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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No, 1 ≠ 0.999...
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Other
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Total votes : 30

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

Postby Magnus Anderson » Wed Jan 22, 2020 5:01 pm

Ecmandu wrote:Now you’re just being weird.

“Ecmandu, you must show me (prove) that you are both looking at a tree and not looking at a tree in order to confirm or deny that you are looking at a tree.”

I don’t know what kind of proof school you went to!


Do you know what logic is?
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Wed Jan 22, 2020 5:52 pm

Magnus Anderson wrote:
Ecmandu wrote:Now you’re just being weird.

“Ecmandu, you must show me (prove) that you are both looking at a tree and not looking at a tree in order to confirm or deny that you are looking at a tree.”

I don’t know what kind of proof school you went to!


Do you know what logic is?


It doesn’t work the way you are explaining it.

I have to see a tree and not see the same tree at the same time in order to prove whether the tree is there or not? That’s not logic, that’s some sort of bizarre psychosis
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Re: Is 1 = 0.999... ? Really?

Postby gib » Wed Jan 22, 2020 7:25 pm

Magnus Anderson wrote:So every number that is not greater than every integer is a finite number?


Yes, boggles the mind doesn't it?!

Magnus Anderson wrote:Infinitesimals are finite numbers?


Yes.

Magnus Anderson wrote:It holds true for every quantity greater than \(0\). And since \(\infty\) is a quantity greater than zero (I hope you agree on that one), it holds true for \(\infty\) just as well.


We never came to an agreement that \(\infty\) is a quantity.

Even if we did, it brings into question whether we can still say \(\sum_{i=1}^{\infty} \frac{9}{10^i} < 1\) since the whole reason one would agree that \(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) for all n > 0 is because this says: if you keep adding 9s after the decimal point but you stop at some number n of 9s, then you will still be below 1. Setting n to \(\infty\) says something very different. It says: if you keep adding 9s and never stop, then you will get a value that [doesn't] equal 1. That's a good reason to question whether it holds for *any* value of n when we allow \(\infty\) as a value.

Magnus Anderson wrote:
It's basic counting, Magnus. I should get a 3 year old to teach you this.


Maybe you should consider the fact that no injection is possible between \(A = \{1, 2, 3\}\) and \(B = \{1, 2, 3\}\).


Huh??? Do you even know what injection is?


You know very well what I'm talking about (or at the very least, you are supposed to know, since all it takes is a little bit of attention.)

There is no injective non-surjective function between the two sets.


:laughing-rolling: I'm "supposed" to know. That's rich. Just like I was "supposed" to know that \(0.\dot9\) is not infinite nor finite. The fact of the matter is, your statement was wrong. There is an injective function between A and B. If there were an injective function between any two sets, it would be A and B. But if you add "non-surjective" then sure. What does that prove?
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Wed Jan 22, 2020 11:57 pm

Ecmandu wrote:It doesn’t work the way you are explaining it.

I have to see a tree and not see the same tree at the same time in order to prove whether the tree is there or not? That’s not logic, that’s some sort of bizarre psychosis


It is indeed psychosis but your own (: That's not what I said.
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Re: Is 1 = 0.999... ? Really?

Postby Fixed Cross » Thu Jan 23, 2020 12:02 am

The first part of the disagreements comes down to whether infinite also means indefinite.
If it does, then it is not a quantity but rather something like a condition, of a set or whatever.

In as far as the logic itself goes, unless there is some change of the rules after a definite number of decimals, Magnus' number will always be less than 1. I have no idea how you guys are using all this code, I cant even quote it.

So: "keep adding 9s after the decimal point but you stop at some number n of 9s, then you will still be below 1" is wrong. The formula doesn't provide for a "stop at some number", it rather says to keep going indefinitely.

But, not to keep going indiscriminately. You have to keep going with a specific task which by definition precludes any step from altering the result of the previous step. Which is what would have to happen for 1 to be reached.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Thu Jan 23, 2020 12:10 am

Magnus Anderson wrote:
Ecmandu wrote:It doesn’t work the way you are explaining it.

I have to see a tree and not see the same tree at the same time in order to prove whether the tree is there or not? That’s not logic, that’s some sort of bizarre psychosis


It is indeed psychosis but your own (: That's not what I said.


That’s exactly what you said. It’s on board record.

You said that if I can’t prove that 0.123...0 is not true AND false, then I can’t disprove it.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Thu Jan 23, 2020 12:30 am

gib wrote:Yes, boggles the mind doesn't it?!


Absolutely. It makes no sense.

M wrote:Infinitesimals are finite numbers?


G wrote:Yes.


Well, they actually aren't.

We never came to an agreement that \(\infty\) is a quantity.

Even if we did, it brings into question whether we can still say \(\sum_{i=1}^{\infty} \frac{9}{10^i} < 1\) since the whole reason one would agree that \(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) for all n > 0 is because this says: if you keep adding 9s after the decimal point but you stop at some number n of 9s, then you will still be below 1. Setting n to \(\infty\) says something very different. It says: if you keep adding 9s and never stop, then you will get a value that [doesn't] equal 1. That's a good reason to question whether it holds for *any* value of n when we allow \(\infty\) as a value.


Do you agree that \(\infty\) is greater than every integer?

Do you agree that an infinite number of apples is more than zero apples? and more than one apple? and two apples? and three apples? and so on?

In other words, do you agree that \(0 < 1 < 2 < 3 < \cdots < \infty\)?

If you do, then you have to accept that, since \(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) for every \(n > 0\) and since \(\infty > 0\), that \(\sum_{i=1}^{\infty} \frac{9}{10^i} < 1\).

How do you calculate the result of an infinite sum that never stops?
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Thu Jan 23, 2020 12:43 am

Magnus!!! Honestly dude!!!

Infinity is not a number!!

It is not greater than or less than an integer!!!

You are so fucking confused on so many levels
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Thu Jan 23, 2020 12:44 am

Fixed Cross wrote:So: "keep adding 9s after the decimal point but you stop at some number n of 9s, then you will still be below 1" is wrong. The formula doesn't provide for a "stop at some number", it rather says to keep going indefinitely.


"Keep going indefinitely" in the context of \(0.\dot9\) is the same as "Stop at infinity".
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Thu Jan 23, 2020 12:46 am

Magnus Anderson wrote:
Fixed Cross wrote:So: "keep adding 9s after the decimal point but you stop at some number n of 9s, then you will still be below 1" is wrong. The formula doesn't provide for a "stop at some number", it rather says to keep going indefinitely.


"Keep going indefinitely" is the same as "Stop at infinity".


No it doesn’t! It means “it is an infinity, that which never ends”.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Thu Jan 23, 2020 12:47 am

Ecmandu wrote:Magnus!!! Honestly dude!!!

Infinity is not a number!!

It is not greater than or less than an integer!!!

You are so fucking confused on so many levels



I’m going to repeat this!!!!!!!!!!!!
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Thu Jan 23, 2020 1:19 am

Ecmandu wrote:
Magnus Anderson wrote:
Fixed Cross wrote:So: "keep adding 9s after the decimal point but you stop at some number n of 9s, then you will still be below 1" is wrong. The formula doesn't provide for a "stop at some number", it rather says to keep going indefinitely.


"Keep going indefinitely" is the same as "Stop at infinity".


No it doesn’t! It means “it is an infinity, that which never ends”.


Better yet, "Keep going indefinitely" is the same as "Stop at the largest number".

Let \(L\) denote the largest number. Since \(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) for each \(n > 0\), and since \(L > 0\), it follows that \(\sum_{i=1}^{L} \frac{9}{10^i} < 1\).
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Thu Jan 23, 2020 1:54 am

No Magnus, doesn’t.

With infinity, there is no “largest number”
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Re: Is 1 = 0.999... ? Really?

Postby Fixed Cross » Thu Jan 23, 2020 2:07 am

Ecmandu wrote:
Magnus Anderson wrote:
Fixed Cross wrote:So: "keep adding 9s after the decimal point but you stop at some number n of 9s, then you will still be below 1" is wrong. The formula doesn't provide for a "stop at some number", it rather says to keep going indefinitely.


"Keep going indefinitely" is the same as "Stop at infinity".


No it doesn’t! It means “it is an infinity, that which never ends”.

Then it is not a number, but a ... motion of sorts. It has a dynamism to it.

Numbers neither do or don't 'end', they never even begin.
Even though the range of their decimals might extend infinitely, the number doesn't move.

But "infinity" does seem to be required to be on the move for it to be comprehensible.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Thu Jan 23, 2020 2:12 am

Infinity (\(\infty\)) = a number that is greater than every integer
Largest number (\(L\)) = a number that is greater than every other number that one can think of

\(0 < 1 < 2 < 3 < \dotso < \infty < 2\infty < 3\infty < \dotso < \infty^2 < \infty^3 < \dotso < \infty^\infty < \dotso < L\)

Whatever "repeats indefinitely" cannot have more than \(L\) repetitions.

"Keep adding 9s and then stop at n number of 9s" simply means "Let the number of 9s be an integer".

"Keep adding 9s and never stop" simply means "Let the number of 9s be a number greater than every integer".
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Re: Is 1 = 0.999... ? Really?

Postby gib » Thu Jan 23, 2020 3:38 am

Fixed Cross

Fixed Cross wrote:The first part of the disagreements comes down to whether infinite also means indefinite.
If it does, then it is not a quantity but rather something like a condition, of a set or whatever.


As far as I know, "indefinite" means something like "we don't know when or if it will ever end." Infinite means "we definitely know it won't end." Infinite is the property of being endless, it's not a quantity. Quantities are the things you find on the number line. Infinity is a property of the number line itself. It is where the number line extends to (or more accurately, the property of its extension being unlimited).

Fixed Cross wrote:In as far as the logic itself goes, unless there is some change of the rules after a definite number of decimals, Magnus' number will always be less than 1. I have no idea how you guys are using all this code, I cant even quote it.


It's LaTeX.

You can see how we do it by quoting our posts and looking at in the text editor.

LaTeX.png
LaTeX.png (15.13 KiB) Viewed 308 times


This board doesn't seem to have all LaTeX features enabled though. I know the mars symbol ♂ can't be posted in LaTeX.

Magnus seems to be the real LaTeX guru. He uses it even to say \(n > 0\). I'm not that hardcore. I'd rather just type out n > 0.

Fixed Cross wrote:So: "keep adding 9s after the decimal point but you stop at some number n of 9s, then you will still be below 1" is wrong. The formula doesn't provide for a "stop at some number", it rather says to keep going indefinitely.


That's a bit ambiguous, turning on what exactly is meant by "indefinitely".

Fixed Cross wrote:But, not to keep going indiscriminately. You have to keep going with a specific task which by definition precludes any step from altering the result of the previous step. Which is what would have to happen for 1 to be reached.


What \(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) for any n means is: pick any integer from the number line. You are completely unlimited in which number you pick. But it does not mean: pick infinity. And not just because infinity isn't a number on the number line, but because substituting \(\infty\) for n actually means: don't pick a value for n. Just keep adding forever.

Magnus,

Magnus Anderson wrote:
gib wrote:
Magnus Anderson wrote:So every number that is not greater than every integer is a finite number?



Yes, boggles the mind doesn't it?!


Absolutely. It makes no sense.


Only you could be driven to say something like this. I honestly wonder if you would have said something like this before this conversation.

Magnus Anderson wrote:
gib wrote:
Magnus Anderson wrote:Infinitesimals are finite numbers?


Yes.



Well, they actually aren't.


Persuasive.

Magnus Anderson wrote:Do you agree that \(\infty\) is greater than every integer?

Sure, why not.

Do you agree that an infinite number of apples is more than zero apples? and more than one apple? and two apples? and three apples? and so on?

Yes. But four apples? Hmm...

In other words, do you agree that \(0 < 1 < 2 < 3 < \cdots < \infty\)?

Yep.

If you do, then you have to accept that, since \(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) for every \(n > 0\) and since \(\infty > 0\), that \(\sum_{i=1}^{\infty} \frac{9}{10^i} < 1\).


Magnus, Magnus, Magnus... you should know my response to this by now. You might have read somewhere that, maybe, possibly, infinity is not a quantity? Ring a bell? I'm honestly more astonished at this point that you aren't able to predict this than the fact that you actually believe your own arguments.

Magnus Anderson wrote:How do you calculate the result of an infinite sum that never stops?


Well, this goes way back to a point I made much earlier in this thread (in fact, I think it was the point I jumped into this thread with). You don't have to calculate the sum of anything. You already have the answer: \(\sum_{i=1}^{\infty} \frac{9}{10^i} = 0.\dot9\). All the 9s, as infinite as they are, are already there (you just can't write them out). The only question at this point is: does \(0.\dot9\) = 1? And there happens to be a nice proof online that it does indeed equal 1:

X = \(0.\dot9\)
10X = \(9.\dot9\)
10X = 9 + \(0.\dot9\)
10X = 9 + X
9X = 9
X = 1

As you're so fond of saying, Magnus: where's the flaw?
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Thu Jan 23, 2020 3:55 am

Magnus Anderson wrote:Infinity (\(\infty\)) = a number that is greater than every integer
Largest number (\(L\)) = a number that is greater than every other number that one can think of

\(0 < 1 < 2 < 3 < \dotso < \infty < 2\infty < 3\infty < \dotso < \infty^2 < \infty^3 < \dotso < \infty^\infty < \dotso < L\)

Whatever "repeats indefinitely" cannot have more than \(L\) repetitions.

"Keep adding 9s and then stop at n number of 9s" simply means "Let the number of 9s be an integer".

"Keep adding 9s and never stop" simply means "Let the number of 9s be a number greater than every integer".


Let’s not confuse our two arguments. You, like me, do not believe that 0.999... = 1.

My argument with you Magnus is about your belief in orders of infinity.

For people who belief in convergence, every whole number equals an infinite amount of convergent series (sequences) therefor, every whole number is infinite (contradiction)

That’s not my issue with you Magnus

Stop trying to conflate the two debates and make this conflation an issue between us.

You also conflate infinity as a process and completed infinities.

Basically, you are confused; inarticulate.
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Re: Is 1 = 0.999... ? Really?

Postby gib » Thu Jan 23, 2020 9:19 am

Just a question I've been dying to ask: when we say that a set consists of infinity members, how do we know it's exactly \(\infty\) as opposed to \(\infty\) + 1 or \(\infty\) - 1? Or 2\(\infty\) or \(\infty\)/2?
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Thu Jan 23, 2020 10:27 am

gib wrote:Magnus, Magnus, Magnus... you should know my response to this by now. You might have read somewhere that, maybe, possibly, infinity is not a quantity? Ring a bell? I'm honestly more astonished at this point that you aren't able to predict this than the fact that you actually believe your own arguments.


If you agree that \(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) for every \(n > 0\), and if you agree that \(\infty > 0\), then it necessarily follows that \(\sum_{i=1}^{\infty} \frac{9}{10^i} < 1\). Everything else is irrelevant. The only condition is that \(n\) is greater than \(0\). No need to satisfy Gib's definition of the word "quantity".

What \(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) for any n means is: pick any integer from the number line. You are completely unlimited in which number you pick. But it does not mean: pick infinity. And not just because infinity isn't a number on the number line, but because substituting \(\infty\) for n actually means: don't pick a value for n. Just keep adding forever.


\(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) holds true for every \(n > 0\). It does not only apply to integers. It literally applies to anything greater than zero.

gib wrote:Just a question I've been dying to ask: when we say that a set consists of infinity members, how do we know it's exactly \(\infty\) as opposed to \(\infty\) + 1 or \(\infty\) - 1? Or 2\(\infty\) or \(\infty\)/2?


We declare how big it is.

When we say that a set consists of a finite number of members, how do we know it consists of exactly \(5\) members as opposed to \(100\) or \(200\)? Well, we don't, unless we specify it further.
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Re: Is 1 = 0.999... ? Really?

Postby surreptitious75 » Thu Jan 23, 2020 10:49 am

When we say that a set consists of infinite members all we can say is how relative it is to other infinite sets
So for example the infinite set of primes is a smaller set than the infinite set of integers because there are fewer primes / more integers
Also infinity plus one / infinity minus one are both still infinity for only infinity minus infinity will result in infinity actually being negated
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Re: Is 1 = 0.999... ? Really?

Postby surreptitious75 » Thu Jan 23, 2020 11:05 am


\(\infty\) + I = \(\infty\)
\(\infty\) - I = \(\infty\)

\(\infty\) + \(\infty\) = \(\infty\)
\(\infty\) - \(\infty\) = 0
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Re: Is 1 = 0.999... ? Really?

Postby surreptitious75 » Thu Jan 23, 2020 11:09 am


\(\infty\) + ANY NUMBER = \(\infty\)
\(\infty\) - ANY NUMBER = \(\infty\)
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Thu Jan 23, 2020 11:45 am

Ecmandu wrote:That’s exactly what you said. It’s on board record.

You said that if I can’t prove that 0.123...0 is not true AND false, then I can’t disprove it.


Again, that's not what I said.

What I said is that you need to show me the statement \(S\) and its negation \(\neg S\) that are implied by \((1, 2, 3, \dotso, 0)\).

And you have to take into account this:

The symbol \((1, 2, 3, \dotso, 0)\) represents an endless series of numbers at the end of which is a single number -- zero. "At the end of an endless series of numbers" is basically saying "At the end of something that has no end". It sounds like a contradiction but in reality it is not. It is not because the first occurrence of the word "end" and the second occurrence of the word "end" do not refer to the same thing (i.e. the same end.) They are talking about two different ends -- one that exists and one that does not.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Thu Jan 23, 2020 11:46 am

surreptitious75 wrote:
\(\infty\) + ANY NUMBER = \(\infty\)
\(\infty\) - ANY NUMBER = \(\infty\)


What exactly do you think you can prove by this?
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Thu Jan 23, 2020 12:00 pm

surreptitious75 wrote:\(\infty\) + \(\infty\) = \(\infty\)
\(\infty\) - \(\infty\) = 0


The first expression is true if what it represents is "A number greater than every integer + 1 = a number greater than every integer". Strictly speaking, however, this is not a mathematical expression since "A number greater than every integer" is not a specific number, but rather, a class of numbers (i.e. the number of numbers greater than every integer is greater than \(1\).)

But what if by \(\infty\) we refer to some specific infinite quantity such as "The smallest number greater than every integer"? (Note that there is only \(1\) such number.) In such a case, the first expression is not true because "The smallest number greater than every integer + 1 > The smallest number greater than every integer".

And what about the second expression?

The second expression is true only if \(\infty\) refers to a specific number. Otherwise, it is not true. "A number greater than every integer - a number greater than every integer" is not necessarily equal to \(0\) in the same way that "A number greater than \(3\) - a number greater than \(3\)" is not necessarily equal to \(0\) (e.g. \(5 - 4 = 1 \neq 0\).)
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