Poll: 0.999... = 1

[Gabanuka]

Ugh, maths. But I'm contracted to say my piece so: Most maths people with normal jobs do in there lives 0.999 will equal 1. Very few people will have to go further then that.

 

[Redingold]

I know there is no end to infinity...that's why I called it theoretical 'end'

There's no theoretical end to infinity, either. Infinity is infinite, it has no end. At all. It might have a beginning, but never an end. It might be countable or uncountable, but it never, ever ends. Ever.
There is never a zero, there are only 9s. When you shift the digits left, it's still just 9s going on forever. If there's ever a shortage of digits, you could just shift it left and mine the integer part for a nine, then repeat (forever!).

Math is so simple, yet so easily misunderstood.

No...sorry, but you're wrong. You cannot shift the digits in any number and just call it a day. Infinite or not. If you take 0.9999... and multiply it with 10, there is a 0 at the end, not a 9. Regardless of how many 9s you write in that number. If you want to prove me wrong, you would have to actually write that number with infinite 9s, and even if you can pull that one of, I still assume that there would be a 0 at the end when you multiply it with 10...

Actually, there really is no such thing as infinity. It is a theoretical concept, but it doesn't exist. Every number has an end. Always.

Wrong. Consider multiplying 0.999... by 10. The second decimal place becomes the first decimal place, the third decimal place becomes the second decimal place and so on. The n+1th decimal place becomes the nth decimal. However, the n+1th decimal place is a 9, because every decimal place is a 9. Thus, when we've multiplied out number by 10, every nth decimal place will be a 9.

No...you're understanding infinity incorrectly. There is no such thing. You can write as many 9s as you want (let's call that number n), the n+1th decimal will not be a 9, since there is no n+1th decimal.

Little example: Let n be 2

0.99 * 10 = 9.90 not 9.99

It is the same thing if you assume 'infinite' 9s...although, as I said, there is no such thing. If you want to do maths with infinity, you cannot use traditional maths....they are not meant to be used with that concept. If you do, you can prove errors like 1 = 0 or 0.9999... = 1, since they have no means to correctly symbolize infinity.

There is an n + 1th decimal, because there are an infinite number of decimal places, all of which are 9. Take a number n. The number n + 1 exists, because there is an infinite amount of numbers. Consider then nth decimal place. The n + 1th decimal place exists, because there are an infinite number of decimal places.

http://en.wikipedia.org/wiki/Hilbert's_hotel

Read that and come back when you understand.

Do you understand the difference between countable infinite and uncountable? If you do come back...(Hint: irrational numbers are not countable...)

Also, don't try to prove me wrong with theoretical Gedankenexperiments...I have stated multiple times that the theory of infinity is perfectly fine, but using traditional maths with it and expecting reasonable outcomes is wishful thinking...

The number of decimal places is a countable infinity.

Okok...I'll give you that...there I was wrong...but my interjection regarding gedankenexperiments still stands

Don't know if you read this in my previous edit, so...

shifting the brackets makes an enormous difference.

(3+5)*7 = 56

3+(5*7) = 38

You see?

Do you know the difference between multiplying and adding?

(3 + 5) - 7 = 3 + (5 - 7)

Why can't I use thought experiments? All of advanced mathematics is essentially thought experiments.

As for shifting the brackets, your hypothesis that shifting them never makes a difference to the answer must be wrong, because you've derived something paradoxical from it (proof by reductio ad absurdum).

 

[Piflik]

I didn't say it never makes a difference. But in this case it doesn't (as long as you keep infinity in mind, or else there will be a -1 at the end, but as multiple people here said, there is no end to infinity, so there is no -1 and 1 = 0).

 

[Redingold]

I didn't say it never makes a difference. But in this case it doesn't (as long as you keep infinity in mind, or else there will be a -1 at the end, but as multiple people here said, there is no end to infinity, so there is no -1 and 1 = 0).

But it does make a difference, because 1 + (-1 + 1) + (-1 + 1) + ... = 1

Infinite addition is not necessarily commutative.

 

[Piflik]

I didn't say it never makes a difference. But in this case it doesn't (as long as you keep infinity in mind, or else there will be a -1 at the end, but as multiple people here said, there is no end to infinity, so there is no -1 and 1 = 0).

But it does make a difference, because 1 + (-1 + 1) + (-1 + 1) + ... = 1

Infinite addition is not necessarily commutative.

So you agree that traditional maths is not necessarily valid when dealing with infinity, because in traditional maths additions are always commutative...so for all intends and purposes you agree with me

 

[Athinira]

It isn't equivalent to zero. That was my point, infinity is the concept of "as close as you can get without being it"... Kinda. 1x10^(-infinity) would equal 0.(an infinite number of zero's)1.
Basically, the difference between 0.9 recurring and 1 is infinitesimally, immeasurably small. But they are not equal.

Wrong. "as close as" doesn't exist, because you can always get closer.

"as close as" implies that we, at some point, stop and examine the number as represented, meaning that we use a finite, not an infinite, amount of decimals. Thats the thing with infinity: You can never stop, and you can always get closer.

Or to put it another way: As close as possible actually means that the two things are the same You can't get any "closer" than "equivalent", at least not in the system of Real Numbers, because that number system doesn't allow infinitesimal values [http://en.wikipedia.org/wiki/Infinitesimal].

 

[quhouv vhqwiufv qwiu]

I will post a proof after a few replies, but other people are free to post proofs before that.

We had this thread a while ago, it all hinges on whether or not you consider the numbers absolute.

.999=/=1, if it were 1 it would be represented as 1, representing as .999 implies a difference to 1, but a difference that is beyond your means or comprehension to measure. Being unable to measure a difference/quantity does not mean it does not exist, so it's different.

Also that proof is flawed, 9x would be 8.999 and 9x1 would be 9. 10x would be 9 and 10x1 would be 10, proves they are not equal as the difference grows with multiplication. Were they equal it would not.

Maths is not open to interpretation and opinion, and nothing hinges on how I consider numbers.

 

[SextusMaximus]

How about this one:

The difference between 0.9999.... and 1.0 is 0.0000... and since the difference between them is 0, the two numbers must be equal.

*EPIC FACEPALM*

Unless you were being sarcastic... the difference is 0.0000000...(infinity)...001

 

[Piflik]

You can't get any "closer" than "equivalent", at least not in the system of Real Numbers, because that number system doesn't allow infinitesimal values [http://en.wikipedia.org/wiki/Infinitesimal].

If you want to allow infinity, you also have to allow infinitesimal values, since that is nothing other than 1/infinity.

 

[grammarye]

I didn't say it never makes a difference. But in this case it doesn't (as long as you keep infinity in mind, or else there will be a -1 at the end, but as multiple people here said, there is no end to infinity, so there is no -1 and 1 = 0).

For those not keeping up, this is Piflik's proof:

1-1 = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1) = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+..... = 0
1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+..... = 0
1 = 0

This is quite distinct, conceptually, from 0.999.... That is a real number, expressed in a single instant in time to be infinite. The above is an infinite equation. It too exists in a single instant of time, infinitely (the time reference is not directly relevant, but people seem to keep insisting on dragging time to evaluate stuff into the discussion).

You could probably choose to express 0.999... as an infinite equation if you so chose, but not many people would - something like:

0.9 + 0.09 + 0.009 + 0.(0...)9 + ... = 0.999...
(Edit: This is most likely not correct mathematical notation, but you get the idea I trust - repeat an ever decreasing addition of 0.9 to 10^-n curse the lack of superscripts, for which algebra would of course be more suited)

However, back to the aforementioned infinite equation involving 1 & -1. You can't evaluate or operate on only part of an infinite equation. You must evaluate or operate on all of it or none of it.

This is why line four is wrong. There is no mathematical operation that 'just moves a bracket'. You must apply an operation to both sides of any expression when you mutate that expression and it must be a valid mathematical operation.

In simpler terms, this proof is arguing that the following is possible:

1-1=0
1+(-1+1) = 0

This is simply nonsense algebraically speaking. That +1 has come from nowhere in the original expression, or put another way, there is a -1 missing, if we are speaking infinitely. It should go:

1-1=0
1+(-1+1-1) = 0

which you will notice evaluates to 0. That you can repeat to infinity. If you add on '+ (1-1)' to each side over and over, that is indeed valid, and evaluates to zero. What you can't do is add 1 on one side and not add a corresponding -1.

As I am one of the people stating there is no end to infinity and you can't insert a 0 on the end of 0.999..., allow me to clarify again - the above is an equation. 0.999... is not an equation. It could be presented as such, but in that case must be operated on as an equation, by equation rules.

 

[quhouv vhqwiufv qwiu]

I stand by the opinion that there's no such thing as infinity. Maybe for time, maybe for the multiverse, but not for matter/energy. Therefore, any mathematic equation that uses infinity is not valid in my eyes.

1) Time as the 4th dimention in higher dimensional physics is not necessarily infinite.
2) Can you show that there is a final decimal place to 0.999...? If not you are going to have to accept that some things involve the concept of infinity.

 

[Piflik]

I didn't say it never makes a difference. But in this case it doesn't (as long as you keep infinity in mind, or else there will be a -1 at the end, but as multiple people here said, there is no end to infinity, so there is no -1 and 1 = 0).

For those not keeping up, this is Piflik's proof:

1-1 = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1) = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+..... = 0
1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+..... = 0
1 = 0

In simpler terms, this proof is arguing that the following is possible:

1-1=0
1+(-1+1) = 0

This is simply nonsense algebraically speaking. That +1 has come from nowhere in the original expression, or put another way, there is a -1 missing, if we are speaking infinitely. It should go:

1-1=0
1+(-1+1-1) = 0

which you will notice evaluates to 0. That you can repeat to infinity. If you add on '+ (1-1)' to each side over and over, that is indeed valid, and evaluates to zero. What you can't do is add 1 on one side and not add a corresponding -1.

As I am one of the people stating there is no end to infinity and you can't insert a 0 on the end of 0.999..., allow me to clarify again - the above is an equation. 0.999... is not an equation. Different rules apply.

I didn't add anything, I shifted brackets (that is no mathematical operation, but additions are commutative, so the order is irrelevant). And I already said that this 'proof' is nonsensical (twice), but if you take infinity into consideration, this 'proof' is just as valid as 0.9999... = 1. I know there is a -1 missing at the end that I omitted, but you (among others) argue that there is no end to infinity, so this -1 doesn't exist.

The difference between 0.99999... and one might be infinitesimal (or infinitely) small, but it is there.

 

[quhouv vhqwiufv qwiu]

You can't get any "closer" than "equivalent", at least not in the system of Real Numbers, because that number system doesn't allow infinitesimal values [http://en.wikipedia.org/wiki/Infinitesimal].

If you want to allow infinity, you also have to allow infinitesimal values, since that is nothing other than 1/infinity.

How about this one:

The difference between 0.9999.... and 1.0 is 0.0000... and since the difference between them is 0, the two numbers must be equal.

*EPIC FACEPALM*

Unless you were being sarcastic... the difference is 0.0000000...(infinity)...001

Epic facepalm to you too. To write infinity in between two decimal places is illogical because you will never reach the decimals after "infinity".

Also I have another proof which I don't think has been posted yet:



A and B are two real numbers. The midpoint (M) of A and B (i.e. the value between them such that AM = MB) can be found with the formula M = (A + B)/2.

Let A = 0.999... and let B = 1.

M = (1 + 0.999...)/2
∴M = 1.999.../2
∴ M = 0.999... (check that on a calculator if you must)

Hence A = 0.999... = M
∴A = M

Therefore in terms of the original equation:

(A + B)/2 = M = A
∴(A + B)/2 = A
∴A + B = 2A
∴B = 2A - A
∴B = A

And substituting the original values:
0.999... = 1


*Note, at no point in that proof before the last line did I use 1 = 0.999... so the proof is valid.

 

[BabyRaptor]

Math makes my brain asplode. I'm suing all you people for all your cookies for making my brain break!

 

[Sikachu]

If x = 0.999999...
Then 10x = 9.9999...
Therefore, 10x - x = 9
Which implies 9x = 9
Thus, x = 1
x also = 0.99999...

In conclusion, I have just proven 1 = 0.9999...

Except of course you're skipping out a shit load of clever and interesting maths about sets and series, but don't let that get in the way of your sense of smug satisfaction. That comment goes for all of you on here who don't know that you don't know why this is the case.

http://en.wikipedia.org/wiki/0.999...
Enjoy trying to wrap your heads around some of those.

 

[Redingold]

I didn't say it never makes a difference. But in this case it doesn't (as long as you keep infinity in mind, or else there will be a -1 at the end, but as multiple people here said, there is no end to infinity, so there is no -1 and 1 = 0).

But it does make a difference, because 1 + (-1 + 1) + (-1 + 1) + ... = 1

Infinite addition is not necessarily commutative.

So you agree that traditional maths is not necessarily valid when dealing with infinity, because in traditional maths additions are always commutative...so for all intends and purposes you agree with me

Of course traditional maths isn't always relevant when we're dealing with infinity. After all, 2 * infinity is infinity, which implies infinity = 0, or that 1 = 2. However, 0.999... is still equal to 1.

M'kay. The number 0.999... is equal to an infinite series 0.9 + 0.09 + 0.009 + 0.0009 and so on. If you know anything about slightly advanced maths, you'll know that the sum of an infinite geometric series is equal to a/(1-r) when |r| < 1 (explained below for those who aren't so good at maths)

In our example here, a, the first term, is 0.9, and r, the common ratio, is 0.1 (because each term is the previous term multiplied by 0.1).

So we have 0.9/(1-0.1) which equals 0.9/0.9 which equals 1.

Explanation of maths involved:

A geometric sequence is one where each term is the previous term multiplied by some number r. The first term is a, the second term is ar, the third term is ar[sup]2[/sup] and so on. The nth term is ar[sup]n-1[/sup].

The sum of a geometric series to n terms, which we shall call S[sub]n[/sub], is therefore equal to a + ar + ar[sup]2[/sup]...+ ar[sup]n-2[/sup] + ar[sup]n-1[/sup]

Multiplying by r, we get rS[sub]n[/sub] = ar + ar[sup]2[/sup] + ar[sup]3[/sup]...+ ar[sup]n-1[/sup] + ar[sup]n[/sup]

Subracting rS[sub]n[/sub] from S[sub]n[/sub] leads to S[sub]n[/sub] - rS[sub]n[/sub] = a - ar[sup]n[/sup]

This means S[sub]n[/sub](1-r) = a(1 - r[sup]n[/sup])

And S[sub]n[/sub] = a(1 - r[sup]n[/sup])/(1-r)

Now, to find the sum to infinity, n must be equal to infinity. If |r| > 1, r[sup]infinity[/sup] is infinite. If |r| < 1, r[sup]infinity[/sup] is equal to zero.

Thus, S[sub]infinity[/sub] = a(1 - r[sup]infinity[/sup])/(1-r) = a(1-0)/(1-r) = a/(1-r) when |r| < 1

Satisfied now?

See?

If you can find a flaw in that, well done.

 

[Piflik]

You can't get any "closer" than "equivalent", at least not in the system of Real Numbers, because that number system doesn't allow infinitesimal values [http://en.wikipedia.org/wiki/Infinitesimal].

If you want to allow infinity, you also have to allow infinitesimal values, since that is nothing other than 1/infinity.

How about this one:

The difference between 0.9999.... and 1.0 is 0.0000... and since the difference between them is 0, the two numbers must be equal.

*EPIC FACEPALM*

Unless you were being sarcastic... the difference is 0.0000000...(infinity)...001

Epic facepalm to you too. To write infinity in between two decimal places is illogical because you will never reach the decimals after "infinity".

Also I have another proof which I don't think has been posted yet:



A and B are two real numbers. The midpoint (M) of A and B (i.e. the value between them such that M-A = B-M) can be found with the formula M = (A + B)/2.

Let A = 0.999... and let B = 1.

M = (1 + 0.999...)/2
&#8756;M = 1.999.../2
&#8756; M = 0.999... (check that on a calculator if you must)

A calculator rounds...and 0.99999....*2 = 1.99999....98...

1.9999999.../2 = 0.9999...95 (note that the 0.9999....95 is 0.0000...05 more than 0.99999...)

so the following is invalid

Hence A = 0.999... = M
&#8756;A = M

Therefore in terms of the original equation:

(A + B)/2 = M = A
&#8756;(A + B)/2 = A
&#8756;A + B = 2A
&#8756;B = 2A - A
&#8756;B = A

And substituting the original values:
0.999... = 1


*Note, at no point in that proof before the last line did I use 1 = 0.999... so the proof is valid.

It is not.

 

[Coldie]

I didn't add anything, I shifted brackets (that is no mathematical operation, but additions are commutative, so the order is irrelevant). And I already said that this 'proof' is nonsensical (twice), but if you take infinity into consideration, this 'proof' is just as valid as 0.9999... = 1. I know there is a -1 missing at the end that I omitted, but you (among others) argue that there is no end to infinity, so this -1 doesn't exist.

The difference between 0.99999... and one might be infinitesimal (or infinitely) small, but it is there.

As you say, there's no operation "shift brackets". What you did is redefine the sequence. You, essentially, created a new, independent sum. With an incorrect result, but oh well - there's no -1 missing after the infinity, it's missing from the right part.

inf.sum(1 - 1) = 0
1 + inf.sum(-1 + 1) = 1
The result of Sum[n=0..inf](-1)[sup]n[/sup] is, of course, {1 for even n, 0 for odd n} and is undetermined when n = infinity, because infinity is both odd and even.

And the difference between 0.(9) and 1 is 0.(0), which is to say, exactly 0.

 

[grammarye]

I didn't add anything, I shifted brackets (that is no mathematical operation, but additions are commutative, so the order is irrelevant). And I already said that this 'proof' is nonsensical (twice), but if you take infinity into consideration, this 'proof' is just as valid as 0.9999... = 1. I know there is a -1 missing at the end that I omitted, but you (among others) argue that there is no end to infinity, so this -1 doesn't exist.

The difference between 0.99999... and one might be infinitesimal (or infinitely) small, but it is there.

Sorry, but that is not a logical sequence. Here are the assumptions I am working with:

* Infinite means infinite - no known end. 0.999... means 9s forever.
* Conceptually even in an infinite equation, both sides must balance, or it is no longer an equation. That means there must conceptually be an 'end' for want of a better word to each side of equation.

If you shift the brackets of that equation along, somehow, then infinitely speaking, and assuming infinite addition is commutative, there is a -1 on the infinite end. That that end is a infinite distance away doesn't change that conceptually. You are taking the use of the phrase 'no end' far too literally & broadly. Infinite numbers do have no end - but infinite equations must, otherwise you couldn't work with them. Lets' take my hypothetical & poorly expressed 0.999... as an equation.

0.9 + 0.09 + 0.009 + 0.(0...)9 + ... = 0.999...

Now lets add 2 to both sides.

0.9 + 0.09 + 0.009 + 0.(0...)9 + ... + 2 = 0.999... + 2

That's still an infinite equation, and it has +2 on the end. Lets assume for a moment that addition isn't commutative (if we were working with a more complex operation for example) - you have to be able to have a infinite 'end' for infinite equations to work. Of course, in most cases they either can't be evaluated, tend to 0, tend to infinity itself, or otherwise are not very useful, but they do exist.

None of that has anything to do with your notion that you can add a zero on the infinite end of 0.999... That is just plain incorrect maths. Multiplication by ten is not and has never been 'shift by one and add a zero on the end'.

(Edit: Coldie beat me to it and much more succinctly)

 

[quhouv vhqwiufv qwiu]

You can't get any "closer" than "equivalent", at least not in the system of Real Numbers, because that number system doesn't allow infinitesimal values [http://en.wikipedia.org/wiki/Infinitesimal].

If you want to allow infinity, you also have to allow infinitesimal values, since that is nothing other than 1/infinity.

How about this one:

The difference between 0.9999.... and 1.0 is 0.0000... and since the difference between them is 0, the two numbers must be equal.

*EPIC FACEPALM*

Unless you were being sarcastic... the difference is 0.0000000...(infinity)...001

Epic facepalm to you too. To write infinity in between two decimal places is illogical because you will never reach the decimals after "infinity".

Also I have another proof which I don't think has been posted yet:



A and B are two real numbers. The midpoint (M) of A and B (i.e. the value between them such that M-A = B-M) can be found with the formula M = (A + B)/2.

Let A = 0.999... and let B = 1.

M = (1 + 0.999...)/2
&#8756;M = 1.999.../2
&#8756; M = 0.999... (check that on a calculator if you must)

A calculator rounds...and 0.99999....*2 = 1.99999....98...

1.9999999.../2 = 0.9999...95 (note that the 0.9999....95 is 0.0000...05 more than 0.99999...)

so the following is invalid

Hence A = 0.999... = M
&#8756;A = M

Therefore in terms of the original equation:

(A + B)/2 = M = A
&#8756;(A + B)/2 = A
&#8756;A + B = 2A
&#8756;B = 2A - A
&#8756;B = A

And substituting the original values:
0.999... = 1


*Note, at no point in that proof before the last line did I use 1 = 0.999... so the proof is valid.

It is not.

Except that I did not use a calculator and I did not at any point multiply 0.999... by 2. Because I was establishing 0.999... = 1 I could not say 0.999... + 1 = 2*0.999... in my proof.