "whole of the remaining" because that breaks the rule set forth in the initial exercise. so even at infinity, or beyond infinity in the case of "hyperreal" blah blah "extended irrationals" or whatever you want the sum STILL doesn't equal the limit.

the same is true of .999~ every digit tacked onto the end brings it closer and closer to 1, but no digit is ever a 10 completing a whole "step"

but it really doesn't matter... whatever :D

 

[chroot]

So there's an integer that is larger than every other integer, eh? Could you do us a favor and show us which integer is the biggest?

- Warren

 

[ram2048]

the one closest to infinity? :O

 

[chroot]

That's the stupidest thing I've ever heard.

- Warren

 

[ram2048]

haha my point is proven

not only can you NOT get to infinity, ever. you can NOT get to the largest integer that is not infinity.

so what hurkyl said about every step being considered is kinda out the window eh?

in any case .999 can never be = to 1 because every step is 9/10's the remaining step towards 1 and "the destination is never in consideration"

there is no step covered that is 10/10ths the remaining distance

sums to infinity are a good approximation

 

[Hurkyl]

What is one plus the "largest integer that is not infinity"?
What is two plus the "largest integer that is not infinity"?

(plus, here, means integer addition)

the same is true of .999~ every digit tacked onto the end brings it closer and closer to 1

Why (and how) are you tacking digits onto .999~?

not only can you NOT get to infinity, ever. you can NOT get to the largest integer that is not infinity.

But you forget, you can "get" to every positive integer. It's a simple proof by induction:

I can get to 1.
If I can get to n, then I can get to n+1 by adding 1 to n.
Thus, by induction, I can get to any positive integer.

If the number system about which you are speaking has numbers to which you cannot "get", then you're not speaking about the integers.

in any case .999 can never be = to 1 because every step is 9/10's the remaining step towards 1 and "the destination is never in consideration"

.999 can never be equal to 1 because it is equal to 999/1000. Of what steps are you speaking?

 

[Hurkyl]

sums to infinity are a good approximation

Approximation of what?

 

[lvlastermind]

ram doesn't understand that you can't have 1/infinity because infinity=0 and you can't divide by zero.

 

[ram2048]

Why (and how) are you tacking digits onto .999~?

doesn't matter why and how, but if you must know it's by process of summation where each "step" resolves into one "digit".

9/10 + 9/100 + 9/1000 ...

.999 can never be equal to 1 because it is equal to 999/1000. Of what steps are you speaking?

you knew what i meant.. :P

if you accept Zeno's conjecture as true, within the confines of the problem set forth you can not reach the destination then you also accept that within mathematics .999~ can never equal 1 simply because they are the same problem with a different curve Zeno's is 1/2 and .999~ is 9/10ths.

 

[lvlastermind]

if you accept Zeno's conjecture as true, within the confines of the problem set forth you can not reach the destination then you also accept that within mathematics .999~ can never equal 1 simply because they are the same problem with a different curve Zeno's is 1/2 and .999~ is 9/10ths.

but you will reach your destination with an infinite number of halfs. This is one example of how infinity doesn't act like a real number. Beacause it doesn't have a value but it still has meaning.

 

[Zurtex]

haha my point is proven

not only can you NOT get to infinity, ever. you can NOT get to the largest integer that is not infinity.

We don't add the things one at a time.

 

[ram2048]

but you will reach your destination with an infinite number of halfs. This is one example of how infinity doesn't act like a real number. Beacause it doesn't have a value but it still has meaning.

no this is an example of how calculus uses "infinity" to approximate.

if something cannot logically EVER be something then infinity and forever it will not be it.

how can you possibly reason that the destination is reached? it means that the last step you took wasn't a half-step but a whole one.

i have described earlier how calculus accepts 1/∞ and 2/∞ as the same number because both would be equal to 0.

indeed if 1/∞ = 0 then it is not actually a step at all, so the runner MUST have reached the destination in step 1/2^nint(∞) where nint defines the nearest integer to infinity.

but then we would have the paradox of being able to halve that distance YET AGAIN. such that 1/2^[nint(∞)+1]

and so on... so somehow the conclusion that the destination CAN be reached has to be wrong. or 1/∞ = 0 is wrong. or both are.

i say both personally ;D

 

[Zurtex]

Your argument is wrong.

1/n > 0

Where n is a positive integer. There is NO nearest integer to infinity, that's just silly. Your problem is that you do not seem to be able to grasp a concept of infinity and that it is not on the real number line.

 

[ram2048]

excuse me, but you guys sum n to infinity ALL THE TIME

that means it's on the same number line, it may not be included in your set of reals, but it's still on the same line.

that means somewhere along the line increasingly greater numbers become infinite. if NOT sum n to infinity has NO meaning. like saying sum n to cow or sum n to vacuum cleaner...

the definition of infinity provides a clear relation of "the concept" to known reals such that even though it's not a number it's a function of numbers so much so that it is possible to use it in calculations.

if you're saying a set of reals can never increase to infinity then you're basically accepting that given the infinite number of steps in Zeno's problem, the man will NEVER reach his destination. clear now?

if you understand that much go back to my other post and read the logical explanation on how even at infinity the destination cannot be reached, so it makes no real difference whether a "largest integer" is real or not. the outcome is still the same

 

[ram2048]

sums to infinity are a good approximation

Approximation of what?

approximation of the actual value that would come out of the calculation if you calculated it out longhand. :|

logically

 

[Zurtex]

excuse me, but you guys sum n to infinity ALL THE TIME

that means it's on the same number line, it may not be included in your set of reals, but it's still on the same line.

that means somewhere along the line increasingly greater numbers become infinite. if NOT sum n to infinity has NO meaning. like saying sum n to cow or sum n to vacuum cleaner...

the definition of infinity provides a clear relation of "the concept" to known reals such that even though it's not a number it's a function of numbers so much so that it is possible to use it in calculations.

if you're saying a set of reals can never increase to infinity then you're basically accepting that given the infinite number of steps in Zeno's problem, the man will NEVER reach his destination. clear now?

if you understand that much go back to my other post and read the logical explanation on how even at infinity the destination cannot be reached, so it makes no real difference whether a "largest integer" is real or not. the outcome is still the same

No, your totally wrong as has been proven and shown on this thread many times.

 

[Hurkyl]

doesn't matter why and how, but if you must know it's by process of summation where each "step" resolves into one "digit".

9/10 + 9/100 + 9/1000 ...

This notation looks like an infinite sum; I don't see any "steps".

(To keep things moving)

I'm presuming by "steps" you are first considering 9/10, then 9/10+9/100, then 9/10+9/100+9/1000, and so on. But, of course, none of these are 9/10 + 9/100 + 9/1000 ... (though the limit of these "steps" is)

if you accept Zeno's conjecture as true, within the confines of the problem set forth you can not reach the destination

I have no problem with that. Zeno only considers the motion up to (but not including) reaching the destination. Thus, it would be silly to think that the destination would be reached in the period he analyzes.

then you also accept that within mathematics .999~ can never equal 1 simply because they are the same problem with a different curve Zeno's is 1/2 and .999~ is 9/10ths.

How do you figure? If I do it with 9/10s, then Zeno considers each of these intervals of position [0, 9/10], [9/10, 99/100], [99/100, 999/1000] ...
Putting all of these intervals together yields the interval [0, 1).

excuse me, but you guys sum n to infinity ALL THE TIME

Right, and sums to infinity are defined by

<br /> \sum_{i=1}^{\infty} a_i = \lim_{m \rightarrow \infty} \sum_{i=1}^m a_i<br />
which can be further resolved (by applying the definition of limit)
\sum_{i=1}^{\infty} a_i = L if and only if for every positive \epsilon there exists an integer N such that for any integer m greater than N, we have
<br /> |L - \sum_{i=1}^m a_i| &lt; \epsilon<br /> [/itex]<br /> <br /> Notice, in particular, that this is not logically equivallent to saying that you keep adding terms one by one until you&#039;ve reached an infinite number of terms. (Though, IMHO, it&#039;s for the most part conceptually equivalent)<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> that means it&#039;s on the same number line, it may not be included in your set of reals, but it&#039;s still on the same line. </div> </div> </blockquote><br /> The number line only has real numbers on it, thus it doesn&#039;t have infinity on it. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f61b.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":-p" title="Stick Out Tongue :-p" data-smilie="7"data-shortname=":-p" /><br /> <br /> But, as mentioned before, mathematicians do use an extension of the real numbers which has a positive and negative infinity on each endpoint. But...<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> that means somewhere along the line increasingly greater numbers become infinite. </div> </div> </blockquote><br /> No it doesn&#039;t. A sequence of increasingly greater numbers can <b>converge</b> to infinity, but none of the individual numbers need be infinite... just like a sequence of numbers can converge to zero, but none of them need to be zero. (e.g. 1, -1, 1/2, -1/2, 1/4, -1/4, 1/8, -1/8, ...)<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> if NOT sum n to infinity has NO meaning. </div> </div> </blockquote><br /> I&#039;ll say it again, sum to infinity has this meaning:<br /> <br /> &lt;br /&gt; \sum_{i=1}^{\infty} a_i = \lim_{m \rightarrow \infty} \sum_{i=1}^m a_i&lt;br /&gt;<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> if you&#039;re saying a set of reals can never increase to infinity </div> </div> </blockquote><br /> I&#039;m saying no real number may be infinite, an entirely different statement.<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> you&#039;re basically accepting that given the infinite number of steps in Zeno&#039;s problem, the man will NEVER reach his destination. clear now? </div> </div> </blockquote><br /> I accept that; the destination is reached <i>after</i> the &quot;steps&quot; contemplated by Zeno.<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> approximation of the actual value that would come out of the calculation if you calculated it out longhand. :|<br /> <br /> logically </div> </div> </blockquote><br /> How do you plan to add an infinite number of terms by longhand?