Is 0.999... Really Equal to 1? Exploring the Mathematical Proofs

[steersman]

1/3 is a non-mathematical representation of a potential infinity: 0.333...etc. There is no such thing as an actual infinity. I refer to the Hilbert's hotel paradox. Potential infinities exist in mathematics because it is a theoretical tool.

Let me put it this way:
does 0.9 = 1? No
does 0.99 = 1? No
does 0.999999999999999999 = 1? No

In each case we are getting closer to 1, but will we ever get there? No.

 

[Erazman]

0.9~ is not 1.
If you can think of space as infinately large,
then just imagine the concept of infinately small.
Distance in space can always be cut in half. And in half again.
Our mind just has a hard time comprehending it...

 

[Integral]

There is no such thing as an actual infinity.

Mathematically speaking there is and it is carefully defined. Maybe you should take a few minutes to read some of the posts in this thread.

0.9~ is not 1.

Certainly is! I guess your concept of the universe does not apply to the real number line. Perhaps you should modify your concept of the universe.

 

[Hurkyl]

Let me put it this way:
does 0.9 = 1? No
does 0.99 = 1? No
does 0.999999999999999999 = 1? No

Does 0.9 = 0.9~? No.
Does 0.99 = 0.9~? No.
Does 0.999999999999999999 = 0.9~? No.

In each case, we are getting closer to 0.9~ but will we ever get there? No.

Why do you think your observation has any bearing on whether 0.9~ = 1?


Have either of you, steersman and Erazman, read through this thread?

 

[steersman]

Have either of you, steersman and Erazman, read through this thread?

Do I have to?

Mathematically speaking there is and it is carefully defined.

So what? Mathematics uses potential infinities not actual ones. This experiment...getting to 1 is a problem because in maths potential infinities exist whereas in reality they don't. This experiment is reality based, despite its mathematical content. It involves the quantization of measurement. You need to take it out of context to see it has no meaning. You can divide a ruler into an infinite of potential parts - does this mean that the ruler itself is infinite? No.

 

[e(ho0n3]

Is .4999... + .4999... = 1? I mean, how does one define an addition or any arithmetic process on these kinds of numbers without first changing them into rational form. Are we stuck using the rational form?

 

[jcsd]

Yes 0.4999... + 0.4999... = 1.

I'm not sure what you mean by rational form as any repeating decimal number is rational

 

[selfAdjoint]

Is .4999... + .4999... = 1? I mean, how does one define an addition or any arithmetic process on these kinds of numbers without first changing them into rational form. Are we stuck using the rational form?

You can define them rigorously via modern methods, or by Eudoxos' theory of proportions, which occupies Book X of Euclid's Geometry, together with a completeness axiom.

 

[Hurkyl]

Do I have to?

No; but you may find it enlightening. And it might save others from being irritable that they have to make the same responses over and over.

 

[balrog]

It's sort of like how no matter how many times you cut a number in half it will never equil zero... but it will get pretty damn close... same thing: no matter how many 9s you add to the end of the decimal you will always get closer to one, but you could never ever reach it! Either I'm an idiot, or you will never ever reach it.

 

[AKG]

It's sort of like how no matter how many times you cut a number in half it will never equil zero... but it will get pretty damn close... same thing: no matter how many 9s you add to the end of the decimal you will always get closer to one, but you could never ever reach it! Either I'm an idiot, or you will never ever reach it.

First of all, you aren't approaching 1, one digit at a time. You have an infinite number of digits. You can make sort of an inductive proof that for all natural numbers n, if 0.999... has n digits, it's still not going to be 1. So with that think you'll never reach 1. But the point is that although for no n will 0.999... (with n digits) equal 1, 0.999... doesn't have a natural number of digits, it has infinite digits. Second of all, 0.99... = 1 for the very strange reason that 0.99... is defined as the infinite sum of the terms t_k, where t_k = 9/10^k. Then, this infinite sum is defined as the limit as X approaches infinity of k=1 to k=X of the sum of t_k. It can be proven that this infinite sum cannot be any real number other than 1, and we simply define the sum to be the real number approached as X approaches infinity. It is useful to define this sum as a real number, and we choose the limit, i.e. the only number it can be, to be that real number. I say this is strange because 0.999... is something you don't use past grade 4, and you don't learn infinite sums and limits until high school.

So, in summary, we choose to define an infinite sum to be a real number (assuming it represents a sequence of converging partial sums), and the number we choose is called the limit, and it's a decent choice because we can prove that it can be no other number.

 

[balrog]

sure there are an infinite number of placeholders after the decimal, I accept that, but even infinity isn't enough to make it 1. As the number of nines after the decimal approach infinity the value will come pretty damned close to 1... but there is no end. It's still not one, it's just undeterminably close to one.

 

[Erazman]

Maybe the real debate here is: What's infinity?

The dictionary describes it is "An indefinitely large number or amount"
the term "indefinate" means having NO distinct limits. Undefined. It's simply a term made up by man to describe an indefinate value.

0.99~ is not definate. If we all of a sudden give a DEFINED VALUE to infinite by calling 0.999~ 1 instead of 0.999~, then we have just destroyed the whole concept of infinity, because infinity HAS NO DEFINED VALUE.

Not only that but...

When i look at 0.999~ i see a 0 before the decimals. the 0 means it cannot possibly be 1. If the number is LESS THAN 1, then a 0 will indicate it, in this case it does...

 

[Hurkyl]

same thing: no matter how many 9s you add to the end of the decimal you will always get closer to one, but you could never ever reach it!

Also, no matter how many 9s you add to the end of the decimal, you will always get closer to 0.9~ but you could never reach it!

As the number of nines after the decimal approach infinity the value will come pretty damned close to 1... but there is no end. It's still not one, it's just undeterminably close to one.

Ah, but the number of 9's in 0.9~ is infinite; it is not some nebulous thing that "approaches infinity".

And it is equal to one; the distance between 0.9~ and 1 is zero.

Maybe the real debate here is: What's infinity?

The dictionary describes it is "An indefinitely large number or amount"
the term "indefinate" means having NO distinct limits. Undefined. It's simply a term made up by man to describe an indefinate value.

And mathematicians are somewhat more precise. In particular... it is mathematically specified that each of the places after the decimal place corresponds to a unique positive integer.

When i look at 0.999~ i see a 0 before the decimals. the 0 means it cannot possibly be 1. If the number is LESS THAN 1, then a 0 will indicate it, in this case it does...

And if that's how < worked, you'd be right.

 

[AKG]

sure there are an infinite number of placeholders after the decimal, I accept that, but even infinity isn't enough to make it 1. As the number of nines after the decimal approach infinity the value will come pretty damned close to 1... but there is no end. It's still not one, it's just undeterminably close to one.

Well, then you didn't read what I wrote. For one, the addition of an infinite number of terms is defined as the limit of the sequence of partial sums, this is proven to be 1 in this case. Second of all, you have no reason to believe that it is not 1 if it has an infinite number of digits after the decimal. Because that means adding an infinite number of terms, and that's something different from adding a finite number of terms. In short, you have no reason to say "infinity isn't enough to make it 1." There's a difference between going on interminably/infinitely and actual infinity. For example, dividing a number by 2 infinitely would not get you zero, because you can keep dividing "forever" and not reach zero. But even if you divided forever, you wouldn't have divided an infinite number of times. I suppose the wording's confusing. What does it mean to cut a number in half infinity times? You can continue cutting it in half for infinity or forever, and never reach zero, but you can never have actually cut it an infinity number of times. Similarly, you can add 9's on to the end of 0.9999 forever and never reach 1, but that's different from having an actual infinite number of 9's. If some number x is divded by 2 n times, then it becomes $x/2^n$. Now, what the heck is $x/2^\infty$? We don't have a way to deal with that, we can only calculate limits, any other answer is meaningless (it's not that it's not zero, it's nothing, at not a real number because reals don't deal with infinities, maybe a surreal number though).

 

[Erazman]

Also, no matter how many 9s you add to the end of the decimal, you will always get closer to 0.9~ but you could never reach it!




Ah, but the number of 9's in 0.9~ is infinite; it is not some nebulous thing that "approaches infinity".

And it is equal to one; the distance between 0.9~ and 1 is zero.




And mathematicians are somewhat more precise. In particular... it is mathematically specified that each of the places after the decimal place corresponds to a unique positive integer.




And if that's how < worked, you'd be right.

sorry, LESS THAN 1 AND GREATER THAN ZERO. i messed up. and this "unique positive integer" your talking about is exactly what shows up in the number itself on paper. 0.9999~ shows the 9999~ corresponding to a ZERO. If it was truly 1, then that number before the decimal place would NOT be zero.

 

[Erazman]

Well, then you didn't read what I wrote. For one, the addition of an infinite number of terms is defined as the limit of the sequence of partial sums, this is proven to be 1 in this case. Second of all, you have no reason to believe that it is not 1 if it has an infinite number of digits after the decimal. Because that means adding an infinite number of terms, and that's something different from adding a finite number of terms. In short, you have no reason to say "infinity isn't enough to make it 1." There's a difference between going on interminably/infinitely and actual infinity. For example, dividing a number by 2 infinitely would not get you zero, because you can keep dividing "forever" and not reach zero. But even if you divided forever, you wouldn't have divided an infinite number of times. I suppose the wording's confusing. What does it mean to cut a number in half infinity times? You can continue cutting it in half for infinity or forever, and never reach zero, but you can never have actually cut it an infinity number of times. Similarly, you can add 9's on to the end of 0.9999 forever and never reach 1, but that's different from having an actual infinite number of 9's. If some number x is divded by 2 n times, then it becomes $x/2^n$. Now, what the heck is $x/2^\infty$? We don't have a way to deal with that, we can only calculate limits, any other answer is meaningless (it's not that it's not zero, it's nothing, at not a real number because reals don't deal with infinities, maybe a surreal number though).

so if 1 / 2~ = not a real number, a surreal number
then 0.99~ = not a real number, a surreal number
adding .x9 to .999~ and calling it "1" is the same thing as 1 / 2~ and calling it "0"
right? therefore 0.99~ doesn't equal 1. It doesn't equal anything other than what it is. 0.99~.

 

[AKG]

so if 1 / 2~ = not a real number, a surreal number
then 0.99~ = not a real number, a surreal number

No. 0.999... is defined to be a real number, the limit of the converging sequence of partial sums. $1/2^\infty$ is not defined to be a real number. I suppose it depends on how you interpret the notation. $\sum _{n=1} ^\infty t_n$ looks like nonsense except when you realize it's a compact notation for $\lim _{x \rightarrow \infty} \sum _{n=1} ^x t_n$. Similarly, if you wanted, you could say that $1/2^\infty$ is just compact notation for $\lim _{x \rightarrow \infty} 1/2^x$. Both the sum and the $1/2^\infty$ have no meaning as real numbers unless you clarify that it's shorthand notation for some limit, as the real numbers don't deal with infinities, i.e. $\infty \notin \mathbb{R}$. It might be more sensible to have an alternate definition for the infinite sum and $1/2^\infty$ as surreal numbers because they deal with infinites and infinitessimals, but for now, we have a logically consistent, useful definition for infinite sums as a limit of a converging sequence of partial sums. Perhaps, don't think of it as proving what it means to add an infinite number of terms together, it is simply defining it to be a certain way.

 

[Integral]

When the real line is extented to include infinity you must include definitions for the arithemetic operations. The ususal definition is

$$\frac 1 {\infty} = 0$$
and
$$x^ {\infty} = \infty \ \forall \ x>0 \ x \in \ R$$
so when these definitions are applied you have:

$$\frac 1 {2^ {\infty}} = \frac 1 {\infty} = 0$$

Just for the record, when extending the real line to include infinity the definition is something like.

$$\infty > x \ \forall \ x \ \in R$$

With a similar definition for negitve infinity.

 

[AKG]

When the real line is extented to include infinity you must include definitions for the arithemetic operations.

Affinely Extended Real Numbers
Projectivly Extended Real Number
Pretty interesting, but notice that with these systems, by adding infinities, you lose some of the properties that the real numbers have.

 

[Integral]

You are absolutly correct, the extended Reals are not a field because of the defined properties of infinity.

 

[Hurkyl]

nd this "unique positive integer" your talking about is exactly what shows up in the number itself on paper. 0.9999~ shows the 9999~ corresponding to a ZERO. If it was truly 1, then that number before the decimal place would NOT be zero.

You entirely misunderstood.

The correspondense is that there is a 1st digit, a 2nd digit, a 3rd digit, and so on; there is an n-th digit for each positive integer n.

Furthermore, every digit can be labelled in this way; for every digit to the right of the decimal place, there is a positive integer n for which you can say that it's the n-th digit.

0.9~ is a sequence such that for each integer n, the n-th digit to the right of the decimal point is a 9.


And, as a sequence of digits, 0.9~ is, indeed, inequal to 1. Furthermore, 0.9~ < 1. according to the lexical ordering of sequences of digits. But 0.9~, as a decimal number, is equal to 1..

 

[e(ho0n3]

I don't know if this has been said already, but the reason I think of why a lot of people don't 'buy' that 0.999... = 1 is simply because these two numbers are syntactically different. In other words, if the numbers don't look the same then they are not the same. I think people should just realize that numbers aren't as straightforward as they think they are. So far nobody has given a convincing logical argument that 0.999... and 1 are different numbers so why is this thread getting so large.

 

[Hurkyl]

Yet, oddly, nobody seems to have any trouble accepting that $\frac{4}{3} = \frac{8}{6} = 1\frac{1}{3}$, and most people can understand that $1 = 13 \mod 12$. I still haven't understood why so many have trouble with $0.\bar{9} = 1$.

 

[Erazman]

alright I've changed sides...
makes more sense now

 

[steersman]

Yet, oddly, nobody seems to have any trouble accepting that , and most people can understand that . I still haven't understood why so many have trouble with

1 and 1/3 are different forms of representation. The example you gave is correct, 1/3 is predicated on there is a 1/3 that can be found and definitely measured. It is a potential representation. This is a predicate in mathematics not reality. The case is different with 0.9~=1. The number 1 is a reality representation.

See, the thing is 0.9 maybe the same as 0.9~. It just depends on how far you are willing to measure. So in a sense 0.92 is more than 0.9 in the way you are justifying your case, despite this being just a matter of measurement.

It is not possible to travel at the speed of light, why is this? because you must expend an infinite amount of energy - which is not possible, because actual infinities do not exist. You would need an actual infinity in 0.9~=1.

This is a philosophical problem not a mathematical problem. It is to do with the philosophical meaning of infinity not any mathematical definition of it.

It's like saying: well if this were possible then this would equal 1. But its not possible.

 

[Hurkyl]

Measurement? Reality? Where did you get any of this stuff?


0.9~ does not depend on any choices whatsoever. It has a 9 in the n-th place for ALL positive integers n, and there exist no other places to the right of the decimal places.

It is not formed by starting with 0. and adding 9's one by one.

 

[steersman]

It is not formed by starting with 0. and adding 9's one by one.

Who said it was?

 

[Hurkyl]

You sort of implied it with "See, the thing is 0.9 maybe the same as 0.9~. It just depends on how far you are willing to measure."

 

[steersman]

You sort of implied it with "See, the thing is 0.9 maybe the same as 0.9~. It just depends on how far you are willing to measure."

No no. The point I was making was that every number that isn't whole is already (potentially) infinite in terms of the infinite regress involved when trying to quantize something.

 

[Integral]

1 and 1/3 are different forms of representation. The example you gave is correct, 1/3 is predicated on there is a 1/3 that can be found and definitely measured. It is a potential representation. This is a predicate in mathematics not reality.

The sentence I have bolded makes no sense to me, could you provide a translation into relevant language?

The case is different with 0.9~=1. The number 1 is a reality representation.

Just what is a reality representation? Is there a mathematical definition for that?

See, the thing is 0.9 maybe the same as 0.9~. It just depends on how far you are willing to measure. [\quote]
I don't care how far you measure, as long as we are talking bout Real numbers 0.9 in NEVER = .9~

So in a sense 0.92 is more than 0.9 in the way you are justifying your case, despite this being just a matter of measurement.

Could you please demonstrate a "sense" were 0.92 is NOT greater then 0.9?

It is not possible to travel at the speed of light, why is this? because you must expend an infinite amount of energy - which is not possible, because actual infinities do not exist.

Wrong thread, discuss Relativity and the speed of light in the correct forum. (BTW you are not exactly on the mark with that statement either!) This is logic applied to Math, no Physics needed or wanted.

You would need an actual infinity in 0.9~=1.

Just so happens that Math HAS a definition of infinity and it means that 0.9~ =1. Perhaps if you knew even the basics of Real Analysis you would have known this.

This is a philosophical problem not a mathematical problem. It is to do with the philosophical meaning of infinity not any mathematical definition of it.

You could not be more wrong. This is a property of the Real Number system, This system has been very carefully constructed on well known and understood axioms followed by careful and through proofs of each and every theorem.

It's like saying: well if this were possible then this would equal 1. But its not possible.

News to me, and every mathematician in the world, perhaps you know something we don't or... Just maybe...

You don't know a LOT that Mathematicians DO KNOW.

 

[steersman]

You could not be more wrong. This is a property of the Real Number system, This system has been very carefully constructed on well known and understood axioms followed by careful and through proofs of each and every theorem.

Maths has limited utility in answering this problem because it's answer does not correspond with reality. That's why I call it a philosophical problem. I do concede though that in maths 0.9~=1. But if that's all you care about then you'll never learn anything.

 

[Integral]

The problem has meaning ONLY in Math. Out side of the definitions of math the the string of symbols .999... has no meaning what so ever.

 

[steersman]

Noone said this was a math problem. It could easily be a philosophical problem. Indeed, many philosophers have pondered this very question in different words and symbols

 

[Hurkyl]

Well, the fact that 0.9~ = 1 (and everything about the decimals!) is chosen so that they're a model of the real numbers. (*sigh* "real" was a poor choice of name, but anyways...)

The real numbers are defined to be a complete ordered field.

To put it loosely, "ordered field" simply means that +, -, *, /, and < all work "properly". The definition is "complete" is more difficult, and is not needed for what follows.


The first thing to notice is that there cannot be any numbers between 0.9~ and 1; if I change any of the digits of 0.9~ into something other than a 9, I'll get a smaller number.

The next thing to notice is that, if we assume $0.\bar{9} \neq 1$:

$$0.\bar{9} < \frac{0.\bar{9} + 1}{2} < 1$$

which contradicts the fact that there are no numbers between 0.9~ and 1.


Thus, in order for the decimal numbers to fulfill the purpose for which they are created... that is, to be a model of the real numbers... it cannot be the case that $0.\bar{9} \neq 1$. In other words, 0.9~ = 1 must be true.



There are, indeed, a lot of interesting philosophical questions that relate to mathematics, but they aren't about whether 0.9~ = 1 in the decimal numbers.