## Is 1 = 0.999... ? Really?

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## Is it true that 1 = 0.999...? And Exactly Why or Why Not?

Yes, 1 = 0.999...
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41%
No, 1 ≠ 0.999...
16
50%
Other
3
9%

### Re: Is 1 = 0.999... ? Really?

Silhouette wrote:
Magnus Anderson wrote:
Silhouette wrote:Every partial product is greater than zero, yes - because it's only partial! Thus the same logic carried over to any "non-partial-product" of some kind of "infinite product" is invalid. Again - poor logic skills on your part.

Every partial product of $$0 \times 0 \times 0 \times \cdots$$ is equal to $$0$$ and that's precisely why its result is equal to $$0$$.

The same does not apply to $$\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \cdots$$.

I can repeat myself too if you want.

That's not getting us anywhere though is it.

I know every partial product is greater than 0. This just panders to your appeal to finitude that you keep denying you're doing.
INFINITES. We're dealing with them, not "partial, finitely somewhere along the way of an infinite series" which you're pretending still counts as valid for evaluating an infinite series.

$$0 \times 0 \times 0 \times \cdots$$ is an infinite product. It's a product made out of an infinite number of terms. It has no end.
Agree?

Even though $$0 \times 0 \times 0 \times \cdots$$ is a product made out of an infinite number of terms, its result is equal to a finite number that is $$0$$.
Agree?

You are certainly not telling us that the result of an infinite product can never be calculated because due to the number of terms being endless there is no end to the process of calculation? I hope that's not what you're trying to tell us.

How do we know that the result of this infinite product is $$0$$? We know it's $$0$$ because we know that all of its partial products are equal to $$0$$. What this means is that we can calculate the result of an infinite product by looking at its partial products.

So yes, I know full well we're dealing with an infinite product (and not merely its partial products.) But I also know that the way to calculate the result of an infinite product (or at least, to calculate its bounds) is by analyzing its partial products.

The insight is pretty basic: regardless of how many terms there are in a product of the form $$\Pi^{n} 0$$, the result will always be $$0$$. Now matter how big the number of terms is (indeed, even if its bigger than every finite number), the result will be $$0$$.

Something similar applies to $$\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \cdots$$. Here, we can observe that every partial product has a different value. It doesn't matter how large the number of terms is (even if its larger than every finite number), the result is always different from every other partial product. This alone tells us that the result is not equal to $$0$$. But there is one more thing we can observe: no matter how large the number of terms is, the result is never equal to or below $$0$$. And this is what tells us that the result is greater than $$0$$.

You, on the other hand, are telling us that we can get $$0$$ by taking $$\frac{1}{10}$$ and raising it to a sufficiently large number (namely, that of infinity.)

The point is that any product of the form $$(\frac{1}{10})^n$$ where $$n$$ is any number greater than $$0$$ is greater than $$0$$. The only condition for $$n$$ is to be greater than $$0$$. It can be a finite number but it can also be a number larger than every finite number (i.e. an infinite number.) Either way, the result is greater than $$0$$.
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### Re: Is 1 = 0.999... ? Really?

Silhouette wrote:If "One day Joe ate them all", and "There is an infinite line of apples", there would be always be more left to eat if the line was infinite. A contradiction.

If he ate them all, this means he left no apple behind.

You can say that he can't eat them all because, by description, the line resists being completely destroyed. So he can eat some (a lot of them, perhaps even an infinite number of them) but he can't eat all of them. But the two statements, "There is an infinite line of apples" and "One day Joe at them all", claim no such thing. The word "infinite" certainly does not mean "incapable of being completely changed or destroyed".

There is no final bound to endlessness that allows the completion of "all".

Let's take a simpler example.

"Once upon a time, there was an infinite number of apples at some point in the universe. One day, every single one of them disappeared."

Are you telling us that the process of an infinite number of apples disappearing cannot be completed within a finite period of time (not to mention instantly)?

"Once upon a time, there was an infinite number of green apples at some point in the universe. One day, every single one of them changed to red."
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### Re: Is 1 = 0.999... ? Really?

Magnus, you need to understand when someone is honestly trying to help you.

As a process 1/10 always leaves a 1 at the end, as an infinity it does not!

I’ve told you twice in this thread already that you treat infinities as objects OR processes as it suits your needs in the current post.

The thing is:

You can’t do this shit, you need to pick a real side and debate it. Thus far, your posts have been non debates.
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### Re: Is 1 = 0.999... ? Really?

phoneutria wrote:
Silhouette wrote:Eating half an apple, then half of that and half of that, which I believe is what you're saying, leaves partial products of slightly more than nothing at all and the limit of the process approaches zero apple.

approaches half apple

For the life of me I can't make sense of this from your wording as approaching half an apple... must be my fault, right? Demonstration please?

$$\prod_{k=0}^\infty(\frac1{2})^k=0$$
Type "infinite product of (1/2)^k" into Wolfram Alpha.
It's the same for any fraction less than 1 (starting with 1 whole apple and proceding to eat any possible fraction of it) and for any k value as a starting point... - you always get zero.

Eat half of a whole apple, you have half left
Eat half of the remaining half, you have a quarter left
Eat half of the remaining quarter, you have an eighth left
Etc. down to the limit of nothingness...

phoneutria wrote:it's not an original thought
all of scientific research operates with this in mind
utility is a compromise

The scientific method is indeed a process of relative improvement, rather than a sure way to arrive at absolute truth - but like dialectics, even with the realisation of compromise, it's intended to tend towards truth or at least a more truthful representation of reality.
Utility being at odds with truth realises that all knowledge, even that refined by the scientific method is in fact a tendency away from the truth of continuity.

Ecmandu wrote:Yeah! I get to play devils advocate to Magnus for a moment!

Magnus writes a number like:

2,4,6, 1... (1 repeating)

Now!

1,1,1,1.... (repeating) is possibly 8 smaller in VALUE, but not correspondence!

For example:

3.000...

Is smaller in VALUE than PI.

Correspondence however is the same.

I don't think Magnus meant that by his ellipsis, but I agree with you that 1 repeating would amount to infinity all the same.
If the 2 corresponds with the first 1, the 4 with the second, the 6 with the third, and with all 1s matching after that, the total difference would be 1+3+5=9, not 8. Is that what you meant by the difference in value? Either way any apparent "value difference" from looking at only the start would get swallowed up by the undefined essence of finitude-being-opposed (infinity). The superficial appearance of different finite values at the start does nothing.

I assume by representing $$3$$ as $$3.\dot0$$ and corresponding each decimal place with $$\pi$$, you get bijection and therefore the implication that the quantities are the same size - or at least the cardinality of the sets is the same.
If it can be accepted that adding in the appropriate number of 0s to any set, and that you can therefore theoretically induce bijection for everything and anything, this would throw into question the whole notion of set cardinality being evaluated by bijection.
It's a big "if" though, but I'm struggling to see any flaws in it other than what I said about quantities that require the quality of infinity to denote them correctly, and those that don't require it (e.g. recurring 0s either before or after any stated finites).

Magnus Anderson wrote:$$0 \times 0 \times 0 \times \cdots$$ is an infinite product. It's a product made out of an infinite number of terms. It has no end.
Agree?

Even though $$0 \times 0 \times 0 \times \cdots$$ is a product made out of an infinite number of terms, its result is equal to a finite number that is $$0$$.
Agree?

You are certainly not telling us that the result of an infinite product can never be calculated because due to the number of terms being endless there is no end to the process of calculation? I hope that's not what you're trying to tell us.

How do we know that the result of this infinite product is $$0$$? We know it's $$0$$ because we know that all of its partial products are equal to $$0$$. What this means is that we can calculate the result of an infinite product by looking at its partial products.

So yes, I know full well we're dealing with an infinite product (and not merely its partial products.) But I also know that the way to calculate the result of an infinite product (or at least, to calculate its bounds) is by analyzing its partial products.

The insight is pretty basic: regardless of how many terms there are in a product of the form $$\Pi^{n} 0$$, the result will always be $$0$$. Now matter how big the number of terms is (indeed, even if its bigger than every finite number), the result will be $$0$$.

Something similar applies to $$\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \cdots$$. Here, we can observe that every partial product has a different value. It doesn't matter how large the number of terms is (even if its larger than every finite number), the result is always different from every other partial product. This alone tells us that the result is not equal to $$0$$. But there is one more thing we can observe: no matter how large the number of terms is, the result is never equal to or below $$0$$. And this is what tells us that the result is greater than $$0$$.

You, on the other hand, are telling us that we can get $$0$$ by taking $$\frac{1}{10}$$ and raising it to a sufficiently large number (namely, that of infinity.)

The point is that any product of the form $$(\frac{1}{10})^n$$ where $$n$$ is any number greater than $$0$$ is greater than $$0$$. The only condition for $$n$$ is to be greater than $$0$$. It can be a finite number but it can also be a number larger than every finite number (i.e. an infinite number.) Either way, the result is greater than $$0$$.

We know the infinite product of zeroes is zero insofar as we accept that any quantity, whether finite or infinite "zero times" is zero.
The fact that the partial product also comes to zero is a symptom of this, it's not the reason for any quantity zero times being zero.
Anything zero times being zero is just logic - we don't need the partial product to know this - that's just a red herring.

Insofar as we analyse what we get when we get away from partial products to their limit, we are "analysing its partial products" to "calculate the result of an infinite product" - but this "insofar" isn't very far if it's exactly what we're trying to get away from...

Btw your notation $$\Pi^{n} 0$$ made me laugh Try something like $$\prod_{n=1}^\infty{0_n}$$ or probably just $$\prod_{n}^\infty{0n}$$

We know from $$\prod_{n=1}^\infty{\frac1{10_n}}$$ that indeed any partial product has a different value - so that's no help, until we extrapolate the limit that it tends towards. The partial product also tells us that for any partial product there's a smaller product - forever. You never get small enough. Anything greater than zero is too large, and the only value that you can't divide smaller is its limit: zero. So whilst the partial product looks like it'll never ever get to zero "no matter how large the number of terms is", zero is still the only value that could make sense - as well as just so happening to be the limit. This is even though $$\prod_{n=1}^\infty{0_n}=0$$ as a separate infinite sum that just so happens to reach the same result - but doesn't mean the infinite series are the same - only that they "look like" they're different and "look like" they ought to give different answers. That's what your point is - that the partial product makes it "look like" it'll always be greater than zero. But you mistake the same to apply for infinites, which cannot yield an answer small enough unless it reaches its limit of 0. Doesn't matter how much larger you want this "infinite number" to be than any "finite number", it will never be small enough - therefore all numbers greater than zero are invalid. Only zero can be valid - I can only apologise on behalf of appearances for fooling you, sorry.

You're just not appreciating what undefinability does to what's so clearly definable for finites only. It fucks things up at a fundamental level, and you need a certain adaptability in your assumptions to intellectually deal with the consequences of infinities.
That's the only barrier to this discussion breaking out of this otherwise infinite loop.

Magnus Anderson wrote:
Silhouette wrote:If "One day Joe ate them all", and "There is an infinite line of apples", there would be always be more left to eat if the line was infinite. A contradiction.

If he ate them all, this means he left no apple behind.

This is convenient, huh? Simply define infinity as being able to have a completed "all" like finites can. Except infinity is the opposite of that, and can't.

Let's give Joe an infinitely large mouth - he both infinitely overshoots with his bite because of the infinity of his mouth, and infinitely undershoots due to there always being infinitely more apples.
Undefined.
It's incapable of definably being completely changed or destroyed - any notion of a completed "all" is invalid.

Magnus Anderson wrote:Let's take a simpler example.

"Once upon a time, there was an infinite number of apples at some point in the universe. One day, every single one of them disappeared."

Are you telling us that the process of an infinite number of apples disappearing cannot be completed within a finite period of time (not to mention instantly)?

"Once upon a time, there was an infinite number of green apples at some point in the universe. One day, every single one of them changed to red."

I love magic stories!
And then what happened?! Feeling... sleepy....  Silhouette
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### Re: Is 1 = 0.999... ? Really?

I meant to type 9 instead of 8 Silhouette, glad you caught it.

So here’s the deal.

Can you see the difference between value and correspondence?

Does PI not register infinitely more value than 3.000...?

I know they correspond, but what do you make of value?
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### Re: Is 1 = 0.999... ? Really?

Silhouette wrote:
phoneutria wrote:
Silhouette wrote:Eating half an apple, then half of that and half of that, which I believe is what you're saying, leaves partial products of slightly more than nothing at all and the limit of the process approaches zero apple.

approaches half apple

For the life of me I can't make sense of this from your wording as approaching half an apple... must be my fault, right? Demonstration please?

$$\prod_{k=0}^\infty(\frac1{2})^k=0$$
Type "infinite product of (1/2)^k" into Wolfram Alpha.
It's the same for any fraction less than 1 (starting with 1 whole apple and proceding to eat any possible fraction of it) and for any k value as a starting point... - you always get zero.

Eat half of a whole apple, you have half left
Eat half of the remaining half, you have a quarter left
Eat half of the remaining quarter, you have an eighth left
Etc. down to the limit of nothingness...

I stand corrected.
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### Re: Is 1 = 0.999... ? Really?

Silhouette wrote:We know the infinite product of zeroes is zero insofar as we accept that any quantity, whether finite or infinite "zero times" is zero.
The fact that the partial product also comes to zero is a symptom of this, it's not the reason for any quantity zero times being zero.
Anything zero times being zero is just logic - we don't need the partial product to know this - that's just a red herring.

Are you saying that my logic is invalid?

Are you saying that it's not true that we can know that an infinite product of $$0$$ is $$0$$ if we know that $$0$$ raised to any number (whether finite or infinite) is equal to $$0$$?

How about an infinite product such as $$1 \times 1 \times 1 \times \cdots$$? How do you know the result of this product is $$1$$? Is it because we know that $$1$$ times any quantity (whether finite or infinite) is $$1$$? Or is it because we know that $$1$$ raised to any quantity (whether finite or infinite) is equal to $$1$$?

That's probably because you're deeply insecure and have a strong need to see flaws in people around you in order to feel good about yourself. And you're looking for any kind of flaws, so as long they are flaws -- big or small, significant or insignificant, etc.

Normal people don't do that.

We know from $$\prod_{n=1}^\infty{\frac1{10_n}}$$ that indeed any partial product has a different value - so that's no help

It's of no help if what you're doing is looking for a number that does not exist e.g. a finite number that is equal to the result of the infinite product $$\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots$$. But if you're merely trying to figure out whether such a number exists, then it's quite a bit of help. It tells you that such a number does not exist.

until we extrapolate the limit that it tends towards

The limit of an infinite product is not the same thing as its result. They are two different concepts.

The partial product also tells us that for any partial product there's a smaller product - forever. You never get small enough. Anything greater than zero is too large

It tells us that the result of the infinite product is smaller than every real number of the form $$\frac{1}{10^n}$$ where $$n \in N$$. Most importantly, it tells us that no matter how large $$n$$ is, the result is always greater than $$0$$.

Your argument is basically that there are no numbers greater than $$0$$ but smaller than every number of the form $$\frac{1}{10^n}$$ where $$n \in N$$.

That's one of our points of disagreement.

So whilst the partial product looks like it'll never ever get to zero "no matter how large the number of terms is", zero is still the only value that could make sense - as well as just so happening to be the limit.

It does not merely "look like". It will never ever get to zero for the simple reason that there is no number $$n$$ greater than $$0$$ that you can raise $$\frac{1}{10}$$ to and get $$0$$.

You can say that $$0.\dot01$$ is approximately equal to $$0$$, and that is true and noone disputes that, but that misses the point of this thread. We're asking whether the two numbers are exactly equal not merely approximately equal.

You can say that $$\frac{1}{0}$$ can be substituted with $$\infty$$ for practical reasons (given that $$0 \approx \frac{1}{\infty}$$) but you cannot say that $$\frac{1}{0} = \infty$$ given that there is no number that you can multiply by $$0$$ and get anything other than $$0$$.

So from your point of view, the only conclusion that should make sense is that $$0.\dot01$$ is a contradiction in terms, and thus, not equal to any quantity. By accepting such a conclusion, you'd have to agree that $$0.\dot9 \neq 1$$. So at least one point of our disagreement (really, the main point of disagreement) would be resolved.

Still, one point of our disagreement would remain, and that would be your insistence that $$0.\dot01$$ is a contradiction in terms based on the premise that there is no quantity that is greater than $$0$$ but less than every number of the form $$\frac{1}{10^n}, n \in N$$.
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### Re: Is 1 = 0.999... ? Really?

Silhouette wrote:This is convenient, huh? Simply define infinity as being able to have a completed "all" like finites can. Except infinity is the opposite of that, and can't.

You have yet to explain where's the contradiction.

Statement 1: "At some point in time at some point in space, there exists an infinite line of apples."

Statement 2: "At some other point in time, no apples exist anywhere in space."

How do the two statements contradict each other?

Certainly, the word "infinite" does not mean "not being able to be something else at some other point in time".
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### Re: Is 1 = 0.999... ? Really?

Silhouette wrote:
MagsJ wrote:I see.. I’d say, that infinite objects can only exist in an infinite space (universe), but concepts are not bound by these physical limitations.. obviously. So adding more apples to a set, could not then make that set become infinite.. it would just make that set larger.

The term infinite is therefore self-defining, so either something is infinite or it is not.
Yes, if you want to make the thought experimental physical, you'd need an infinite space (universe), and yes, concepts are not bound by these physical limitations.

Adding more apples to a finite set would make that set larger, no issue there at all.
The issue is that adding apples to an infinite set of apples doesn't make "infinity bigger" because expanding the bounds of something to make it bigger only applies to the bounded and not the boundless.

..the exact point at which infinity becomes self-defining, so yes.. anything infinite is not bounded within a defined measurable set.

The term infinite defies definition by the definition of "definition" and of "finite". Either something is infinite or it is not - of course. Infinite is only "definable" insofar as we can easily define finite... and then saying "not that". This says what you don't have and not what you do have. The analogy I used is that this "defines" what's in a hole by defining the boundaries of the hole and what's outside of it (i.e. it doesn't define what's inside the hole at all).

A good definition.. it’s not what you’ve got, it’s what you ain’t got. I like it. Magnus Anderson wrote:Certainly, the word "infinite" does not mean "not being able to be something else at some other point in time".

..but then wouldn’t that simply mean that something is either infinite or not? which I ‘think’ Silhouette (I don’t want to put words in his mouth) is also saying.
The possibility of anything we can imagine existing is endless and infinite.. - MagsJ

I haven't got the time to spend the time reading something that is telling me nothing, as I will never be able to get back that time, and I may need it for something at some point in time.. Wait, What! - MagsJ

You’re suggestions and I, just simply don’t mix.. like oil on water, or a really bad DJ - MagsJ MagsJ
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### Re: Is 1 = 0.999... ? Really?

Max wrote:Certainly, the word "infinite" does not mean "not being able to be something else at some other point in time".

Mags wrote:..but then wouldn’t that simply mean that something is either infinite or not? which I ‘think’ Silhouette.. I don’t want to put words into his mouth, is also saying.

I am not exactly sure what you mean, so I'll have to make a guess.

I suppose what you mean is that something is either finite or infinite i.e. that it cannot be both at the same time.

I agree with that. The number of elements within a set is either finite or it is infinite. There is no third option here.

Unfortunately, I cannot understand how that relates to what I said in the above quote.

The following two statements certainly do not make a claim that a line of apples existing at some point in time at some point in space is both finite and infinite.

Statement 1: "At some point in time at some point in space, there exists an infinite line of apples."

Statement 2: "At some other point in time, no apples exist anywhere in space."

They merely state that at one point in time the line is made out of an infinite number of apples and that at some other point in time the line is made out of zero apples.

As for Silhouette's claim that:

The issue is that adding apples to an infinite set of apples doesn't make "infinity bigger" because expanding the bounds of something to make it bigger only applies to the bounded and not the boundless.

It's not true that you cannot make an infinite quantity bigger. As for the rest, it's difficult to respond to because it's difficult to understand.
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### Re: Is 1 = 0.999... ? Really?

See, the issue I have with you Magnus, is not that 0.666... is larger in value than 0.555...

The issue I take with you, is that there’s ANY infinity that is NOT in correspondence.
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### Re: Is 1 = 0.999... ? Really?

Ecmandu wrote:See, the issue I have with you Magnus, is not that 0.666... is larger in value than 0.555...

The issue I take with you, is that there’s ANY infinity that is NOT in correspondence.
Neither of those numbers is infinite.

0.666... is exactly equal to 2/3

0.555... is exactly equal to 5/9

They're just regular numbers.

You can easily confirm it by doing a long division. (A level of math which is taught in middle school.)
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### Re: Is 1 = 0.999... ? Really?

If every partial product is greater than $$0$$, it does not follow that the complete product is greater than $$0$$.

This is what Silhouette pointed out and he's correct.

Here's an example:

$$\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times 0$$

Partial products:
$$\frac{1}{10} \times \frac{1}{10} = 0.01$$
$$\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} = 0.001$$

All greater than $$0$$.

Complete product:
$$\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times 0 = 0$$

Equal to $$0$$.

So I was wrong when I said that we know that the result of the infinite product $$0 \times 0 \times 0 \times \cdots$$ is equal to $$0$$ because we know that all of its partial products are equal to $$0$$. The real reason why we know the result is $$0$$ is because $$0$$ raised to any quantity (and not merely any quantity less than the number of terms in the sum) is equal to $$0$$; indeed, it's because zero times any product (finite or infinite) is equal to $$0$$.

Interestingly, despite my mistake, the type of reasoning that Silhouette says we use to conclude that the infinite product of $$0 \times 0 \times 0 \times \cdots$$ is equal to $$0$$ can be used to show that $$0.\dot01$$ is greater than $$0$$. Basically, there is no $$n > 0$$ such that $$\frac{1}{10^n}\ = 0$$. Note that $$n$$ can be any number greater than $$0$$ including numbers greater than every integer (i.e. infinite quantities.) And yet, Silhouette keeps insisting that $$0.\dot01$$ is equal to $$0$$.
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### Re: Is 1 = 0.999... ? Really?

phyllo wrote:
Ecmandu wrote:See, the issue I have with you Magnus, is not that 0.666... is larger in value than 0.555...

The issue I take with you, is that there’s ANY infinity that is NOT in correspondence.
Neither of those numbers is infinite.

0.666... is exactly equal to 2/3

0.555... is exactly equal to 5/9

They're just regular numbers.

You can easily confirm it by doing a long division. (A level of math which is taught in middle school.)

I’m not trying to be a butt here, but this means also that:

5+5+5...

Is less than

6+6+6....

Or

2+4+6+ 1 (repeating)

Is greater than

1+1+1+1 (repeating)

That was my point.

The value is greater, but correspondence is not
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### Re: Is 1 = 0.999... ? Really?

Ecmandu wrote:
phyllo wrote:
Ecmandu wrote:See, the issue I have with you Magnus, is not that 0.666... is larger in value than 0.555...

The issue I take with you, is that there’s ANY infinity that is NOT in correspondence.
Neither of those numbers is infinite.

0.666... is exactly equal to 2/3

0.555... is exactly equal to 5/9

They're just regular numbers.

You can easily confirm it by doing a long division. (A level of math which is taught in middle school.)

I’m not trying to be a butt here, but this means also that:

5+5+5...

Is less than

6+6+6....

Or

2+4+6+ 1 (repeating)

Is greater than

1+1+1+1 (repeating)

That was my point.

The value is greater, but correspondence is not

I forgot my concluding sentence (apologies)

In order to prove orders of infinity, you have to prove greater correspondence (not merely value)
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### Re: Is 1 = 0.999... ? Really?

Well no.

5+5+5+... can be expanded as (1+1+1+1+1)+(1+1+1+1+1)+(1+1+1+1+1)+...

which is 1+1+1+... when the brackets are removed.

The same goes for 6+6+6+...

And same goes for 2+4+6+ 1(repeating) which is (1+1)+(1+1+1+1)+(1+1+1+1+1+1)+1 (repeating)

They are all equal to 1+1+1+... because "..." does not represent some fixed number of terms.

One could also go in the 'opposite' direction and accumulate terms :

1+1+1+... could be written (1+1+1)+(1+1+1)+...

resulting in 3+3+3+...

If you think that you have infinity/3 terms in the final equation, then what is the number of terms if not infinite? ie Either infinity/3=infinity or intinity/3= equals some finite number. (What is that number?)
phyllo
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### Re: Is 1 = 0.999... ? Really?

phyllo wrote:Well no.

5+5+5+... can be expanded as (1+1+1+1+1)+(1+1+1+1+1)+(1+1+1+1+1)+...

which is 1+1+1+... when the brackets are removed.

The same goes for 6+6+6+...

And same goes for 2+4+6+ 1(repeating) which is (1+1)+(1+1+1+1)+(1+1+1+1+1+1)+1 (repeating)

They are all equal to 1+1+1+... because "..." does not represent some fixed number of terms.

One could also go in the 'opposite' direction and accumulate terms :

1+1+1+... could be written (1+1+1)+(1+1+1)+...

resulting in 3+3+3+...

If you think that you have infinity/3 terms in the final equation, then what is the number of terms if not infinite? ie Either infinity/3=infinity or intinity/3= equals some finite number. (What is that number?)

And 1+1+1 can also be written as 1/2+ 1/2 + 1/2...

Converging to zero 1/4 + 1/4 + 1/4 .... etc

Do you believe numbers converge?

Because with your logic all infinite decimals equal zero if you do
Ecmandu
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### Re: Is 1 = 0.999... ? Really?

gib wrote:You agreed that $$0.\dot9 \neq \infty$$. And it's not a finite number. What's left?

It's a number that is greater than every number of the form $$\frac{9}{10^n}, n \in \{1, 2, 3, \dotso\}$$ but less than $$1$$.

Note that my claim is that this is precisely the meaning of the symbol $$0.\dot9$$. It represents the infinite sum $$\sum_{n=1}^{\infty} \frac{9}{10^n}$$ which in turn represents a number that is greater than every number of the form $$\frac{9}{10^n}, n \in \{1, 2, 3, \dotso\}$$ but less than $$1$$.

Since it's less than $$1$$, it's clear that it's not a number larger than every integer. But it's also not a number that can be represented as a finite sum of rational numbers.

Sure, but only in the sense that rearranging the bricks in a building doesn't mean the building hasn't changed. But that's irrelevant to comparing it to another building and asking: do they still contain the same number of bricks?

But my point is precisely that rearranging the remaining points in the line does not show that the number of points did not change.

We started with a set $$A = \{P_1, P_2, P_3, P_4, \dotso\}$$. We removed every odd point to get a new set $$A' = \{P_2, P_4, \dotso\}$$. Since we preserved the identity of points, we can see that $$P_1$$ and $$P_3$$ are missing from $$A'$$. Note that we didn't merely change the position of points. If that were the case, $$A'$$ would include $$P_1$$ and $$P_2$$ (as well as all other odd points.) Also note that we're actively disregarding where on the line these points are. $$P_1$$ does not necessarily represent the first point on the line. $$P_1$$ can be ANY point on the line. The only condition is that it's the only point on the line that is represented by this symbol.

Let's go back to your example.

We have two infinitely long lines $$A$$ and $$B$$ each represented by a set of points that constitute them.

$$A = \{a1, a2, a3, \dotso\}$$
$$B = \{b1, b2, b3, \dotso\}$$

$$a1$$ represents the first point on the line $$A$$, $$a2$$ the second point on the line $$A$$, and so on.

Similarly, $$b1$$ represents the first point on the line $$B$$, $$b2$$ the second point on the line $$B$$, and so on.

Then, we pick any point on the line $$A$$ and rename it to $$p1$$. Let's take $$a3$$ and rename it to $$p1$$.

$$A = \{p1, a1, a2, a4, \dotso\}$$

Then, we do the same for $$B$$. Let's pick $$b1$$ and rename it to $$p1$$.

$$B = \{p1, b2, b2, b3, \dotso\}$$

We then repeat this process for all other elements. But before we can do this, we have to specify how big these sets are in relation to each other. Are they equal in size or is one of them larger than the other? In your example, you said that the two parallel lines are equally long, so that means we have to declare that the two sets are equal in size. This gives us the following result:

$$A = \{p1, p2, p3, \dotso\}$$
$$B = \{p1, p2, p3, \dotso\}$$

This allows us to say that there is a one-to-one correspondence between the two sets based on the simple observation that every element in $$A$$ is also present in $$B$$ and vice versa. (We couldn't do this with what we had at the start.)

Now, we take set $$B$$ and take every second point out. What do we get? We get:

$$B' = \{p1, p3, p5, p7, \dotso\}$$

We can now compare $$B'$$ to $$A$$ and conclude that $$A$$ is greater than $$B'$$ because every element in $$B'$$ is present in $$A$$ but not every element in $$A$$ is present in $$B$$.
Magnus Anderson
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### Re: Is 1 = 0.999... ? Really?

And 1+1+1 can also be written as 1/2+ 1/2 + 1/2...

Converging to zero 1/4 + 1/4 + 1/4 .... etc
Those series do not converge to zero. They don't converge at all.
Do you believe numbers converge?

Because with your logic all infinite decimals equal zero if you do
Numbers don't converge ... the word is not applicable to numbers. Series either converge or don't converge or it's unknown if they converge or not.
phyllo
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### Re: Is 1 = 0.999... ? Really?

phyllo wrote:
And 1+1+1 can also be written as 1/2+ 1/2 + 1/2...

Converging to zero 1/4 + 1/4 + 1/4 .... etc
Those series do not converge to zero. They don't converge at all.
Do you believe numbers converge?

Because with your logic all infinite decimals equal zero if you do
Numbers don't converge ... the word is not applicable to numbers. Series either converge or don't converge or it's unknown if they converge or not.

Your being non responsive FINE! Series sometimes converge and not numbers.

That means that 1+1+1... = 0

That means that 10 + 4 + 6 + 9 (repeating) = 0

You refuse to understand the implications of your argument.
Ecmandu
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### Re: Is 1 = 0.999... ? Really?

Your being non responsive FINE! Series sometimes converge and not numbers.
Non-responsive? I did respond.
That means that 1+1+1... = 0

That means that 10 + 4 + 6 + 9 (repeating) = 0

You refuse to understand the implications of your argument.

The first term of both those series is greater than zero and there are no negative terms. The implication is that if the series converges, then it converges to a value which is greater than( or equal to) the first term.

That's the first problem with your post.

The second problem is that you have no idea what convergence means in mathematics.
phyllo
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### Re: Is 1 = 0.999... ? Really?

Returning to an exchange that took place almost 10 pages ago.

viewtopic.php?p=2755763#p2755763

Silhouette wrote:You literally created identical sets, i.e. bijective by definition, $$A = \{1, 2, 3, \dotso\}$$ and $$B = \{1, 2, 3, \dotso\}$$

Max wrote:But they aren't identical. You are merely not listening.

Silhouette wrote:Right. $$A = \{1, 2, 3, \dotso\}\neq{B} = \{1, 2, 3, \dotso\}$$

What am I not listening to when you literally write out the exact same set twice? You saying they're not the same even though they are?
So sorry for not believing you when you literally write out right in front of everyone the exact same set.

Basically, the assumption is that the two representations $$A = \{1, 2, 3, \dotso\}$$ and $$B = \{1, 2, 3, \dotso\}$$ are specifying two different sets of the same size, so you cannot say they differ in size. I'm going to argue that this is yet another example of being fooled by the appearances.

Let's take what some have argued previously:

The set of natural numbers and the set of even natural numbers are the same size.

This is backed up by the claim that there is a bijective function between the set of natural numbers and the set of even natural numbers.

And yes, there is such a function.

$$f(x) = 2x$$

$$1 \mapsto 2$$
$$2 \mapsto 4$$
$$3 \mapsto 6$$
$$\cdots$$

Every member of $$N$$ is uniquely associated with a single member of $$2N$$ and vice versa.

But it's not the only function that exists between the two sets. There are functions that are not bijective.

$$f(x) = 4x$$

$$\hspace{0.83cm} 2$$
$$1 \mapsto 4$$
$$\hspace{0.83cm} 6$$
$$2 \mapsto 8$$
$$\hspace{0.83cm} 10$$
$$3 \mapsto 12$$
$$\hspace{0.83cm} 14$$
$$\cdots$$

So which one is it? Is $$N = \{1, 2, 3, \dotso\}$$ the same size as $$2N = \{2, 4, 6, \dotso\}$$ or is it actually smaller?

The answer is that $$N = \{1, 2, 3, \dotso\}$$ does not specify the size of the set. The size of the set is something that is specified separately (usually merely assumed, without any kind of explicit specification.)
Magnus Anderson
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### Re: Is 1 = 0.999... ? Really?

phyllo wrote:
Your being non responsive FINE! Series sometimes converge and not numbers.
Non-responsive? I did respond.
That means that 1+1+1... = 0

That means that 10 + 4 + 6 + 9 (repeating) = 0

You refuse to understand the implications of your argument.

The first term of both those series is greater than zero and there are no negative terms. The implication is that if the series converges, then it converges to a value which is greater than( or equal to) the first term.

That's the first problem with your post.

The second problem is that you have no idea what convergence means in mathematics.

You assert that 3 is (1+1+1)

By that logic,

1+1+1 = 1/2+1/2+1/2+1/2+1/2+1/2!!!!!!!

And 1/2 = 1/4+1/4 etc...

And 1/4 = 1/8+1/8. Etc....

That means that ALL whole numbers equal zero!!!

Do you want to hear my logic now?
Ecmandu
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:It's a number that is greater than every number of the form $$\frac{9}{10^n}, n \in \{1, 2, 3, \dotso\}$$ but less than $$1$$.

Note that my claim is that this is precisely the meaning of the symbol $$0.\dot9$$. It represents the infinite sum $$\sum_{n=1}^{\infty} \frac{9}{10^n}$$ which in turn represents a number that is greater than every number of the form $$\frac{9}{10^n}, n \in \{1, 2, 3, \dotso\}$$ but less than $$1$$.

Since it's less than $$1$$, it's clear that it's not a number larger than every integer. But it's also not a number that can be represented as a finite sum of rational numbers.

Why do you think it needs to be represented as a finite sum of rational numbers in order to be finite? It still seems like you can't wrap your head around the difference between an infinite number of terms in a sum, and an infinite number.

In any case, you seem to think that numbers that can only be described as approaching a limit without ever attaining the limit are a third kind of number after finite and infinite. Have a name for these kinds of numbers?

Magnus Anderson wrote:
Sure, but only in the sense that rearranging the bricks in a building doesn't mean the building hasn't changed. But that's irrelevant to comparing it to another building and asking: do they still contain the same number of bricks?

But my point is precisely that rearranging the remaining points in the line does not show that the number of points did not change.

We started with a set $$A = \{P_1, P_2, P_3, P_4, \dotso\}$$. We removed every odd point to get a new set $$A' = \{P_2, P_4, \dotso\}$$. Since we preserved the identity of points, we can see that $$P_1$$ and $$P_3$$ are missing from $$A'$$. Note that we didn't merely change the position of points. If that were the case, $$A'$$ would include $$P_1$$ and $$P_2$$ (as well as all other odd points.) Also note that we're actively disregarding where on the line these points are. $$P_1$$ does not necessarily represent the first point on the line. $$P_1$$ can be ANY point on the line. The only condition is that it's the only point on the line that is represented by this symbol.

Let's go back to your example.

We have two infinitely long lines $$A$$ and $$B$$ each represented by a set of points that constitute them.

$$A = \{a1, a2, a3, \dotso\}$$
$$B = \{b1, b2, b3, \dotso\}$$

$$a1$$ represents the first point on the line $$A$$, $$a2$$ the second point on the line $$A$$, and so on.

Similarly, $$b1$$ represents the first point on the line $$B$$, $$b2$$ the second point on the line $$B$$, and so on.

Then, we pick any point on the line $$A$$ and rename it to $$p1$$. Let's take $$a3$$ and rename it to $$p1$$.

$$A = \{p1, a1, a2, a4, \dotso\}$$

Then, we do the same for $$B$$. Let's pick $$b1$$ and rename it to $$p1$$.

$$B = \{p1, b2, b2, b3, \dotso\}$$

We then repeat this process for all other elements. But before we can do this, we have to specify how big these sets are in relation to each other. Are they equal in size or is one of them larger than the other? In your example, you said that the two parallel lines are equally long, so that means we have to declare that the two sets are equal in size. This gives us the following result:

$$A = \{p1, p2, p3, \dotso\}$$
$$B = \{p1, p2, p3, \dotso\}$$

This allows us to say that there is a one-to-one correspondence between the two sets based on the simple observation that every element in $$A$$ is also present in $$B$$ and vice versa. (We couldn't do this with what we had at the start.)

Now, we take set $$B$$ and take every second point out. What do we get? We get:

$$B' = \{p1, p3, p5, p7, \dotso\}$$

We can now compare $$B'$$ to $$A$$ and conclude that $$A$$ is greater than $$B'$$ because every element in $$B'$$ is present in $$A$$ but not every element in $$A$$ is present in $$B$$.

Ooooh, I see. Because every second point in line A no longer maps to points in line B, that obviously means line A must be longer than line B. I get it now! Thanks Magnus!

But wait... I seem to remember this argument recurring. Yeah, I seem to remember it recurring a lot. Over and over and over and oh my god you're repeating yourself!

It doesn't matter how the points are paired. You're allowed to relabel them. Take p3 in B and relabel it p2. Take p5 and relabel it p3. Take p7 and relabel it p4. So B goes from {p1, p3, p5, p7 ...} to {p1, p2, p3, p4 ...}. And voila! A one-to-one correspondence with A again. It's not like you'll run out of points in B. It's infinite!

The reason why you can relabel is because you're using a method of counting whereby you pair natural numbers with elements in the set. The rule is: so long as every natural number starting from 1 (or 0 if you want to consider that the first natural number) and going up in order can be paired with every element of the set, the number of elements in the set is infinite. Note that the rule doesn't say anything about preserving the original pairing that you started out with. So long as the natural numbers can be so paired up (even if you have to relabel or rearrange), you can still apply the rule. If you start pairing elements from 1 to $$\infty$$, and then you remove every second element, leaving you with 1, 3, 5, ... you're allowed to redo the pairing. You can replace 3 with 2, 5 with 3, etc. and you will once again have a one-to-one pair of all the natural numbers to all the elements, showing that it's still infinite. In the example above, you're just using a slightly different labeling system: p1, p2, p3... but obvious it's the same idea. It's the same as using 1, 2, 3...

Here's some vsauce for you:

Starting from 3:45, Michael Stevens explains this concept perfectly.

I realize you're going to say: but of course the line is still infinite, that wasn't my point. My point is that it is now a lesser infinity than before. But I don't know how you're labeling system proves that. The fact that a1 would map to b2, a2 to b4, a3 to b6, etc. proves nothing to me unless you can convince me that what applies to finite sets also applies to infinite sets. Ball's in your court.
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### Re: Is 1 = 0.999... ? Really?

gib wrote:Why do you think it needs to be represented as a finite sum of rational numbers in order to be finite? It still seems like you can't wrap your head around the difference between an infinite number of terms in a sum, and an infinite number.

I understand very well that there are infinite sums of rational numbers that can be represented as a finite sum of rational numbers.

$$2 + 2 + 2 + 0 + 0 + 0 + \cdots$$ is an infinite sum of integers that can be represented as a finite sum e.g. $$2 + 2 + 2$$.

I merely said that $$0.\dot9$$ cannot be represented as a finite sum of rational numbers.

In any case, you seem to think that numbers that can only be described as approaching a limit without ever attaining the limit are a third kind of number after finite and infinite. Have a name for these kinds of numbers?

I don't.

It doesn't matter how the points are paired. You're allowed to relabel them.

That's where we disagree.

You aren't allowed to relabel them. By relabelling them you can literally prove anything you want.

Take $$B' = \{p_1, p_3, p_5, p_7, \dotso\}$$. You can relabel its elements by renaming $$p_3$$ to $$p_2$$, $$p_5$$ to $$p_3$$, $$p_7$$ to $$p_4$$ and so. That's what you're doing, right? By doing so, you can "prove" that $$B'$$ is equal in size to $$A$$. But by relabelling them in a different way, you can "prove" that $$B'$$ is actually bigger than $$A$$. Rename $$p_1$$ to $$p_0.5$$, $$p_3$$ to $$p_1$$, $$p_5$$ to $$p_1.5$$, $$p_7$$ to $$p_2$$, and so on. Voila! You have a set $$B' = \{p_{0.5}, p_1, p_{1.5}, p_2, \dotso\}$$ that is "clearly" bigger than $$A$$!

Take p3 in B and relabel it p2. Take p5 and relabel it p3. Take p7 and relabel it p4. So B goes from {p1, p3, p5, p7 ...} to {p1, p2, p3, p4 ...}. And voila! A one-to-one correspondence with A again. It's not like you'll run out of points in B. It's infinite!

But there's no one-to-one correspondence. You merely created an illusion that there is.

Let's take a look at an example involving a line consisting of a finite number of football players. Each player has a number written on the back of their shirt. Let's say there are only five players in the line numbered 1, 2, 3, 4 and 5. Suppose now that one day you wake up and decide to kill one of them e.g. number 3. By killing number 3, you get something like 1, 2, 4, 5. There's a clear gap in the line, so someone who knows that the line had no gap before this incident can become suspicious. So what you wanna do is you want to erase this gap by changing the numbers. So number 4 becomes 3 and number 5 becomes 4 leading to a line that looks something like this: 1, 2, 3, 4. Still, if there's someone who knows that there was a player number 5 standing in the line, he can become suspicious even though there is no gap. This is the disadvantage of finite sets: it's much more difficult to be deceptive with them.

Now let's take a look at an example involving a line consisting of an infinite number of football players. Each player has a number written on their back but instead of there being 5 football players, there is now an infinite number of them. The line looks something like this: {1, 2, 3, 4, 5, ...}. So one day you wake up and kill one of them e.g. number 3. This leads to {1, 2, 4, 5, ...}. Clearly, there is a gap in the line, so anyone aware of the fact that there was no gap can become suspicious. So what you have to do is conceal this fact by changing the numbers such that 4 becomes 3, 5 becomes 4, 6 becomes 5 and so on. This leads to {1, 2, 3, 4, 5, ...}. Now, noone can see that a guy is missing from the line. The player that is missing is number 3, but since there's a player with that number in the line, and since no other number is missing, it's quite difficult for anyone to become suspicious. This is the disadvantage of infinite sets: it's super easy to be deceptive with them.

The reason why you can relabel is because you're using a method of counting whereby you pair natural numbers with elements in the set.

And that's not a valid method of counting because the set of natural numbers does not have a size on its own.

Let's map $$N = \{1, 2, 3, \dotso\}$$ to $$B' = \{p_1, p_3, p_5, \dotso\}$$.

You can use bijection to do so:

$$f(x) = p_{2x - 1}$$

$$1 \mapsto p_1$$
$$2 \mapsto p_3$$
$$3 \mapsto p_5$$
$$\cdots$$

This makes the two sets equal in size.

But you can also use any other kind of function e.g. injection:

$$f(x) = p_{4x - 1}$$

$$\hspace{0.83cm} p_1$$
$$1 \mapsto p_3$$
$$\hspace{0.83cm} p_5$$
$$2 \mapsto p_7$$
$$\hspace{0.83cm} p_9$$
$$3 \mapsto p_{11}$$
$$\hspace{0.83cm} p_{13}$$
$$\cdots$$

This makes $$A$$ smaller than $$B'$$.

With this kind of "logic", you can literally prove anything you want.

The reason why you can relabel is because you're using a method of counting whereby you pair natural numbers with elements in the set. The rule is: so long as every natural number starting from 1 (or 0 if you want to consider that the first natural number) and going up in order can be paired with every element of the set, the number of elements in the set is infinite. Note that the rule doesn't say anything about preserving the original pairing that you started out with. So long as the natural numbers can be so paired up (even if you have to relabel or rearrange), you can still apply the rule. If you start pairing elements from 1 to $$\infty$$, and then you remove every second element, leaving you with 1, 3, 5, ... you're allowed to redo the pairing. You can replace 3 with 2, 5 with 3, etc. and you will once again have a one-to-one pair of all the natural numbers to all the elements, showing that it's still infinite. In the example above, you're just using a slightly different labeling system: p1, p2, p3... but obvious it's the same idea. It's the same as using 1, 2, 3...

I understand very well how the rule works. I'm simply saying it's not a valid rule.

If people say that a rule is valid, does that mean it's valid?

How do you know it's a valid rule? Based on what?

Starting from 3:45, Michael Stevens explains this concept perfectly.

I realize you're going to say: but of course the line is still infinite, that wasn't my point. My point is that it is now a lesser infinity than before. But I don't know how you're labeling system proves that. The fact that a1 would map to b2, a2 to b4, a3 to b6, etc. proves nothing to me unless you can convince me that what applies to finite sets also applies to infinite sets. Ball's in your court.

I'm certainly not a machine that convinces people (:
Magnus Anderson
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