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#451 2007-02-27 05:23:29

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: 0.9999....(recurring) = 1?

if 0.00..01 with infinite digits means only 0, 0.111... has only finite digits (1 in any infinite digit simply means 0) thus does not equal to 1/9

No, you misunderstand.  It's not that the 1 that appears in 0.000...01 is in the "infinite digit" (whatever that is), it's that 1 never appears in 0.000...01.

There is no such thing as an "infinite digit".  Each digit in a number has a finite position.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#452 2007-02-27 08:33:46

MathsIsFun
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Registered: 2005-01-21
Posts: 7,713

Re: 0.9999....(recurring) = 1?

Don't worry John, your ideas are sound in the real world.

"Points" would be the elementary particles (Quarks and Leptons). This is the difference: in mathematics we can divide things endlessly, but in the real world you end up with indivisible particles. ("Atom" comes from Greek "atomos" meaning indivisible, but of course we now know that atoms are made of electrons etc.)


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#453 2007-02-27 23:35:10

George,Y
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Registered: 2006-03-12
Posts: 1,379

Re: 0.9999....(recurring) = 1?

Maelwys wrote:

2/ 0.111... is a repeating digit, which can repeat for an infinite number of digits without changing. 0.000...01 implies an infinite number of digits that repeat for a while, and then suddenly change. What could possibly produce that change? 1/9 = 0.111... is reached because when you divide 1 / 9 you see that 9 goes into 10 once, leaving a remainder of 1, and then 9 goes into 10 once, leaving a remainder of 1, and then ... etc etc. The same result each time propogating the same result means that the same digit will keep appearing without end.

Uh, yes, so you mean 0.111...simply means finite digits going on and on, well, every finite 0.00001 or 0.0000000000000million01 adds on new value so your 0.111... though possible but varying. I think the discussion cannot be resumed until your 0.111... becomes (become?) constant. Equiting a varying thing to a static 1 makes no sense.


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#454 2007-02-27 23:53:25

Dross
Member
Registered: 2006-08-24
Posts: 325

Re: 0.9999....(recurring) = 1?

George,Y wrote:

I think the discussion cannot be resumed until your 0.111... becomes (become?) constant.

Note that 0.1111... is constant - it has a constant value all the time.

Don't be fooled into thinking that the value of 0.1111... changes value as you go along. What does happen, however, is that as a human you would write down 0.1, then adding another digit you'd have written 0.11, then 0.111, then 0.1111 and so on and so forth.

While it is indeed true that the numbers 0.1, 0.11, 0.111, ... are an increacing sequence and all of them are different, it's important to note that none of them are the number we are concerned with, namely 1.1111... .

The value of 0.1111... is indeed constant - it's only our approximations to it that may change as they get more accurate.


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#455 2007-03-01 00:01:16

George,Y
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Registered: 2006-03-12
Posts: 1,379

Re: 0.9999....(recurring) = 1?

Dross, do you mean you have gotten 0.111... such that any finiteth digit of 1 has already been included in it?


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#456 2007-03-01 00:08:37

Dross
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Registered: 2006-08-24
Posts: 325

Re: 0.9999....(recurring) = 1?

George,Y wrote:

Dross, do you mean you have gotten 0.111... such that any finiteth digit of 1 has already been included in it?

What do you mean "have I gotten"? (assume by "any finiteth digit" you mean "any digit", right?)

Do you mean to ask if I, personally, have written down on some piece of paper the number that includes every digit of 0.1111...?


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#457 2007-03-01 00:11:14

George,Y
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Registered: 2006-03-12
Posts: 1,379

Re: 0.9999....(recurring) = 1?

No, just imaginarily.


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#458 2007-03-01 00:20:59

Dross
Member
Registered: 2006-08-24
Posts: 325

Re: 0.9999....(recurring) = 1?

Well, in that case yes I have - right here:

0.1111...

Ask what the nth digit is, it is possible to give the correct answer of "1".


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#459 2007-03-01 00:25:14

Anthony.R.Brown
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Registered: 2006-11-16
Posts: 516

Re: 0.9999....(recurring) = 1?

Quote:

" So what is 1 - 0.9999... equal to, then? "

A.R.B

1 - 0.9999... has to equal a Number <> 1 and a Number <> 0.9999...

The Magic Number is the Infinite Difference!

So that ( 1 +  the Infinite Difference! ) = 1.1  and ( 0.9999...+  the Infinite Difference! ) = 1

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#460 2007-03-01 00:27:05

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: 0.9999....(recurring) = 1?

Then how can you get any finiteth digit of 1 included in your static 0.111... without  having got already Infinite amount of digits in it ?


X'(y-Xβ)=0

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#461 2007-03-01 01:01:10

Dross
Member
Registered: 2006-08-24
Posts: 325

Re: 0.9999....(recurring) = 1?

Anthony.R.Brown wrote:

Quote:

" So what is 1 - 0.9999... equal to, then? "

A.R.B

1 - 0.9999... has to equal a Number <> 1 and a Number <> 0.9999...

The Magic Number is the Infinite Difference!

So that ( 1 +  the Infinite Difference! ) = 1.1  and ( 0.9999...+  the Infinite Difference! ) = 1

From the bold text, it is then clear that "the Infinite Difference!" = 0.1, which is clearly wrong.


George,Y wrote:

Then how can you get any finiteth digit of 1 included in your static 0.111... without  having got already Infinite amount of digits in it ?

But there are an infinite amound of digits in it. That's what the dots at the end indicate. 0.1111... means that the "1"-digits repeat endlessly - i.e. there are an infinite amount of them.

Last edited by Dross (2007-03-01 01:04:30)


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#462 2007-03-01 01:06:24

Anthony.R.Brown
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Registered: 2006-11-16
Posts: 516

Re: 0.9999....(recurring) = 1?

Quote:

" From the bold text, it is then clear that "the Infinite Difference!" = 0.1, which is clearly wrong. "

A.R.B

Math don't lie! Does it!!

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#463 2007-03-01 01:24:43

Maelwys
Member
Registered: 2007-02-02
Posts: 161

Re: 0.9999....(recurring) = 1?

Anthony.R.Brown wrote:

" From the bold text, it is then clear that "the Infinite Difference!" = 0.1, which is clearly wrong. "

A.R.B

Math don't lie! Does it!!

Math doesn't lie, but it can be wrong if the fomulae used don't make sense.  If 1 + "infinite difference!" = 1.1, then 1.1 - 1 = "infinite difference!", therefore "infinite difference!" = 0.1
That means that by your idea 0.999... + 0.1 = 1? but we know that 0.999... + 0.1 = 1.0999... which is definitely not 1. So there's obviously a problem with your formulae.

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#464 2007-03-01 01:30:45

Anthony.R.Brown
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Posts: 516

Re: 0.9999....(recurring) = 1?

To Maelwys

Quot:
" but we know that 0.999... + 0.1 = 1.0999... which is definitely not 1. "

A.R.B

But wait a minute! according to the way you think! 1 + 0.9999.. must equal 2 does it not?
after all you keep saying 0.999... equals 1

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#465 2007-03-01 01:47:02

Maelwys
Member
Registered: 2007-02-02
Posts: 161

Re: 0.9999....(recurring) = 1?

Anthony.R.Brown wrote:

To Maelwys

Quot:
" but we know that 0.999... + 0.1 = 1.0999... which is definitely not 1. "

A.R.B

But wait a minute! according to the way you think! 1 + 0.9999.. must equal 2 does it not?
after all you keep saying 0.999... equals 1

Yes, I agree. What's the question?

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#466 2007-03-01 01:51:37

Dross
Member
Registered: 2006-08-24
Posts: 325

Re: 0.9999....(recurring) = 1?

Anthony.R.Brown wrote:

To Maelwys

Quot:
" but we know that 0.999... + 0.1 = 1.0999... which is definitely not 1. "

A.R.B

But wait a minute! according to the way you think! 1 + 0.9999.. must equal 2 does it not?
after all you keep saying 0.999... equals 1

Why do you keep coming up with these non-sequiturs? The following things are stated:

1) 0.999... + 0.1 = 1.0999... which is definitely not 1

2) But... 1 + 0.9999.. must equal 2 does it not?

Now, please explain to me how they have anything to do with each other? I agree with both of them - they blatantly do not contradict each other at all.

Please, please don't think you can "baffle" us into agreeing with you.


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#467 2007-03-02 00:48:11

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: 0.9999....(recurring) = 1?

Dross wrote:
George,Y wrote:

Then how can you get any finiteth digit of 1 included in your static 0.111... without  having got already Infinite amount of digits in it ?

But there are an infinite amound of digits in it. That's what the dots at the end indicate. 0.1111... means that the "1"-digits repeat endlessly - i.e. there are an infinite amount of them.

I think  you make some inconsistency, this time your 0.111... has come to grow again? 0.111...repeating.

0.111... either static or growing, OK? You can choose any, but you cannot choose one upon your convienence. (I can choose the opposite at my convience too)

For already infinite digits, refer to the previous page and look up the apple & basket problem, and review Post 96 and Post 107 please. Tell me Exactly where the finiteth 1 and infiniteth 1 are divided, in anyway you like.

Last edited by George,Y (2007-03-02 00:58:28)


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#468 2007-03-02 01:54:10

Anthony.R.Brown
Banned
Registered: 2006-11-16
Posts: 516

Re: 0.9999....(recurring) = 1?

Quote:

" Math doesn't lie, but it can be wrong if the fomulae used don't make sense.  If 1 + "infinite difference!" = 1.1, then 1.1 - 1 = "infinite difference!", therefore "infinite difference!" = 0.1
That means that by your idea 0.999... + 0.1 = 1? but we know that 0.999... + 0.1 = 1.0999... which is definitely not 1. So there's obviously a problem with your formulae. "

A.R.B

Let's make this clearer!

Quot: " but we know that 0.999... + 0.1 = 1.0999... "

A.R.B

even 1 + 0.1 = 1.1 so how can the above 0.999... + 0.1  = 1.0999...

as I have said if you think 0.999.. = 1 then the above must = 2

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#469 2007-03-02 02:36:28

Maelwys
Member
Registered: 2007-02-02
Posts: 161

Re: 0.9999....(recurring) = 1?

Anthony.R.Brown wrote:

Quote:

" Math doesn't lie, but it can be wrong if the fomulae used don't make sense.  If 1 + "infinite difference!" = 1.1, then 1.1 - 1 = "infinite difference!", therefore "infinite difference!" = 0.1
That means that by your idea 0.999... + 0.1 = 1? but we know that 0.999... + 0.1 = 1.0999... which is definitely not 1. So there's obviously a problem with your formulae. "

A.R.B

Let's make this clearer!

Quot: " but we know that 0.999... + 0.1 = 1.0999... "

A.R.B

even 1 + 0.1 = 1.1 so how can the above 0.999... + 0.1  = 1.0999...

as I have said if you think 0.999.. = 1 then the above must = 2

"the above must = 2"? So now you're trying to prove that 0.1 = 1? In what world does "0.999... = 1" mean that "0.999... + 0.1 = 2"? No way, no how. 0.999 + 0.1 = 1.0999..., which for the same reasonings described many times about the odd nature of the repeating 9s, also = 1.1. But it certainly does not = 2, ever.

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#470 2007-03-02 04:41:52

Dross
Member
Registered: 2006-08-24
Posts: 325

Re: 0.9999....(recurring) = 1?

George,Y wrote:

I think  you make some inconsistency, this time your 0.111... has come to grow again? 0.111...repeating.

0.111... either static or growing, OK? You can choose any, but you cannot choose one upon your convienence. (I can choose the opposite at my convience too)

For already infinite digits, refer to the previous page and look up the apple & basket problem, and review Post 96 and Post 107 please. Tell me Exactly where the finiteth 1 and infiniteth 1 are divided, in anyway you like.

0.1111... does not "change value" over time. For example, the value of 0.13 is always the same - it doesn't "start" at 0 and then "become" 0.1 and then "become" 0.13. Numbers do not change with time. 0.1111... always has the same value, and it has an infinite number of digits after the decimal place - what's the problem with this?

Also, what exactly are "finiteth" and "infiniteth" supposed to mean? Please don't make up words.


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#471 2007-03-02 15:56:31

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: 0.9999....(recurring) = 1?

0.1111... does not "change value" over time. For example, the value of 0.13 is always the same - it doesn't "start" at 0 and then "become" 0.1 and then "become" 0.13. Numbers do not change with time. 0.1111... always has the same value, and it has an infinite number of digits after the decimal place - what's the problem with this?

Also, what exactly are "finiteth" and "infiniteth" supposed to mean? Please don't make up words.

Well, I can read from your post that you have chosen the static, and you assume 0.999... as an number with infinite digits.
So what's the problem with the assumption that 0.00...01 is also a number that has the same amount of digits as that of 0.999...(infinite digits)? or 0.111...(infinite digits)?

"Also, what exactly are "finiteth" and "infiniteth" supposed to mean? Please don't make up words."
-so you can make up words like infinite digits and I cannot naturally derive it to the infiniteth digits?
when I mean infiniteth digits, I mean exactly the digit after infinite-1 digits, are you gonna say it is impossible? Well then, your infinite digits is impossible as well since you aren't even able to have infinite-1 digits.(This is indeed a little repeative to Post 107, talking to believers is quite time-consuming)


X'(y-Xβ)=0

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#472 2007-03-02 19:41:33

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: 0.9999....(recurring) = 1?

because for a number like 0.000....1 to exist, you have to have a finite number of digits, because you are putting an end to the zero's, which makes them non infinite


The Beginning Of All Things To End.
The End Of All Things To Come.

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#473 2007-03-02 22:47:47

Dross
Member
Registered: 2006-08-24
Posts: 325

Re: 0.9999....(recurring) = 1?

George,Y wrote:

Well, I can read from your post that you have chosen the static, and you assume 0.999... as an number with infinite digits.
So what's the problem with the assumption that 0.00...01 is also a number that has the same amount of digits as that of 0.999...(infinite digits)? or 0.111...(infinite digits)?

What Luca said. For the number 0.1111..., for every natural number n you give me, I can tell you what digit is at that position, and that digit will always be one. However, there is no natural number that you can give me such that the nth digit of 0.0000...0001* is equal to 1. Thus this number* is equal to 0. Happy with that?
If you want to say that 0.0000... has the same number of digits as 0.1111..., that's fine, I'll agree with you. So long as you recognise that 0.0000...0001 doesn't actually make sense (or at least that it's equivalent to 0).

George,Y wrote:

"Also, what exactly are "finiteth" and "infiniteth" supposed to mean? Please don't make up words."
-so you can make up words like infinite digits and I cannot naturally derive it to the infiniteth digits?
when I mean infiniteth digits, I mean exactly the digit after infinite-1 digits, are you gonna say it is impossible? Well then, your infinite digits is impossible as well since you aren't even able to have infinite-1 digits.(This is indeed a little repeative to Post 107, talking to believers is quite time-consuming)

Right - I think what you're trying to do is ask me what the second to last digit is - there is no such digit because there is not a last digit. The string of digits never terminates. Note that this does not mean that the number is "non-static" or changing or anything like that, it just means that trying to find the last digit is like trying to find the largest integer - you can't do it. Not because you're not good enough, but because there isn't one.


And we are able to have an infinite number of digits, which is exactly the same number of digits as "infinite - 1" digits, since if we take a way a finite number of elements from a set with cardinality
, we're still left with
digits.



* Assuming for a second that this number even makes sense.


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#474 2007-03-03 00:28:57

Anthony.R.Brown
Banned
Registered: 2006-11-16
Posts: 516

Re: 0.9999....(recurring) = 1?

Quote:

" the above must = 2"? So now you're trying to prove that 0.1 = 1? In what world does "0.999... = 1" mean that "0.999... + 0.1 = 2"? No way, no how. 0.999 + 0.1 = 1.0999..., which for the same reasonings described many times about the odd nature of the repeating 9s, also = 1.1. But it certainly does not = 2, ever."

A.R.B

1 - 0.9 = 0.1 and so if 0.999... also equals 1 then 0.999... - 0.1 also = 0.1

if 1 and 0.999... are the same as you both keep saying they are?

then 1 + 0.1 = 1.1 and so does 0.999... + 0.1 = 1.1 and not 0.999... + 0.1 = 1.0999...,

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#475 2007-03-03 00:36:35

Maelwys
Member
Registered: 2007-02-02
Posts: 161

Re: 0.9999....(recurring) = 1?

Anthony.R.Brown wrote:

Quote:

" the above must = 2"? So now you're trying to prove that 0.1 = 1? In what world does "0.999... = 1" mean that "0.999... + 0.1 = 2"? No way, no how. 0.999 + 0.1 = 1.0999..., which for the same reasonings described many times about the odd nature of the repeating 9s, also = 1.1. But it certainly does not = 2, ever."

A.R.B

1 - 0.9 = 0.1 and so if 0.999... also equals 1 then 0.999... - 0.1 also = 0.1

if 1 and 0.999... are the same as you both keep saying they are?

then 1 + 0.1 = 1.1 and so does 0.999... + 0.1 = 1.1 and not 0.999... + 0.1 = 1.0999...,

I assume you mean 0.999... 0.9 = 0.1
And yes, 0.999... + 0.1 = 1.1, because as I said above, 1.0999... = 1.1, by the same principles that we used to describe 0.999... = 1

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