Is 0.999... Truly Equal to 1?

[TENYEARS]

I don't care what proof you come up with.

1 - I take a geometric object a cicle or a square and divide it into three parts and then add the decimal values the number is .999999 but is 1 relative to the object as a whole. This is correct for it is the totality of the object.

2 - I take the number .9999... out of the blue with no reference to a geometric representation of an object this is not 1. If I take a geometric reference to the universe, divide it into three equadistant rays starting from a central point and extending into infinity running a wall of ray for the length of extension so you have three defined parts, make that a decimal value of .333... added together, then I will equate .99999... to one with repsect to the universe.

If there is no geometric reference, and it is just a number, it is not equal to 1.

 

[NateTG]

Really, this seems like a funky philosophy of math question but, there is an obvious subset of \Re that is isomorphic to the integers with the appropriate operations. As long as you think of it more as an instatiation of the integer 1 rather than as the only integer 1, you should be fine.

I have only seen two contructions of R (Dedekind cuts & Cauchy sequences) and both construct R from Q. And Q is defined by Z and Z by N. So I would think that is rigorous logical foundation, but what do I know?

You can also construct the real numbers from things like the set of all countable sequences of zeroes and ones that do not end in reapeating 1's, or something, but a construction like that has a PITA factor while Cauchy sequences and Dedekind cuts can readily be shown to have the desired properties. I bring this up because it allows for 0.\bar{1}_2 <binary notation>, not to be a real number.

I expect that, the (mistaken) notion that 0.\bar{9} and 1 are distinct is a result of the mistaken assumption that decimal representations are unique.

 

[honestrosewater]

I don't care what proof you come up with.

I think you are giving ahkron too much credit.
If ahkron has seen further it is by standing on the shoulders of Giants.

Now, if you cut a giant into 3 parts... only kidding, in good fun

Okay, now I can't help myself ;)

"If I have not seen as far as others, it is because giants were standing on my shoulders." -- HalAbelson

"In the sciences, we are now uniquely privileged to sit side by side with the giants on whose shoulders we stand." -- GeraldHolton?

"If I have not seen as far as others, it is because I was standing in the footprints of giants"

"If I have seen farther than others, it is because I was standing on a really big heap of midgets." -- EricDrexler (Nice for those of us who believe the inspiration of giants isn't the only engine of progress.)

"If I have seen further than others, it is because I was surrounded by dwarves." -- attributed to MurrayGellMann?, possibly maliciously.

"I cannot see very far, because my eyes are full of midgets."

 

[honestrosewater]

You can also construct the real numbers from things like the set of all countable sequences of zeroes and ones that do not end in reapeating 1's, or something, but a construction like that has a PITA factor while Cauchy sequences and Dedekind cuts can readily be shown to have the desired properties. I bring this up because it allows for 0.\bar{1}_2 <binary notation>, not to be a real number.

Great, now I'm confused too Is there a quick way to explain how that construction proceeds? Oh, countable is a clue methinks. No, the set is countable? Or the sequences are countable? Yeah, What is this PITA factor you speak of? (I get the PITA, but what is it?)

 

[Hurkyl]

Basically, you define all the arithmetic operations by the method of elementary school arithmetic, but the trick is that you have to perform addition from left to right. The "PITA" factor is in cases where 'ambiguous'
whether you should have a carry or a borrow when doing an operation.

E.G. when adding 0.1100... and 0.011000..., you can "look ahead" to see that the second place to the right of the decimal point generates a carry, and it's propagated through the next place, so you can set the one's digit to be a 1, and then so on.

However, when adding 0.101010... and 0.010101..., there's nothing to say that there should or should not be a carry. Thus, you make a definition; you either say that in this situation you will always consider there to be a carry, never consider there to be a carry, or define both options as being equal.

 

[honestrosewater]

However, when adding 0.101010... and 0.010101..., there's nothing to say that there should or should not be a carry. Thus, you make a definition; you either say that in this situation you will always consider there to be a carry, never consider there to be a carry, or define both options as being equal.

1) carry-> 1.000...
2) no carry-> 0.111...
3) equal-> 1.000...=0.111...

Yes? No? So why can .111... not be real? I am still missing something; hopefully I will see clearer after I some shuteye.

Happy thoughts
Rachel

EDIT- Oh duh- if you decide to carry, you cannot ever get the "noncarry" number, and vice versa.

 

[Swigs]

My understanding of infinite numbers is that you can not treat them like any other number. Like what 'tenyear" was talking about.

Infinite numbers are just a mathimatical idea and don't really exist in the real universe as we know it. I thought that is why "limits" were created anyway.

 

[mikesvenson]

infinity, as I've been thinking about it, is whithin every real number as a result of mathamatical processes. e.g, divide to infinity, add to infinity, subtract to infinity, ...etc. Even whithin 0, there is infinity.

Infinity is the true nature of all things from greatly massive to micro. "Limits", are just easier ways of looking at things and are nothing more than a generalization of reality.

 

[lvlastermind]

infinity, as I've been thinking about it, is whithin every real number as a result of mathamatical processes. e.g, divide to infinity, add to infinity, subtract to infinity, ...etc. Even whithin 0, there is infinity.

every real number has an infinite number of digits but no real number has an infinite magnitude. e.g. ...000000001.0000000...

 

[mikesvenson]

it has an infinite divisible magnatude

 

[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&gt;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 &gt; 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.