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...
12
32%
No, 1 ≠ 0.999...
22
58%
Other
4
11%
 
Total votes : 38

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

Postby obsrvr524 » Mon Feb 08, 2021 6:16 pm

phyllo wrote:
Wikipedia wrote:
Countable
In mathematics, a countable set is a set with the same cardinality as some subset of the set of natural numbers. A countable set is either a finite set or a countably infinite set.
I don't think that you are using it in that sense. Basically by writing infA-1, you're counting backwards from infA. That seems like a no no. You know, counting back from where exactly??

I made that argument myself much earlier. I don't use \(infA-1\) as an index for that reason - but Magnus does.
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Re: Is 1 = 0.999... ? Really?

Postby phyllo » Mon Feb 08, 2021 6:58 pm

Your definition for 999... has infinity+1 digits.

Whatever that means. :-k

I tend to think that 999... is just infinity. (As is any number represented as n...)
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Mon Feb 08, 2021 7:44 pm

obsrvr524 wrote::lol:
But I was hoping that you would skip that obvious argument.

If that is where you want to go and have nothing else to offer then you have expired your defense on that issue by not convincing me of your premise (4) - so the rest of that argument (version 4) is moot.

Do you have a different argument to make addressing the proposal? If so - it's time to post it. O:)


PoO #2
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Mon Feb 08, 2021 8:34 pm

wtf wrote:Once you stipulate that your expressions "stand for a positive number," you are limited to finite-length strings of decimal digits. Those are the rules for positive numbers.

Of course you can make up your own notation if you can make sense of it (you haven't yet done so) but you can't call them positive numbers as if they're natural numbers. They're not.


I used the term "positive number" to mean "a number greater than or equal to \(0\)". That does not have to be a natural number. For example, \(1.5\) is not a natural number but it is a positive number.

The conventional meaning of the term "positive number" is not relevant though I do think that it is the same as the one assigned by me.
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Re: Is 1 = 0.999... ? Really?

Postby wtf » Mon Feb 08, 2021 9:02 pm

Magnus Anderson wrote:
I used the term "positive number" to mean "a number greater than or equal to \(0\)". That does not have to be a natural number. For example, \(1.5\) is not a natural number but it is a positive number.


Perhaps you already forgot what you wrote. Perfectly normal as we age, think nothing of it. Let me help you out.

Magnus Anderson wrote:
1) Let \(X\) be a class of decimal / base-10 numerals that have no fractional part and that stand for a positive number.


Emphasis mine.

I reiterate: If you are making up your own kind of numbers, you are free to do so if you can make them logically coherent, which you have so far failed to do. But if you are talking about the standard natural numbers, as you clearly seem to be, each and every one of them has finite length.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Mon Feb 08, 2021 9:25 pm

Wtf wrote:Perhaps you already forgot what you wrote. Perfectly normal as we age, think nothing of it. Let me help you out.


I didn't forget what I wrote.

Yes, I spoke of base-10 numerals that stand for positive numbers and that have no fractional part.

But if you are talking about the standard natural numbers, as you clearly seem to be, each and every one of them has finite length.


If a base-10 numeral has no fractional part, it does not mean it represents a natural number.

I reiterate: If you are making up your own kind of numbers, you are free to do so if you can make them logically coherent, which you have so far failed to do.


Can you explain the bolded part?
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Re: Is 1 = 0.999... ? Really?

Postby wtf » Mon Feb 08, 2021 9:29 pm

Magnus Anderson wrote:it does not mean it represents a natural number.

Can you explain the bolded part?




You agree that you are NOT talking about natural numbers, and you haven't given a coherent definition of "9999999..."

But you know if you are really interested in making sense of numbers with infinitely many digits to the left of the decimal point, you might be interested in the p-adic numbers.

https://en.wikipedia.org/wiki/P-adic_number
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Mon Feb 08, 2021 9:43 pm

wtf wrote:But you know if you are really interested in making sense of numbers with infinitely many digits to the left of the decimal point, you might be interested in the p-adic numbers.

https://en.wikipedia.org/wiki/P-adic_number


I am not interested in p-adic numbers. I am interested in you explaining what's wrong with what I am saying (if you are interested, that is.) And note that "explaining what's wrong with what I am saying" has a very specific meaning.

You agree that you are NOT talking about natural numbers, and you haven't given a coherent definition of "9999999..."


Yes, I am not talking about natural numbers. What about that?

As for \(99\dot9\), it's a numeral that involves an infinite number of \(9\)s in the direction of the least significant digit. The weight of the first digit (and by extension, the weight of all other digits) is not specified and neither is the exact number of \(9\)s (we only know the number is greater than every integer.) That makes it a class of numbers rather than some specific number. What's incoherent about that?
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Re: Is 1 = 0.999... ? Really?

Postby wtf » Mon Feb 08, 2021 9:48 pm

Magnus Anderson wrote:I am not interested in p-adic numbers. I am interested in you explaining what's wrong with what I am saying (if you are interested, that is.) And note that "explaining what's wrong with what I am saying" has a very specific meaning.


I couldn't find your definition but I seem to recall it involves expressions like \(10^\infty\) which makes no sense. If you have a more current definition I did not take the trouble to read back to find it. If all you mean is that it's infinite, then not much harm is done, but you seem to mean something more specific than that.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Mon Feb 08, 2021 10:03 pm

Magnus Anderson wrote:There are implicit \(0\)s everywhere. For example, there's an implicit digit right before the first digit in \(99\dot9\). That digit is \(0\). There is also an implicit digit at \(-1\) (which is the index of the first digit after the decimal point.) That digit is also \(0\).


obsrvr524 wrote:But your index can't reach either of those positions - so they are not implied.
    1) You restricted the number to be non-fractional and also
    2) your "..." restricts you from exceeding your upper "9" to any supposed preceding "0" (which wouldn't be relevant anyway).


An implicit digit is a digit that is implied by the symbol rather than explicitly stated by it.

For example, in the case of \(1\), there is exactly one explicit digit -- that digit is \(1\). But there are lots of implicit digits. For example, the digit immediately before it is \(0\). The digit associated with \(10^{-1}\) is \(0\) too. And so on.

Similarly, in the case of \(999\dotso\), there are three explicit digits (the three \(9\)s), many implicit \(0\)s and lots of \(9\)s implied by the ellipsis.

Magnus Anderson wrote:I assume what you're saying here is that you do not think that the digit associated with \(10^0\) is \(0\) but rather \(9\).


obsrvr524 wrote:
    3) Certainly -- \((9x10^0) = 9\).


The problem with that is that the ellipsis in \(999\dotso\) indicates that the digits expand endlessly in the direction of the least significant digit. This means that they can't stop at \(10^0\). The \(9\) associated with \(10^0\) CAN'T be the last \(9\) because there is NO last \(9\). That's what the symbol is saying: no last \(9\).
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Mon Feb 08, 2021 10:05 pm

wtf wrote:I couldn't find your definition but I seem to recall it involves expressions like \(10^\infty\) which makes no sense. If you have a more current definition I did not take the trouble to read back to find it. If all you mean is that it's infinite, then not much harm is done, but you seem to mean something more specific than that.


Cool. But why do you think it makes no sense? It merely stands for \(10 \times 10 \times 10 \times \cdots\). It's \(10\) multiplied by itself infinitely.
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Re: Is 1 = 0.999... ? Really?

Postby wtf » Mon Feb 08, 2021 10:20 pm

Magnus Anderson wrote:
Cool. But why do you think it makes no sense? It merely stands for \(10 \times 10 \times 10 \times \cdots\). It's \(10\) multiplied by itself infinitely.


Because "10 multiplied by itself infinitely" is not defined anywhere.

Multiplication is defined as a binary operation. It inputs two numbers and outputs a third. 3 x 5 = 15. Two inputs, one output.

By induction we can extend the definition to any finite collection of input values. To show that 2 x 3 x 5 makes sense we note that it's (2 x 3) x 5. And 2 x 3 is defined because it's two inputs, so we get 6 x 5, which is again two inputs, so we get 30. We can use the associative law to note that we could have done it as 2 x (3 x 5) and the answer would be the same.

You have not defined "10 multiplied by itself infinitely." All we can do with the standard rules of math is show by induction that any FINITE number of numbers multiplied together can ultimately be reduced to a sequence of binary operations.

So if you want to extend the definition to infinitely many multiplicands, you have to make a definition. And this you have not done. Which is why I say you haven't done it.

You have an intuitive idea of what you mean, but you haven't formalized it logically. And till you do that, you haven't got a definition.
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Re: Is 1 = 0.999... ? Really?

Postby obsrvr524 » Mon Feb 08, 2021 10:38 pm

I agree that \(\infty \) cannot be used as a number because it is not well defined.
\(infA\) is a different story.
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Re: Is 1 = 0.999... ? Really?

Postby obsrvr524 » Mon Feb 08, 2021 10:49 pm

Magnus Anderson wrote:
Magnus Anderson wrote:There are implicit \(0\)s everywhere. For example, there's an implicit digit right before the first digit in \(99\dot9\). That digit is \(0\). There is also an implicit digit at \(-1\) (which is the index of the first digit after the decimal point.) That digit is also \(0\).
obsrvr524 wrote:But your index can't reach either of those positions - so they are not implied.
    1) You restricted the number to be non-fractional and also
    2) your "..." restricts you from exceeding your upper "9" to any supposed preceding "0" (which wouldn't be relevant anyway).
Magnus Anderson wrote:in the case of \(999\dotso\), there are three explicit digits (the three \(9\)s), many implicit \(0\)s and lots of \(9\)s implied by the ellipsis.

I explained why that red clause is not true in the partial post you just quoted. If you have some rebuttal to that explanation - give it.

Magnus Anderson wrote:
Magnus Anderson wrote:I assume what you're saying here is that you do not think that the digit associated with \(10^0\) is \(0\) but rather \(9\).
obsrvr524 wrote:
    3) Certainly -- \((9x10^0) = 9\).

The problem with that is that the ellipsis in \(999\dotso\) indicates that the digits expand endlessly in the direction of the least significant digit. This means that they can't stop at \(10^0\). The \(9\) associated with \(10^0\) CAN'T be the last \(9\) because there is NO last \(9\). That's what the symbol is saying: no last \(9\).

That might be true except that you specified "NO FRACTIONAL DIGITS" - that means there is no "9.###" but merely "9." And your "..." actually bumped your "999" upward from 0 to make it an infinite count. If the "..." is allowed to go below zero then your number "999..." is undefined and can't be rationally discussed. There is no way to know how big it is. It could just stand for "9999.9999..." or "999.999..." or any other amount. But in no case is there a 0 digit involved.
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              You have been observed.
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    • the light from the dark, and
    • blame each for the sins of the other
    • - until they beg you to take charge.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Mon Feb 08, 2021 11:29 pm

obsrvr524 wrote:I agree that \(\infty \) cannot be used as a number because it is not well defined.


I disagree.

It can be used but it has its limits.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Mon Feb 08, 2021 11:35 pm

wtf wrote:Because "10 multiplied by itself infinitely" is not defined anywhere.

Multiplication is defined as a binary operation. It inputs two numbers and outputs a third. 3 x 5 = 15. Two inputs, one output.

By induction we can extend the definition to any finite collection of input values. To show that 2 x 3 x 5 makes sense we note that it's (2 x 3) x 5. And 2 x 3 is defined because it's two inputs, so we get 6 x 5, which is again two inputs, so we get 30. We can use the associative law to note that we could have done it as 2 x (3 x 5) and the answer would be the same.

You have not defined "10 multiplied by itself infinitely." All we can do with the standard rules of math is show by induction that any FINITE number of numbers multiplied together can ultimately be reduced to a sequence of binary operations.

So if you want to extend the definition to infinitely many multiplicands, you have to make a definition. And this you have not done. Which is why I say you haven't done it.

You have an intuitive idea of what you mean, but you haven't formalized it logically. And till you do that, you haven't got a definition.


I don't think that "10 multiplied by itself infinitely" is undefined. And I also do not think the rest of your post shows otherwise. What makes you think that an infinite number of numbers multiplied together cannot be reduced to an infinite sequence of binary operations? And what do you want me to do to prove that an infinite product such as \(10^{\infty}\) is not undefined?
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Re: Is 1 = 0.999... ? Really?

Postby obsrvr524 » Mon Feb 08, 2021 11:42 pm

Magnus Anderson wrote:
obsrvr524 wrote:I agree that \(\infty \) cannot be used as a number because it is not well defined.


I disagree.

It can be used but it has its limits.

It is an ambiguous term. You don't know if it means merely the size of the natural numbers (such as \(infA\)) or 10 times that or that squared. If you try to use it in maths, you find that you can prove almost anything is equal to almost anything else.

\(\infty + 10 = \infty + 3 = \infty^2 = \infty - 40\)

Because what is infinite is infinite - especially if you add anything to it.
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    • the light from the dark, and
    • blame each for the sins of the other
    • - until they beg you to take charge.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Feb 09, 2021 12:00 am

obsrvr524 wrote:I explained why that red clause is not true in the partial post you just quoted. If you have some rebuttal to that explanation - give it.


I thought I already did.

That might be true except that you specified "NO FRACTIONAL DIGITS" - that means there is no "9.###" but merely "9."


And your "..." actually bumped your "999" upward from 0 to make it an infinite count.


The ellipsis is to the right side of the digits which indicates that the digits are expanding to the right, which is to say, in the direction of the least significant digit.

Let \(x\) be the index of the first (the leftmost) \(9\) in \(99\dot9\). With that in mind, the index of the second is \(x - 1\), the index of the third is \(x - 2\), the index of the fourth is \(x - 3\) and so on. And since the ellipsis indicates that this process continues without an end, there is no last digit, which means, there is no \(9\) with the lowest index ("the least significant \(9\)").

In the case of YOUR number, which is \(\sum_{i=0}^{i->\infty} 9\times10^i\), the least significant \(9\) exists.

That alone tells you there is a difference between the two numbers.

If the "..." is allowed to go below zero then your number "999..." is undefined and can't be rationally discussed.


If it's allowed to go below zero, the concept becomes contradictory, which means, it cannot stand for anything real, but that does not mean we can't talk about it and/or use it in other ways.

And note that I didn't say that \(999\dotso\) necessarily represents that concept. It can be used to represent that concept but it can also be used to represent other concepts. It all depends on the index of the first digit. If the index of the first digit is \(infA^2\), and the number of \(9\)s is \(infA\), then "..." does not extend beyond the decimal point.

Normally, decimal numerals have "the rightmost digit before the decimal point". \(150\) is an example. The rightmost digit before the decimal point is \(0\). In such a case, we have a rule for determining the index of any digit within the numeral. The index of the rightmost digit, for example, is \(0\). The index of any other digit is based on how far away it is from the rightmost digit and whether it is to the right or to the left of it. For example, the index of \(5\) is \(1\) and the index of \(1\) is \(2\).

But \(999\dotso\) does not have "the rightmost digit", so we cannot use this rule. And since no other rule exists, there's no way to deduce the index of any digit.
Last edited by Magnus Anderson on Tue Feb 09, 2021 12:04 am, edited 1 time in total.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Feb 09, 2021 12:02 am

obsrvr524 wrote:It is an ambiguous term. You don't know if it means merely the size of the natural numbers (such as \(infA\)) or 10 times that or that squared. If you try to use it in maths, you find that you can prove almost anything is equal to almost anything else.

\(\infty + 10 = \infty + 3 = \infty^2 = \infty - 40\)

Because what is infinite is infinite - especially if you add anything to it.


You can do logic with ambiguous terms -- if you know how to do it. And you can't prove just about anything with \(\infty\) unless you're using it improperly.
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Re: Is 1 = 0.999... ? Really?

Postby obsrvr524 » Tue Feb 09, 2021 12:35 am

Magnus Anderson wrote:
obsrvr524 wrote:I explained why that red clause is not true in the partial post you just quoted. If you have some rebuttal to that explanation - give it.


I thought I already did.

You merely repeated what I had already rebutted. Repetition doesn't count as a rebuttal - it indicates that there is no more argument to be presented and the end of that encounter.

Magnus Anderson wrote:
That might be true except that you specified "NO FRACTIONAL DIGITS" - that means there is no "9.###" but merely "9."
And your "..." actually bumped your "999" upward from 0 to make it an infinite count.

The ellipsis is to the right side of the digits which indicates that the digits are expanding to the right, which is to say, in the direction of the least significant digit.

Let \(x\) be the index of the first (the leftmost) \(9\) in \(99\dot9\). With that in mind, the index of the second is \(x - 1\), the index of the third is \(x - 2\), the index of the fourth is \(x - 3\) and so on. And since the ellipsis indicates that this process continues without an end, there is no last digit, which means, there is no \(9\) with the lowest index ("the least significant \(9\)").

Again, you are repeating yourself. I already explained why that is not acceptable -
obsrvr524 wrote:That might be true except that you specified "NO FRACTIONAL DIGITS" - that means there is no "9.###" but merely "9." And your "..." actually bumped your "999" upward from 0 to make it an infinite count. If the "..." is allowed to go below zero [the decimal point] then your number "999..." is undefined and can't be rationally discussed. There is no way to know how big it is. It could just stand for "9999.9999..." or "999.999..." or any other amount. But in no case is there a 0 digit involved.
Magnus Anderson wrote:If it's allowed to go below zero, the concept becomes contradictory, which means, it cannot stand for anything real, but that does not mean we can't talk about it and/or use it in other ways.

I see that I misspoke - I meant to say that it cannot go below the decimal point.

Magnus Anderson wrote:And note that I didn't say that \(999\dotso\) necessarily represents that concept. It can be but not necessarily.

It cannot because of the reasons given above.

Magnus Anderson wrote:If the index of the first digit is \(infA^2\), and the number of \(9\)s is \(infA\), then "..." does not extend beyond the decimal point.

Again you are repeating what I have already explained away - your number cannot be of a higher cardinality than the natural numbers because that is all we are dealing with and the ellipsis is only about the natural "countable" number set.

Magnus Anderson wrote:Normally, decimal numerals have "the rightmost digit before the decimal point". \(150\) is an example. The rightmost digit before the decimal point is \(0\). In such a case, we have a rule for determining the index of any digit within the numeral. The index of the rightmost digit, for example, is \(0\). The index of any other digit is based on how far away it is from the rightmost digit and whether it is to the right or to the left of it. For example, the index of \(5\) is \(1\) and the index of \(1\) is \(2\).

That is counter to how you were indexing originally but I agree that an index should begin AT the decimal point - and that means that your first "9" of your "999..." figure is AT index \(infA\) (or \(infA-1\) if you prefer), else the entire number doesn't make sense.

Magnus Anderson wrote:And since no other rule exists, there's no way to deduce the index of any digit.

I explained that "other rule" in my first rebuttal in our first attempt -
obsrvr524 wrote:
Magnus Anderson wrote:4) Therefore, \(999\dotso\) and \(\sum_{i=0}^{i->\infty} 9\times10^i\) represent two different numbers.

But now disagreed.

1) The same numerical value can be represented in two different ways

2) The representation - "999..." declares a top digit but no bottom digit.

3) The representation - \(\sum_{i=0}^{i->\infty} 9\times10^i\) declares a bottom digit but no top digit.

4) An infinitely large number has no top digit (endlessly large).

5) An infinitely large number (non-decimal) does have a bottom digit.

9) When a representation of an infinitely large value is presented as topping out with a "999" it can be acceptably understood that 9's are the entirety of the infinite value (despite the indication of a proposed top).

10) When a representation of an infinitely large value is presented as having a bottom of "999.0" it can be acceptably understood that 9's are the entirety of the infinite value (maintaining that there is no top).

11) Both representations can be accepted as representative of a value that is all 9's
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              You have been observed.
    Though often tempted to encourage a dog to distinguish color I refuse to argue with him about it
    It's just same Satanism as always -
    • separate the bottom from the top,
    • the left from the right,
    • the light from the dark, and
    • blame each for the sins of the other
    • - until they beg you to take charge.
    • -- but "you" have been observed --
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Re: Is 1 = 0.999... ? Really?

Postby wtf » Tue Feb 09, 2021 12:45 am

Magnus Anderson wrote:I don't think that "10 multiplied by itself infinitely" is undefined.


Where is it defined? If you have defined it, I missed it, so please repeat the definition. But if you haven't defined it, then where is it defined? There is of course a notion of an infinite product in math, but it has a technical definition, and your usage doesn't fit that case. At best we can say that your notation is an infinite product that diverges. If that's what you mean, I'll accept that. I am not sure though that this is what you mean.
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Re: Is 1 = 0.999... ? Really?

Postby obsrvr524 » Tue Feb 09, 2021 12:56 am

wtf wrote:If you claim it's defined, what's the definition? How can I show something's not defined anywhere, by copying every math book ever written to show it's not there? On the contrary, you have to show it's defined if you claim it's defined. And if you claim it's defined, just tell me the definition.

Magnus prefers - "if you claim that it is not there - you should prove that it is not". :-"

Been through that with him already. O:)
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              You have been observed.
    Though often tempted to encourage a dog to distinguish color I refuse to argue with him about it
    It's just same Satanism as always -
    • separate the bottom from the top,
    • the left from the right,
    • the light from the dark, and
    • blame each for the sins of the other
    • - until they beg you to take charge.
    • -- but "you" have been observed --
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Feb 09, 2021 12:59 am

obsrvr524 wrote:You merely repeated what I had already rebutted. Repetition doesn't count as a rebuttal - it indicates that there is no more argument to be presented and the end of that encounter.


I don't think I repeated myself. I stated something I never stated before -- what I mean by "implicit digit". I thought you did not understand the term.

obsrvr524 wrote:Again, you are repeating yourself. I already explained why that is not acceptable


if the "..." is allowed to go below zero [the decimal point] then your number "999..." is undefined


You seem to think that if a meaning is contradictory that it cannot possibly be associated with a word.

It's like saying the meaning of "square-circle" is not "a shape that is both a circle and a shape" because a shape cannot be both a circle and a shape.

I see that I misspoke - I meant to say that it cannot go below the decimal point.


I understood you correctly.

Again you are repeating what I have already explained away - your number cannot be of a higher cardinality than the natural numbers because that is all we are dealing with and the ellipsis is only about the natural "countable" number set.


Yes, that's what you think. I disagree with that.

That is counter to how you were indexing originally


The only difference is that I'm starting with \(0\).

but I agree that an index should begin AT the decimal point


That's how I did it before:
https://www.ilovephilosophy.com/viewtop ... 5#p2803118

and that means that your first "9" of your "999..." figure is AT index \(infA\) (or \(infA - 1\) if you prefer), else the entire number doesn't make sense.


That does not follow. That's merely you stitching things together so that they make sense to you. That's like me saying "square-circle" means "square" because otherwise the concept is contradictory. You don't get to decide the meaning of an expression based on what makes sense to you and what doesn't. It means what it means.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Feb 09, 2021 1:00 am

obsrvr524 wrote:
wtf wrote:If you claim it's defined, what's the definition? How can I show something's not defined anywhere, by copying every math book ever written to show it's not there? On the contrary, you have to show it's defined if you claim it's defined. And if you claim it's defined, just tell me the definition.

Magnus prefers - "if you claim that it is not there - you should prove that it is not". :-"

Been through that with him already. O:)


The onus of proof is ALWAYS on the one making the claim (whether that claim is positive or negative is irrelevant.)
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Feb 09, 2021 1:04 am

wtf wrote:
Magnus Anderson wrote:I don't think that "10 multiplied by itself infinitely" is undefined.


Where is it defined? If you have defined it, I missed it, so please repeat the definition. But if you haven't defined it, then where is it defined? There is of course a notion of an infinite product in math, but it has a technical definition, and your usage doesn't fit that case. At best we can say that your notation is an infinite product that diverges. If that's what you mean, I'll accept that. I am not sure though that this is what you mean.


"10 multiplied by itself infinitely" is another expression for \(10^{\infty}\).

And if you want me to define the entire expression, I don't think that's how it works. The meaning of a statement is derived from the meaning of its words (among other things.)

So what exactly do you want me to define?

What exactly is unclear?
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