# 0.999… revisited

The original question was: 0.999… = 1 does not make sense with respect to my conception of the number line, I do not know much about number classes but for the number line from lets say 0 to 1 there is an infinite number of points, so there is a number right next to 1 we cant write it in entirety because the decimal expansion is infinite but the difference between that 1 and that number is 1*10^(-infinity) (sorry if I am abusing notation). so that number to me should be 0.999… but it is not where am I missing the point?

Physicist: In the language of mathematics there are “dialects” (sets of axioms), and in the most standard, commonly-used dialect you can prove that 0.999… = 1.  The system that’s generally taught now is used because it’s useful (in a lot of profound ways), and in it we can prove that 0.99999… = 1.  If you want to do math where 1/infinity is a definable and non-zero value, you can, but it makes math unnecessarily complicated (for most tasks).  The way the number system is generally taught (at the math-major level, where the differences become important) is that the real numbers are defined such that (very long story short) 1/infinity = 0 and there isn’t a “next number” for any number.  That is, if you think you’ve found a number, x, that’s closer to 1 than any other number, then I can find a number half way between it and 1, (1+x)/2, that’s even closer.  That’s not a trivial statement.  In the system of integer numbers there is a next number; for 3 it’s 4, for 26 it’s 27, etc..  In the system of real numbers every number can be added, subtracted, multiplied, and divided without “leaving” the real numbers.  That leads to the fact that we can squeeze a new number between any two different numbers.  In particular, there’s no greatest number less than one.  If there were, then you couldn’t fit another number between it and one, and that would make it a big weird exception.  Point is: it’s tempting to say that 0.999… is the “first number below 1”, but that’s not a thing.

The term “real numbers” is just a name for a “sand box” of mathematical tools that have become standard because they’re useful.  However!  There are other systems where “very very very slightly less than 1” , or more precisely “less than one, but greater than every number that’s less than one”, makes mathematical sense.  These systems aren’t invalid or wrong, they’re just… not as pretty and fluid as the simple (as it reasonably can be), solid, dull as dishwater, real number system.

It is a fact, immutable and True, that every rook is safe for the next move according to the most widely accepted rules for chess.  In other games that may not be the case.  0.999…=1 in the most widely accepted rules for real numbers.

In the set of “real numbers” (as used today) a number can be defined as the limit of the decimal expansion taken one digit at a time.  For example, the number “2” is {2, 2.0, 2.00, 2.000, …}.  The “square root of 2” is {1, 1.4, 1.41, 1.414, 1.4142, …}.  The number, and everything you might ever want to do with it (as a real number), can be done with this sequence of ever-longer decimals (although, in practice, there are usually more sophisticated methods).

These sequences are “equivalent” and describe the same number if they get (arbitrarily) closer and closer to that same number forever.  Two sequences don’t need to be identical to be equivalent.  The sequences {1, 1.0, 1.00, 1.000, …} and {0, 0.9, 0.99, 0.999, …} both get closer and closer to each other and to the value “1” forever, so they’re equivalent.  In absolutely every way that counts (in terms of the real numbers), the number “0.99999…” and the number “1” or “1.0000…” are exactly the same.

It does seem very bizarre that two numbers that look different can be the same, but there it is.  This is basically the only exception; you can write things like “0.5 = 0.49999…”, but the same thing is going on.

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### 39 Responses to 0.999… revisited

1. Flavian Popa says:

I love this posts about Mathematics. Number theory and topology, one of the best fields if not the very one, in modern Mathematics, since it has numerous applications in Quantum Mechanics…If you could post also about manifolds, it would be terrific!

All my respect for your work and keep it up!

2. The Cool Dude says:

Certainly floor(.9999…)=0.

3. Anthony Rose says:

Interesting. It has always ‘offended’ 🙂 my mathematical sensibilities that two numbers, intuitively different by definition of what writing means (since 0.999… is always less than 1.000… if you stop writing 9s and 0s at any limit of decimal digits), should be regarded as the same. But after reading this today it has just dawned on me that at some point the difference becomes vanishingly small in *reality* – literally vanishingly. For example, if you cut a cake in half, and have 0.4999… of it on one side, then (assuming no crumbs adhere to the knife) you have actually cut it perfectly in half to the very last molecule, which is way bigger than the difference between 0.4999… and 0.5. So in practice 0.4999… is really the same thing as 0.5 because you can’t, in practice, cut real things small enough to see the difference. In fact, because the number is recurring, it can always stay ahead of you, even if you went sub-atomic. Maybe this is a dumb layman’s way of viewing things – probably is. I just thought I’d share my light-bulb moment anyway because if it helps me stay sane, maybe it’ll help someone else.

4. Brendan Hayden says:

If I find a number x = 0.99999999….. so that it is one planck length away from 1 on the number line can you find a number that is closer? remembering that half a planck length is not something that makes sense. Sure you can do it mathematically (the above mentioned (1+x)/2 ) but that is not a valid length in this scenario.
So in any real world scenario a number closest to 1 does exist and if maths doesn’t have to relate to real word scenarios we may as well study unicorn poetry…

5. The Cool Dude says:

That logic only makes sense if you’re using math only to specifically represent distance, which is an infinitesimal sub-sect of applying mathematics to reality.

6. Waytfm says:

@ Brenden Hayden

You’re mixing physical reality with number systems, which doesn’t really work. The real line isn’t something physical. You can’t apply something like the planck length to it.

The real world may be discrete, but the real line certainly isn’t. There is no smallest unit of length on the real number line.

7. Waytfm says:

@ Brenden Hayden

That’s an extremely simplistic and anti-intellectual view on what math should be. You have no way of knowing what topics in math may have applications in the future. Look at applications of group theory in physics, or knot theory’s applications to string theory or biology. Neither one of those fields had those applications when they were first theorized, but they sure turned out to be useful.

Furthermore, I think that the pursuit of knowledge for knowledge’s sake is a perfectly worthwhile goal.

8. Waytfm says:

@ TheCoolDude

Nope. In the real numbers, 0.999… is exactly one. So floor[0.999…] = 1. Remember, the floor function is typically defined as a function from the reals to the integers. Now, if you were fiddling around in Non-Standard Analysis, where you have number systems that can deal with non-zero infinitesimals. You might could define a function from that number system to the integers such that floor[0.999…] = 0. But it’d probably be really messy.

There’s a reason why we stick with the reals for everything.

9. Anshul says:

@Brendan – Real world is quantized. The real number line is not. The planck length is a result of quantum theory, which by definition requires quantization of values. The real number line is an abstract concept and is continuous. So you’re literally comparing opposite concepts.

Also, your point about if math doesn’t reflect the real world, we may as well study unicorn poetry is flawed, because math is an abstraction of the real world, and studying abstractions carefully may sometimes reveal things about the real world that we don’t see in daily observations.

10. Tre says:

Perhaps this can make it even more plausible:

$\;\frac{1}{3}=0.333333333\ldots$
$+\frac{2}{3}=0.666666666\ldots$
————————————–
$\;1=0.999999999\ldots$

11. The Cool Dude says:

Thanks @Waytfm

12. Stan says:

From the point of view of the pure theory o.99999… is not 1. But there is a problem. In reality infinitisimals do not exist. In reality the nines in 0.99999… have to end. Things in reality could be ever changing but they are defined at every single moment.

Rational periodic numbers are periodic only because of the number system we use. We can always find another number system where rational numbers will not be periodic. 1/3 is not 0.3333333… It is more. There will always be residual of 1 which is the cause for the ever repeating threes.

I am not sure if distinction between 1 and 0.999999… is needed in maths. For now – seems not. But in theory there is a difference – purely abstract, because all manifested things are finite.

13. Waytfm says:

@ Stan

“From the point of view of the pure theory o.99999… is not 1.”

This isn’t really the case. “Pure theory” isn’t a very well defined term. There are many different number systems. You can’t just point to “pure theory” to make your claim. In the real numbers 0.999… is exactly equal to one. The only infinitesimal in the real numbers is 0.

Unless you’re doing Non-Standard Analysis, 0.999… is exactly equal to one.

14. kyrbis says:

Pythagoras of Samos resurrections

http://oi62.tinypic.com/km8mf.jpg

15. Brendan Hayden says:

@Waytfm

I certainly didn’t mean to be anti intellectual and I accept that the pursuit of knowledge for knowledge’s sake is perfectly valid.

My unicorn poetry comment was a little tounge in cheek.

I’m just interested in the interaction of pure maths and the real world and what you do when there seems to be a conflict.

16. The Physicist says:

@Brendan Hayden
The real world wins.
But that’s more of a problem for physicists, since math is a strictly theoretical pursuit. Physical reality is the concern of every other kind of scientist.

17. Anthony Rose says:

Oh boy… NOW I get it! We don’t have to apply it to the real world, so we define it whichever way we like. In this instance, we have DEFINED 0.999… = 1.0. In other systems, we could define it – just like the Mathematician said (how could I have missed this?!) – as “less than one, but greater than every other number that is less than one”. If we wanted to. But we don’t. When it comes to applying maths to the real world, we just need to think about whether our definitions work for reality or not.

18. Mark Schaal says:

@Anthony

Not really. No mathematician would define 0.999… = 1.0. They define what the real numbers are and how they can be manipulated, and as a consequence of those definitions we can prove that 0.999… = 1.

There are lots and lots and lots of ways of writing the real number 1 (an infinite number of them). Some are more obvious (1, 1.0, 1.0000, 2/2, 9-8, 1+0, 0+1) and some are less obvious (-8 – (-(-(-9))), 0.999…, sin(pi/2), 1/2 + 1/4 + 1/8…), but they are all equally valid expressions for the same number.

I think it is a combination of our brains not working well with infinity (and why would they?) and using base-10 that makes this one particular representation cause confusion. Would writing it as 9/10 + 9/100 + 9/1000… make any difference?

Even in number systems with infinitesimals, 0.999… = 1. They do have numbers that are “less than one, but greater than every [real] number that is less than one”, but they would be written differently.

19. Stan says:

As I remember the real numbers include sq., cube, forth … n-th roots of numbers. N-th root (n->infinity) of any positive number is 1, if we accept that 1/inf=0 (or that 0.99999…. and 1.0000….1 = 1). Now let’s just try to invert the calculations. It’s impossible. Example: inf.-th root of 0.5 = 1. But 1 to the power of infinity is 1 and not 0.5. So there are infinitisimals not equal to zero even in the real numbers. For me the only reason o.99999… to be equal to one is because it is more convenient practicaly.

20. The Cool Dude says:

In the most simple terms possible, the abstraction we’re speaking of is equal to
lim(n→∞) of the sum(9*10^-i) from i=1→n.
Basically, .9+.09+.009+.0009, forever.
Using calculus and geometric series, you can easily show that the limit is 1, the limit being the thing it never reaches but gets infinitely close to. The sum of that series for any finite number of terms never reaches 1, but the limit is exactly equal to 1, because that’s what a limit is, the thing something never reaches but gets very close to. The difference can also be shown by using the floor function, such that, if you take the floor of the limit of the sum, it’s 1, but if you take the limit of the floor of the sum, it’s always zero. Hence why it isn’t a matter of definition, it’s a matter of thinking you can reverse the order of the limit and some other function.

@Stan
The conundrum you’re considering is a result of the fact that while lim(n→∞) of c^(1/n) is always 1, lim(n→∞) of 1^n is also always 1, and it comes from the order of the limits. If you operate one limit after another, those two statements will always be true, but if you do them at the same time, the result could be anything.
Since 1=n^(1/∞), if you raise 1^∞, it’s the same as n^(∞/∞). Since ∞/∞ is indeterminate, it produces a universal set, thus 1^∞ is technically anything and everything, so long as you do both limits at the same time. That is, lim(n→∞) of c^(n/f(n)), such that lim(n→∞) of f(n) is infinite as well. The result will depend on how much faster or slower f(n) raises to infinity in comparison to ‘n’.
In Calculus, there are various cases in which something like this happens, and it’s part of solving limits. It usually happens whenever you can show that the solution to a function at a certain point must, in some variation, contain the entity 0/0. Other common variations of this entity that show that a limit can’t be directly evaluated include 0^0, 1^∞, ln(0)*0, ∞*ln(1), (1/0)^(1/0), ln(∞)*0.

21. Stan says:

Using powers and radicals you can not get from 1 a different number. n^(inf/inf)=1^inf only if there is difference between n^(1/inf) and 1. That infinitesimal difference raised to an infinite power will give us an indeterminate result other than one. In other words, if 1=0.999… , 1^inf=n^(inf/inf), but in the same time there has to be a difference between 1 and 0.999… for that to happen. It’s a contradiction for me. On the other hand if 1 is not equal to 0.999…, n^(1/inf), (0<n<1) will be 0.999… or 1-1/inf, but not one, and one will be only a limit. 1^(inf/inf) will be 1 and not n^(inf/inf). And in this case I see no contradictions.

22. The Cool Dude says:

@Stan
I think you mixed them up. ∞/∞ is already indeterminate, so f(∞/∞) is also indeterminate so long as f(x) isn’t constant, and in this case, f(x)=n^x.

1^∞ is n^(∞/∞) only if there ISN’T a difference between n^(1/∞) and 1: Raise both sides to ∞, n^(∞/∞)=1^∞. Since there is no difference, 1^∞ is indeterminate.

Back to limits, if you have a function f(x)= g(x)^h(x), such that g(∞)=1, and h(∞)=∞, then the limit can not be directly evaluated, because 1^∞ is indeterminate. A good example of this would be the limit definition of ‘e’.
e=lim(x→∞) of (1+1/x)^x
As x approaches infinity, 1+1/x approaches 1, thus the function approaches 1^∞. If you multiply the exponent by a constant ‘c’, it changes to 1^(c*∞), but if c is positive and nonzero, it still evaluates to 1^∞ at x=∞, but the actual limit is e^c. This can only happen because the actual evaluation of 1^∞ is indeterminate.

23. Stan says:

Yes, but you have to prove that 1+1/inf is one. This is just what we are discussing here – is 1+1/inf equal to one.

And I have a question, because I’ve read several explanation why 1^inf is indeterminate and everywhere 1 is shown as function in one way or another (above it is 1+1/x). Why is that and is it needed?

Yes, infinity is not a number in the way the finite numbers are. But in 1^inf case we have a well defined action (1*1) which is repeated infinite times. As the result of the action is always one, and nobody ever witnessed anything else, I think a proof is needed, that the infinite repeating of the action can result in something different from one and the reason for the difference has to be explained. Where does it comes from?

I know that inf/inf is indeterminate because there are different infinity sets. But sometimes we have more information. I’ve read some explanations of why is 1^inf indeterminate like: 1^inf=(2/2)^inf=(2^inf)/(2^inf)=inf/inf, which is indeterminate. But is it? Since it is the same infinity we apply to the numerator and the denumerator? Yes, infinity is not a well defined number, but in this case we have additional information that infinity in the numerator is exactly the same infinity in the denominator. We can even use it as argument that 1^inf=1

24. The Cool Dude says:

@Stan
You kinda lost me there, I’m not sure what subject you’re actually talking about right now.
If you want an informal proof that 1^∞ is indeterminate, though, I can give you one. If you use infinity as though it were a number, you’re essentially agreeing to the statement that
lim(n→∞) of f(n)=f(∞), for all f(x).
Given that basic definition, there are more than enough anomalies to say that when infinity appears twice in a single function, the result could be practically anything, simply determined by how you take each limit. Aside from that, however, I can use the above definition on the function |1/x|, where x→0. It can be shown (easily) that:
lim(x→0) of |1/x|=∞ from all directions.
However, the ratio between ‘1/x’ and |1/x| is equal to the sign of x:
1/x=|1/x|*sign(1/x)
And when you combine those two things, you find that when used as if infinity could be a number, it is true that:
1/0=∞*sign(1/0)
Or, more usefully put:
|1/0|=∞
This is a pretty interesting and seemingly logical statement on its own, which I think is pretty neat. To expand this to 1^∞, you just do three simple replaces, and simplify:
1^∞=
(Replace ∞ with |1/0|)
1^|1/0|=
(Replace 1 with n^0)
(n^0)^|1/0|=
n^(0*|1/0|)=
(Replace x with |x|*sign(x), where x=0)
n^(sign(0)*|0|*|1/0|)=
n^(sign(0)*|0/0|)
Since any function which operates on the universal set must also equate to the universal set, 1^∞ equates to the universal set, which essentially means that it’s indeterminate.

Your reasoning is not completely flawed though. The same reasoning can be used on 0^∞ and it is not incorrect. A simple explanation for why this happens is that, if you take ANY number vaguely around zero, and raise it to the power of infinity, it still ends up equaling zero. However, if you take a number vaguely around one, and raise it to the power of infinity, for numbers of a finite difference, it either limits to zero or infinity, but for numbers infinitesimally close to one, it could be anything. To demonstrate what I mean, I’ll use a graphic: http://prntscr.com/4t4yjm
The RGB line represents the function f(x)=x^n, (n→∞). When n is even, the function limits to the blue and green lines, and when n is odd, it limits to the red and green lines. The vertical lines are at x=1, and as you can see, 1^∞ and (-1)^∞ limits to all numbers via this graph. You can’t easily prove that directly, so instead, think of the inverse graph, g(x)=x^(1/n), (n→∞). For all positive x, while n is odd, g(x) limits to 1, and for even n, g(x) is both positive AND negative. For negative x, while n is odd, g(x) limits to -1, and while n is even, g(x) is imaginary. Since ∞ is both/neither even and/or odd, both cases for classification of n are true. I won’t get too much into complex numbers, but if you were to plot the function in the complex plane, including ALL possible solutions, then the inverse function, at n→∞, would have g(x)=e^(iθ) for all x AND all θ. (Where |e^(iθ)|=1, for reference)
Since f(g(x))=x for any x (By definition of g):
f(g(x))=f(e^(iθ));
f(e^(iθ))=x
Thus, for any number ‘Y’ that has an absolute value of one, Y^∞=x, for all x.

25. Stan says:

I’m talking about 1^inf. Because if it is 1 (exactly and only) as a consequence 1 will not equal 0.9999… I asked why we have to take limit of (1+1/x)^x, x->inf, since we have one exactly raised to the power of infinity. Limit of 1^x, x->inf is one. Why do mathematicians represent the base as function (1 as 1+1/x, x->inf), since it is not. This changes entirely the subject of the problem and the result is not valid for the original problem. It is only my opinion of course.

26. Mark Schaal says:

@Stan, I’ll take a crack at an explanation.

Keep in mind that infinity is not a number. When talking about infinity in a numerical context, we are always talking about limits and the behavior of something as it approaches a limit. Infinity is not a number, very very important!

When talking about 1^∞ it is not a number but an expression called an “indeterminate form” and represents how such expressions behave as they approach a limit. The more complete way of writing it out would be:
f(x)^g(x)
where
f(x) -> 1 and g(x) -> ∞ in the limit as x -> something
but that’s a lot of words so we usually use the shorthand abbreviation.

One specific example of f(x) would be the constant expression 1, so f(x) = 1, and in that specific case as you observe the overall expression will also be 1.

I’m not clear on why you think this affects the original topic. 1 = 0.999… = (0.999…)^2 = (0.999…)^ 3 etc. Raising 0.999… to a power doesn’t suddenly make it less than one.

In your first post you seem to be concerned with something like “0.5^(1/∞) = 1, but algebraically raise both sides to ∞ then you get 0.5 = 1^∞ which is bogus.” The problem with that is that infinity is not a number and you cannot do algebra with it. For a simpler example consider ∞ – ∞ (another indefinite form). If you could do algebra with infinity, this would equal zero. However it is easy to make this equal to any number at all. Let’s pick 13 for fun.
Let f(x) = x + 13, g(x) = x.
Then as x -> ∞, f(x) -> ∞ and g(x) -> ∞ s
So f(x) – g(x) is of the form ∞ – ∞.
But if we do the math, f(x) – g(x) = x + 13 – x = 13, a constant expression, and clearly the value of 13 as x->∞ is going to be 13 and not zero. Doing algebra with infinity just doesn’t work.

27. The Cool Dude says:

@Stan
You are 100% right that 1^n as (n→∞) is 1; if 1 is a constant with respect to n, and n is an integer, 1^n as (n→∞) is exclusively 1. The discrepancy comes from when you treat ∞ as a number and a constant, because now you have to define the unit infinity, and how it works with respect to finite numbers. Is it even? Is it odd? Is it divisible by five? Is it divisible by all integers? Does that mean it’s divisible by its self plus one? Is it divisible by twenty three? I hope so, twenty three is a cool number. But anyway, If and ONLY if ∞ is an integer can 1^∞ exclusively be 1. In this way, you have defined the specific infinity you are talking about, and can be different from other infinite quantities by degree, ratio, AND finite difference. If ∞ is different from ∞+.5, then it must also be different from ∞+1, and so on, though that presents an additional problem, that you need more information to define infinity, since 1^∞=1 only shows that ∞ is an integer. In short, by deciding that ∞ is a number and a constant, you’ve now got to come up with a specific method of measure for what the “unit” infinity is, but you can’t compare an infinity to finite numbers without using other infinities or infinitesimals, all of which are composed of that unit infinity, which means you can’t define the unit infinity without using its self, which defies definition. In that way, you can’t particular say that infinities are numbers like finites are, since you can only compare an infinity to a function of its self, and nothing else.

It IS good to notice the differences though; that will surely make you a good mathematician.

28. Stan says:

The question “is infinity an integer?” is interesting. But what’s the integer? We can get 0.1 as a base unit and still have an integer in the case of 15.6 for example. Even if infinity is irrational there still in theory has to exist a base unit, if the existance of the irrational number (composed of base units) is a fact. But even if infinity is not an integer there are at maximum 2 options: 1)1^inf = 1 or 2) 1^inf =+/-1. If we work with complex numbers we will have n outcomes for 1^(1/n), but for real numbers they are one or two, depending on is n odd or even. If n is irrational we’ll not know if -1 is possible outcome, but still we’ll know that 1 is for sure. But main thing is that result other of +/-1 is impossible, and -1 is possible only if infinity is not an integer, irrational in the part after the decimal point and under the condition it is not compound of any kind of base units. Sad thing is we can’t have all that information and will never know is -1 is a possible outcome. But this is not important. The important thing is 1^inf can never be let’s say 0.4. Which mean that 0.4^(1/inf) can never be 1. And that means there is a difference between one and infinitely close to one.

By the way 0.9999… is irrational and 1 is rational. It has to be that way, because if 0.9999… was rational 1/inf would also have been rational. And the last is impossible.

29. Mark Schaal says:

‘The question “is infinity an integer?” is interesting.’
No, it isn’t. Infinity is not a number, it is a statement about limits. Since infinity isn’t even a number, it obviously isn’t an integer. It also obviously isn’t rational or irrational.

“By the way 0.9999… is irrational”
No, it isn’t. 0.999… is 1, 1 is rational, therefore 0.999… is rational.

“1/inf would also have been rational. And the last is impossible.”
No, not impossible. 1/inf -> 0 and 0 is rational.

Stan, sorry to be so blunt, but you are just stringing together false statements and using those as justifications for other false statements. That’s not going to get you anywhere mathematically. Infinity is not a number, so any reasoning starting with a stated or unstated “assume infinity is a number…” is unreliable.

30. Stan says:

Actually the problem is not mathematical, but here is not the place for other analyses. Whether we’ll have infinitely repeating nines or infinetely repeating 1*1 – where is the difference?

It is truth that we have to be very careful when operating with infinity. For example:
(1-1/inf)*10 – (1-1/inf) = 10-1/inf – 1 + 1/inf = 9
but we can have
(1-1/inf)*10 – (1-1/inf) = (1-1/inf)(10-1)=9-1/inf
Yes infinity is not a real number. But what’s above is the well known proof:
x=0.999999…
10*x=9.99999….
10*x-x=9.999…. – 0.9999…
9*x=9
->0.9999… = 1
I just represented 0.9999… as 1-1/inf. And my calculations above are not less correct than 9.9999…-0.9999…=9. (I’m not saying they are correct)

31. betaneptune says:

TheCoolDude asks about the floor function. He says that floor(*0.999…)=0. Excellent question. Actually it equals 1, because 0.999…=1 and floor(1)=1.

But consider the sequence

floor(0.9), floor(0.99), floor(0.999) –> 0

So the limit of the floors is not equal to the floor of the limit. Basically it matters how you take the limit. A bit subtle, but it makes all the difference.

I suppose you could rewrite the limit of the floors as

floor(0.9), floor(0.99), floor(0.999), . . .

So if the ellipsis is inside the parentheses, the answer is one. If it’s outside, as in the sequence above, the answer is zero.

32. David Martin says:

I would be interested in any comments, particularly from the ‘Ask a’ guy, who knows this general area better than I do. But I think I can show that the two numbers are different.

If instead of approaching 1, you approach zero, there are plenty of equations that have a series of connected results for 0.0001, 0.0000001, 0.0000000001, and so on. It seems that with an infinite one, ie 0.000—- etc — 0001, that is still part of that series.

The equation I’m thinking about is simply an angle in degrees divided by its sine, so
x/(sin x). The smaller the angle, the closer it approaches 1 radian in degrees, or
180/[pi]. and at x = 0.000—- etc — 0001, it is exactly equal to 180/[pi].

So the equation has an output value that approaches a particular known number when you go along that series, so this known number is actually at the infinite one at the end of the series, 0.000—- etc — 0001.

But at x = zero, the equation gives zero, a completely different output number. Therefore these two numbers are different. Any thoughts? Thanks.

33. Andres says:

Easy proof!
x = 0.99999… =>
10x = 9.99999… =>
10x – x = 9.99999 – 0.99999 =>
9x = 9 =>
x=1
this 1 = 0.9999…

34. Drak says:

So what’s the difference between [-1;1] and (-1;1) intervals if 0,99… equals 1?
Or what’s the point of asymptote, why is it so complicated if -1+0,99… = 0?

35. Drak says:

Also what if you round these numbers down to the nearest natural number?

36. Angel says:

@Drak: the difference between [-1,1] and (-1,1) is that (-1,1) DOES NOT include 0.999….. while [-1,1] does. The point of asymptote is less than 0.999….. . Remember that 1 = 0.9999……. .Thus, you cannot round 0.999…… because you cannot round 1. The nearest natural number to 1 is 1 itself.

37. betaneptune says:

If 0.333… is 1/3, then 0.999… is 1. A repeating decimal is defined as the limit of the sequence of adding one more decimal to the number for each step. It’s as simple as that. My analysis book says it’s defined as the least upper bound of the sequence. As best as I can tell, that’s the same thing, at least in this case.

It’s a matter of definition.

You could say that the sequence 0.9, 0.99, 0.999, … never gets to 1, but that doesn’t matter because when you add the … the definition tells you to take the limit. You could say that about any series. E.g., 1 + x + x^2/2! + x^3/3! + . . . never makes it to e, but the sum of the series _is_ e, nonetheless. So the definition is a useful one.

One more word about 0.999…: No matter where you stop, you can always get closer by adding another digit. So you can’t say it’s less than 1.

If you don’t believe 0.999… = 1, then you’ll have to mess up all other series, too. And that would be bad.

38. Howard Ludwig says:

Two points:

1. 0.999…, where the … means to repeat 9’s forever, is a rational number. A number in standard decimal form is rational if and only if the standard decimal form terminates on the right or it has a fixed block of digits that repeats forever. In the case of 0.999…, there is a repeating block that is 1 digit in length, namely 9.

2. I suspect many people get confused on this subject because they are taught, correctly, that for every number written in standard decimal form, perhaps an unending string of digits to the right of the decimal marker, whether with or without a repeating fixed block of digits (or, equivalently, every real number) corresponds to a point on the real number line, and that point is unique. Teachers will also say, again correctly, that every point on the real number line has a standard decimal form representation. [Remember that in mathematics the expression “has a” means “has at least one” rather than exactly one.] Sometimes the teacher will extend this statement–this time incorrectly–or students will assume incorrectly that uniqueness works in this direction as well, namely that each point on the real number line has one and only one standard decimal form representation. Some real numbers, like 1/7 and √2 do indeed have only one standard decimal representation, but some, like the number one, have more than one standard decimal representation (1; 1.0; 1.00; 1.000; …; 1.000…; 0.999…, i.e., any finite count of 0’s to the right of the decimal marker in “1.”, an infinite string of 0’s to the right of the decimal marker in “1.”, or an infinite string of 9’s after the decimal marker in “0.”). It can be very difficult for people to overcome that assumption if they have believed it for a while: How can every value represented in standard decimal form correspond to one and only one point on the real number line, while every point on the real number line has at least one standard decimal form representation, not necessarily exactly one? It is a reasonable question to ponder, but it is a provable fact for the standard system of real numbers. Mathematics is often not intuitive, so do not try to impose false properties in mathematics to make the concepts more intuitive and palatable, as doing so will lead to even more unpleasant conclusions downstream than 0.999… = 1 is.

39. netzweltler says:

If 0.999… stands for infinitely many commands