Is 0.999... equal to 0?

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The difference is that you are taking the sum of the elements in the series, not the limit of the series itself. The limit is the value that the series approaches, not the sum of its elements. In the series 0.1, 0.01, 0.001, ..., the limit is 0, but the sum of the elements is not. In the series 0.9, 0.09, 0.009, ..., the limit is 0.999..., but the sum of the elements is not. It is important to understand the difference between the two.
  • #1
Seasons
Hi folks, my first post here. I was researching that guy who claims to be the smartest person in the world (Chris Langan) and found thread here… seemed like a good place to ask my question, or submit my observation.

So… I'm thinking about 1/9, and I visualize a graph where the ones are slowly converging to zero as they become smaller through the "expanding" regress. And I sat and pondered the argument on wikipedia. 9*1/9 = 1. The convergence proof, right?

But I thought, if you're going to assume convergence using the same rules, then why is not the first step, 1/9 = 0, before you multiply it nine times. There isn't a ratio that gives you 0.9…, you have to multiply it to get there, so by order of operations, the steps that get you to .9… should actually get you to zero, if you use the assumption of convergence.

How am I wrong?
 
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  • #2
If you are going to reference outside material, it is customary to provide a link so that people have a chance to know what you are talking about. Please provide the appropriate link for context.

Seasons said:
then why is not the first step, 1/9 = 0
Probably because 1/9 is not equal to zero. Without knowing what "first step" you are referring to, it is impossible to answer in any other way.
 
  • #3
Seasons said:
9*1/9 = 1. The convergence proof, right?
There is no convergence to study here. 9*1/9=1 by definition of 1/9.
Seasons said:
then why is not the first step, 1/9 = 0
Why should 1/9 be 0? It cannot be 0.Decimal representations are ways to write a number. They are not what defines a number!
0.1=1/10=5/50 is all the same number - written in three different ways.
 
  • #4
1/9 has to be zero, because it converges to zero, ONLY IF .999… converges to 1. It's using the same exact logic. Simply saying 0.1… doesn't equal zero is not an argument, especially when you argue that 0.9… converges to 1 using the same exact logic.

I stated that the argument in question was found among the 20 or so arguments on the wikipedia page: https://en.wikipedia.org/wiki/0.999…

All you have to do is think through the steps… 0.1 gets smaller and smaller and smaller, getting closer and closer and closer to zero if you plot the regress on a graph, so if you assume convergence, as you do with other sequences, 0.1… converges at zero. It seems obvious to me.

Mentor note for the benefit of readers of this thread. The OP, @Seasons, has many misconceptions about how real numbers work, almost too many to count.
"1/9 has to be zero" -- It should be obvious that this can't possibly be true.
"...because it converges to zero" -- 1/9 is a number, a constant that doesn't change -- convergence doesn't enter into the discussion.
"ONLY IF .999… converges to 1" -- .999... is also a constant, that happens to be exactly equal to 1. No convergence here, either.
"Simply saying 0.1… doesn't equal zero is not an argument" -- It is a fact that is (almost) universally recognized that 0.1... is not equal to zero.
"0.1 gets smaller and smaller" -- No, 0.1 is a constant that doesn't change.

Several members have responded in rest of this thread, attempting to correct the OP's misconceptions, but the misconceptions are so numerous, not all of them have been addressed.
 
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  • #5
Seasons said:
1/9 has to be zero, because it converges to zero
No it doesn't. 1/9 is a well defined rational number. Assuming that you are at high-school level, I would say that this error is mainly based on lack of knowledge of what "converges to" means.

Seasons said:
Simply saying 0.1… doesn't equal zero is not an argument, especially when you argue that 0.9… converges to 1 using the same exact logic.
It is not the same logic. You can easily find a number between 0.1... and 0 (take 0.05 for example) and so the limit cannot be zero. The corresponding thing would be 0.0... = 0, which is obvious.

Seasons said:
0.1 gets smaller and smaller and smaller, getting closer and closer and closer to zero
No it doesn't. You are thinking of the series 0.1, 0.01, 0.001, ... which has the limit zero but is not the same thing as 0.11..., which is the limit of the series 0.1, 0.11, 0.111, ...
 
  • #6
That's where my potential confusion is!

if you add .1, .01, .001 = .111 correct?

so how are they different? I don't understand this part
 
  • #7
Seasons said:
if you add .1, .01, .001 = .111 correct?
Yes, but this is not what the series is. You are confusing the limit of the series with the sum of the elements in the series. The series 0.9, 0.09, 0.009, ... converges to zero. If you make a series where the nth entry is the sum of that series' first n elements you get the series 0.9, 0.99, 0.999, ... - which is a different series that converges to one.
 
  • #8
Seasons said:
That's where my potential confusion is!

if you add .1, .01, .001 = .111 correct?

so how are they different? I don't understand this part
What is different from what?
0.1+0.01+0.001=0.111, correct.
Note that 0.11 is larger than 0.1 and 0.111 is larger than 0.11.
 
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  • #9
Orodruin said:
Yes, but this is not what the series is. You are confusing the limit of the series with the sum of the elements in the series. The series 0.9, 0.09, 0.009, ... converges to zero. If you make a series where the nth entry is the sum of that series' first n elements you get the series 0.9, 0.99, 0.999, ... - which is a different series that converges to one.
Ok, this is my issue. There is no such thing as a ratio that gives you 0.999…

It can ONLY be arrived upon by a sum. Either 0.333…*3 0r 0.111…*9

So in one sentence you said I'm confusing the limit with the sum, yet, you can only use a sum to get to the limit. I'm not fully out of my mental problem with this.
 
  • #10
mfb said:
What is different from what?
0.1+0.01+0.001=0.111, correct.
Note that 0.11 is larger than 0.1 and 0.111 is larger than 0.11.

Let me think about that… it seems obvious, but I think the rules that make that number might cause a contradiction - which are the rules of summing.

What I mean to suggest, is that the same rules used to make 0.999… equal to 1, appear to make 0.111… equal to 0. Almost as if, the mental hurdle that people have with the 0.999… issue, might be something that facilitates an additional mental hurdle for the implications of those rules applied to the number 1; namely, summing and convergence.

Is it the case that 0.9, 0.09, 0.009… equals zero in the same sense 0.1, 0.01, 0.001… equals zero, as was also stated earlier. Argghh… its hurting my brain. Maybe I need to take all this in!
 
  • #11
Seasons said:
There is no such thing as a ratio that gives you 0.999…
Yes there is: 2/2 = 0.999... because 0.999... = 1.

You have to understand which real number is intended when we write it on decimal form. By a number on the form ##0.a_1a_2a_3\ldots## where the ##a_k## are digits, we mean the limit of the series ##0.a_1##, ##0.a_1a_2##, ##0.a_1a_2a_3##, ##\ldots## or, equivalently, the sum
$$
\sum_{n = 1}^\infty \frac{a_n}{10^n}.
$$
In the case of 0.999..., this sum is equal to one.

A different way of stating this is that for any distinct real numbers, it is possible to find a number (even a rational number) which is between them. You can find no such number for 0.999... and 1 and therefore they must be the different representations of the same real number.
 
  • #12
Seasons said:
What I mean to suggest, is that the same rules used to make 0.999… equal to 1, appear to make 0.111… equal to 0.
No they don't. You are applying them differently even if you do not see it. The corresponding thing in the 0.999... case would be to claim that it converges to zero because the series 0.9, 0.09, 0.009, ... converges to zero - which is not true. The value of 0.999... is the limit of the series 0.9, 0.99, 0.999, ... not the limit of the series 0.9, 0.09, 0.009. Just as 0.111... is the limit of the series 0.1, 0.11, 0.111, ... not the limit of the series 0.1, 0.01, 0.001, ...
 
  • #13
Seasons said:
What I mean to suggest, is that the same rules used to make 0.999… equal to 1, appear to make 0.111… equal to 0.
They do not. They make 0.000... = 0.

0.1111... is well above 0.
0.1 is larger than 0 and the sequence 0.1, 0.11, 0.111 ... is increasing. How can you expect to end up at 0?
 
  • #14
Orodruin said:
Yes there is: 2/2 = 0.999... because 0.999... = 1.

You have to understand which real number is intended when we write it on decimal form. By a number on the form ##0.a_1a_2a_3\ldots## where the ##a_k## are digits, we mean the limit of the series ##0.a_1##, ##0.a_1a_2##, ##0.a_1a_2a_3##, ##\ldots## or, equivalently, the sum
$$
\sum_{n = 1}^\infty \frac{a_n}{10^n}.
$$
In the case of 0.999..., this sum is equal to one.

A different way of stating this is that for any distinct real numbers, it is possible to find a number (even a rational number) which is between them. You can find no such number for 0.999... and 1 and therefore they must be the different representations of the same real number.
Ok here! You're using a lot of implied steps that I'm puzzling about to make the claim that 2/2 = 0.999… Everyone knows that there is no actual ratio that spits out repeating 9's, the only method actually used is summing lower ratios in decimal form, and then making the equality from there.

So the question is, if you're going to sum those ratios, why can't you sum other ratios?

There's also no ratio that gives you 0.000…1, only a summing sequence, just like 0.999…

So, if you're going to use summing sequences to make decimals that have no ratios, then it begs the question of why 0.111… (ratio) doesn't equal the summing sequence of 0.1, 0.01, 0.001… (no ratio) The sum gets greater (0.111…) as the summing sequence gets smaller (0.1, 0.01, 0.001) - how are they not equal? You're doing the same thing for 1 (ratio) and 0.999… (no ratio). ?? Because you've proven a ratioless number equals a ratio number here, how is not the ratioless number of 0.000…1 not equal to the sum of 0.1, 0.01, 0.001… not equal to 0.111?

I'm also thinking about the argument that there is no number between 0.999… and 1, but the is between say, 0 and 0.111… : I want to ponder this further, because I think I might be able to make the argument that anything between them solves as them (one or the other / both beings equal)
 
  • #15
Seasons said:
So, if you're going to use summing sequences to make decimals that have no ratios, then it begs the question of why 0.111… (ratio) doesn't equal the summing sequence of 0.1, 0.01, 0.001… (no ratio)
It does equal the sum of the sequence. The sum of the sequence is not equal to the limit of the sequence!
 
  • #16
Orodruin said:
It does equal the sum of the sequence. The sum of the sequence is not equal to the limit of the sequence!

Ok! This is the second time you've said this to me. So I should consider this quite seriously. The point I was trying to make is that this method is defining sums as limits! The method of 0.999… = 1 is using only sums to define the limit. Yet you make the distinction that I'm being confused when I do this for numbers like 1/9. Defining sums as limits that is.
 
  • #17
The limit of a (Cauchy) series ##\{x_n\}## is defined as the number ##x## such that ##|x-x_n|## is arbitrarily small as long as ##n## is large enough. An infinite sum can be defined by letting
$$
y_n = \sum_{k=1}^n x_n
$$
and taking the limit of the series ##\{y_n\}##. If ##\{x_n\}## is a series, there is no reason to believe that it converges to the same number as the series ##\{y_n\}## as defined above, since the series are different.
 
  • #18
Orodruin said:
Yes, but this is not what the series is. You are confusing the limit of the series with the sum of the elements in the series.
@Orodruin, you are misusing some standard terminology here. "Limit of the series" is usually stated as the "limit of the sequence of partial sums of the series." This limit, if it exists, is the sum of the series.
Infinite series: ##\sum_{n = 0}^\infty a_n##
Sequence of terms: ##\{a_0, a_1, a_2, \dots, a_n, \dots \}##
Sequence of partial sums: ##\{S_0, S_1, S_2, \dots, S_N, \dots \}##, where ##S_N = \sum_{n = 0}^N a_n##
If the limit as N grows without bound exists in the sequence of partial sums, this limit is the sum of the series.
Orodruin said:
The series 0.9, 0.09, 0.009, ... converges to zero.
What you wrote is a sequence. This sequence (not series) converges to zero, but the series (.9 + .09 + .009 + ...) converges to 1.
Orodruin said:
If you make a series where the nth entry is the sum of that series' first n elements you get the series 0.9, 0.99, 0.999, ... - which is a different series that converges to one.
Again, what you wrote here is a sequence.
 
  • #19
Mark44 said:
you are misusing some standard terminology here
Yes, sorry. I meant to write Cauchy sequence and sequence ##\{x_n\}## etc.

Edit: Come to think of this. I think this is a translation issue I have been misusing for quite some time. Always room to teach old dogs to sit.
 
  • #20
Seasons said:
There's also no ratio that gives you 0.000…1, only a summing sequence, just like 0.999…
0.000...1 is not a number. The ellipsis ('...') means that the 0's continue indefinitely, which implies that there can't be a 1 digit "out there" somewhere. If you specify that the 1 is at a specific position, then that is meaningful, but without other context, 0.000...1 is not.
Seasons said:
So, if you're going to use summing sequences to make decimals that have no ratios, then it begs the question of why 0.111… (ratio) doesn't equal the summing sequence of 0.1, 0.01, 0.001… (no ratio) The sum gets greater (0.111…) as the summing sequence gets smaller (0.1, 0.01, 0.001) - how are they not equal?
There are two sequences involved here.
1) {.1, .01, .001, ...} -- This is the sequence of terms.
2) (.1, .11, .111, ...} -- This is the sequence of partial sums of the series ##\sum_{k = 1}^\infty \frac 1 {10^k}##
The second sequence is produced by adding the terms in the first sequence. I.e., .1 = .1, .11 = .1 + .01, .111 = .1 + .01 + .001, and so on.
 
  • #21
Orodruin said:
The limit of a (Cauchy) series ##\{x_n\}## is defined as the number ##x## such that ##|x-x_n|## is arbitrarily small as long as ##n## is large enough. An infinite sum can be defined by letting
$$
y_n = \sum_{k=1}^n x_n
$$
and taking the limit of the series ##\{y_n\}##. If ##\{x_n\}## is a series, there is no reason to believe that it converges to the same number as the series ##\{y_n\}## as defined above, since the series are different.

I'm super-sorry that this means nothing to me, I literally don't understand what all that even means to be able to deconstruct it. I posted this as basic math, and you're replying with advanced logical symbols that I have no referent for as your argument. If you can please try to speak in plain english, it would help, or if someone else can. One thing I think you are saying is that 0.111… is a series that is a ratio, and is different than a series 0.1, 0.01, 0.001… that has no ratio. But you do agree that they sum together to be equal. Something that has no ratio can equal something that has a ratio. 0.999… has no ratio, yet equals 1, which has a ratio. You do this by summing the sequence. So why can't I sum a non-ratio sequence to equal a ratio?

Let me address the earlier point raised.

0.010101… (a ratio) is greater than 0.000…1 (derived by sequence - not ratio), but less than 0.111…; because 1 is greater than zero, therefor, as someone else stated, 0.111… cannot equal 0, or cannot be less than 0.0(1…), but only greater. 1/100, 1/10,000, 1/1,000,000… is less than 1/10, 1/100, 1/1000… ? However, let me posit this in terms of convergence - there are an infinite number of one's and zeroes in both sequences, equal to each other, one set is not larger or smaller than the other when it is infinity. What this means is that the sequences before the adding have exactly the same number of zeroes and 1's; after they are converted, 0.111… obviously has no zeroes at all. So anything with a zero would be less. So you have an instance where the sequences yield the same number of zeroes and ones before conversion; hypothetically equal… yet you say it is the sum that defines the convergence and reveals the flaw, while get this, complaining to me that I'm using sums as conversions and confusing the two. It seems to me, that you are summing to get the conversions, not me, in this instance, and so I am confused, why you're using sums to converge, then telling me that you're not, and then telling me that I'm using sums to converge and am thus confusing myself.

Does that make any sense?
 
  • #22
First, see Mark's comments about nomenclature.

Seasons said:
But you do agree that they sum together to be equal.

No. You have gotten this wrong. One of the sequences is the sequence of partial sums of the other sequence. The sum of one sequence is the limit of the other and the sum of the other is divergent.

Seasons said:
I literally don't understand what all that even means to be able to deconstruct it.
I understand that it is difficult to understand this on a B level. However, to understand it deeper you might have to accept that it is currently beyond you and trust in the result stated by people who have studied more maths. The remedy is of course to revisit this when you know more about how the real numbers and their representations are constructed.

Seasons said:
0.000…1
Again, as Mark said, this number does not exist.
 
  • #23
Also, if you have problems with that particular derivation, try one of the others. What about the statement that there are no numbers between 0.999... and 1? Whatever number you pick it is going to be either smaller or larger than both of them. Does that help?
 
  • #24
I see that 0.000…1 is not a number, however 0.1, 0.001, 0.0001… is a sequence. So, you can have sequences that aren't numbers as well? There are implications to that question, I'm curious how you answer it.
 
  • #25
Seasons said:
So, you can have sequences that aren't numbers as well?
A sequence is never a number. A number can be the limit of a sequence. 0.999... is the limit of the sequence 0.9, 0.99, 0.999, ...
 
  • #26
Seasons said:
I'm super-sorry that this means nothing to me, I literally don't understand what all that even means to be able to deconstruct it. I posted this as basic math, and you're replying with advanced logical symbols
The question you asked was about infinite series, which aren't part of "basic math." You also used terms such as "converges" which has a precise mathematical meaning. When you start talking about adding an infinite number of terms, you are no longer in basic math, and the explanations require more than plain vanilla elementary school arithmetic.
Seasons said:
that I have no referent for as your argument. If you can please try to speak in plain english, it would help, or if someone else can. One thing I think you are saying is that 0.111… is a series that is a ratio
0.111... is a ratio (1/9), and if you like, we can call it an infinite sum, which is the same as an infinite series.
The sum is 0.1 + 0.01 + 0.001 + ... ad infinitum
As you add more and more terms of this sum, you get a number that gets closer and closer to 1/9.
A sequence is nothing more than a list of numbers -- no addition or other operation is performed. The sequence of terms in this series is {.1, .01, .001, .0001, ...}. As you go farther out in this sequence, you get numbers that are closer and closer to zero, meaning that this sequence converges to zero. It is a necessary condition for an infinite series to converge, that its sequence of terms converges to zero. In other words, if the sequence did not converge, then the series also would not converge.
Seasons said:
, and is different than a series 0.1, 0.01, 0.001… that has no ratio.
This is NOT a series (sum). It's a sequence of numbers.
Seasons said:
But you do agree that they sum together to be equal. Something that has no ratio can equal something that has a ratio. 0.999… has no ratio, yet equals 1, which has a ratio. You do this by summing the sequence. So why can't I sum a non-ratio sequence to equal a ratio?
So what if there is no division expression that produces 0.999...?
This is a consequence that most numbers have two decimal representations, and has nothing to do with whether or not the decimal representation was produced by a division operation.

Here are a few examples of numbers with two decimal representations:
1 and 0.9999...
.5 and 0.49999...
.1 and 0.09999 ...
.25 and .24999...
etc.
Seasons said:
Let me address the earlier point raised.

0.010101… (a ratio) is greater than 0.000…1 (derived by sequence - not ratio), but less than 0.111…; because 1 is greater than zero, therefor, as someone else stated, 0.111… cannot equal 0, or cannot be less than 0.0(1…), but only greater. 1/100, 1/10,000, 1/1,000,000… is less than 1/10, 1/100, 1/1000… ? However, let me posit this in terms of convergence - there are an infinite number of one's and zeroes in both sequences, equal to each other, one set is not larger or smaller than the other when it is infinity. What this means is that the sequences before the adding have exactly the same number of zeroes and 1's; after they are converted, 0.111… obviously has no zeroes at all. So anything with a zero would be less. So you have an instance where the sequences yield the same number of zeroes and ones before conversion; hypothetically equal… yet you say it is the sum that defines the convergence and reveals the flaw, while get this, complaining to me that I'm using sums as conversions and confusing the two. It seems to me, that you are summing to get the conversions, not me, in this instance, and so I am confused, why you're using sums to converge, then telling me that you're not, and then telling me that I'm using sums to converge and am thus confusing myself.

Does that make any sense?
 
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  • #27
I'm even using words wrong! And was corrected that this isn't basic math. I just assumed division was basic math, since infinity is implied when they teach it in second grade, apparently not! Why is division so bizarre compared to addition, subtraction and multiplication, that people need all these bizarre rules that require PhD's to understand? I think I should study what You've all thus far given me, and probably stop asking more at this point. Thank you.
 
  • #28
Actually, I do have one more question(s)!

If you can't get a number from a fraction, isn't it irrational by definition?
Does that mean every irrational number equals a ratio? If so, wouldn't you be able to order all numbers?
 
  • #29
Seasons said:
Why is division so bizarre compared to addition, subtraction and multiplication, that people need all these bizarre rules that require PhD's to understand?
It is not division that is complicated, it is the formal definition of real numbers and their representation as a sequence of decimals that is. In order to grasp what it actually means you need to know the definition. Again, I suggest that you consider a different approach to the question, such as there not being any number between 0.999... and 1.

Seasons said:
I'm even using words wrong!
Yes, so was I. There is no reason to get upset about that. Those words have a precise mathematical meaning and if you don't use them properly people will generally not understand you. This is why Mark corrected me and why he corrected you. A more constructive response would be
Orodruin said:
Yes, sorry. I meant to write Cauchy sequence and sequence ##\{x_n\}## etc.

Edit: Come to think of this. I think this is a translation issue I have been misusing for quite some time. Always room to teach old dogs to sit.

Seasons said:
If you can't get a number from a fraction, isn't it irrational by definition?
Yes. But as already stated, 0.999... can be written as a ratio because it is equal to 1. An example of such a ratio is 9/9. You are confusing "getting the decimal expression for a number using a certain algorithm" with "being equal to a ratio". Since 0.999... = 1, it holds that 9/9 = 1 = 0.999... and therefore 0.999... can be written as a ratio.
Seasons said:
Does that mean every irrational number equals a ratio?
No. By definition irrational numbers cannot be written as a ratio.
Seasons said:
If so, wouldn't you be able to order all numbers?
You can order all numbers. For any two numbers ##x## and ##y## you either have ##x < y##, ##y < x## or ##x = y##.
 
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  • #30
Seasons said:
I just assumed division was basic math
It is. Division is one of the four arithmetic operations, so is part of arithmetic, AKA basic math.
Seasons said:
, since infinity is implied when they teach it in second grade, apparently not!
I don't think I was taught division until about 4th grade, and no one mentioned infinity to me for a long time after that. The long division algorithm that you learned is just an iterative method that generates the decimal digits of a division operation. I don't see that infinity is implied in this process.
Seasons said:
Why is division so bizarre compared to addition, subtraction and multiplication, that people need all these bizarre rules that require PhD's to understand?
A PhD is not required (I don't have one). What bizarre rules? There really is only one rule in division -- you can't divide by zero.
 
  • #31
OK!

I've been thinking more about this, what you all said, which has been super helpful, thank you! I think I forgot a couple thumbs ups, I'll get to those.

So, I realized something interesting to me about what you all discussed and seemed to all agree upon.

If you expand a sequence from the inside out, rather than just adding to the end, the end number gets dropped.

And I started thinking: If the end number gets dropped, then there is a new end number, and that gets dropped as well, because everything, again, before it, is infinite. So for example: 0.1, 0.01, 0.001… is not an addition as mark pointed out, it is a subtraction. It was made clear by everyone that what this implies, is an infinite amount of zeroes… so that the 1 gets dropped. but actually, the .01, then gets dropped, and then the .001 gets dropped, so it ends up just equalling 0. Now let's do this sequence for 0.4, 0.14, 0.114… since there's always a new end number being added which needs to be dropped (at infinity), this sequence also equals 0! Anytime, you make a sequence y, xy, xxy, xxxy, … it always equals zero. Correct?
 
  • #32
I also want to add, to Marks' reply so that Mark can understand what I meant, that division seems baffling compared to addition, subtraction and multiplication. If you missed my last post, I hope you all read that as well.

The reason this bothers me so much, is that even though the numbers are rational, you can separate these rational numbers into three categories.

Terminating decimals without a repeat
Repeating decimals (though, because they regress, technically, they're not repeating)
and finally
Decimals that are not created through long division, but rather by summing decimals lower than them.

Now, all these categories seem extremely odd to me, considering how neat multiplication, addition and subtraction are.

I think people would reply, "Well, they repeat, because they don't go into the number equally"

Ok, but why don't they go into the number equally? why

I actually don't think anyone knows the answer to that. And, I speculate, because nobody knows the answer to that, their other theorems are simply guesses to that extent. We all know why 1 apple and 1 apple is 2 apples. Because there is an identity copy. Addition isn't weird like division. And two of you in this thread seemed totally perplexed why this totally perplexes me.
 
  • #33
This is a problem of how numbers are represented, not of numbers in themselves. If you use base 3, then 1/3 would be written as 0.1. Also note that 0.999... and 1 (and, as Mark noted 0.001251999... and 0.001252) are just different ways of writing the same number. You can also write 1 as 3/3 if you would choose to do so.

As we have pointed out, a decimal digits are a representation of a number, just as 3/3 is a representation of a number, it is not that number and all numbers can in general be represented in different ways.
 
  • #34
Orodruin said:
This is a problem of how numbers are represented, not of numbers in themselves. If you use base 3, then 1/3 would be written as 0.1. Also note that 0.999... and 1 (and, as Mark noted 0.001251999... and 0.001252) are just different ways of writing the same number. You can also write 1 as 3/3 if you would choose to do so.

As we have pointed out, a decimal digits are a representation of a number, just as 3/3 is a representation of a number, it is not that number and all numbers can in general be represented in different ways.
That's a great reply. I would point out with only division (and other types of divisory functions), when you operate on whole numbers can you get these bizarre read outs. Changing the base, just shifts which numbers do and don't give these bizarre read outs, it doesn't make them go away. You accept that it's just a different way numbers are represented, without really explaining why it's so relatively convoluted compared to addition, subtraction and multiplication. When I see something so inexplicably bizarre like this, I can understand why people are confused enough to challenge these notions, particularly when people can't seem to explain why, but then they have all types of rules for this phenomenon nobody understands. I do think to myself when I ponder this stuff, that maybe we aren't doing division correctly. I'm not sure what doing it correctly entails, but the methods we try to explain around, may actually be a form of psychosis.
 
  • #35
Seasons said:
I do think to myself when I ponder this stuff, that maybe we aren't doing division correctly.
I believe that when you say "doing division" you are again referring to the process of getting from the representations of two numbers ##x## and ##y## to the number ##z## such that ##x = yz##. This is again a problem of representations and of the methods used to find the representation of a new unknown number, not a problem of division itself, which is very well defined.

The thing is that when it comes to representing numbers there is no "correct" way. There are some ways that are more convenient than others in some circumstances, but then again may be worse in others. For example, representing rational numbers as fractions can be very convenient when you are multiplying and dividing. However, decimal representation can be preferable in other circumstances.
 
<h2>1. Is 0.999... equal to 0?</h2><p>No, 0.999... is not equal to 0. It is infinitely close to 1, but never actually reaches 1.</p><h2>2. How can 0.999... be equal to 1 if there are an infinite number of 9s?</h2><p>The concept of infinity can be difficult to grasp, but in this case, it means that the number 0.999... continues on forever without ever reaching an endpoint. Even though there are an infinite number of 9s, the number is still less than 1.</p><h2>3. Can you prove that 0.999... is equal to 1?</h2><p>Yes, there are multiple mathematical proofs that show 0.999... is equal to 1. One example is using the infinite geometric series formula, which states that the sum of an infinite geometric series is equal to a/(1-r), where a is the first term and r is the common ratio. In this case, a=0.9 and r=0.1, so the sum is 0.9/(1-0.1) = 0.9/0.9 = 1.</p><h2>4. Why is it important to understand that 0.999... is equal to 1?</h2><p>Understanding that 0.999... is equal to 1 is important because it helps us understand the concept of infinity and the limitations of our numerical system. It also has practical applications in fields such as calculus and number theory.</p><h2>5. Can there be other numbers that are infinitely close to a whole number like 0.999... is to 1?</h2><p>Yes, there can be other numbers that are infinitely close to a whole number. For example, 0.111... is infinitely close to 0.1, and 0.888... is infinitely close to 0.9. However, it is important to note that these numbers are not equal to the whole number, just as 0.999... is not equal to 1.</p>

1. Is 0.999... equal to 0?

No, 0.999... is not equal to 0. It is infinitely close to 1, but never actually reaches 1.

2. How can 0.999... be equal to 1 if there are an infinite number of 9s?

The concept of infinity can be difficult to grasp, but in this case, it means that the number 0.999... continues on forever without ever reaching an endpoint. Even though there are an infinite number of 9s, the number is still less than 1.

3. Can you prove that 0.999... is equal to 1?

Yes, there are multiple mathematical proofs that show 0.999... is equal to 1. One example is using the infinite geometric series formula, which states that the sum of an infinite geometric series is equal to a/(1-r), where a is the first term and r is the common ratio. In this case, a=0.9 and r=0.1, so the sum is 0.9/(1-0.1) = 0.9/0.9 = 1.

4. Why is it important to understand that 0.999... is equal to 1?

Understanding that 0.999... is equal to 1 is important because it helps us understand the concept of infinity and the limitations of our numerical system. It also has practical applications in fields such as calculus and number theory.

5. Can there be other numbers that are infinitely close to a whole number like 0.999... is to 1?

Yes, there can be other numbers that are infinitely close to a whole number. For example, 0.111... is infinitely close to 0.1, and 0.888... is infinitely close to 0.9. However, it is important to note that these numbers are not equal to the whole number, just as 0.999... is not equal to 1.

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