**Moderator:** Flannel Jesus

An equals sign requires a balance. Nothing but one equals one. It doesn't matter how far you carry the decimal.

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

That's question begging, because if \(0.\dot9 = 1\), then it points to the same quantity as the words in "nothing but one equals one". \(\frac{2}{2} = 1\), and that doesn't challenge the claim that "nothing but one equals one".

But the second third sentence is just wrong: we think of \(1\) as a single digit, but it absolutely matters that the decimal expansion is \(1.0000000...\) with infinite \(0\)s. If there were not infinite zeros following the \(1\), it would be greater than \(1\).

In the same way that we can think of \(0.\dot9\) as an infinite sum that approaches 1 from below, \(1.\dot0\) can be thought of as the infinite sum that approaches 1 from above. Both equal 1 in the limit.

But the second third sentence is just wrong: we think of \(1\) as a single digit, but it absolutely matters that the decimal expansion is \(1.0000000...\) with infinite \(0\)s. If there were not infinite zeros following the \(1\), it would be greater than \(1\).

In the same way that we can think of \(0.\dot9\) as an infinite sum that approaches 1 from below, \(1.\dot0\) can be thought of as the infinite sum that approaches 1 from above. Both equal 1 in the limit.

User Control Panel > Board preference > Edit display options > Display signatures: No.

- Carleas
- Magister Ludi
**Posts:**6084**Joined:**Wed Feb 02, 2005 8:10 pm**Location:**Washington DC, USA

Both equal 1 in the limit.

But it is non-finite. An equation requires a balance and .9 how ever many times will never balance equal 1. With each decimal an infinity smaller amount must be added to balance. A smallest number, beyond which?

Anything more or less then one would require the decimal places. 1 does not require them. Some amount other then 1. (give or take, does.) O requires no decimal places to distinguish it. At what point does 0.000.... magically become .000...1 cause that seems like the only way that equation is going to balance.

1.000...-0.999... is not equal to 0.000...

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

But the infinite decimal place is implied. There are in fact an infinite number of \(0\)s following the decimal point following \(1\). We don't write them because we don't need to, we imply them by convention. But we need every single one of them, and it's only in the limit that we get the number the convention implies.

It's the same with \(0.\dot9\). You describe each \(9\) as "adding to the balance", but that's not right: it assumes that each decimal place is a \(0\) until we examine it and discover a \(9\), which is understandable because that's what the convention implies. Instead, think of each decimal place as uncertain, and each \(9\) adding certainty, narrowing the scope of possible values. When we explore the implications of infinite \(9\)s, we have to conclude that there is no uncertainty, and there is no distance between \(0.\dot9\) and \(1.\dot0\): they point to the same number.

It's the same with \(0.\dot9\). You describe each \(9\) as "adding to the balance", but that's not right: it assumes that each decimal place is a \(0\) until we examine it and discover a \(9\), which is understandable because that's what the convention implies. Instead, think of each decimal place as uncertain, and each \(9\) adding certainty, narrowing the scope of possible values. When we explore the implications of infinite \(9\)s, we have to conclude that there is no uncertainty, and there is no distance between \(0.\dot9\) and \(1.\dot0\): they point to the same number.

User Control Panel > Board preference > Edit display options > Display signatures: No.

- Carleas
- Magister Ludi
**Posts:**6084**Joined:**Wed Feb 02, 2005 8:10 pm**Location:**Washington DC, USA

We don't write them because we don't need to

We don't need to because an infinite string of zero's doesn't change the unit or value of 1, it is some other number near one that requires the decimal places to distinguish itself from 1, as having a different value. Two numbers that can be distinguished as representing different values from each other can not also be equal.

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

Mowk wrote:We don't need to because an infinite string of zero's doesn't change the unit or value of 1

It does if we don't assume they're there already. If we aren't assuming that every decimal place is zero, then when we say "\(1\)" we can narrow it down to a number between \(1.\dot0\) and \(2.\dot0\). When we say \(1.0000000\), we can narrow that down to any number between \(1.000000\dot0\) and \(1.0000001\).

You're implicitly assuming all zeroes, and so you see specifying additional zeros as not changing a value you've already implicitly assumed. But if we explicitly assume that every decimal places is a \(9\), then each additional \(9\) doesn't change the value either, we've already explicitly assumed it's a \(9\), so pointing out that it's a \(9\) doesn't change the value.

User Control Panel > Board preference > Edit display options > Display signatures: No.

- Carleas
- Magister Ludi
**Posts:**6084**Joined:**Wed Feb 02, 2005 8:10 pm**Location:**Washington DC, USA

But if we explicitly assume that every decimal places is a 9, then each additional 9 doesn't change the value either,

Each additional 9 to either side of the decimal point does change the value. .90 does not equal .99 while .9 =.90000... and 99. is not equal to 9. if it doesn't balance on one side of the decimal place it won't balance on the other.

You're implicitly assuming all zeroes, and so you see specifying additional zeros as not changing a value you've already implicitly assumed.

That is because of where they are being added in relation to the decimal. Numbers to the left of the decimal are whole numbers, numbers to the right are fractions. a whole number will always be greater than any fraction of that whole number. A number greater than or less than another number can not be equal to it. 1 > .9 (infinitely recurring.)

.99 > .90, .999 > .990, .999... < 1

Last edited by Mowk on Tue Feb 11, 2020 9:32 pm, edited 1 time in total.

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

As I said, we're explicitly assuming that every decimal place is a \(9\). That's the starting point. Proceeding from there, what value is changing?

This is parallel to the implicit assumption that all the decimal places following \(1\) are \(0\)s, and then, proceeding from that assumption, you say that "an infinite string of [\(0\)s] doesn't change the unit or value of \(1\)". I agree that it doesn't, because I agree with the implicit assumption that they were already \(0\)s. In the case of \(0.\dot9\), I'm proceeding from a different, explicit assumption.

Just remove either assumption and see what happens. Do you agree that, without any implicit or explicit assumption about what numbers follow the decimal point, we can only say that a number given as "\(1\)" is between \(1.\dot0\) and \(2.\dot0\)? With no explicit or implicit assumption, it's tantamount to writing \(1.x_1 x_2 x_3 x_4...\); each decimal place is an unknown.

This is parallel to the implicit assumption that all the decimal places following \(1\) are \(0\)s, and then, proceeding from that assumption, you say that "an infinite string of [\(0\)s] doesn't change the unit or value of \(1\)". I agree that it doesn't, because I agree with the implicit assumption that they were already \(0\)s. In the case of \(0.\dot9\), I'm proceeding from a different, explicit assumption.

Just remove either assumption and see what happens. Do you agree that, without any implicit or explicit assumption about what numbers follow the decimal point, we can only say that a number given as "\(1\)" is between \(1.\dot0\) and \(2.\dot0\)? With no explicit or implicit assumption, it's tantamount to writing \(1.x_1 x_2 x_3 x_4...\); each decimal place is an unknown.

User Control Panel > Board preference > Edit display options > Display signatures: No.

- Carleas
- Magister Ludi
**Posts:**6084**Joined:**Wed Feb 02, 2005 8:10 pm**Location:**Washington DC, USA

we can only say that a number given as "1" is between 1.˙0 and 2.˙0?

We can't say that...1 is not between them. It is equal to the smallest of the numbers in the range specified. 1 = 1.0 recursive and would be less then all the other numbers between the range specified.

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

Mowk wrote:We can't say that...1 is not between them. It is equal to the smallest of the numbers in the range specified.

You're still assuming that there are infinite zeros.

User Control Panel > Board preference > Edit display options > Display signatures: No.

- Carleas
- Magister Ludi
**Posts:**6084**Joined:**Wed Feb 02, 2005 8:10 pm**Location:**Washington DC, USA

It was your notation that placed them there. 1 is not between 1 and 2.

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

Again, you are making an implicit assumption that 1 = 1.0 = 1.00000 = 1.000000000000000000000000000000000000

That's true by convention.

But if you don't make that assumption, if we drop that convention, then \(1.\dot0 \leq 1 \leq 2.\dot0 \)

In that inequality, I'm explicitly indicating where there are infinite zeros. Anywhere there aren't explicitly infinite zeros, we don't know what lies beyond the decimal point, because we are not applying the convention of implicitly assuming infinite zeros.

Without the convention, \(1 = 1.????????\)

That's true by convention.

But if you don't make that assumption, if we drop that convention, then \(1.\dot0 \leq 1 \leq 2.\dot0 \)

In that inequality, I'm explicitly indicating where there are infinite zeros. Anywhere there aren't explicitly infinite zeros, we don't know what lies beyond the decimal point, because we are not applying the convention of implicitly assuming infinite zeros.

Without the convention, \(1 = 1.????????\)

User Control Panel > Board preference > Edit display options > Display signatures: No.

- Carleas
- Magister Ludi
**Posts:**6084**Joined:**Wed Feb 02, 2005 8:10 pm**Location:**Washington DC, USA

A whole number can not be less then itself. That's not an assumption. 1 is equal to 1.? if and only if ? = 0. If "?" were any other value 1 would not be equal to 1.? The equation .9 (recursive) = .9 (recursive) .9 (recursive) can not be equal to 1 as it represents a fraction and 1 represents a whole number.

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

What fraction does \(0.\dot9\) represent?

User Control Panel > Board preference > Edit display options > Display signatures: No.

- Carleas
- Magister Ludi
**Posts:**6084**Joined:**Wed Feb 02, 2005 8:10 pm**Location:**Washington DC, USA

9 (recurring) over 10 (recurring)

Last edited by Mowk on Wed Feb 12, 2020 4:02 pm, edited 1 time in total.

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

Interesting, although it doesn't move us forward; \(1\dot0 - \dot9 = 0\), so that fraction equals \(1 = 1/1 = 0.\dot9\)

Let me go back to this:

So we have '1', the symbol, which points to 1, the concept. We can say of the concept that there is no decimal part, it's an integer. By convention, we treat the symbol '1' as pointing to that concept, and we assume that where '1' doesn't explicitly specify the decimal expansion, it implies it. But as you say, if there is any decimal part that is not 0, we agree that it would not be the concept 1. By convention, the symbol '1' implies that there is no decimal part.

We can point to the same concept in a number of different ways: \(\frac{2}{2}, x^0, 0!, -e^{i\pi}\) -- those are all the same concept.

So the claim is that \(0.\dot9\) is another symbol for that concept. To show that it doesn't, you need to show that they are conceptually distinct. Similar to what I say above, \(1 - 0.\dot9 = 0\), and there's no other value we can coherently put on the right side of the equality. Magnus' approach, if I understand it correctly, seems to be to posit \(0.\dot01\), but that too is equal to the concept \(0\), because by its construction there's no difference between them.

Related: do you agree that \(\infty+1 = \infty\) (or rather, that the cardinality of the set of integers is the same as the cardinality of the union between the set of integers and the set {1.234})?

Let me go back to this:

Mowk wrote:1 is equal to 1.? if and only if ? = 0. If "?" were any other value 1 would not be equal to 1.?

So we have '1', the symbol, which points to 1, the concept. We can say of the concept that there is no decimal part, it's an integer. By convention, we treat the symbol '1' as pointing to that concept, and we assume that where '1' doesn't explicitly specify the decimal expansion, it implies it. But as you say, if there is any decimal part that is not 0, we agree that it would not be the concept 1. By convention, the symbol '1' implies that there is no decimal part.

We can point to the same concept in a number of different ways: \(\frac{2}{2}, x^0, 0!, -e^{i\pi}\) -- those are all the same concept.

So the claim is that \(0.\dot9\) is another symbol for that concept. To show that it doesn't, you need to show that they are conceptually distinct. Similar to what I say above, \(1 - 0.\dot9 = 0\), and there's no other value we can coherently put on the right side of the equality. Magnus' approach, if I understand it correctly, seems to be to posit \(0.\dot01\), but that too is equal to the concept \(0\), because by its construction there's no difference between them.

Related: do you agree that \(\infty+1 = \infty\) (or rather, that the cardinality of the set of integers is the same as the cardinality of the union between the set of integers and the set {1.234})?

User Control Panel > Board preference > Edit display options > Display signatures: No.

- Carleas
- Magister Ludi
**Posts:**6084**Joined:**Wed Feb 02, 2005 8:10 pm**Location:**Washington DC, USA

Wow. the site is slow today, painfully.

This is taking me no where. I can't even copy and paste what you have written. I disagree with the language. 1 as an integer does not require the concept of division or exponents, or ... Your examples include them as part of the concept of the integer 1.

I'm no Georg Cantor. The notion of infinity has messed with better heads then mine.

You are a more informed mathematician then I.

We can point to the same concept in a number of different ways: 22,x0,0!,−eiπ -- those are all the same concept.

This is taking me no where. I can't even copy and paste what you have written. I disagree with the language. 1 as an integer does not require the concept of division or exponents, or ... Your examples include them as part of the concept of the integer 1.

I'm no Georg Cantor. The notion of infinity has messed with better heads then mine.

You are a more informed mathematician then I.

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

The proof involving let x=.9 (recurring) introduces a rounding error in the equation. The presence of the function of the infinitely recurring property is lost in the process and therefore the equation isn't balanced. At best the result is an approximation. That's my best guess. Rather like rounding up .6 (recurring) to .67 or truncating π to a specific decimal place. That's what my gut tells me, but i can't prove it. An infinity long line and a finite length 180 degree arc share the same number of points along their paths because a finite length path can be divided infinitely. All infinities are not created equally. Thank you Georg Cantor. Perhaps this "proof" ignores this detail, but I'm not the mathematician to prove it.

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

Mowk wrote: I can't even copy and paste what you have written.

Using quote from the normal view should show the LaTeX markup. The quote buttons in the "Topic Review" below the reply box are kind of misleading, they only tack on the poster name, they don't actually quote from the post they're attached to (you can see this by selecting arbitrary text anywhere on the page and then clicking the quote button: it will format the selected text as a quote by the author of the post that the button is attached to).

Mowk wrote:1 as an integer does not require the concept of division or exponents, or ... Your examples include them as part of the concept of the integer 1.

What does "=" mean? If \(\frac{2}{2} = 1\), what do we mean by "=" there? I wouldn't say that \(\frac{2}{2}\) is "part of" the concept of the integer 1, but it is equivalent to the integer 1, and it's certainly an important part of the concept that it is the multiplicative identity, of which \(\frac{2}{2} = 1\) is an example.

Mowk wrote:I'm no Georg Cantor. The notion of infinity has messed with better heads then mine.

I think it messed with Cantor's head as well. He seems to have somewhat agreed with Magnus, in that he thought there was a greatest number, though he also seems to have recognized that the concept leads to contradiction. His first quote on that page is downright mystical, though, basically saying that "absolute infinity" is tantamount to god.

User Control Panel > Board preference > Edit display options > Display signatures: No.

- Carleas
- Magister Ludi
**Posts:**6084**Joined:**Wed Feb 02, 2005 8:10 pm**Location:**Washington DC, USA

Sorry, I got use to the select, copy, paste. The quote button always grabs the whole post, but it does work better to see the code that makes it all pretty.

I guess equals means more to me then you, huh? And to add insult to injury, the extra meaning is likely invalid. I get side tracked easily, had to come up with some grip of this LaTeX thingy.

Anyway, 1, the integer is a counting unit. It requires no 'function' to be performed on it to be it. In the case of 2 divided by 2 an element has been added to the units as counting digits that requires a division function to be performed. So when you concluded "-- those are all the same concept." I thought there was more then just one concept in play. So yes, the value resulting from the division function results in 1 which is equal to 1. Yet there is something 'more' taking place on the left (in your example) of the equals sign then on the right. Not arguing what you mean just how you said it; as if the two, as a package, didn't make the transit. Likely the confusion was on my end.

Anyway, I am more likely to suspect that some quality is not being taken into account, or there is yet more at play then meets these eyes. But if the question comes up in a quiz I'll be better prepared to answer counter intuitively.

I guess equals means more to me then you, huh? And to add insult to injury, the extra meaning is likely invalid. I get side tracked easily, had to come up with some grip of this LaTeX thingy.

Anyway, 1, the integer is a counting unit. It requires no 'function' to be performed on it to be it. In the case of 2 divided by 2 an element has been added to the units as counting digits that requires a division function to be performed. So when you concluded "-- those are all the same concept." I thought there was more then just one concept in play. So yes, the value resulting from the division function results in 1 which is equal to 1. Yet there is something 'more' taking place on the left (in your example) of the equals sign then on the right. Not arguing what you mean just how you said it; as if the two, as a package, didn't make the transit. Likely the confusion was on my end.

Anyway, I am more likely to suspect that some quality is not being taken into account, or there is yet more at play then meets these eyes. But if the question comes up in a quiz I'll be better prepared to answer counter intuitively.

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

If on a math test I came upon the question x equals two minus one, or x plus one equals 2, solve for x. If I answered .9 (recurring) would my answer be considered correct? Or if presented the question; two plus two equals x, solve for x, and my answer was 3.9 (recurring)?

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

I deeply regret that I did not think of testing this when I was in school. There are infinite answers to any "solve for x" problem, and I now realized the many missed opportunities to be a precocious wise ass.

The repeating decimal is an interesting case here. By comparison, if the question is \(2-1=x\), then \(x=2^0\) is true, but I suspect the examiner would say that it's not in its most "reduced" form, and that seems legitimate since there's an unresolved operation baked in. But I don't think that objection works as well for \(0.\dot0\): it's not really any less "reduced" than \(1\), at least not as obviously as an answer that contains an operation.

Ultimately, "is it correct" and "would it be considered correct" are different questions. It is correct, but I suspect many teachers wouldn't accepted it. Lots of teachers punish wiseassery, even when it obeys the letter of the law, and appealing to mathematical trivia to provide a technically correct but counterintuitive and less-clear answer would be read as being a wise ass.

I haven't read much at the intersection of math and philosophy of language, but my impression is that popular conceptions give it more meaning than it needs. I think it's possible to build math as a purely formal language, with no real-world analogues to the transformations and relationships necessary to specify how its objects are interconnected. Define rules about how we can manipulate symbols, and then use the rules to show that two symbols satisfy the relationship "=".

But that's beyond me, and I'm only about 80% sure it's true.

The repeating decimal is an interesting case here. By comparison, if the question is \(2-1=x\), then \(x=2^0\) is true, but I suspect the examiner would say that it's not in its most "reduced" form, and that seems legitimate since there's an unresolved operation baked in. But I don't think that objection works as well for \(0.\dot0\): it's not really any less "reduced" than \(1\), at least not as obviously as an answer that contains an operation.

Ultimately, "is it correct" and "would it be considered correct" are different questions. It is correct, but I suspect many teachers wouldn't accepted it. Lots of teachers punish wiseassery, even when it obeys the letter of the law, and appealing to mathematical trivia to provide a technically correct but counterintuitive and less-clear answer would be read as being a wise ass.

Mowk wrote:I guess equals means more to me then you, huh?

I haven't read much at the intersection of math and philosophy of language, but my impression is that popular conceptions give it more meaning than it needs. I think it's possible to build math as a purely formal language, with no real-world analogues to the transformations and relationships necessary to specify how its objects are interconnected. Define rules about how we can manipulate symbols, and then use the rules to show that two symbols satisfy the relationship "=".

But that's beyond me, and I'm only about 80% sure it's true.

User Control Panel > Board preference > Edit display options > Display signatures: No.

- Carleas
- Magister Ludi
**Posts:**6084**Joined:**Wed Feb 02, 2005 8:10 pm**Location:**Washington DC, USA

Funny, in my previous post I was going to ask if it would be correct or I'd get sent to the office for being a wise ass, or maybe both. I was being precocious asking it.

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

It's the decimal. 0. means something different then .0, following or preceding yet, on each own they mean the same thing. And within the infinity between zero and one there is an infinite opportunity for counting. Save for the first from no thing to some thing. Save between counting of something and counting nothing. How can no thing be counted? Absence? the decimal on it's own, merely a divider, a quantifier of a whole or a part. That is how much difference exists between .9 recurring and one, the smallest thing remaining countable is as infinite as .9 recurring.

- Mowk
- Philosopher
**Posts:**1914**Joined:**Thu Feb 02, 2012 8:17 pm**Location:**In a state of excessive consumption

This might be a pedantic over-emphasis of your specific wording, but the real numbers aren't countable. There's no next real number. For any two distinct real numbers, there are infinitely many real numbers between them. 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).

User Control Panel > Board preference > Edit display options > Display signatures: No.

- Carleas
- Magister Ludi
**Posts:**6084**Joined:**Wed Feb 02, 2005 8:10 pm**Location:**Washington DC, USA

Return to Science, Technology, and Math

Users browsing this forum: Google [Bot]