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0.999...= 1

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Re: 0.999...= 1

#631
post #551

Earlier quoted context omitted.

It does exist. The other poster just clearly showed that it exists by referring to it. The problem is that if we include such a number in our formal system of math, we quickly find contradictions and the whole system falls apart. So such a number is incompatible with any formal system of math (though I guess you could start building one which does include such a number and see what properties it has). Herein lies the…

In math, when assuming the existence of something proved a contradiction, we conclude that the thing does not exist. The description may exist "integer between 3 and 4", but there is not described object. A description names a set or a class, and that class can have 0,1, or more numbers.

>In math, when assuming the existence of something proved a contradiction, we conclude that the thing does not exist.

Well only to the extent that you don't want to throw away any of the other axioms. Sometimes you do and there are some fun systems of math, but few have any practicality and those that do are often so advanced that even someone with an undergraduate focus in math can't appreciate those systems.

It is much the same with computer science. I personally enjoyed playing around with formal concepts of computation and adding some extras to see what happens. For example, what happens to a Turing machine if part of the machine can time travel or has access to an oracle. Does this make concepts like time travel inherently contradictory to our notion of computation?

But the practicality of these exercises does not exceed their entertainment value.

Re: 0.999...= 1

#632

Earlier quoted context omitted.

A countable set is one where you can reach any element in a finite amount of steps. See: https://en.wikipedia.org/wiki/Countable_set

If you're saying we never get from the initial 0 to the trailing 1 in a finite number of steps, that's true and that's what DavidVoid is saying. But I haven't made the list of digits uncountably large by adding one element. Instead I put that element in a transfinite position in the ordering and I have to watch out for weird consequences. I'm no expert, but countability doesn't depend on how the set is ordered. It de…

It's not the set of digits you use which is uncountable, it's the representation itself.

Re: 0.999...= 1

#633

There is no proof that will ever satisfy a person dead-set against this. Ever since I brought this home from school as a child, my whole family ribbed me mercilessly for it. If you tell a person that 3/6 = 1/2, they'll believe you - because they have been taught from an early age that fractions can have multiple "representations" for the same underlying amount. People mistakenly believe that decimal numbers don't hav…

Maybe try expressing it in the form of money? Let's say a gallon of gas is 99 cents with infinitely repeating 9's. 0.99999999999 cents. You're still going to end up paying a dollar a gallon for it because eventually it's going to get rounded off. No gas store operator is going to try to cut a penny for you and give you a fraction of a penny. They could argue that the fraction of a penny becomes infinitely small and that giving you a shaving off the side of a copper penny would be infinitely too large.

Now don't mind me while I open up a store where every price tag ends in 0.99...repeating and have a poor college student at the checkout lane with a penny shaver to calm down any rowdy customers he or she can't explain away.

Re: 0.999...= 1

#634
post #192

A formally rigorous proof of this (in Metamath) is here: http://us.metamath.org/mpeuni/0.999....html Unlike typical math proofs, which hint at the underlying steps, every step in this proof only uses precisely an axiom or previously-proven theorem, and you can click on the step to see it. The same is true for all the other theorems. In the end it only depends on predicate logic and ZFC set theory. All the proofs have…

It depends on more than just ZFC, also on the definitions of the real/complex numbers. The crux of the proof is that 0.99999... is being constructed within the real/complex numbers, and in that system it is equal to 1. And at the point where students see this, the whole concept of real numbers and infinity is usually ill-defined. I actually understand the scepsis for this theorem and where it comes from. The proof re…

It's true that it depends on the definitions of real/complex numbers. Many other things turn out to be provable from ZFC. A discussion about this, from the viewpoint of Metamath, is here: http://us.metamath.org/mpeuni/mmcomplex.html

Re: 0.999...= 1

#636
post #564

Earlier quoted context omitted.

I expect mjd is thinking of irrational bases. The number might still be written as 10, in digits that look decimal.

That is not what I meant at all. (My phrasing was unclear. Sorry for the confusion.) Jordi understood what I meant though.

It was unclear; I understand rational numbers to be ratios, and irrationals to be inexpressible as fractions, and I see an almost direct connection between that and digital representation in rational bases, so it seemed deeply confusing to see a consequence of the definition of rationals being refuted. The only way out was an irrational base.

Re: 0.999...= 1

#637
post #625

Earlier quoted context omitted.

Can something be true and not obvious?

Obviously not. Which is the point in using the word obvious, obviously. Namely, using it to feel superior or to not provide a better argument.

When I use the word, the point is to call out the fact that I think it's obvious, so if others don't, they can explain why. Not to forestall any discussion.

Anyone who uses the word differently is doing it wrong.

Re: 0.999...= 1

#638
post #594

Earlier quoted context omitted.

> > What is an "infinitely small" number? > What is an infinitely large number? Neither is a well-defined concept within the standard reals, and completely unnecessary for understanding that 0.999…=1. > > What does it mean to say X is a number, if you can't subtract it from another number and get a number as an answer? > By that logic, 0.99 repeating isn't a number at all, and therefore can't be equivalent to 1, beca…

0.9 is not equal to 1, 0.99 is not equal to 1, 0.999 is not equal to 1, 0.9999 is not equal to 1, 0.99999 is not equal to 1, 0.999999 is not equal to 1, and so on, ad infinitum. Saying that if you add enough "9"s it suddenly equals 1.0 makes absolutely no sense to me, and I seriously doubt that anyone will be able to convince me that it does make sense. I've read every single post in this thread and none of you have…

> 0.9 is not equal to 1,

> 0.99 is not equal to 1,

> 0.999 is not equal to 1,

> 0.9999 is not equal to 1,

> 0.99999 is not equal to 1,

> 0.999999 is not equal to 1,

> and so on, ad infinitum.

You are correct about all of these, and all finite strings of the above form.

> Saying that if you add enough "9"s it suddenly equals 1.0 makes absolutely no sense to me, and I seriously doubt that anyone will be able to convince me that it does make sense. I've read every single post in this thread and none of you have gotten me any closer at all to believing or understanding that 0.9 repeating equals 1.

I think it's because you, and a lot of other people in this thread, are turning the question on its head. The difficulty does not so much lie in figuring out whether 0.999… is equal to 1 or not, but rather in what we mean when we write 0.999….

I know I'm repeating myself from elsewhere in the thread, but I'll try again. Try to go through these step by step, and feel free to let me know where you lose the thread.

DEFINITION: A finite decimal representation of a real number is a finite string of the form `a_m a_{m-1} … a_0 . b_1 b_2 … b_n` where each `a_i` and each `b_i` is a natural number between 0 and 9 inclusive (a digit). We say that this finite decimal representation represents the real number

    a_m*10^m + a_{m-1}*10^{m-1} + … + a_0 + b_1*10^{-1} + b_2*10^{-2} + … + b_n*10^{-n}.
Note: The previous definition deals with finite strings and finite sums. I hope we can agree that these are well-defined and unambiguous concepts.

EXAMPLE: The string `12.98` has `m=1`, `n=2` with `a_1=1`, `a_0=2`, `b_1=9` and `b_2=8`. It therefore represents the real number

    1*10^1 + 2*10^0 + 9*10^{-1} + 8*10^{-2}
(duh!).

Within this standard framework, there is no way to ask "what is 0.999…?. It is not yet defined, because we have only defined what finite strings mean. The standard definition for what one means by 0.999… follows. (One can obviously also define these things 0.888…, 1.999…, etc., but let's stick to one case here).

DEFINITION: Let `(c_n)_{n natural}` be a sequence of real numbers (let me know if you need a definition of sequences!). We say that the sequence has the limit x as n tends to infinity (these are words, you don't have to ascribe meaning to "infinity" in that sentence – it's just a word, like "gnarf"!) if, given any real eps>0, there exists an M such that for all m > M, |c_m - x| Definition (this is the definition you have to wrap your head around before continuing): Consider the sequence `(c_n)_{n natural}` where `c_n` is the finite sum

    9*10^{-1} + 9*10^{-2} + … + 9*10^{-n}
The string `0.999…` (which we colloquially speak of as "zero point nine nine nine with nines repeating forever") denotes the limit of the sequence `(c_n)_{n natural}` as n tends to infinity (if it exists).

"THEOREM": The limit defining `0.999…` does exist. It is `1`.

PROOF: You can fill this in. If you can't, I'm happy to do it.

As you can see, at no point in the above did feelings or beliefs matter :-)

Re: 0.999...= 1

#639
post #624

Earlier quoted context omitted.

Why? https://news.ycombinator.com/item?id=23009160 posited that we were in a situation where (1) holds, so that's where we're starting. If we assume (1) holds, then (2) cannot hold. We can abandon (1) but then we're no longer replying to that specific comment, now we're trying to prove something else.

My point is that it cannot be clear what assuming (1) entails when you aren't properly defining the quantities involved. You have to answer in clear and mathematical language what the quantity in (1) is defined as. What is the definition of "0.666…7"? As it currently stands, assumption (1) is similar in nature to me saying "gnarfgnarf is an imaginary number". It's completely meaningles unless I define what I mean by…

No, it isn't. The idea that saying "gnarfgnarf is an imaginary number is completely meaningless" is the opposite of true: if you assert that gnarfgnarf is an imaginary number, that is the definition we'll be using for the remainder of whatever proof we use that in. Anywhere the proof now talks about gnarfgnarf, we're talking about something that is an imaginary number, and has to follow all the rules that imaginary numbers have to follow, without ever having to say which imaginary number it is, or further define it. It's "any" imaginary number, we just call it "gnarfgnarf" instead of "x" or "a + bi" or the like.

Same here: we have a number written as 0.666...7 using conventional mathematical notation. The comment that is being replied to asserts that this can be treated as a sequence, and so we start the proof with that definition: "0.666...7 is a sequence", and now we're done. You, as reader of the proof, have been informed that those nine symbols, in that order, for the rest the proof, represent a sequence. Not "a specific sequence", but "any sequence", and it must follow all the rules that sequences follow.

We then show that simply by being "a sequence", due to the properties of sequences, we get a contradiction. Our first assertion is the definition for the purpose of this proof, and is sufficient.

Re: 0.999...= 1

#640
post #534

Earlier quoted context omitted.

0.000...1 can be written as 1/inf, which has sense in Surreal Numbers math.

You could just as easily say that 0.000…54234 can be written as 1/∞. Surreal Numbers is a bit of a detour in this case. The premise of the idea that 1 - 0.999… could be written 0.000…1 is the mistaken concept that there is some point "after an infinite number of steps" where the expansion of 0.999… stops and you can leave the remaining 1. The expansion never stops and there is no final remainder. The result is 0.000……

    x = 0.999...
    x = 9/10 + 9/10^2 + 9/10^3 ... 9/10^inf
    x = 9/10 + (9/10 + 9/10^2 + 9/10^3 ... 9/10^(inf-1))/10
    x = 9/10 + (x - 9/10^inf)/10
    x - 9/10 = (x - 9/10^inf)/10
    10x - 9 = x - 9/10^inf
    9x - 9 + 9/10^inf = 0
    9x = 9 - 9/10^inf
    x = 1 - 1/10^inf
    x = 1 - 0.000...1
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