Is 1 = 0.999... ? Really?

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)?

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.

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.

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.

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).

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.

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.

I consider the extents in a range as not included in the subset of the range defined by them. Key word being “between”.

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.

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.

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?

phyllo nailed it and didn’t.

Differing sets of convention.

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

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:

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?

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.

To recap:

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:

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.

“Number”

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.

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.

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

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.

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?).

(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?

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.)

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?)

(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).

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.