Poll: 0.999... = 1

[grammarye]

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 still haven't answered my question - write out for me in decimal the value you get when you divide 1 by 3, and then explain to me how, having divided 1 into 3 pieces, how you put them back together?

 

[Piflik]

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?

Yes exactly. 1/3 can only be accurately represented by 1/3. I know that it is not practical for usage and we will always have to use some sort of approximation, but 1/3 is 1/3 and nothing else... Just like pi is pi and e is e...you cannot ever write one of those in any other way without being inaccurate.

 

[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.

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...

Yes, I know. So what? How is that relevant to my proof?
My proof used the limits themselves, so it doesn't matter that the sequences are different. Because the limits are the same, even though the sequences are certainly different.

 

[Redingold]

So, Piflik, have you come up with a reasonable explanation as to why the sum of ALL negative integer powers of two is not one? Why can I not replace all values of one with the sum of this series? What errors would this cause?

 

[Maze1125]

Just like pi is pi and e is e...you cannot ever write one of those in any other way without being inaccurate.

That's because pi and e are irrational numbers. 1/3 is a rational number, and so it can be accurately be represented by a decimal, as all rational numbers can. And the decimal in question is 0.333...

 

[grammarye]

SNIP

Already told you, love: there's a problem with your maths, not the question.

There was a problem with our globes once, too: they were flat.

So, you wouldn't particularly mind if we dropped the decimal system then, it being fundamentally flawed and all... Got a replacement in mind?

 

[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...

Yes, I know. So what? How is that relevant to my proof?
My proof used the limits themselves, so it doesn't matter that the sequences are different. Because the limits are the same, even though the sequences are certainly different.

The point is, that the limits are limits...I know the squeeze theorem and argued with my Professor back then, too...I couldn't convince him, he couldn't convince me...The upper one will not reach the limit, the lower one will not reach the limit, so the middle one can be between the infinitely small deviations from that value...that's all I am saying...o.33333... might be the closest representation of 1/3 without typing 1/3, but I still say it is inaccurate.

So, Piflik, have you come up with a reasonable explanation as to why the sum of ALL negative integer powers of two is not one? Why can I not replace all values of one with the sum of this series? What errors would this cause?

Yes, and just for you I will post it a third time

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. And there will always be a next term, since we are talking infinity here.

Just like pi is pi and e is e...you cannot ever write one of those in any other way without being inaccurate.

That's because pi and e are irrational numbers. 1/3 is a rational number, and so it can be accurately be represented by a decimal, as all rational numbers can. And the decimal in question is 0.333...

And how is that accurate? You are an missing infinite amount of decimals in that number.

 

[Athinira]

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 there 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~

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

Wrong.

Irrational numbers are numbers that can't be represented with a terminating or repeating decimal. 0.9r is represented with repeating decimals, and is not an irrational number.

>> Whist infinitely small it is still relevant <<

No, because it can't exist in the Real Number system in the first place.

I refer to my posts about linear continuum again. For x < y in the real number system, there must exist a real number z, so that x < z < y. If z doesn't exist, then x = y.

>> If X = 1 AND .9~ the difference would be .0~1 <<

That number doesn't exist. You are trying to put a number behind an infinite string of zeroes, which is impossible since the string is infinite. It's just as impossible as finding the ending of a circle. The number doesn't exist in the real number system.

You didn't disprove anything. You just failed at understanding infinity (and irrational numbers).

 

[Vanaron]

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

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

Same reason I believe 1 and 2 are different numbers.

*shrug*

That's like saying

1/2 != 0.5

because they're represented differently...

 

[Maze1125]

SNIPPY

That's different, hun. You aren't adding a 1 to 0.999..., only another .9.

So 1 + 1 + 1 + 1 + 0.999... != 5.

It's not different in regards to your previous post.

Just because 1 =/= 2 and 1.9 =/= 2, doesn't mean that 1.999... =/= 2.
Equally, just because 1 =/= 5 and 1 + 1 =/= 5, doesn't mean that 1 + 1 + 1 + 1 + 1 =/= 5.

As far as I can tell, your only reason for thinking they're different numbers is because they're written differently on the page.
But 1/2 and 0.5 are written differently, they still represent exactly the same number, so why can't 1 and 0.999... do the same?

 

[Ishnuvalok]

SNIP

i. She.

ii. Yes I do.

iii. You would need an infinite bottle of water to test that.

iv. No, it doesn't terminate. But no matter how many 9's you add, it will always just be 1.999...

Alright. Lets try this again.

We will define 0.999... as an infinite series.

0.999...=0+9(1/10) + 9(1/10)^2 + 9(1/10)^3 + 9(1/10)^4 + ...

Now we use a convergence theorem.

ar+ar^2+ar^3+...=ar/(1-r)

So

0+9(1/10) + 9(1/10)^2 + 9(1/10)^3 + ... = 9(1/10)/1-(1-10)

Which of course = 1.

 

[Redingold]

So, Piflik, have you come up with a reasonable explanation as to why the sum of ALL negative integer powers of two is not one? Why can I not replace all values of one with the sum of this series? What errors would this cause?

Yes, and just for you I will post it a third time

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. And there will always be a next term, since we are talking infinity here.

Yes, but just because it never actually reaches that value, does not mean we cannot do maths with said value. We can work out what the value is, and in the case of 0.9 + 0.09 + 0.009 + 0.0009 + ..., it turns out to be 1.

Mathematicians use infinities all the time. Integral calculus uses it to find the area under graphs, geometric series can be used to solve Zeno's paradox. Just because in practice it never reaches that limit, does not mean it will do the same in theory.

 

[Maze1125]

so the middle one can be between the infinitely small deviations from that value...

So you believe that infinitely small, but non-zero, real number exist?

And how is that accurate? You are an missing infinite amount of decimals in that number.

No I'm not, they're all there, represented by the "...". You don't have to write down every number for it to be accurate, you only need to find a way to unambiguously represent them.

I can't write 3.141... for pi, because it would be ambiguous as to what the ... meant, but when we're talking about 1/3. It's clear that the ... represents 3s all the way.

 

[Addendum_Forthcoming]

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 there 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~

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

Wrong.

Irrational numbers are numbers that can't be represented with a terminating or repeating decimal. 0.9r is represented with repeating decimals, and is not an irrational number.

>> Whist infinitely small it is still relevant <<

No, because it can't exist in the Real Number system in the first place.

I refer to my posts about linear continuum again. For x < y in the real number system, there must exist a real number z, so that x < z < y. If z doesn't exist, then x = y.

>> If X = 1 AND .9~ the difference would be .0~1 <<

That number doesn't exist. You are trying to put a number behind an infinite string of zeroes, which is impossible since the string is infinite. It's just as impossible as finding the ending of a circle. The number doesn't exist in the real number system.

You didn't disprove anything. You just failed at understanding infinity (and irrational numbers).

Whatever. As I said in the post that infinitely small, it's still relevant

.9~ / 1 =/= 1

.9~ / 1 = .9~

If .9~ = 1, then both would be interchangeable. They aren't.

Then can't substitude .9~ for one because it is not properly divisible by 1.

Does 12 x .9~ = 12? No.

Therefore it's not 1 <.<

 

[Maze1125]

SNIP

I'm not saying you're wrong; using your maths, you are right.

I'm saying our maths is wrong.

Just like our maps were wrong when they were missing countries and our globes when we 'knew' Earth was flat and was orbited by the sun.

The facts don't change. Only our way of seeing does.

Why do you think it's a fact that 0.999... even exists as a number separate to 1?

The only reason we can write 0.999... down is because our representation of the numbers gives us some redundancy. 0.999... doesn't exist in reality. It only exists on paper due to the way we write numbers.

 

[Maze1125]

If .9~ = 1, then both would be interchangeable. They aren't.

Yes they are.

The can't substitude .9~ for one because it is not properly divisible by 1.

Yes it is.
0.999... / 1 = 0.999... = 1

Does 12 x .9~ = 12? No.

Yes it does.

 

[Piflik]

So, Piflik, have you come up with a reasonable explanation as to why the sum of ALL negative integer powers of two is not one? Why can I not replace all values of one with the sum of this series? What errors would this cause?

Yes, and just for you I will post it a third time

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. And there will always be a next term, since we are talking infinity here.

Yes, but just because it never actually reaches that value, does not mean we cannot do maths with said value. We can work out what the value is, and in the case of 0.9 + 0.09 + 0.009 + 0.0009 + ..., it turns out to be 1.

Mathematicians use infinities all the time. Integral calculus uses it to find the area under graphs, geometric series can be used to solve Zeno's paradox. Just because in practice it never reaches that limit, does not mean it will do the same in theory.

You can do math with it all you want...I do it myself...it is useful. Inaccurate, but useful and when dealing with such small inaccuracies not working with it would be exceptionally stupid. I am not arguing that...for all practical intents and purposes 0.99999... = 1, but when you want to delve into the unambiguous beauty of maths, these two numbers have to be different.

so the middle one can be between the infinitely small deviations from that value...

So you believe that infinitely small, but non-zero, real number exist?

And how is that accurate? You are an missing infinite amount of decimals in that number.

No I'm not, they're all there, represented by the "...". You don't have to write down every number for it to be accurate, you only need to find a way to unambiguously represent them.

I can't write 3.141... for pi, because it would be ambiguous as to what the ... meant, but when we're talking about 1/3. It's clear that the ... represents 3s all the way.

Yes, yes...turtles all the way down.

Of course there are infinitely small non-zero numbers.