Is 1 = 0.999... ? Really?

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

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 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…

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.

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.

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.

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.

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

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.

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.

You’re still assuming that there are infinite zeros.

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

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.???)

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.

What fraction does (0.\dot9) represent?

9 (recurring) over 10 (recurring)

[attachment=0]fraction.jpg[/attachment]

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})?

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.

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.

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).

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.

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.

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.