Math Is Fun Forum

Anthony.R.Brown

Banned

Registered: 2006-11-16

Posts: 516

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

Quote:

" because as I said above, 1.0999... = 1.1 "

A.R.B

Ok let's seperate your answer above!

we have the Value 1 above

and we have the Value 0.999... above

you say 1 = 1

you say 0.999... also = 1

so 1 + 1 must = 2

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

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

It's not the value of 0.999... he has, it's 0.0999... move the decimal place one to the right... it happens to equal 0.1, and that added to 1 is 1.1.

Anthony.R.Brown

Banned

Registered: 2006-11-16

Posts: 516

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

Qote:"
It's not the value of 0.999... he has, it's 0.0999... move the decimal place one to the right... it happens to equal 0.1, and that added to 1 is 1.1."

A.R.B

0.999... the .9's never end!!

0.099... THE .9's never end!!

Maelwys

Member

Registered: 2007-02-02

Posts: 161

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

It's amazing, I've actually agreed with everything you've said in your last 2 posts.
1 = 1
0.999... = 1
1 + 1 = 2
0.999... the 9s never end
0.0999... the 9s never end

I agree with all of these statements. I'm not sure I understand your point... but I at least agree with your statements. It's a breakthrough!

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

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

Dross, please answer the question I proposed in 461. Just tell me whether your 0.111... has gotten an infinite amount of digits or not.

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.

But how can you have 0.999... without a "last" digit? The question, Dross, is not whether we have the ability to write or to find the "last" digit, but whether such a digit is already included in the imaginary model of "0.111..." . However, I don't demand it to be the last digit only, the infiniteth digit is enough.

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George,Y

Member

Registered: 2006-03-12

Posts: 1,379

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

Actually when you have "all integers" at the same time, you admit you have got infinity.

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X'(y-Xβ)=0

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

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

how can you have 0.999.... without a last digit? are you saying you can't have a number without a last digit?

so i guess pi isnt a number either then? since it doesnt have a list digit

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George,Y

Member

Registered: 2006-03-12

Posts: 1,379

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

No, I don't demand the last digit. I only demand the infiniteth digit as guarantee to have any finiteth (the 3rd, the 3millionth, etc) digit included in the structure of 0.999... you can read Post 96 and Post 107 for more details.

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luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

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

theres no such thing as the infiniteth digit, that makes no sense.

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George,Y

Member

Registered: 2006-03-12

Posts: 1,379

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

So positive... I can just say 0.999... doesn't exist either.

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

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

No George.  You can't just say that without reason.  The reason there is no infiniteth digit is because digits in real number are countable, and thus all have a finite distance from the decimal place.  I've said this before, but you just ignored it.  Perhaps you'll do so again.

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"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..."

Anthony.R.Brown

Banned

Registered: 2006-11-16

Posts: 516

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

---------------------------------------------------------------------------------------------------------------
                        RECURRING DEFINITION PROOF 03/03/07  by Anthony.R.Brown.
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The Definition for Recurring is given below..............................................................................

“ A Number that repeats itself Endlessly! Continuously repeating the same Number! “

Below are examples of Recurring Numbers! They are of course Endless! Length is example only! 

Recurring ( 0.1 ) = .111111111111111111111111111111111111111111111111111111111111111

Recurring ( 0.2 ) = .222222222222222222222222222222222222222222222222222222222222222

Recurring ( 0.3 ) = .333333333333333333333333333333333333333333333333333333333333333

Recurring ( 0.4 ) = .444444444444444444444444444444444444444444444444444444444444444

Recurring ( 0.5 ) = .555555555555555555555555555555555555555555555555555555555555555

Recurring ( 0.6 ) = .666666666666666666666666666666666666666666666666666666666666666

Recurring ( 0.7 ) = .777777777777777777777777777777777777777777777777777777777777777

Recurring ( 0.8 ) = .888888888888888888888888888888888888888888888888888888888888888

Recurring ( 0.9 ) = .999999999999999999999999999999999999999999999999999999999999999

The above examples are shown to be True! in the Calculations below! According to the Definition for Recurring!

( 0.1 ) Recurring < 1 and <> 1

( 0.2 ) Recurring < 1 and <> 1

( 0.3 ) Recurring < 1 and <> 1

( 0.4 ) Recurring < 1 and <> 1

( 0.5 ) Recurring < 1 and <> 1

( 0.6 ) Recurring < 1 and <> 1

( 0.7 ) Recurring < 1 and <> 1

( 0.8 ) Recurring < 1 and <> 1

( 0.9 ) Recurring < 1 and <> 1

---------------------------------------------------------------------------------------------------------------

Maelwys

Member

Registered: 2007-02-02

Posts: 161

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

( 0.1 ) Recurring < 1 and <> 1

( 0.2 ) Recurring < 1 and <> 1

( 0.3 ) Recurring < 1 and <> 1

( 0.4 ) Recurring < 1 and <> 1

( 0.5 ) Recurring < 1 and <> 1

( 0.6 ) Recurring < 1 and <> 1

( 0.7 ) Recurring < 1 and <> 1

( 0.8 ) Recurring < 1 and <> 1

( 0.9 ) Recurring < 1 and <> 1

I'm not sure I see your logic here... just because 1 < 9, 2 < 9, 3 < 9,  4 < 9, 5 < 9, 6 < 9, 7 < 9, 8 < 9, doesn't mean 9 < 9... you can't just create an arbitrary logic pattern and assume that it continues. Sorry, you need better logic than that.

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

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

No George.  You can't just say that without reason.  The reason there is no infiniteth digit is because digits in real number are countable, and thus all have a finite distance from the decimal place.  I've said this before, but you just ignored it.  Perhaps you'll do so again.

Well, after so many posts you still don't get it Ricky. How many digits do you have in 0.999... so long as you have "All" finiteth digits included in it?

The problem is not that you check one digit, finiteth, another, finiteth, a third, finiteth,...(you cannot check over but that doesn't mean the check is OK) The problem is that how you can have "all" of them together in one "decimal expansion". If you have forgotten what I wrote, check Post 107 on
The 5th page of this thread, whose url differs from this page by p=5 instead of p=20

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X'(y-Xβ)=0

Dross

Member

Registered: 2006-08-24

Posts: 325

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

No, I don't demand the last digit. I only demand the infiniteth digit as guarantee to have any finiteth (the 3rd, the 3millionth, etc) digit included in the structure of 0.999... you can read Post 96 and Post 107 for more details.

1) 0.9999... has an infinite number of digits.
2) (1) => 0.9999... has no last digit.
3) (2) => 0.9999... has no digit at position "∞ - 1"*
4) (1) and (3) do not contradict each other. (however, you seem to think that they do - or at least seem to be trying to get me to say that they do)

There is no problem at all with having an infinite number of digits after the decimal place.

Also, consider this - 0.3333... in base-10 is equal to 0.1 in base-3. So:

Now, there is clearly no doubt at all that:

(right? - remember

)

and that:

Which should be enough to lead you to the conclusion that

because equations (a) and (b) are exactly the same**, just written in a different base. (okay - so there are umpteen other proofs in this thread that have also shown this, so I don't expect you to pay attention, I'm just hoping you might)

* which is what I understand you mean by the "infinitieth digit", right?
** yes - I mean exactly the same

Last edited by Dross (2007-03-04 23:28:21)

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Bad speling makes me [sic]

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

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

1) 0.9999... has an infinite number of digits.
2) (1) => 0.9999... has no last digit.

This time you interpret "infinite" as finite step goes on and on, (1,2,3,4,...) Dross. So this time you change the view on the terminology "infinite" and give its the property of "growing".

Base system is already discussed earlier, it intends to express a given number by a sum of exponents of a certain integer (10, 2, 16 etc). But whether the exponent can be infinite or whether the sum can have infinite entries are still the same question as this one. Do I have to list my previous post against this "proof"?

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George,Y

Member

Registered: 2006-03-12

Posts: 1,379

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

Here,
Post 40

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X'(y-Xβ)=0

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

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

Infinity isn't a number, it's a concept. As such, you can't label it on a number line or point to an "infinitieth" digit. However, just because that's not possible doesn't mean that you can't use infinity validly in arguments.

Also, 0.999... doesn't grow, it just doesn't have an end to its digits, and it never did. If it never had an end, then the end can't grow further away.

Your example of 1, 2, 3, 4... is a sequence that could be interpreted as growing, but it doesn't damage our argument because infinity is not mentioned in that sequence, and it would be incorrect to do so.

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Dross

Member

Registered: 2006-08-24

Posts: 325

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

1) 0.9999... has an infinite number of digits.
2) (1) => 0.9999... has no last digit.

This time you interpret "infinite" as finite step goes on and on, (1,2,3,4,...) Dross. So this time you change the view on the terminology "infinite" and give its the property of "growing".

I second what Mathsy said - as has already been stated, just because the series is infinitely long does not mean it "grows" over time. 0.9999... has exactly the same number of digits at this moment in time as it does in any other moment in time.

Also, your post is quite vague - could you please highlight exactly what you are unhappy with about the way base-3 arithmemtic (or any other arithmetic) is used in my last post? If you can't, there must be nothing wrong with it, and it must be correct.

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George,Y

Member

Registered: 2006-03-12

Posts: 1,379

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

"argument because infinity is not mentioned in that sequence, and it would be incorrect to do so."

--Exactly, it is hided rather than mentioned.
1,2,3,...,1000,...,1000000,.... doesn't cause the problem.

However, the "number" 0.999... containing digts growing or 0.999... containing some digits enough to accommodate "all" of 1,2,3,...,1000,...,1000000,....  does.

I am not rejecting 1/3 interpreted as 0.1 in 3-base because it is indeed 0+1×3^(-1) and can be expressed in finite digits. However, 1/3 interpreted as 0.333... in 10-base means adding 3×10(-1)+3×10(-2)+3×10(-3)+3×10(-4)+... and has the same problem with the validity of infinite digits.

Just stating there is already a system embeding infinite digits or infinite sum in it doesn't prove the action of embeding is right, does it?!!!!!!!!!!!!!!!!!!!!!!!!!!

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George,Y

Member

Registered: 2006-03-12

Posts: 1,379

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

See this post and answer me whether the time is 19

By the way, you are not the only person that knows base system.

Last edited by George,Y (2007-03-05 03:00:34)

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

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

Well, after so many posts you still don't get it Ricky. How many digits do you have in 0.999... so long as you have "All" finiteth digits included in it?

There are an infinite amount of digits, but every digit has a finite distance from the decimal point.  That is the basic definition of countable, George.  Tell me what natural number is the "infinitith" number?  Yet, there are an infinite amount of them.

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"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..."

lightning

Real Member

Registered: 2007-02-26

Posts: 2,060

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

nice avatar

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George,Y

Member

Registered: 2006-03-12

Posts: 1,379

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

Well, after so many posts you still don't get it Ricky. How many digits do you have in 0.999... so long as you have "All" finiteth digits included in it?

There are an infinite amount of digits, but every digit has a finite distance from the decimal point.  That is the basic definition of countable, George.  Tell me what natural number is the "infinitith" number?  Yet, there are an infinite amount of them.

That's only an evasive statement only to ban me from pointing the delima. You take advantage of the inability to say infiniteth from the finite side. I can demonstrate this in the next post.

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

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

That's only an evasive statement only to ban me from pointing the delima. You take advantage of the inability to say infiniteth from the finite side. I can demonstrate this in the next post.

George, the ability to create a list (even if its an infinite list) of numbers is the definition of countable.  That's all decimal expansion is, a list of numbers.  Countable means that we can map the natural numbers to them.  As every natural number is finite, we can say that if n (a natural number) maps to some object, its the nth object.  Thus, as there is no such thing as the infinitith natural number, there is no such thing as the infinitith object.

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"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..."