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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Total votes : 30

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

Postby Mowk » Thu Feb 20, 2020 4:01 pm

If it isn't zero; the decimal point on it's own, it is counting by default. Any unit recognized as not zero is a form of counting.
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Thu Feb 20, 2020 6:42 pm

Carleas wrote: So the term "the smallest thing remaining countable" is undefined, there can be no smallest real number distinct from zero ("smallest" in absolute value, I assume that is your intent as well).


It seems we are getting hung up on this notion of size. Zero is not the "smallest" in absolute value, it represents a value as non existent regardless of size, sequence and series. Any unit recognized as existing can not be zero. This is perhaps more an existential question then one of math. If there were not things to count we would have no reason to create a symbol representing the absence of the countable things. We wouldn't have a zero as a symbol for no value if we didn't have things we place values upon.

Zero represents nothing to count. Zero is a symbol given to not counting. As soon as any other symbol shows up counting has taken place. You've recognized a thing and things as distinguishable from the absence of the thing and between things. It is sort of binary, zero or one. You've presented me with a range of an infinite set of numbers between 0 and 1 which are real numbers. Given this information, there are three values involved. A) No thing represented by zero (0) B) "a" thing represented by (1) where X is not equal to 1 and 1 is not equal to 0 And as a given C) there is an infinite subset of "somethings" between them N is not (0) AND n is not (1). I can count there is at least one thing between 0 and 1, N. All real numbers can be counted just not in an infinite sequence that is dependent on an infinite number of digits. There is/are N of them between 0 and 1. They can be counted in sequence as 0, N, 1. Where N can not be less then or equal to zero or equal to or more then 1.
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Thu Feb 20, 2020 7:18 pm

I consider the extents in a range as not included in the subset of the range defined by them. Key word being "between".
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Thu Feb 20, 2020 7:29 pm

There is a persistent but unfounded notion that between can be used only of two items and that among must be used for more than two. Between has been used of more than two since Old English; it is especially appropriate to denote a one-to-one relationship, regardless of the number of items. It can be used when the number is unspecified economic cooperation between nations , when more than two are enumerated between you and me and the lamppost partitioned between Austria, Prussia, and Russia — Nathaniel Benchley , and even when only one item is mentioned (but repetition is implied). pausing between every sentence to rap the floor — George Eliot Among is more appropriate where the emphasis is on distribution rather than individual relationships. discontent among the peasants When among is automatically chosen for more than two, English idiom may be strained. a worthy book that nevertheless falls among many stools — John Simon
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Re: Is 1 = 0.999... ? Really?

Postby Carleas » Thu Feb 20, 2020 9:01 pm

Sure, but between \(1\) and \(0\), there are a bunch of other numbers; we can point to any of them (0.123, for example) and prove that \(1\) and \(0\) are distinct numbers. There are no numbers between \(1\) and \(0.\dot9\), which isn't possible if they are distinct real numbers.
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Thu Feb 20, 2020 10:59 pm

the decimal point. It is not, to the best of my understanding, a number in and of itself. But is does reside between 0.9 and 1.0 in our numbering system.

Count to 1.0 by .9s recurring? What is used to move the decimal point and assign the next most significant digit? The application of the convention of counting natural numbers. So we have all these different descriptions of groups of numbers; natural, whole, fractions, integers, rational, irrational, prime, positive, negative, real and complex and imaginary.

Not done.
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Fri Feb 21, 2020 2:54 am

Darn, composed for too long and my service provider killed the connection.

I will agree that within a hypothetical convention where infinity exists 1 = 0.9 recurring. I would also argue that using the natural numbers as a convention the numbers 0 and recurring decimals 9's don't exist to have equivalence.

It is a matter of convention and not all conventions are created equal. Yet it is a postulate that .9 recurring using an infinite convention is equal to 1 limited by another convention of the infinite. Different sets that are not inclusive.

We "let" it be equal in an infinite set as an axiom, while it doesn't exist within the convention of natural numbers as other example of an infinite set. We have already determined all infinite sets aren't created equal such that a .9 recurring in a set does not require it be equal to a 1 from every other case of an infinite set. 1 can not be equal to .9 recurring in a set that does not include .9 recurring as a member of it's set.

It is presented as a one or the other case while there is an "and" option in place of "either/or". Just what exactly are we "letting" the unequal non-equivalent sets include axiomatically?

OK. this 'seems' to "sum" up the argument as reasoning thus far. (don't know, could be other members of an infinite set, axiomatically). I don't know maybe the first draft was a member from a set that was a more accurate representation. What are we going to "let" axiomatically?
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Fri Feb 21, 2020 3:55 am

phyllo nailed it and didn't.

This is one of those issues that display the clear distinction between a good philosopher and a expert mathematician.

A good engineer will tell them that they are wasting their time.

Reminds me of this joke:

A mathematician and an engineer are sitting at a table drinking when a very beautiful woman walks in and sits down at the bar.

The mathematician sighs. "I'd like to talk to her, but first I have to cover half the distance between where we are and where she is, then half of the distance that remains, then half of that distance, and so on. The series is infinite. There will always be some finite distance between us."

The engineer gets up and starts walking. "Ah, well, I figure I can get close enough for all practical purposes."


Differing sets of convention.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Fri Mar 13, 2020 4:49 am

Carleas wrote:That phrase is pretty vague


What exactly is vague about the statement "A number larger than every other number"?

the way you are using it, it is a contradiction of much of standard math, e.g. the associative property of multiplication on the real numbers, the property that the set of real numbers is closed under addition and multiplication, etc


Closure

If you add two real numbers, you will get another real number. What makes you think that \(L\) is a real number and that what applies to real numbers also applies to it?

Of course, you're aware of the possibility that \(L\) is not a real number. Earlier you said "So, \(L\) is some special type of number, it doesn't follow normal rules of arithmetic. So it's possible that \(L-1 = L\), we can't really say because we don't really have a definition of what \(L\) is." But we do have the definition of \(L\). You're merely choosing to ignore this for some reason.

Associativity

Earlier, you said "Right, so \(L + (1-1) \neq (L+1)-1\). That means \(L\) isn't a real number, because addition and multiplication aren't associative on it." I never confirmed or denied this inequality since it doesn't strike me as particularly necessary. What makes you think that \(L + (1 - 1) \neq (L + 1) - 1\) can't be the case?

I'm now going to return to this:

Carleas wrote:
Magnus Anderson wrote:\(L + 1 - 1\) is fine.

How can that be? L+1 is undefined! You can't subtract 1 from an undefined quantity and get a defined quantity.


What about \(\sqrt{-1}\)? Is it undefined? Is there a real number the square of which is equal to \(-1\)? If the answer is "no", does that mean you reject complex numbers? If you do not reject complex numbers, why do you reject pseudo-numbers such as numbers larger than the largest number? And why do we even have to speak of such numbers?
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Fri Mar 13, 2020 5:09 am

Carleas wrote:Sure, but between \(1\) and \(0\), there are a bunch of other numbers; we can point to any of them (0.123, for example) and prove that \(1\) and \(0\) are distinct numbers. There are no numbers between \(1\) and \(0.\dot9\), which isn't possible if they are distinct real numbers.


What makes you think there are no numbers between \(1\) and \(0.\dot9\)? You can't say it's because there are no standard real numbers in between them because that would be like saying there are no numbers between \(0\) and \(1\) because there are no integers in between them.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Fri Mar 13, 2020 6:20 am

To recap:

Image

The reason this proof is wrong is because the two underlined numbers are not equal and they are not equal because the number of nines in the first is lower than the number of nines in the second.

An image representing \(10 \times 0.\dot9\) operation:

Image

The green rectangle (representing the first number) is shorter than the purple rectangle (representing the second number).

This is based on the premise that \(0.\dot9\) represents a specific infinite quantity. However, most people do not see it this way. So it might not be the best counter-argument.

So let us define \(0.\dot9\) to be a non-specific infinite number instead.

A number \(x\) is said to be specific if \(x - x = 0\). If \(x - x \neq 0\) then \(x\) is not a specific number.

According to some, \(\infty - \infty = \infty\). This is fine, nothing wrong with it, but let's see where it leads.

If \(\infty - \infty = \infty\) then \(0.\dot9 - 0.\dot9 \neq 0\). This is because the number of non-zero terms in the first sum is not the same as the number of non-zero terms in the second -- there's always an infinite difference. Hence, \(10x - x \neq 9x\).

Indeed, the same conclusion follows for any case where \(\infty - \infty\) is non-zero.
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Re: Is 1 = 0.999... ? Really?

Postby Carleas » Tue Mar 24, 2020 1:49 am

Magnus Anderson wrote:What exactly is vague about the statement "A number larger than every other number"?

"Number"
Magnus Anderson wrote:What makes you think that \(L\) is a real number and that what applies to real numbers also applies to it?

If memory serves, this is the first time you've suggested otherwise. But that's sort of my point, \(L\) can't be a real number for all the reasons I've pointed out, so if it's not a reason then we agree there. I just think that's a big problem for your argument, because you're treating \(L\) as though it's a real number.
Magnus Anderson wrote:we do have the definition of \(L\). You're merely choosing to ignore this for some reason.

I'm not ignoring it, I'm pointing out that it's vague. \(L\) is clearly not a number like the ones we usually work with, and so I'm not making assumptions about what you mean by "number". You haven't clarified what happens when we subtract 1 from \(L\), what kind of number do we get? If it's a real number, than that number plus 1 must be \(L\), in which case \(L\) is the sum of two real numbers. If it's some number less than \(L\) but of the same type, then does \(L-1 = L\)? Is \(L\) hyperreal?

Failing to say explicitly what you mean lets you play calvinball around \(L\): no conclusion that could defeat your claim can be argued for, because you offer attributes of \(L\) only if and when they become necessary to undermine some argument.

Magnus Anderson wrote:I never confirmed or denied this inequality since it doesn't strike me as particularly necessary.

Here is a great example: just confirm or deny the inequality!

Magnus Anderson wrote:why do you reject pseudo-numbers such as numbers larger than the largest number?

I have yet to see a coherent definition of the "the largest number", and this question makes me think whatever you have in mind has some problems. But I guess we'll see when you tell me what you have in mind.

Magnus Anderson wrote:According to some, \(\infty - \infty = \infty\).

This here is treating \(\infty\) like a number, and it isn't a number. I would say that that equation is meaningless. I've probably been sloppy in my language around this, and I apologize. When I say that two infinities are equal, I mean that every element of one set can be mapped to exactly one element of the other set. So, for example, \(.\dot9\) and \(9.\dot9\) have the same number of decimal places, in the sense that each decimal place in one corresponds to exactly one decimal place in the other.

Somewhat as an aside: in your system, \(.\dot9\) is ambiguous, since if I just say \(.\dot9\) we don't know if it's just the simple repeating decimal, or some different version of the repeating decimal obtained by multiplication by a power of 10 and subtracting the integer portion. This strikes me as a larger problem than you seem to acknowledge (how might 1 fewer zeros on the end of an integer's decimal expansion change its value?).
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Wed Mar 25, 2020 12:54 am

Carleas wrote:\(L\) is clearly not a number like the ones we usually work with, and so I'm not making assumptions about what you mean by "number". You haven't clarified what happens when we subtract 1 from \(L\), what kind of number do we get? If it's a real number, than that number plus 1 must be \(L\), in which case \(L\) is the sum of two real numbers. If it's some number less than \(L\) but of the same type, then does \(L-1 = L\)? Is \(L\) hyperreal?


\(L - 1\) is not a real number. It's also not \(L\). Even if we say that \(L\) and \(L - 1\) are numbers of the same type (i.e. if we put them in the same category), it does not mean they are the same number, that they are equal.

I am not sure why you're insisting on categorizing \(L\). What exactly do we gain by placing it in a category?

Here is a great example: just confirm or deny the inequality!


I don't have to confirm or deny irrelevant claims. Not answering irrelevant questions isn't evasion in the negative sense of the word. If it is an evasion, it is a positive kind of evasion (one that tries to push or pull the discussion in the right direction.)

This here is treating \(\infty\) like a number, and it isn't a number. I would say that that equation is meaningless.


If I recall correctly, your claim is that \(0.999\dotso = 1\). If this is true, it means that you think that an infinite number of non-zero terms (which is what \(0.\dot9\) is) can be equal to a number (such as \(1\).) In other words, it would imply that at least SOME infinite quantities are numbers (since \(1\) is a number, right?)
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Wed Mar 25, 2020 1:13 am

When I say that two infinities are equal, I mean that every element of one set can be mapped to exactly one element of the other set. So, for example, \(0.\dot9\) and \(9.\dot9\) have the same number of decimal places, in the sense that each decimal place in one corresponds to exactly one decimal place in the other.


\(0.\dot9\) and \(9.\dot9\) have the same number of decimal places because you said so. It's an arbitrary decision. Nothing is stopping you from picking a different position e.g. that \(0.\dot9\) has fewer decimal places than \(9.\dot9\).
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Thu Mar 26, 2020 3:32 pm

And the result of dividing .9 recurring in half?

The question is are they significant decimal places? As example 125 = (1x100)+(2x10)+(5x1). further 3076 = (3x1000)+(0x100)+(7x10)+(6x1)

In the case of 0.9, 0.9 = (0x1)+(9/10) the number in decimal form is already telling you there are zero 1s. If there are zero 1s .9 recurring can't be equal to 1 because there are zero 1s in the expression.
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Re: Is 1 = 0.999... ? Really?

Postby Carleas » Thu Mar 26, 2020 8:05 pm

Magnus Anderson wrote:I am not sure why you're insisting on categorizing \(L\). What exactly do we gain by placing it in a category?

We gain the ability to have a conversation! If your argument depends on the properties of \(L\), then we need a full description of those properties (e.g. by placing \(L\) into an existing category) to evaluate your claims.

Magnus Anderson wrote:I don't have to confirm or deny irrelevant claims.

Do you not know the answer? You don't have to do anything, but if you know the answer, it would be awfully kind of you to just say it. Or if the question doesn't make sense I'd love to hear why. Or if the answer is completely toxic to understanding and also irrelevant, then make that case.

Magnus Anderson wrote:If I recall correctly, your claim is that \(0.999…=1\). If this is true, it means that you think that an infinite number of non-zero terms (which is what \(0.\dot9\) is) can be equal to a number (such as \(1.\)) In other words, it would imply that at least SOME infinite quantities are numbers (since \(1\) is a number, right?)

Your conclusion doesn't follow. \(0.\dot9\) has infinite decimal places. It is not itself infinite, and infinity isn't used as a number in defining that.

Magnus Anderson wrote:\(0.\dot9\) and \(9.\dot9\) have the same number of decimal places because you said so.

It follows from the definition of a repeating decimal. If you have some other definition, please specify (and really, you need to clarify what it means if you're going to hold that \(9+0.\dot9 = 9.\dot9 \neq 9.\dot9 = 10 * 0.\dot9\))
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Re: Is 1 = 0.999... ? Really?

Postby Carleas » Thu Mar 26, 2020 8:08 pm

Mowk wrote:And the result of dividing .9 recurring in half?

The long division makes sense here, and doesn't really inform the question.

\(\frac{0.\dot9}{2}= .4\dot9 = .5\)
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Re: Is 1 = 0.999... ? Really?

Postby iambiguous » Thu Mar 26, 2020 8:12 pm

Next up: Does 1 covid 19 infection = 0.999... covid 19 infection?
He was like a man who wanted to change all; and could not; so burned with his impotence; and had only me, an infinitely small microcosm to convert or detest. John Fowles

Start here: viewtopic.php?f=1&t=176529
Then here: viewtopic.php?f=15&t=185296
And here: viewtopic.php?f=1&t=194382
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Thu Mar 26, 2020 9:03 pm

Carleas wrote:
Mowk wrote:And the result of dividing .9 recurring in half?

The long division makes sense here, and doesn't really inform the question.

\(\frac{0.\dot9}{2}= .4\dot9 = .5\)


Interesting but you are not expressing the fraction .9 recurring. divided by 2 as a fraction.

In mathematics the set of all numbers that can be expressed in the form a/b, where a and b are integers and b is not zero.

Your example is invalid as .9 recurring is not an integer. the representation .9 can be written as (0x1) + (9/10) Does it matter how many 9's follow the expression as the expression itself clearly declares there are zero 1s.
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Re: Is 1 = 0.999... ? Really?

Postby Carleas » Thu Mar 26, 2020 9:18 pm

Mowk wrote:Interesting but you are not expressing the fraction .9 recurring. divided by 2 as a fraction.

My point was that the division works as expected: the long division algorithm produces \(0.499999...\)

So, I don't think it answers the question either way because if \(0.\dot9 = 1\), then \(0.4\dot9 = 0.5\), and if it doesn't it doesn't.

Mowk wrote:Does it matter how many 9's follow the expression as the expression itself clearly declares there are zero 1s. How can a description of a number that has zero 1's be equal to an expression that has one?

There's a lot of addition in there too, and I don't think anyone's denying that 0x1+5/10+5/10 =1
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Thu Mar 26, 2020 10:49 pm

Carleas wrote:
Mowk wrote:Interesting but you are not expressing the fraction .9 recurring. divided by 2 as a fraction.

My point was that the division works as expected: the long division algorithm produces \(0.499999...\)

So, I don't think it answers the question either way because if \(0.\dot9 = 1\), then \(0.4\dot9 = 0.5\), and if it doesn't it doesn't.

Mowk wrote:Does it matter how many 9's follow the expression as the expression itself clearly declares there are zero 1s. How can a description of a number that has zero 1's be equal to an expression that has one?

There's a lot of addition in there too, and I don't think anyone's denying that 0x1+5/10+5/10 =1


Yeah but you haven't respected the notion of decimal formatting in your example either. You have added two fractions together from the same decimal position. 0.9 recurring still states there are zero 1s within it's notation. Adding .5 + .5 is the math required to move the decimal position, .9 recurring leaves the decimal point where it is and leaves it as zero 1s.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Fri Mar 27, 2020 1:11 am

Careleas wrote:Your conclusion doesn't follow. \(0.\dot9\) has infinite decimal places. It is not itself infinite, and infinity isn't used as a number in defining that.


Somehow, you forgot the fact that \(0.\dot9\) is an infinite sum, which makes it an infinite number of, non-zero terms. Are you going to tell me it's a finite sum? a finite number? Is it finite or is it infinite?

Are you suddenly denying that \(0.\dot9 = 0.9 + 0.09 + 0.009 + \dotso\)? What's at the right side of the expression if not an infinite number of things?

It's pretty clear that infinites and finites overlap. An infinite number of zeroes is a zero. One raised to infinity is one. And so on. If you can calculate the result of \(\infty \times 1\) what's the reason you can't calculate the result of \(\infty - \infty\)?

If you say that \(9.\dot9\) and \(0.\dot9\) have the same number of \(9\)'s following the dot, then it follows that \(9.\dot9 - 0.\dot9 = 9\). That's doing arithmetic with infinites.

If you say that \(9.\dot9\) and \(9.\dot9\) have the same number of \(9\)'s after the dot, then it follows that \(9.\dot9 - 9.\dot9 = 0\).

If you say that \(\infty\) and \(\infty\) refer to the same number, then it follows that \(\infty - \infty = 0\).

There's ABSOLUTELY NOTHING meaningless about \(\infty - \infty = \infty\). That's just you not willing to play the game. You don't win the game by not playing the game.

We gain the ability to have a conversation! If your argument depends on the properties of \(L\), then we need a full description of those properties (e.g. by placing \(L\) into an existing category) to evaluate your claims.


How about you place \(i\) in one of the categories that existed before people invented the category of complex numbers? How useful is that? Very unuseful, isn't it? What's the point of placing something where it doesn't belong?

How about placing rationals in the category of integers?

And why is any of this relevant? I don't even remeber why I mentioned \(L\). That's how relevant it is. It served some very limited purpose, one I can't even remember, precisely because it's so limited.
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Fri Mar 27, 2020 1:14 am

The question; Is 1 = 0.999...?

Another way of asking the same question might sound something like; is an infinite number of smaller and smaller "parts" equal to 1 finite "whole" ?

If I had 1 volume of space and I divided it, equally in half, an infinite number of times, the result would be an infinite number of parts. I guess if I could cut a volume of space an infinite number of times, it would still have come from the 1 original volume.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Fri Mar 27, 2020 1:34 am

@Carleas

Your argument rests on this: infinity minus infinity is meaningless but \(0.\dot9 - 0.\dot9\) is equal to zero. Not sure how to say it but don't you realize that infinity is equivalent to a sum of ones? Like so: \(\infty = 1 + 1 + 1 + \cdots\). And that \(0.\dot9\) is also equivalent to a sum of numbers? Like so: \(0.\dot9 = 0.9 + 0.09 + 0.009 + \dotso\). Now, you take \(0.9 + 0.09 + 0.009 + \cdots\) and you subtract from that \(0.9 + 0.09 + 0.009 + \cdots\) AND YOU HAVE NO PROBLEM WITH THAT. You go like "Hm, okay, take first numbers from both sums, subtract them, we get -- hehe -- zero! Now, let us take second number, what do we get? oh yeah baby, we get yet another zero! and so on and so forth!" But you take \(1 + 1 + 1 + \cdots\) and subtract from that \(1 + 1 + 1 + \cdots\) and you tell me it's MEANINGLESS? As in meaning-fucking-less no less. Caps lock doesn't have big enough letters to communicate the bewilderment I am currently suffering through :) What is going on in here?
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Fri Mar 27, 2020 1:38 am

Mowk wrote:The question; Is 1 = 0.999...?

Another way of asking the same question might sound something like; is an infinite number of smaller and smaller "parts" equal to 1 finite "whole" ?

If I had 1 volume of space and I divided it, equally in half, an infinite number of times, the result would be an infinite number of parts. I guess if I could cut a volume of space an infinite number of times, it would still have come from the 1 original volume. Seems like the argument could be made from there, that infinity = 1.


Infinity (\(\infty\)) times infinitesimal (\(\frac{1}{\infty}\)) equals \(1\). No doubt about that. But \(0.\dot9\) is a different beast -- one that never adds up to \(1\).
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