Is 1 = 0.999... ? Really?

I suppose it has something to do with the fact that you can subtract (1) from (1) to get (0)?

Not sure what you mean when you say that you expect that “Multiplying (L) by (2) is impossible”.

The point is that it’s a logical contradiction to speak of a number that is twice the size the largest number. It’s not impossible to speak of such a number (indeed, it’s not impossible to contradict yourself) but it’s a logical contradiction to do so.

I can say that Socrates is a man and then later on say that he’s a woman. Nothing impossible about that. But it’s a logical contradiction to do so. Either Socrates is a man or he’s a woman. You can’t have it both ways. Either there is no number larger than (L) or there is a number larger than (L). You can’t have it both ways.

(2L) is a reference to a number larger than the largest number. Nothing impossible about saying such a thing. But it’s a logical contradiction to do so.

It is not my argument that (9 + 0.\dot9 \neq 9.\dot9). Rather, my argument is that (10 \times 0.\dot9 \neq 9.\dot9).

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.

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.

I think we could say the same thing about the idea of the largest number itself: it’s not impossible to speak of such a number, but it contradicts a significant part of standard math.

Let’s say that there are (L) (9)s following the decimal point in (0.\dot9). How many digits are there in (10 \times 0.\dot9) ? (L+1)? If there are (L), are there only (L-1) decimal places? What’s in the (L)th decimal place?

What if we go the other way:
(\frac{9 + 0.\dot9}{10} \stackrel{?}{=} .\dot9)
(\frac{9.\dot9}{10} \stackrel{?}{=} 9.\dot9 - 9)
If so, it seems we can proceed and conclude (0.\dot9 = 1). If not, why not? One of the (9)s gets pushed of the end, but we don’t have the (L+1) problem.

A further thought on this: I think this breaks the associative property of multiplication for all numbers, not just (L). Because
((0.\dot9\ * 10) * \frac{1}{10} \neq 0.\dot9\ * (10 * \frac{1}{10}))

Magnus, Carleas…

I’m glad Carleas can speak in your latex language, because you’ll be more likely to understand it.

I already gave you my disproof Magnus, for “completed infinities” (I am the originator of this disproof and you ignored it)

If you take any real number… (I’ll use the number 1)

If you take the number 1 and divide it by 1/2, it equals 1/2+1/2!

If you continue the sequence … 1/2+1/2=1

If you continue the sequence 1/2+1/2=1/4+1/4+1/4+1/4

If you continue this sequence to convergence!!!

1=0+0+0…

1=0

You can do this for ANY real number if completed infinities exist.

Proof through contradiction:

Completed infinities don’t exist.

Actually, (L - 1 < L). And we do have a definition of (L). It’s a number larger than every other number.

Except that “A number larger than every other number” is not a contradiction in terms.

Let’s say that the number of (9)s in (0.\dot9) is some infinite number (a). The number of (9)s in (10 \times 0.\dot9) would be (a - 1).

Carleas was trying to show you through proof contradiction that infinity cannot act upon real numbers. He was not saying that L-1 was less that L.

He was showing you the absurdity of adding operators to infinity.

Anyways… answer my last post. You believe orders of infinity exist because of convergence (completed infinities) take down that argument !!

viewtopic.php?p=2759527#p2759527

That phrase is pretty vague, but 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.

I’ve given you a lot of examples of how it screws with standard math. You haven’t provided any response to those. Can you?

What operations work on the “infinite number[s]”? What are the properties of those operations? Can you add real and infinite numbers? Is the sum or difference always an infinite number? Can you add real numbers together and eventually get an infinite number? What does it mean for an infinite number to be larger than another infinite number? Is L larger than all infinite numbers?

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.