Is 0.9999... really the same as 1?

What is the basis for reason? And mathematics?

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Is 0.9999... really the same as 1?

Yes 0.9999... = 1
5
31%
No 0.9999... is slightly less than 1
7
44%
Other
4
25%
 
Total votes: 16

wtf
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Re: Is 0.9999... really the same as 1?

Post by wtf » Wed Mar 08, 2017 3:43 am

Philosophy Explorer wrote: Here's a good question to ask. There are different foundations proposed for mathematics. Which one would you support (if any?)
PhilX
It's not necessary to quote my entire shaggy dog story just to type in one sentence. It wasn't THAT good :-) Feel free to delete it to make the page more readable.

Why do I have to support or oppose a particular foundation? It's not a war. The foundations are not the math. If I didn't make that clear then I failed to communicate.

Philosophy Explorer
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Re: Is 0.9999... really the same as 1?

Post by Philosophy Explorer » Wed Mar 08, 2017 3:54 am

wtf wrote:
Philosophy Explorer wrote: Here's a good question to ask. There are different foundations proposed for mathematics. Which one would you support (if any?)
PhilX
It's not necessary to quote my entire shaggy dog story just to type in one sentence. It wasn't THAT good :-) Feel free to delete it to make the page more readable.

Why do I have to support or oppose a particular foundation? It's not a war. The foundations are not the math. If I didn't make that clear then I failed to communicate.
The reason why the foundations are important is that mathematicians are looking for the right road to follow. Follow the wrong road and mathematicians will say you may go astray. Another way of putting it is that mathematicians are looking for the most consistent system to follow. Godel shocked the math world showing that math may be consistent with certain axioms (later followed by Richard Cohen who showed that math may be just as consistent without those axioms).

PhilX

wtf
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Re: Is 0.9999... really the same as 1?

Post by wtf » Wed Mar 08, 2017 4:54 am

Philosophy Explorer wrote: The reason why the foundations are important is that mathematicians are looking for the right road to follow. Follow the wrong road and mathematicians will say you may go astray.
That's true. But the collective opinion of the right road is a matter of fashion and historical contingency. Math has gone in and out of periods of rigor versus loosey-goosy anything goes. When standards of rigor were low there was much creativity ending in eventually getting stuck in the logic problems. Too much rigor is stifling too, it emphasizes technique at the expense of ideas.

Newton knew calculus worked and his writings show that he himself understood the logical problems in his definition of the derivative. Those problems took 200 years to sort out. Should math have put calculus on hold for 200 years till they could nail down the right axioms for it? Of course not.

Just ask a physicist. They think mathematicians have gone off the deep end with abstraction. The physicists think the mathematicians have already gone astray a long time ago!
Philosophy Explorer wrote: Another way of putting it is that mathematicians are looking for the most consistent system to follow. Godel shocked the math world showing that math may be consistent with certain axioms (later followed by Richard Cohen who showed that math may be just as consistent without those axioms).
Paul Cohen might have had something to to with it too :-)

I don't agree that the average working mathematician is worried about a consistent foundation. They're not. If set theory were discovered to be inconsistent tomorrow morning nobody would care. The group theorists and algebraic geometers and differential topologists would just keep doing their work, and the foundationalists would get busy patching the problems.

A group theorist studies the class of mathematical objects that satisfy the axioms of group theory. One foundation is as good as another. Only the defining properties of groups matter. They're also called axioms but here an axiom is a defining property and not a "statement accepted without proof."

In fact if we turn it around and reframe set theory as the study of mathematical objects that satisfy the axioms of sets, we get a beautiful area of math without any of the ontological or foundational baggage.

That's the modern view as I understand it.

marsh8472
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Re: Is 0.9999... really the same as 1?

Post by marsh8472 » Wed Mar 08, 2017 8:39 am

Given that 0.9999... can be rewritten as lim x->infinity of f(x) where f(x) = 1 - 10^(-x) and the definition of limit being for every real number ε > 0, there exists a natural number N such that for all x > N, we have | f(x) − L | < ε. By the definition of limit and by the definition of "0.9999..." expressed as a limit, 0.9999... = 1.

But the limit implies that the series approaches 1 without being 1. We know that f(x) < 1 for any real number and f(x) will never equal 1 for any real number or ∀x∈ℝ f(x)<1 . The Archimedean property in real numbers may negate the point about any perceived infinitesimal non-zero difference between 1 and 0.9999... but the Archimedean property also implies that there are no infinitely large numbers as well, f(infinity) = 1 does not work out because infinity is not a real number. If I were to say X is an infinitesimal amount less than Y and then ask which is larger X or Y, it does not make intuitive sense to say X and Y are both equal because infintesimal numbers do not exist in the real number system and therefore their infinitesimal difference does not exist either. In reality if an infintesimal difference does exist, their sameness would not exist either by law of non-contradiction.

I see a separation between math and linguistics here that's causing a problem too. In common language when people ask whether 0.9999... equals 1 they're asking whether they're precisely equal or not. Asking whether this series 0.9, 0.99, 0.999, 0.9999, ... approaches 1 is supposed to be a different question than asking whether 0.99999... = 1. It's similar to asking whether a line will ever intersect its asymptote. In that case people are inquiring about the line, not the asymptote. The line is supposed to never have the same value as the asymptote by definition. Likewise, a series that converges to a value is different than the value it's converging to by the definitions in our language making 0.9999... and 1 not equal by that definition of equal.

wtf
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Re: Is 0.9999... really the same as 1?

Post by wtf » Wed Mar 08, 2017 8:08 pm

marsh8472 wrote:Given that 0.9999... can be rewritten as lim x->infinity of f(x) where f(x) = 1 - 10^(-x)
It's closer to the definition to say .999... = lim(n->infinity) of sum(1 to n) 9/10^n, but the difference isn't important for our purposes. I'll accept what you wrote.
marsh8472 wrote: and the definition of limit being for every real number ε > 0, there exists a natural number N such that for all x > N, we have | f(x) − L | < ε. By the definition of limit and by the definition of "0.9999..." expressed as a limit, 0.9999... = 1.
Yes. If you know this, I don't understand the rest of your post. You agree that by the definition of the limit of a sequence, .999... = 1 and the equal sign is what an equal sign always is: a statement of equality. The two different expressions point to the same abstract mathematical object.

Or if you are a formalist and don't believe in abstract mathematical objects, .999... = 1 by a formal derivation, the key step of which is the exact definition you just gave.

Honestly I could not have explained this any better than you just did.

[Edit -- You are slightly confusing the definition of the limit of a sequence with the limit of a function, but no great harm is done to the discussion].
marsh8472 wrote: But the limit implies that the series approaches 1 without being 1.
If you said this and hadn't just written down the formal definition of the limit of a sequence, I'd explain to you the definition. But since you yourself just wrote down the definition, I don't understand how you can now pretend to deny that it says exactly what it says.
marsh8472 wrote: We know that f(x) < 1 for any real number and f(x) will never equal 1 for any real number or ∀x∈ℝ f(x)<1 .
True. That's why limits are cool. We formalize the notion of "arbitrarily close." But again I am baffled that I have to explain this to someone who already understands that! Perhaps I'm totally missing your point. You already gave the perfect answer as to why .999... = 1, namely that it follows directly from the formal definition of a limit. You can explain this just as well as I can.
marsh8472 wrote: The Archimedean property in real numbers may negate the point about any perceived infinitesimal non-zero difference between 1 and 0.9999... but the Archimedean property also implies that there are no infinitely large numbers as well, f(infinity) = 1 does not work out because infinity is not a real number. If I were to say X is an infinitesimal amount less than Y and then ask which is larger X or Y, it does not make intuitive sense to say X and Y are both equal because infintesimal numbers do not exist in the real number system and therefore their infinitesimal difference does not exist either. In reality if an infintesimal difference does exist, their sameness would not exist either by law of non-contradiction.
Maybe, I don't know. I don't think so. In any model of the first-order theory of the real numbers, .999... = 1 is a theorem. In other words if you want to work in the hyperreals of nonstandard analysis, a model of the first-order theory of the reals that contains infinitesimals and infinite quantities, it is still true that .999... = 1. This is a point not often appreciated. Infinitesimals do not help the argument against .999... = 1 in the least.
marsh8472 wrote: I see a separation between math and linguistics here that's causing a problem too.
In my own experience, the biggest doubters of .999... = 1 are not linguists or average everyday people, who frankly have no interest in the subject at all. The biggest doubters are usually career engineers. People who mastered calculus but didn't take Real Analysis. People highly skilled in working with numbers in a practical sense, with years of experience using tight approximations, but without knowledge of the underlying formal theory of the real numbers.

That's my impression. The average person on the street wouldn't care one way or another.
marsh8472 wrote: In common language when people ask whether 0.9999... equals 1 they're asking whether they're precisely equal or not. Asking whether this series 0.9, 0.99, 0.999, 0.9999, ... approaches 1 is supposed to be a different question than asking whether 0.99999... = 1. It's similar to asking whether a line will ever intersect its asymptote. In that case people are inquiring about the line, not the asymptote. The line is supposed to never have the same value as the asymptote by definition. Likewise, a series that converges to a value is different than the value it's converging to by the definitions in our language making 0.9999... and 1 not equal by that definition of equal.
What does common language have to do with a technical field? In common language, heavy objects fall down. In the language of physics, it's all Higgs bosons and the relativistic theory of gravity. It matters not what the person on the street says, since their opinion is irrelevant. I wish math education taught people to think more precisely about the real numbers, but I'm not in charge of the curriculum.

But the bottom line is that I simply don't understand anything you wrote. If you hadn't given a perfect account of the definition of a limit, I'd explain it to you. But since you understand it just as well as I do, I don't understand where you're coming from.

Perhaps you are trying to explain why some people are confused. If so, I agree. We can send someone to calculus class and they can go on to a career as an engineer or physicist and never know the definition of a real number or a limit. I think that's sad but the education establishment is what it is.

On the other hand, perhaps you're saying that you KNOW that .999... = 1 but you don't believe it. In which case I'd say just put on your formalist hat and forget naive intuitions that lead you astray.

marsh8472
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Re: Is 0.9999... really the same as 1?

Post by marsh8472 » Wed Mar 08, 2017 9:53 pm

wtf wrote:
marsh8472 wrote:Given that 0.9999... can be rewritten as lim x->infinity of f(x) where f(x) = 1 - 10^(-x)
It's closer to the definition to say .999... = lim(n->infinity) of sum(1 to n) 9/10^n, but the difference isn't important for our purposes. I'll accept what you wrote.
marsh8472 wrote: and the definition of limit being for every real number ε > 0, there exists a natural number N such that for all x > N, we have | f(x) − L | < ε. By the definition of limit and by the definition of "0.9999..." expressed as a limit, 0.9999... = 1.
Yes. If you know this, I don't understand the rest of your post. You agree that by the definition of the limit of a sequence, .999... = 1 and the equal sign is what an equal sign always is: a statement of equality. The two different expressions point to the same abstract mathematical object.

Or if you are a formalist and don't believe in abstract mathematical objects, .999... = 1 by a formal derivation, the key step of which is the exact definition you just gave.

Honestly I could not have explained this any better than you just did.

[Edit -- You are slightly confusing the definition of the limit of a sequence with the limit of a function, but no great harm is done to the discussion].
marsh8472 wrote: But the limit implies that the series approaches 1 without being 1.
If you said this and hadn't just written down the formal definition of the limit of a sequence, I'd explain to you the definition. But since you yourself just wrote down the definition, I don't understand how you can now pretend to deny that it says exactly what it says.
marsh8472 wrote: We know that f(x) < 1 for any real number and f(x) will never equal 1 for any real number or ∀x∈ℝ f(x)<1 .
True. That's why limits are cool. We formalize the notion of "arbitrarily close." But again I am baffled that I have to explain this to someone who already understands that! Perhaps I'm totally missing your point. You already gave the perfect answer as to why .999... = 1, namely that it follows directly from the formal definition of a limit. You can explain this just as well as I can.
marsh8472 wrote: The Archimedean property in real numbers may negate the point about any perceived infinitesimal non-zero difference between 1 and 0.9999... but the Archimedean property also implies that there are no infinitely large numbers as well, f(infinity) = 1 does not work out because infinity is not a real number. If I were to say X is an infinitesimal amount less than Y and then ask which is larger X or Y, it does not make intuitive sense to say X and Y are both equal because infintesimal numbers do not exist in the real number system and therefore their infinitesimal difference does not exist either. In reality if an infintesimal difference does exist, their sameness would not exist either by law of non-contradiction.
Maybe, I don't know. I don't think so. In any model of the first-order theory of the real numbers, .999... = 1 is a theorem. In other words if you want to work in the hyperreals of nonstandard analysis, a model of the first-order theory of the reals that contains infinitesimals and infinite quantities, it is still true that .999... = 1. This is a point not often appreciated. Infinitesimals do not help the argument against .999... = 1 in the least.
marsh8472 wrote: I see a separation between math and linguistics here that's causing a problem too.
In my own experience, the biggest doubters of .999... = 1 are not linguists or average everyday people, who frankly have no interest in the subject at all. The biggest doubters are usually career engineers. People who mastered calculus but didn't take Real Analysis. People highly skilled in working with numbers in a practical sense, with years of experience using tight approximations, but without knowledge of the underlying formal theory of the real numbers.

That's my impression. The average person on the street wouldn't care one way or another.
marsh8472 wrote: In common language when people ask whether 0.9999... equals 1 they're asking whether they're precisely equal or not. Asking whether this series 0.9, 0.99, 0.999, 0.9999, ... approaches 1 is supposed to be a different question than asking whether 0.99999... = 1. It's similar to asking whether a line will ever intersect its asymptote. In that case people are inquiring about the line, not the asymptote. The line is supposed to never have the same value as the asymptote by definition. Likewise, a series that converges to a value is different than the value it's converging to by the definitions in our language making 0.9999... and 1 not equal by that definition of equal.
What does common language have to do with a technical field? In common language, heavy objects fall down. In the language of physics, it's all Higgs bosons and the relativistic theory of gravity. It matters not what the person on the street says, since their opinion is irrelevant. I wish math education taught people to think more precisely about the real numbers, but I'm not in charge of the curriculum.

But the bottom line is that I simply don't understand anything you wrote. If you hadn't given a perfect account of the definition of a limit, I'd explain it to you. But since you understand it just as well as I do, I don't understand where you're coming from.

Perhaps you are trying to explain why some people are confused. If so, I agree. We can send someone to calculus class and they can go on to a career as an engineer or physicist and never know the definition of a real number or a limit. I think that's sad but the education establishment is what it is.

On the other hand, perhaps you're saying that you KNOW that .999... = 1 but you don't believe it. In which case I'd say just put on your formalist hat and forget naive intuitions that lead you astray.
I'm on the fence about whether they are equal, not equal, or I just do not know. I never messed with hyper reals but the answer here explains how its unclear whether 0.9999...=1 if you want to evaluate their explanation https://www.quora.com/Is-0-999-dots-1-in-the-hyperreals I kind of fall in the middle of your spectrum too. I have a degree in applied math, took all the math courses and got as far as intro to real analysis which I do remember going through the proofs for calculus but it's been a while.

From a limits interpretation I do agree that 0.9999... = 1. But if we interpret this problem as a word problem from others who asked it might not accept the definition of "= 1" to be "approaches 1". I tried looking for a formal definition of the equal sign to help resolve this as well as the common usage of the term equals.

There could be a difference between the spirit of the question and literal interpretation of the question. Also the reality of numbers vs the real number system may not correlate with this problem. Maybe an axiom is wrong. Is an infinitesimal amount the same thing as 0? For instance if I were to select a random real number between 0 and 1 in a uniform distribution, I can show using limits that the probability of any real number being selected equals 0 and yet a real number is still selected. This would suggest that equals is actually being used as an approximate comparison instead of a precise comparison between two values.

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Greta
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Re: Is 0.9999... really the same as 1?

Post by Greta » Wed Mar 08, 2017 11:07 pm

With help earlier in the thread I see the situation as this:

0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

is definitely less than 1.

However 0.999... must be equal to 1 because the 9s are infinite. Infinity is not only probably not present in the physical world (unless nothingness is infinite) but infinity is also beyond mathematics, so the standard mathematical logic of "it must be just a little less than one" cannot work. Any such number that is less than one cannot be 0.999.... It instead would be a finite decimal, as above, with a finite number of repeated 9s.

wtf
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Re: Is 0.9999... really the same as 1?

Post by wtf » Thu Mar 09, 2017 12:55 am

Greta wrote: However 0.999... must be equal to 1 because the 9s are infinite.
I disagree with this logic.

First, you are looking for a reason. But there is no reason. .999... = 1 is a legal position in the formal game of the modern theory of the real numbers. End of story.

But now you say (I assume) that you are wondering about the "moral" reason. That's what we're really interested in.

But the claim that .999... = 1 is true "because" the number of 9's is infinite, can be falsified.

The Quora link marsh8472 gave is a clear and simple technical overview of the question of whether .999... = 1 in the hyperreals. I'm going to talk about this a little when I respond to marsh8472's post. Normally I don't like Quora, but in this case there is an answer by Alon Amit, who in my opinion is God when it comes to expositing math on Quora. In this case he did a beautiful job of showing an example where we can interpret a particular hyperreal notation as containing infinitely many 9's, yet the number so expressed would not be 1. The problem is that it's not "ALL" nines, which turns out to not be a number in the hyperreals. I will say something about all this later, it's not actually that difficult.

But the bottom line here is that we have to make the case that .999... = 1 in the hyperreals. It's not given just because "there are infinitely many 9's."

To sum up:

* There is no "reason." .999... = 1. Two centuries of formalization conspired to make it true. Whether it's true in some moral sense is a totally different question. But there is no reason for anything in math. Of course we are always trying to find deeper levels of explanation, perhaps that's what you mean by reason. If so ...

* Even if we did try to find the underlying property that makes .999... = 1, having an infinite number of 9's doesn't help. In that Quora link you will see a reference to a construction of a hyperreal number in a special hyperreal notation, such that there are infinitely many 9's yet the expression is strictly less than 1. Even in this model .999... = 1, because you need a different notation to get this example to work. But it involves infinitely many 9's.

Does this make sense? You want to "understand "it. John von Neumann once said, "In math you don't understand things. You just get used to them." .999... = 1 is exactly the kind of thing he's talking about.


Greta wrote: Infinity is not only probably not present in the physical world (unless nothingness is infinite)
Right. And -- yet another often underappreciated point -- even if the world turns out to be infinite, there is no reason to think 20th century set theory applies to it! That's the thing. Even if the universe does turn out to be infinite, there's no evidence that our fancy mathematical theory of infinity refers to it! More evidence for my thesis that .999... = 1 is purely a question of mathematics and isn't even a well-defined question outside of formal math.

Greta wrote: but infinity is also beyond mathematics,
Math has a very well developed and beautiful theory of the infinite. Modern set theory is really wild, it's all about "large cardinals" that are so big their existence can't be proved in standard set theory. So they keep adding all these crazy new kinds of infinity and then figuring out whether their new theory is still consistent. Or at least as consistent as the previous one, since consistency proofs are always relative.

We agree that math doesn't have a theory of the "worldly" infinite. The type of infinity that exists in physics ... if such a thing is ever proven to exist.

But math isn't designed to study the world. That's the job of physics. Math is only designed to study patterns, real or not. So you can't fault math for not answering the questions that the physicists are supposed to answer.
Greta wrote: so the standard mathematical logic of "it must be just a little less than one" cannot work.
The idea of limits formalizes the intuitive notion of "must be juuuuust a little bit less" than 1. And the idea does work. It's been very successful.
Greta wrote: Any such number that is less than one cannot be 0.999.... It instead would be a finite decimal, as above, with a finite number of repeated 9s.
Yes that's the proof! That's more precise than "infinitely many 9's." But if that's what you meant, then you are absolutely right! If we claim that .999... is strictly less than 1, then it's less than 1 by some number ε > 0. But we can always find a finite string of 9's that's closer to 1 than ε, showing that we were mistaken when we said that .999... differed from 1 by ε. It's closer than that. It's closer than ANY ε you could name. Therefore by definition it's equal to 1.

This is the way in which the theory of limits finesses the question of infinity. We replace vague notions of infinity with the precise logic of "epsilonics" as this technique is often called. (That's a term with a little snark attached, it's not an official word in math).

Is any of this helpful or am I just on some kind of obscure hobby horse about all this?

marsh8472
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Re: Is 0.9999... really the same as 1?

Post by marsh8472 » Thu Mar 09, 2017 2:53 am

Greta wrote:With help earlier in the thread I see the situation as this:

0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

is definitely less than 1.

However 0.999... must be equal to 1 because the 9s are infinite. Infinity is not only probably not present in the physical world (unless nothingness is infinite) but infinity is also beyond mathematics, so the standard mathematical logic of "it must be just a little less than one" cannot work. Any such number that is less than one cannot be 0.999.... It instead would be a finite decimal, as above, with a finite number of repeated 9s.
I can prove they're not equal with bases:
Assume we're working in base 16
0.99999... in base 16 = 9/16 + 9/256 + 9/4096 + .... in base 10 = (9/16) / ( 1 - 1/16 ) in base 10 = (9/16) / ( 15/ 16) in base 10 = 144 / 240 = 0.6 in base 10 which does not equal 1 in base 10
There's some interpretation involved when answering the question particularly pinning down what equals and real numbers are.

I'm still trying to figure out whether 0.9999... is a real number or not. I know the order properties of real numbers states adding two real numbers or multiplying two real numbers produces a real number. In that case:

9/10 is real, 9/10 + 9/100 + 9/1000 is real, but I don't know about an infinite series like 9/10 + 9/100 + 9/1000 + 9/1000 + .... this doesn't seem to help since we know an infinite series like 1+2+3+4+... does not produce a real number either.

Any theorems of axioms that I've looked at to determine whether 0.9999... = 1 or whether 0.9999... is a real number I run into circularity issues like here

https://en.wikipedia.org/wiki/Construct ... _sequences
Synthetic approach

The synthetic approach axiomatically defines the real number system as a complete ordered field. Precisely, this means the following. A model for the real number system consists of a set R, two distinct elements 0 and 1 of R, two binary operations + and × on R (called addition and multiplication, respectively), and a binary relation ≤ on R, satisfying the following properties.

(R, +, ×) forms a field. In other words,
For all x, y, and z in R, x + (y + z) = (x + y) + z and x(yz) = (xy)z. (associativity of addition and multiplication)
For all x and y in R, x + y = y + x and xy = yx. (commutativity of addition and multiplication)
For all x, y, and z in R, x(y + z) = (xy) + (xz). (distributivity of multiplication over addition)
For all x in R, x + 0 = x. (existence of additive identity)
0 is not equal to 1, and for all x in R, x*1 = x. (existence of multiplicative identity)
For every x in R, there exists an element −x in R, such that x + (−x) = 0. (existence of additive inverses)
For every x ≠ 0 in R, there exists an element x^−1 in R, such that x*x^−1 = 1. (existence of multiplicative inverses)
(R, ≤) forms a totally ordered set. In other words,
For all x in R, x ≤ x. (reflexivity)
For all x and y in R, if x ≤ y and y ≤ x, then x = y. (antisymmetry)
For all x, y, and z in R, if x ≤ y and y ≤ z, then x ≤ z. (transitivity)
For all x and y in R, x ≤ y or y ≤ x. (totalness)
The field operations + and × on R are compatible with the order ≤. In other words,
For all x, y and z in R, if x ≤ y, then x + z ≤ y + z. (preservation of order under addition)
For all x and y in R, if 0 ≤ x and 0 ≤ y, then 0 ≤ xy (preservation of order under multiplication)
The order ≤ is complete in the following sense: every non-empty subset of R bounded above has a least upper bound. In other words,
If A is a non-empty subset of R, and if A has an upper bound, then A has a least upper bound u, such that for every upper bound v of A, u ≤ v.

The rational numbers Q satisfy the first three axioms (i.e. Q is totally ordered field) but Q does not satisfy axiom 4. So axiom 4, which requires the order to be Dedekind-complete, is crucial. Axiom 4 implies the Archimedean property. Several models for axioms 1-4 are given below. Any two models for axioms 1-4 are isomorphic, and so up to isomorphism, there is only one complete ordered Archimedean field.
Using "For all x and y in R, if x ≤ y and y ≤ x, then x = y. (antisymmetry)" as an example
if 0.9999... <= 1 and 1 <= 0.9999... then 0.9999... = 1. 0.9999... <= 1 is true but proving 1<=0.9999... would involve proving 0.9999... = 1 first, it's circular

My real analysis book tries to incorporate the concept of limits with the equal sign called infimum or greatest lower bound without saying whether they are the same thing as far as I can see.

it says:

If S is bounded below, then w is said to be an infimum (or greatest lower bound) of S if it satisfies the conditions:
1) w is a lower bound of S, and
2) if t is any lower bound S, then t<=w

For example if I picked the subset S of all reals greater than 0, every negative would qualify as a lower bound but 0 would be the greatest lower bound even though that bound is not part of the subset. The same would be true if subset S of all reals greater than or equal to 0.

I found this theorem a Corollary of the Archimedean Property

If S := {1/n : n∈N}, then inf S = 0

Which at first I thought could have been helpful to prove whether an 0.00000...1 = 0 or 1 - 9/10 - 9/100 - 9/1000 + ... = 0, a similar problem to the 0.99999... problem. But because "2) if t is any lower bound S, then t<=w" both 0.00000...1 < 0 or 0.00000...1 = 0 could be true and still not contradicts this Corollary

Another question that I thought could distinguish 0.9999... and 1 is the rounding down function

Does floor(0.9999...) = 0 or floor(0.9999...) = 1? It's pretty much the same dilemma.

Here's the definition of the floor function:
floor(x) = max{m∈Z | m<=x}

but floor(0.99999...) cannot be evaluated without first knowing whether 0.99999...>=1 aka whether 0.99999...=1 beforehand. I suspect I'll run into this same problem of not being able to pin down "=" in real numbers and both interpretations may work within the reals unless someone can point out something that I haven't looked at?

wtf
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Re: Is 0.9999... really the same as 1?

Post by wtf » Thu Mar 09, 2017 3:12 am

marsh8472 wrote: I'm still trying to figure out whether 0.9999... is a real number or not.
A nonempty set of reals bounded above has a least upper bound. That's the defining property of the real numbers, which differentiates it from dense ordered fields such as the rational numbers.

https://en.wikipedia.org/wiki/Completen ... al_numbers

They told you this in real analysis class :-)

Example: The set of rational numbers whose square is less than 2 does NOT have a least upper bound in the rationals. That shows the rationals are not complete. They don't have the least upper bound property.

Example: The set of real numbers {.9, .99, .999, ...} is nonempty and bounded above (by 47, say). Therefore it has a least upper bound, which we can prove must be exactly 1. That's because the real numbers are complete. There are no "holes" in them, whereas the rationals have a hole at sqrt(2), for example.

Note in passing that an ordered field containing infinitesimals cannot be complete. (Can be proved but I'm not proving it here). The hyperreals are shot full of holes hence make a poor model of anybody's idea of a continuum.

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Re: Is 0.9999... really the same as 1?

Post by marsh8472 » Thu Mar 09, 2017 4:58 am

wtf wrote:
Greta wrote: Any such number that is less than one cannot be 0.999.... It instead would be a finite decimal, as above, with a finite number of repeated 9s.
Yes that's the proof! That's more precise than "infinitely many 9's." But if that's what you meant, then you are absolutely right! If we claim that .999... is strictly less than 1, then it's less than 1 by some number ε > 0. But we can always find a finite string of 9's that's closer to 1 than ε, showing that we were mistaken when we said that .999... differed from 1 by ε. It's closer than that. It's closer than ANY ε you could name. Therefore by definition it's equal to 1.

This is the way in which the theory of limits finesses the question of infinity. We replace vague notions of infinity with the precise logic of "epsilonics" as this technique is often called. (That's a term with a little snark attached, it's not an official word in math).
This last sentence is the part I'm not certain of.

Whether f(x)=y is defined as: for every real number ε > 0, there exists a natural number N such that for all x > N, we have | f(x) − y | < ε
or whether this more narrow definition of x=y should be the correct one: There exists a natural number N such that for all x > N, we have | f(x) − y | = 0

Also 1 - 0.99999... = 1-0.9-0.09-0.009 - ... = 0.00000...1

Based on this part
"If we claim that .999... is strictly less than 1, then it's less than 1 by some number ε > 0"
If we say 0.00000...1>0 and 0.999... is less than 1 because 1 and 0.999... differ by this amount then it could be strictly less than.

But I assume the reply to this would either be 0.00000...1 is not greater than 0 for similar reasons as why 0.99999... does not less than 1. But reapplying this strictly less than definition to 0 and 0.00000...1 results in circularity. Otherwise another reply could be x would have to be less than y by a real number for x<y and 0.0000...1 is not a real number based on the Archimedean property. Then if 0.0000...1 is not a real number why shouldn't 0.99999... be excluded from real numbers as well?

It would seem like if we're working within a set of integers and ask whether 3 = pi, 3 and pi do differ by an amount but not an integer amount. Since pi-3 is an irrational number and irrational numbers are not integers therefore 3=pi. That's how the 0.9999...=1 proof appears.

wtf wrote:
marsh8472 wrote: I'm still trying to figure out whether 0.9999... is a real number or not.
A nonempty set of reals bounded above has a least upper bound. That's the defining property of the real numbers, which differentiates it from dense ordered fields such as the rational numbers.

https://en.wikipedia.org/wiki/Completen ... al_numbers

They told you this in real analysis class :-)

Example: The set of rational numbers whose square is less than 2 does NOT have a least upper bound in the rationals. That shows the rationals are not complete. They don't have the least upper bound property.

Example: The set of real numbers {.9, .99, .999, ...} is nonempty and bounded above (by 47, say). Therefore it has a least upper bound, which we can prove must be exactly 1. That's because the real numbers are complete. There are no "holes" in them, whereas the rationals have a hole at sqrt(2), for example.

Note in passing that an ordered field containing infinitesimals cannot be complete. (Can be proved but I'm not proving it here). The hyperreals are shot full of holes hence make a poor model of anybody's idea of a continuum.
Yes but that does not prove to me that 0.9999... = 1. 0.9999... is not element of this set {0.9, 0.99, 0.999, ...}, per completeness it has a least upper bound within the set of real numbers as defined here https://en.wikipedia.org/wiki/Infimum_and_supremum

An upper bound b of S is called a supremum (or least upper bound, or join) of S if
for all upper bounds z of S in P, z ≥ b (b is less than any other upper bound).

If 0.9999... is not a real number then 1 can still be the least upper bound without contradicting the idea of 0.9999... < 1 while {0.9, 0.99, 0.999, ...} converges arbitrarily closer to both 0.9999... and 1. This would actually be true regardless of whether 0.9999... = 1 as I see it.

The problem with the "Non-uniqueness of decimal representation" section in https://en.wikipedia.org/wiki/Decimal is that it's using the limit of an infinite geometric series to prove that every real number (except 0) has a second decimal representation of infinite 9's. I'm still trying to justify that it's appropriate to include that within the definition of "=".

Is does say no gaps are implied in reals as an axiom but really because another axiom denies their possible existence.
https://en.wikipedia.org/wiki/Completen ... al_numbers
Intuitively, completeness implies that there are not any “gaps” (in Dedekind's terminology) or “missing points” in the real number line.
https://en.wikipedia.org/wiki/Archimedean_property
Roughly speaking, it is the property of having no infinitely large or infinitely small elements
If consecutive points do not exist on a number line, could this not also imply that gaps exist somewhere between the nearest non-consecutive points?

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Re: Is 0.9999... really the same as 1?

Post by Greta » Thu Mar 09, 2017 5:11 am

marsh8472 wrote:
Greta wrote:With help earlier in the thread I see the situation as this:

0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

is definitely less than 1.

However 0.999... must be equal to 1 because the 9s are infinite. Infinity is not only probably not present in the physical world (unless nothingness is infinite) but infinity is also beyond mathematics, so the standard mathematical logic of "it must be just a little less than one" cannot work. Any such number that is less than one cannot be 0.999.... It instead would be a finite decimal, as above, with a finite number of repeated 9s.
I can prove they're not equal with bases:
Assume we're working in base 16
0.99999... in base 16 = 9/16 + 9/256 + 9/4096 + .... in base 10 = (9/16) / ( 1 - 1/16 ) in base 10 = (9/16) / ( 15/ 16) in base 10 = 144 / 240 = 0.6 in base 10 which does not equal 1 in base 10
There's some interpretation involved when answering the question particularly pinning down what equals and real numbers are.

I'm still trying to figure out whether 0.9999... is a real number or not. I know the order properties of real numbers states adding two real numbers or multiplying two real numbers produces a real number. In that case:

9/10 is real, 9/10 + 9/100 + 9/1000 is real, but I don't know about an infinite series like 9/10 + 9/100 + 9/1000 + 9/1000 + .... this doesn't seem to help since we know an infinite series like 1+2+3+4+... does not produce a real number either.

Any theorems of axioms that I've looked at to determine whether 0.9999... = 1 or whether 0.9999... is a real number I run into circularity issues like here
Wtf has been doing a job on us and I think you are the final straw that exploded my brain to the point of mixed metaphor!

This has brought back some memories of school confusion with repeated decimals due to practicality and rounding up with one third, or 0.67 as rounded up at school being "twice" one third or 0.333..., 3 x 0.333... equalling 0.999..... It all came across as not-quite-logical to me and so I did as wtf spoke about earlier - just accepted it for what it was.

Anyway, that's the view from the "ground floor". By all means continue moving up towards the penthouse, gents :)

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Re: Is 0.9999... really the same as 1?

Post by wtf » Thu Mar 09, 2017 5:29 am

Greta wrote: Wtf has been doing a job on us
I'm sorry, how should I take that remark? You think I'm bs-ing or flimflamming in some way?

I know the mathematical side of this very well and I'm doing my best to explain it. If you think I'm "doing a job" on you I can see I should save my keystrokes.

Is this a language issue or perhaps my misunderstanding your meaning?

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Re: Is 0.9999... really the same as 1?

Post by marsh8472 » Thu Mar 09, 2017 5:41 am

I read this over, they pretty much said all the things I've said already but probably explain it better https://betterexplained.com/articles/a- ... er-0-999-1

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Re: Is 0.9999... really the same as 1?

Post by marsh8472 » Thu Mar 09, 2017 5:52 am

wtf wrote:
Greta wrote: Wtf has been doing a job on us
I'm sorry, how should I take that remark? You think I'm bs-ing or flimflamming in some way?

I know the mathematical side of this very well and I'm doing my best to explain it. If you think I'm "doing a job" on you I can see I should save my keystrokes.

Is this a language issue or perhaps my misunderstanding your meaning?
I've understood everything you've said. I think he's just saying it's interesting. Within real numbers and with limits 0.9999... = 1. But on another level is 0.9999...=1 true too? Using criteria like these https://en.wikipedia.org/wiki/Criteria_of_truth

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