Poll: 0.999... = 1

[grammarye]

SNIP

There's a problem with your maths, not the question.

Please do explain. I'm always keen to know where I've gone wrong. Where is the fundamental flaw? Please, before you answer, reference the numerous actual mathematical proofs in this page alone.

 

[Redingold]

Hey, let's try something.

Let n = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... and so on, forever, involving all negative integer powers of 2.

Now, 2n = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... and so on.

This is 1 more than n.

2n = n + 1

Subtract n...

n = 1

So what you're saying is that n is actually infinitesimally smaller than 1, huh?

Would that make it equal to 0.999..., since that is also infinitesimally smaller than 1?

But I've just shown it's one.

What am I doing wrong? Where is the flaw that shows that when n + 1 = 2n, n =/= 1?

Again your geometric series...I said it before and will say it again. A limit is not a value. It is a limit and will never reach that value.

The limit of the sum 1/(2[sub]n[/sub]) for n -> infinity is 1. That much is true, but the function will approach this value asymptotically and never reach it.

It will never reach it...in a finite amount of steps. But the ... indicates that I have used an infinite amount of sets. Also, I did just show that it's sum was 1. Maybe you'd like to actually explain where my maths is flawed, rather than just repeatedly stating that it will never actually reach 1. In sums to infinity, you can use an infinite number of terms. That's rather the point of sums to infinity.

Please explain why the limit 1 does not equal the value 1.

 

[Piflik]

Because the limit for 1/n for n->0 does not equal 1/0.

The point of converging infinite sums is, that you can always add another term and still not reach the limit. That's why it is called a limit. It will never go above this limit, no matter how many of these terms you add, and thus it can also never reach that limit, because then the next term you add will take it above the limit.

 

[Maze1125]

0.333... * 3 =/= 1, since 0.33333... =/= 1/3

0.333... is a flawed representation of 1/3. No matter how far you go, will always be an infinitesimal difference between 1/3 and 0.333...

Let a[sub]n[/sub] be the nth term of the sequence: 0.4, 0.34, 0.334, 0.3334, ...
Let b[sub]n[/sub] be the nth term of the sequence: 1/3, 1/3, 1/3, 1/3, ...
Let c[sub]n[/sub] be the nth term of the sequence: 0.3, 0.33, 0.333, 0.3333, ...

Now, the limits of a[sub]n[/sub] and c[sub]n[/sub], as n tends to infinity, are both obviously 0.333...
The limit of b[sub]n[/sub] is also clearly 1/3.
But, for all n, a[sub]n[/sub] > b[sub]n[/sub] > c[sub]n[/sub], and the limits of a[sub]n[/sub] and c[sub]n[/sub] are equal. So, by the squeeze theorem, the limit of b[sub]n[/sub] is equal to the limits of a[sub]n[/sub] and c[sub]n[/sub].

So, the limits of b[sub]n[/sub] are both 1/3 and 0.333..., and limits are unique. Therefore 0.333... = 1/3
QED

Refer to my previous posts...Limits are not values. Both sequences will approach 1/3 asymptotically and never reach it. Ever.

Please re-read my proof, because I never made that fallacy.
Yes, I used limits, but I used them correctly.

 

[Maze1125]

SNIP

There's a problem with your maths, not the question.

Please do explain. I'm always keen to know where I've gone wrong. Where is the fundamental flaw? Please, before you answer, reference the numerous actual mathematical proofs in this page alone.

It's pretty simple, hun.

If my maths told me red equaled blue, I'd be getting a new maths.

What next? 2 = 5? Madness!

Why do you believe that 0.999... and 1 are different numbers?

 

[grammarye]

SNIP

There's a problem with your maths, not the question.

Please do explain. I'm always keen to know where I've gone wrong. Where is the fundamental flaw? Please, before you answer, reference the numerous actual mathematical proofs in this page alone.

It's pretty simple, hun.

If my maths told me red equaled blue, I'd be getting a new maths.

What next? 2 = 5? Madness!

It is rather more complex than that. Ignoring for a moment the fun little debate over whether something is a value or tends to it that Redingold & Piflik are having, there's a small problem with decimal maths if you consider them entirely different. I'm just going to quote the following, to indicate that this is not merely my minor personal delusion (ignoring the big red text in the image, it is actually informative):

I don't understand why people are still arguing this

Shouldn't common sense prevail soon?

Not if your "common sense" is wrong (or as a clever man once said: "Common sense is not so common.").

Math is a complicated branch of science and doesn't make sense if you don't take your time to understand it's rules. Specifically, you haven't taken your time to understand that one value can be represented in several (or rather, infinite) ways. 1 = 2/2 = 2 - 1 = (1/2) * 2. And yes, 1 = 0.999... as well.

What's common sense to you is laughable in the eyes of math Ph.d.'s. It's commonly accepted that 0.999... equals 1 amongst educated mathmaticians, even if it sounds like bollocks to people who doesn't understand this branch of math properly.

 

[Athinira]

Refer to my previous posts...Limits are not values. Both sequences will approach 1/3 asymptotically and never reach it. Ever.

Refer to my previous post about continuum being the proof (a concept you ironically mentioned yourself), and how you are confusing the Real Number system with the Extended Real Number system.

I'll repeat again, if two numbers doesn't have an average in the real number system (and that average still can't be infinitesimal), then they are the same number. Period.

 

[Piflik]

0.333... * 3 =/= 1, since 0.33333... =/= 1/3

0.333... is a flawed representation of 1/3. No matter how far you go, will always be an infinitesimal difference between 1/3 and 0.333...

Let a[sub]n[/sub] be the nth term of the sequence: 0.4, 0.34, 0.334, 0.3334, ...
Let b[sub]n[/sub] be the nth term of the sequence: 1/3, 1/3, 1/3, 1/3, ...
Let c[sub]n[/sub] be the nth term of the sequence: 0.3, 0.33, 0.333, 0.3333, ...

Now, the limits of a[sub]n[/sub] and c[sub]n[/sub], as n tends to infinity, are both obviously 0.333...
The limit of b[sub]n[/sub] is also clearly 1/3.
But, for all n, a[sub]n[/sub] > b[sub]n[/sub] > c[sub]n[/sub], and the limits of a[sub]n[/sub] and c[sub]n[/sub] are equal. So, by the squeeze theorem, the limit of b[sub]n[/sub] is equal to the limits of a[sub]n[/sub] and c[sub]n[/sub].

So, the limits of b[sub]n[/sub] are both 1/3 and 0.333..., and limits are unique. Therefore 0.333... = 1/3
QED

Refer to my previous posts...Limits are not values. Both sequences will approach 1/3 asymptotically and never reach it. Ever.

Please re-read my proof, because I never made that fallacy.
Yes, I used limits, but I used them correctly.

The first Sequence approaches 1/3 from above without reaching it, the last one approaches it from below without reaching it (SINCE THEY ARE LIMITS!!!! THEY CANNOT EVER REACH THAT VALUE)

Edit: Posted this to the geometric sum argument, but the essence still stays correct here...

The point of converging infinite sums is, that you can always add another term and still not reach the limit. That's why it is called a limit. It will never go above this limit, no matter how many of these terms you add, and thus it can also never reach that limit, because then the next term you add will take it above the limit.

 

[Maze1125]

Why do you believe that 0.999... and 1 are different numbers?

'Cause one is 1 and the other is 0.999...

And why can't those be the same thing? Just represented in a different way.

1 and 2 can't be the same thing, because 2 - 1 = 1. So there is a difference so 1 between them. But 1 - 0.999... = 0.000... = 0, so there is no mathematical difference between them.

 

[Addendum_Forthcoming]

I suck at maths but technically wouldn't 10x (if x is .9~) - x = 8.9999999~?

If 10x = 9.999... and 9x = 8.999... then x = 1 since the difference between 8.9... and 9.9... is is 1, which still proves that 0.999... = 1

It's not though o.o

The difference is .9~,

.9~(x) + 8.9~ (9x) = 9.9~

.'. x = .9~

Actually no.

If 0.9r ISN't equal to 1, then ".9~(x) + 8.9~ (9x) = 9.9~" implies that the number 8 must exist somewhere in the end result (9.9r).

You bare basically saying that the difference between 8.9r and 9.9r is 0.9r. That implies that 0.9r equals 1, since 9.9r - 8.9r = 1.

If X = 1 AND .9~ the difference would be .0~1 ... .9~ is not 1 n.n

Whilst the same can not be said of 3/3 = 1 ... and all numbers are divisible by themselves into real numbers.... 3/3 = 1 is still mathematically correct.

That which cannot be said about .9~/1 = 1, as cannot equal .9~ =/= 1.

1 is a rational number, the other irrational (.9~).

Whist infinitely small it is still relevant, if .9~ and 1 were the same then they could mutually interchangeable, but they're not x.x 12 x .9~ will always be 11.9~.

This is what makes them irrational numbers.

For all that simplisticd theory I can disprove it in one sentence.

.9~(x) / 1 = .9~

.'. x = .9~

 

[Ishnuvalok]

Why do you believe that 0.999... and 1 are different numbers?

I think he/she doesn't realize that 0.999... doesn't terminate. It goes on forever.

A practical example. Lets say you have a bottle of water and you pour the water into three glasses so that the amount of water is the same in each glass.

1/3 = 0.33333...

1/3+1/3+1/3=1

0.33333...+0.33333...+0.33333...=0.99999...

0.99999...=1

 

[Maze1125]

0.333... * 3 =/= 1, since 0.33333... =/= 1/3

0.333... is a flawed representation of 1/3. No matter how far you go, will always be an infinitesimal difference between 1/3 and 0.333...

Let a[sub]n[/sub] be the nth term of the sequence: 0.4, 0.34, 0.334, 0.3334, ...
Let b[sub]n[/sub] be the nth term of the sequence: 1/3, 1/3, 1/3, 1/3, ...
Let c[sub]n[/sub] be the nth term of the sequence: 0.3, 0.33, 0.333, 0.3333, ...

Now, the limits of a[sub]n[/sub] and c[sub]n[/sub], as n tends to infinity, are both obviously 0.333...
The limit of b[sub]n[/sub] is also clearly 1/3.
But, for all n, a[sub]n[/sub] > b[sub]n[/sub] > c[sub]n[/sub], and the limits of a[sub]n[/sub] and c[sub]n[/sub] are equal. So, by the squeeze theorem, the limit of b[sub]n[/sub] is equal to the limits of a[sub]n[/sub] and c[sub]n[/sub].

So, the limits of b[sub]n[/sub] are both 1/3 and 0.333..., and limits are unique. Therefore 0.333... = 1/3
QED

Refer to my previous posts...Limits are not values. Both sequences will approach 1/3 asymptotically and never reach it. Ever.

Please re-read my proof, because I never made that fallacy.
Yes, I used limits, but I used them correctly.

The first Sequence approaches 1/3 from above without reaching it, the last one approaches it from below without reaching it (SINCE THEY ARE LIMITS!!!! THEY CANNOT EVER REACH THAT VALUE)

Yes, I know, and as I said, I never tried to claim they could in my proof. My proof works perfectly well even though that is true.

Yes, I used limits, but I used them correctly. You're not telling me anything that I didn't already account for in my proof.

But, if you're so sure I did make that fallacy, then please point out where so I can explain what I really meant to you.

Edit: And that is still true with your edit.

 

[grammarye]

The point of converging infinite sums is, that you can always add another term and still not reach the limit. That's why it is called a limit. It will never go above this limit, no matter how many of these terms you add, and thus it can also never reach that limit, because then the next term you add will take it above the limit.

Correct me if I am wrong, but aren't you essentially stating that, to pick our controversial value, the only possible way to define 1/3 is as 1/3. It can't be a decimal, it can't be an equation that relates to that because those use limits - so aren't we pretty much stuck with purely 1/3 as the only way to define it?

I'm not saying this is mathematically wrong or anything - just it strikes me that for all intents and purposes this precise rigour may not be that practical for usage... Is there a way you can define 1/3 in your precise definition that might be, say, encoded in a PC or whatever? Aren't we stuck using some form of approximation (in your mind at any rate, even if it's a matter of debate) for almost any purpose?

 

[Piflik]

0.333... * 3 =/= 1, since 0.33333... =/= 1/3

0.333... is a flawed representation of 1/3. No matter how far you go, will always be an infinitesimal difference between 1/3 and 0.333...

Let a[sub]n[/sub] be the nth term of the sequence: 0.4, 0.34, 0.334, 0.3334, ...
Let b[sub]n[/sub] be the nth term of the sequence: 1/3, 1/3, 1/3, 1/3, ...
Let c[sub]n[/sub] be the nth term of the sequence: 0.3, 0.33, 0.333, 0.3333, ...

Now, the limits of a[sub]n[/sub] and c[sub]n[/sub], as n tends to infinity, are both obviously 0.333...
The limit of b[sub]n[/sub] is also clearly 1/3.
But, for all n, a[sub]n[/sub] > b[sub]n[/sub] > c[sub]n[/sub], and the limits of a[sub]n[/sub] and c[sub]n[/sub] are equal. So, by the squeeze theorem, the limit of b[sub]n[/sub] is equal to the limits of a[sub]n[/sub] and c[sub]n[/sub].

So, the limits of b[sub]n[/sub] are both 1/3 and 0.333..., and limits are unique. Therefore 0.333... = 1/3
QED

Refer to my previous posts...Limits are not values. Both sequences will approach 1/3 asymptotically and never reach it. Ever.

Please re-read my proof, because I never made that fallacy.
Yes, I used limits, but I used them correctly.

The first Sequence approaches 1/3 from above without reaching it, the last one approaches it from below without reaching it (SINCE THEY ARE LIMITS!!!! THEY CANNOT EVER REACH THAT VALUE)

Yes, I know, and as I said, I never tried to claim they could in my proof. My proof works perfectly well even though that is true.

Yes, I used limits, but I used them correctly. You're not telling me anything that I didn't already account for in my proof.

But, if you're so sure I did make that fallacy, then please point out where so I can explain what I really meant to you.

Ok...again...the first sequence approaches 0.3333... from above, but it will never reach it, so it will always be bigger than 0.3333... (if it would somewhere reach 0.3333..., the next member in the sequence would be smaller than 0.3333... but that is not possible, since 0.3333... is the limit...and there will always be a next member when you deal with infinity)

The last sequence has the same problem, but it will always stay smaller than 0.33333... thus 1/3 would be somewhere between 0.333333....- and 0.3333333....+ (+ and - represent infinitesimal small deviations from the exact value), but not exactly 0.3333...

 

[Ishnuvalok]

SNIP

1.9 != 2.

1.99 != 2.

1.999 != 2.

1.9999 != 2.

1.99999 != 2.

1.999999 != 2.

1.9999999 1= 2.

Can you guess what 1.9999999 equals?

Spoiler: You guessed it!

1.9999999

You're making the same mistake over and over and over again.
0.999... does NOT TERMINATE.

 

[Maze1125]

SNIP

1.9 != 2.

1.99 != 2.

1.999 != 2.

1.9999 != 2.

1.99999 != 2.

1.999999 != 2.

1.9999999 1= 2.

Can you guess what 1.9999999 equals?

Spoiler: You guessed it!

1.9999999

1 =/= 5

1 + 1 =/= 5

1 + 1 + 1 =/= 5

1 + 1 + 1 + 1 =/= 5

Can you guess what 1 + 1 + 1 + 1 + 1 equals?

Spoiler: You guessed it!

1 + 1 + 1 + 1 + 1