## 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...
13
42%
No, 1 ≠ 0.999...
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48%
Other
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10%

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

Magnus Anderson wrote:
Ecmandu wrote:Every sequence is an algorithm

1,2,3,4,5 ... is the plus 1 algorithm

"Not really" to both.

The following is an algorithm:

1) Set $$a$$ to $$0$$
2) Increase $$a$$ by $$1$$
3) Go to step 2

That's a sequence of instructions (an algorithm), whereas $$1, 2, 3, \dotso$$ is a sequence of numbers (not an algorithm.)

And no, they are not always equal to 1 at convergence of limits. One stays the same and the other gets smaller and smaller and smaller.

I have no idea what you're talking about. Unless you make an effort to clarify your position, this is the end of the discussion between the two of us. You have no right to complain at this point.

The algorithm is implied. Without the algorithm, the sequence is not possible.

For the second part:

I honestly have no clue why you can’t see that one side stays the same as the other side grows infinitely smaller.

1/263784

Is smaller than 1/1!

So... imagine 1/263782 multiplied 263784 times...

Equals one? Yes?

Now imagine the “convergence” at infinity...!!!

It becomes zero times zero! It gets smaller and smaller and smaller! It is a sequence!
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### Re: Is 1 = 0.999... ? Really?

Ecmandu wrote:The algorithm is implied. Without the algorithm, the sequence is not possible.

That doesn't ring a bell.

For the second part:

I honestly have no clue why you can’t see that one side stays the same as the other side grows infinitely smaller.

1/263784

Is smaller than 1/1!

So... imagine 1/263782 multiplied 263784 times...

Equals one? Yes?

Now imagine the “convergence” at infinity...!!!

It becomes zero times zero! It gets smaller and smaller and smaller! It is a sequence!

That's NOT the other side.

Take this equation:

$$1 = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}$$

$$\frac{1}{8}$$ is NOT the right side of that equation, it's merely one small part of it. The right side is $$\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}$$ and it's equal to $$1$$. Magnus Anderson
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:
Ecmandu wrote:The algorithm is implied. Without the algorithm, the sequence is not possible.

That doesn't ring a bell.

For the second part:

I honestly have no clue why you can’t see that one side stays the same as the other side grows infinitely smaller.

1/263784

Is smaller than 1/1!

So... imagine 1/263782 multiplied 263784 times...

Equals one? Yes?

Now imagine the “convergence” at infinity...!!!

It becomes zero times zero! It gets smaller and smaller and smaller! It is a sequence!

That's NOT the other side.

Take this equation:

$$1 = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}$$

$$\frac{1}{8}$$ is NOT the right side of that equation, it's merely one small part of it. The right side is $$\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}$$ and it's equal to $$1$$.

Magnus, I don’t deny that 8*1/8 equals 1.

We’re talking about infinite sequences here... convergent series.

Look at the title of the thread to gain context!
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### Re: Is 1 = 0.999... ? Really?

I know what this thread is about. What I don't know is what YOU are talking about. Magnus Anderson
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:I know what this thread is about. What I don't know is what YOU are talking about.

And that’s why you lost credibility in this thread. Like I stated before. You have your fancy symbols... but plain English is something you fail to grasp.

I’m not speaking this in complex ways.

I wish I could dumb it down for you, but I’m not sure that I can.

I’ll try again....

1=1/1
1=0.5+0.5
1=0.25+0.25+0.25+0.25

Etc....

On the right side, the numbers get lower and lower and lower.

For example: the next step is:

1=1/8+1/8+1/8+1/8+1/8+1/8+1/8+1/8

....

Eventually, in this series, if limits converge, every number to the right will equal zero.

I honestly cannot make it simpler than that. I’m sorry, it’s impossible to me to make it simpler!
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### Re: Is 1 = 0.999... ? Really?

You complain when people ignore you, you complain when people ask you questions. All in all, Ecmandu complains. Magnus Anderson
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:You complain when people ignore you, you complain when people ask you questions. All in all, Ecmandu complains.

I complain because we live in a zero sum world. Where there’s a winner, there’s a loser.

I complain about my losses and I complain about my victories. I hate this world. If you had any sense, could put just 2 neurons together in that brain of yours, you’d hate it too.
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### Re: Is 1 = 0.999... ? Really?

Ecmandu wrote:Eventually, in this series, if limits converge, every number to the right will equal zero.

I honestly cannot make it simpler than that. I’m sorry, it’s impossible to me to make it simpler!

It makes perfect sense and I understand the argument, and you've explained it fine for anyone with a brain.

The limit of each number on the right is indeed 0, as all the denominators tend towards infinity. Add all those limits together to get 0, when the left hand side remains 1 throughout.

The problem is that you have an infinite number of fractions on the right hand side.
This undefined element is what you're not respecting, because you can't only pay attention to "any number of zeroes added together is zero". You also have to pay attention to "an infinite number of positive numbers added together is infinite" and also "any number of that same number's reciprocal added together is one" etc.

1="0" isn't the only answer you can get, but it's the only answer you're paying attention to.

I'm not being biased when I say "look at all the possible solutions, not just one". That's literally the opposite of being biased.
I'm sorry, but it's just not a valid proof that there's something wrong with math. Nice try, but no cigar. Silhouette
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### Re: Is 1 = 0.999... ? Really?

Silhouette wrote:
Ecmandu wrote:Eventually, in this series, if limits converge, every number to the right will equal zero.

I honestly cannot make it simpler than that. I’m sorry, it’s impossible to me to make it simpler!

It makes perfect sense and I understand the argument, and you've explained it fine for anyone with a brain.

The limit of each number on the right is indeed 0, as all the denominators tend towards infinity. Add all those limits together to get 0, when the left hand side remains 1 throughout.

The problem is that you have an infinite number of fractions on the right hand side.
This undefined element is what you're not respecting, because you can't only pay attention to "any number of zeroes added together is zero". You also have to pay attention to "an infinite number of positive numbers added together is infinite" and also "any number of that same number's reciprocal added together is one" etc.

1="0" isn't the only answer you can get, but it's the only answer you're paying attention to.

I'm not being biased when I say "look at all the possible solutions, not just one". That's literally the opposite of being biased.
I'm sorry, but it's just not a valid proof that there's something wrong with math. Nice try, but no cigar.

What’s interesting to me here silhouette, is that if this series doesn’t converge at zero (as you state) then 0.9... cannot equal 1. You’re arguing against yourself here!

I’m creating a very specific box with this proof:

If my proof is false, YOU’RE the one destroying math, not me!

Math goes on just fine if convergence is false! If convergence is true ... that’s the end of all math!

Do you understand that message?
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### Re: Is 1 = 0.999... ? Really?

Ecmandu wrote:What’s interesting to me here silhouette, is that if this series doesn’t converge at zero (as you state) then 0.9... cannot equal 1. You’re arguing against yourself here!

Not quite, if I'm understanding you right.

Are you trying to suggest that:
1) if the fractions do actually reach their limit of 0, then the right hand side properly equals 0 when the left hand side still equals 1, and the difference between $$0.\dot9$$ and $$1$$ can also genuinely reach its limit of 0, but math is therefore broken? And,
2) if the fractions on the right hand side of your equations don't reach their limit of 0, leaving something to work with to maintain equality with the left hand side of 1, then there is the same kind of "something" between $$0.\dot9$$ and $$1$$, and mathematical consistency is maintained?

You let me know if I understand the message.

What I'm trying to say is that those aren't the only two options.
The limit of the equations you're presenting leads to an undefined number of fractions, such that whether each fraction reaches their limit of 0 or not, the right hand side of the equation isn't neatly "0" (or anything else) until you do more work and fully narrow down a singular defined answer only.

By contrast, there's no extra work to do to get the difference between $$0.\dot9$$ and $$1$$. There is no other number that it can be than the singular defined answer of zero. You could arbitrarily phrase the difference as $$\lim_{n\to\infty}(n\times\frac1n)=0$$ just like in your argument to try and force the same complication, because the the difference may as well be $$\sum_{n=0}^\infty\frac0{10^n}=0$$, but we already know the singular defined answer never gets to a point where it's larger than 0 no matter how far you go down the decimal expansion.

It's simply not the same situation, so the complication you stop at with your argument, instead of working it through to get beyond the undefined element and finding a way to fully get to a singular defined answer, doesn't apply. It doesn't prove anything by itself, never mind anything about $$1=0.\dot9$$, because you stopped short and concluded a singular defined answer without doing more work to resolve the undefined element that could just as easily yield a different answer for you to stop short at instead, if you wanted.

Let me know if you, in turn, understand my message. Silhouette
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### Re: Is 1 = 0.999... ? Really?

Silhouette wrote:The limit of each number on the right is indeed 0

Not the limit of every number on the right side (that makes no sense) but the limit of the sequence of numbers that is formed by picking one number from the right side of every equation. That sequence is $$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dotso$$ and its limit is indeed $$0$$.

Add all those limits together to get 0, when the left hand side remains 1 throughout.

Why are we adding those limits together?

Let's say that his starting assumption is that an infinite sum of $$\lim_{n\to\infty} \frac{1}{n}$$ is equal to $$1$$ (even though that doesn't appear to be the starting point of his argument) then what he is saying is the following:

$$1 = \lim_{n\to\infty} \frac{1}{n} + \lim_{n\to\infty} \frac{1}{n} + \lim_{n\to\infty} \frac{1}{n} + \cdots$$

$$1 = 0 + 0 + 0 + \cdots$$

$$1 = 0$$

The conclusion logically follows from the premises.

But since we know the conclusion is wrong, it follows that some of our premises are wrong. Ecmandu is claiming that the premise that is false is the premise that infinite series converge. I disagree. What's actually wrong is the premise that $$1$$ is equal to an infinite sum of limits of the sequence $$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dotso$$. Magnus Anderson
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:Not the limit of every number on the right side (that makes no sense) but the limit of the sequence of numbers that is formed by picking one number from the right side of every equation.

You're doing that thing again.

Mathematician: <mathematical fact>
Non-mathematician: <no questions asked, just straight to...> "no, that makes no sense".

Yes it does make sense. The number that's the outcome of each term on the right hand side tends towards a limit of 0 as the denominator tends towards infinity. This is a fact, so just stop with the premature ejaculation and for once try and figure out why it's a fact first before blurting out that it doesn't yet make sense TO YOU and is therefore wrong.

Magnus Anderson wrote:That sequence is $$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dotso$$ and its limit is indeed $$0$$.

If you mean the sequence of each fraction on the right hand side, then yes - that's what I literally just said.
If you mean those are the terms on the right hand side of the equation, then no - that's not what he said.
Either way, shh, listen, learn, and stop pretending.

Magnus Anderson wrote:
Silhouette wrote:Add all those limits together to get 0, when the left hand side remains 1 throughout.

Why are we adding those limits together?

Because that's literally the pattern he's trying to communicate and the whole point of his argument maybe?

Magnus Anderson wrote:Let's say that his starting assumption is that an infinite sum of $$\lim_{n\to\infty} \frac{1}{n}$$ is equal to $$1$$ (even though that doesn't appear to be the starting point of his argument)

It is an incomplete part of the starting point of his argument.
Multiply that limit by "n", and it equals 1.
This is clearly the case for all finite values of "n" on the way to infinity, but not so clear for when "n" is infinite.

Magnus Anderson wrote:then what he is saying is the following:

$$1 = \lim_{n\to\infty} \frac{1}{n} + \lim_{n\to\infty} \frac{1}{n} + \lim_{n\to\infty} \frac{1}{n} + \cdots$$

$$1 = 0 + 0 + 0 + \cdots$$

$$1 = 0$$

The conclusion logically follows from the premises.

The logic is incomplete, as I literally just described.
As I literally just described there are various patterns to use to get to various different conclusions.

Magnus Anderson wrote:But since we know the conclusion is wrong, it follows that some of our premises are wrong. Ecmandu is claiming that the premise that is false is the premise that infinite series converge. I disagree. What's actually wrong is the premise that $$1$$ is equal to an infinite sum of limits of the sequence $$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dotso$$.

The premises are fine, the incomplete logic is wrong. As I literally just described.

He's not even saying that this sequence that you keep repeating is 1.
And infinite series absolutely can converge (either that or they diverge).

Why do you keep talking to me when you keep complaining that I keep calling you out on your bullshit? What did your post just achieve? It's not even clear you understand his simple argument from several things that you're saying and you're still asserting your conclusions as fact without a hint of self-questioning. Some things you're "telling me" when I literally said them just before. And the one question you actually did ask, I literally just answered it in my last post.
Like seriously, what the fuck was that post of yours? A waste of everyone's time, STILL no indication whatsoever that you've adjusted your approach in the slightest to reflect the reality of your expertise on the subject, or that you've learned a single thing, only an attempt to talk to me again to hear what I keep telling you you need to do before this whole thing can even begin to be remotely constructive. Silhouette
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### Re: Is 1 = 0.999... ? Really?

Silhouette wrote:If you mean the sequence of each fraction on the right hand side, then yes - that's what I literally just said.

That's not what you LITERALLY said even though that might be (and probably is) what you WANTED to say.

What you literally said is:

The limit of each number on the right is indeed 0

Do numbers have limits?
(The answer is, I believe, no.)

What's the limit of the number $$0.5$$?
(The question is, I believe, non-sensical.)

Multiply that limit by "n", and it equals 1.

What limit? Do you mean $$\lim_{n\to\infty} \frac{1}{n}$$? If so, that limit is $$0$$, so when you multiply it by $$n$$, you get $$0 \times n$$ which is $$0$$.

The premises are fine

So you think that $$1 = \lim_{n\to\infty} \frac{1}{n} + \lim_{n\to\infty} \frac{1}{n} + \lim_{n\to\infty} \frac{1}{n} + \cdots$$ is true?

How can that be the case if the limit of the sequence $$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dotso$$ is $$0$$? Magnus Anderson
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:
Silhouette wrote:The limit of each number on the right is indeed 0

Not the limit of every number on the right side (that makes no sense) but the limit of the sequence of numbers that is formed by picking one number from the right side of every equation. That sequence is $$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dotso$$ and its limit is indeed $$0$$.

Add all those limits together to get 0, when the left hand side remains 1 throughout.

Why are we adding those limits together?

Let's say that his starting assumption is that an infinite sum of $$\lim_{n\to\infty} \frac{1}{n}$$ is equal to $$1$$ (even though that doesn't appear to be the starting point of his argument) then what he is saying is the following:

$$1 = \lim_{n\to\infty} \frac{1}{n} + \lim_{n\to\infty} \frac{1}{n} + \lim_{n\to\infty} \frac{1}{n} + \cdots$$

$$1 = 0 + 0 + 0 + \cdots$$

$$1 = 0$$

The conclusion logically follows from the premises.

But since we know the conclusion is wrong, it follows that some of our premises are wrong. Ecmandu is claiming that the premise that is false is the premise that infinite series converge. I disagree. What's actually wrong is the premise that $$1$$ is equal to an infinite sum of limits of the sequence $$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dotso$$.

Magnus, I just want to correct you a moment on your interpretation of my argument.

I’m not ONLY stating the series:

1/2, 1/4, 1/8, 1/16. Etc....

I’m stating!

1/2*2, 1/4*4, 1/8*8, 1/16*16. Etc...

That will ALSO equal zero!
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### Re: Is 1 = 0.999... ? Really?

Ecmandu wrote:I’m stating!

1/2*2, 1/4*4, 1/8*8, 1/16*16. Etc...

That will ALSO equal zero!

How can a sequence be equal to zero (or any other number)?

Are you, perhaps, trying to say that the limit of the sequence $$2 \times \frac{1}{2}, 4 \times \frac{1}{4}, 8 \times \frac{1}{8}, \dotso$$ is equal to zero?

But how can that be the case? That sequence is LITERALLY a sequence of $$1$$'s i.e. $$1, 1, 1, \dotso$$

This goes on to show how much TROUBLE you have expressing your thoughts. Magnus Anderson
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:
Ecmandu wrote:I’m stating!

1/2*2, 1/4*4, 1/8*8, 1/16*16. Etc...

That will ALSO equal zero!

How can a sequence be equal to zero (or any other number)?

Are you, perhaps, trying to say that the limit of the sequence $$2 \times \frac{1}{2}, 4 \times \frac{1}{4}, 8 \times \frac{1}{8}, \dotso$$ is equal to zero?

But how can that be the case? That sequence is LITERALLY a sequence of $$1$$'s i.e. $$1, 1, 1, \dotso$$

This goes on to show how much TROUBLE you have expressing your thoughts.

I forget to use the catchphrase “limit” from time to time. Sue me.

Yes it always equals 1 until the limit!

Infinity over zero!

It’s funny to me that we’re all still taking jabs at each other. =)

Not in a bad way, I just find it funny.
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### Re: Is 1 = 0.999... ? Really?

Ecmandu wrote:
Magnus Anderson wrote:
Ecmandu wrote:I’m stating!

1/2*2, 1/4*4, 1/8*8, 1/16*16. Etc...

That will ALSO equal zero!

How can a sequence be equal to zero (or any other number)?

Are you, perhaps, trying to say that the limit of the sequence $$2 \times \frac{1}{2}, 4 \times \frac{1}{4}, 8 \times \frac{1}{8}, \dotso$$ is equal to zero?

But how can that be the case? That sequence is LITERALLY a sequence of $$1$$'s i.e. $$1, 1, 1, \dotso$$

This goes on to show how much TROUBLE you have expressing your thoughts.

I forget to use the catchphrase “limit” from time to time. Sue me.

Yes it always equals 1 until the limit!

Infinity over zero!

It’s funny to me that we’re all still taking jabs at each other. =)

Not in a bad way, I just find it funny.

I know what you’re thinking right now...

1/1, 2/2, 4/4, 8/8, 16/16. Etc...

Cannot possibly equal infinity over zero!

You’re half correct!

In this formation you always have 2* 1/2, 4*1/4... that’s an equality!

2,4,6,8... reaches infinity at convergence!

1/2, 1,4, 1/8, 1/16... equals zero at convergence.

Thus: following transitive laws of math, the series is:

Infinity over zero!
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### Re: Is 1 = 0.999... ? Really?

Ok fine! I know what you’re thinking as well!!

It’s not infinity over zero, it’s infinity times zero. Infinity zero times is zero.

My proof still stands.
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### Re: Is 1 = 0.999... ? Really?

Ecmandu wrote:Read my last two posts:

Ok fine! I know what you’re thinking as well!!

It’s not infinity over zero, it’s infinity times zero. Infinity zero times is zero.

My proof still stands.

You're not acknowledging the problem.

Anything times zero is zero, but anything times infinity is infinity, so what's zero times infinity? Zero AND infinity? Zero and not infinity?
And anything times its reciprocal is one, so why is it zero and not infinity or one?
Why are you just focusing on one of these general rules? There's a conflict here which you're just ignoring.

The proof doesn't stand. Why is nobody getting this? Silhouette
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### Re: Is 1 = 0.999... ? Really?

Silhouette wrote:
Ecmandu wrote:Read my last two posts:

Ok fine! I know what you’re thinking as well!!

It’s not infinity over zero, it’s infinity times zero. Infinity zero times is zero.

My proof still stands.

You're not acknowledging the problem.

Anything times zero is zero, but anything times infinity is infinity, so what's zero times infinity? Zero AND infinity? Zero and not infinity?
And anything times its reciprocal is one, so why is it zero and not infinity or one?
Why are you just focusing on one of these general rules? There's a conflict here which you're just ignoring.

The proof doesn't stand. Why is nobody getting this?

I get what you’re saying from your last posts silhouette: it’s undefined

When something is undefined, 3 different interpretations (as you presented them) are EQUALLY viable! I get that!

So, I posit you this:

If it’s undefined, how can anyone assert a positive, neutral or negative argument?!

This is a non-starter for any conversation.
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### Re: Is 1 = 0.999... ? Really?

Silhouette wrote:Anything times zero is zero, but anything times infinity is infinity, so what's zero times infinity? Zero AND infinity? Zero and not infinity?

You can't accept both statements ("Anything times zero is zero" and "Anything times infinity is infinity") since they are in opposition to each other (accepting them both would be a logical contradiction) so you have to pick one.

If you accept that "Anything times zero is zero" then zero times infinity is zero.

If you accept that "Anything times infinity is infinity" then zero times infinity is infinity.

Either way, $$lim_{n\to\infty} \frac{1}{n} + lim_{n\to\infty} \frac{1}{n} + lim_{n\to\infty} \frac{1}{n} + \cdots$$ is not equal to $$1$$. It's either $$0$$ or it's infinity.

And the correct answer is $$0$$ (since infinity times zero, properly speaking, without deviating from the standard meaning of the term zero and/or the standard meaning of the term multiplication, is zero, not infinity.)

But that's no proof that infinite series that converge do not exist (Ecmandu's point.) Magnus Anderson
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### Re: Is 1 = 0.999... ? Really?

Ecmandu wrote:So, I posit you this:

If it’s undefined, how can anyone assert a positive, neutral or negative argument?!

This is a non-starter for any conversation.

You do get it, good.

For an undefined, there isn't really a positive or negative argument, maybe a neutral one.
But for the thread topic as a whole, there's a singular defined answer, so you can start conversation about it. Somebody did and math ended it several times over throughout this thread.

Magnus Anderson wrote:You can't accept both statements ("Anything times zero is zero" and "Anything times infinity is infinity") since they are in opposition to each other (accepting them both would be a logical contradiction) so you have to pick one.

You accept one preliminary statement: it's undefined.
You accept this because yes - you can't pick more than one defined conclusion when they're mutually incompatible, even when they're seemingly just as valid as each other.
"Picking one" singular defined conclusion -> invalid (no more or less valid than any other, for now at least).
"Picking one" undefined conclusion is the only option left at this point.

Magnus Anderson wrote:If you accept that "Anything times zero is zero" then zero times infinity is zero.

If you accept that "Anything times infinity is infinity" then zero times infinity is infinity.

Either way, $$\lim_{n→\infty}\frac1n+\lim_{n→\infty}\frac1n+\lim_{n→\infty}\frac1n+⋯$$ is not equal to $$1$$. It's either $$0$$ or it's infinity.

And the correct answer is 0 (since infinity times zero, properly speaking, without deviating from the standard meaning of the term zero and/or the standard meaning of the term multiplication, is zero, not infinity.)

Wrong.

You only mention two of the possible "general rules" that I mentioned. Another one was that any number multiplied by its reciprocal is 1. So yes, there is a perfectly valid rule that can also be seen as applying to this situation, which equals 1 like the tendency did all the way through all possible finite numbers. This is all the extra work you need to do in this case to actually resolve what would otherwise be "undefined" as the answer, and it's just the same as you would do in calculus (as I explained in a previous post to iambiguous). Just like in calculus where limits of 0 and infinity seem to pose conceptual issues along the way to an answer, the answers you end up with are still perfectly valid, exact and unique, you can get past the same conceptual issues in the same way for Ecmandu's argument as well.

So you begin rejecting 1 singular defined answer over any other that's seemingly just as valid, investigate this undefined answer to see if more work can yield a singular defined answer, it can - so that's the correct answer: 1. That's the correct, full process. You don't stop short and conclude there's a problem with math as soon as you come across what might seem like a problem with math.

The step of converting the problem to the form "$$\lim_{n→\infty}\frac1n+\lim_{n→\infty}\frac1n+\lim_{n→\infty}\frac1n+⋯$$", or "$$\lim_{n\to\infty}(n)\times\lim_{n\to\infty}(\frac1n)$$" just turns out to be a red herring, when you could have just simplified the problem to $$\lim_{n\to\infty}(\frac{n}n)$$ or just $$\lim_{n\to\infty}(1)$$, which is quite obviously 1 like it was for all finite numbers for n. Trying to turn it into something that looked like it would suddenly switch to equalling 0 at its limit just created more work, which at best was only useful for showing how easy it is to manipulate infinity to giving arbitrary undefined answers. If you're careful with them, you can often get past undefineds in a legitimate way.

Don't simply claim this is wrong. Calculus has been doing this and getting correct answers for 350 years.

Magnus Anderson wrote:But that's no proof that infinite series that converge do not exist (Ecmandu's point.)

So in Ecmandu's example, his infinite series does converge - to 1.
In the thread topic, there's also a convergence of what can be represented as an infinite series ($$0.\dot9$$) also to 1. Silhouette
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### Re: Is 1 = 0.999... ? Really?

So... Magnus ...

I did reverse engineer the problem. If convergence is true, it’s impossible for math to work (proof through contradiction)

Let’s set aside my proof that 1 always equals zero, let’s look at it a different way!

1=1

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

1/4+1/4+1/4+1/4=1/4+1/4+1/4+1/4

Etc... this is a series!!

If sums converge at the limit in infinite sets, what I wrote above HAS to equal:

0=0!!!

Am I wrong? No, I’m not wrong.

In order for every possible number to not equal zero, thus contradicting all of math, Infinite series cannot converge at limits.
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### Re: Is 1 = 0.999... ? Really?

Silhouette,

Let’s assume there’s a series that converges at the number 2!

My proof is flawless; it also means it converges at zero!

What do we make of this?

You say it’s undefined.

Do you understand the implications of this!?!?

That means that ALL infinite series are undefined!!!

Ahh... now you get it!
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### Re: Is 1 = 0.999... ? Really?

Magnus wrote:You can't accept both statements ("Anything times zero is zero" and "Anything times infinity is infinity") since they are in opposition to each other (accepting them both would be a logical contradiction) so you have to pick one.

Silhouette wrote:You accept one preliminary statement: it's undefined.
You accept this because yes - you can't pick more than one defined conclusion when they're mutually incompatible, even when they're seemingly just as valid as each other.
"Picking one" singular defined conclusion -> invalid (no more or less valid than any other, for now at least).
"Picking one" undefined conclusion is the only option left at this point.

I think you missed the part where I said:

You can't accept both statements ("Anything times zero is zero" and "Anything times infinity is infinity") since they are in opposition to each other (accepting them both would be a logical contradiction) so you have to pick one.

It's a logical contradiction to say that both statements are true. They can't be both true. So which one is true? You didn't answer this question. Is it "Anything times zero is zero" or is it "Anything times infinity is infinity"? It's either one or the other, you can't have it both ways.

You only mention two of the possible "general rules" that I mentioned. Another one was that any number multiplied by its reciprocal is 1.

Yes. The third rule says that $$n \times \frac{1}{n} = 1$$ for any number $$n$$. I ignored it because it didn't look particularly relevant.

The argument put forward by Ecmandu is, at least as I see it, based on the premise that $$1 = \lim_{n\to\infty} \frac{1}{n} + \lim_{n\to\infty} \frac{1}{n} + \lim_{n\to\infty} \frac{1}{n} + \cdots$$. This premise can be restated as $$1 = \infty \times \lim_{n\to\infty} \frac{1}{n}$$.

The three general rules you mention are relevant because they decide how we're going to evaluate the right side of the equation.

Rule #1: Anything times zero is zero.

If we accept it then the right side of the equation is $$0$$ which is how both me and Ecmandu evaluate it. This leads to $$1 = 0$$ which is a contradiction.

Rule #2: Anything times infinity is infinity.

If we accept it then the right side of the equation is $$\infty$$. This is not how we evaluate it, but even if we did evaluate it this way, we'd end up with $$1 = \infty$$ which is still a logical contradiction.

Rule #3: Anything times its reciprocal is one.

This rule doesn't seem relevant. This is because there is an underlying, unspoken, assumption that zero is the reciprocal of infinity. So perhaps, it is better stated as "The reciprocal of infinity is zero".

Thus, $$1 = \infty \times \lim_{n\to\infty} \frac{1}{n}$$ is equal to $$1 = \infty \times 0$$ and since $$0$$ is the reciprocal of infinity this leads to $$1 = \infty \times \frac{1}{\infty}$$ which is $$1 = 1$$. Thus, no logical contradiction.

The question is:
Do we accept the premise? I personally don't. Zero is not the reciprocal of infinity and this is evident in the fact that zero times infinity isn't $$1$$.

(The third rule can also be stated as "Zero times infinity is one" which makes it more similar to the other rules.) Magnus Anderson
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