0.9999....(recurring) = 1?

kylekatarn · 2007-02-10 01:47:03

Of course there is always another 9 (-__-)' it's obvious. That's not the question.

IS 0.(9) = 1 ? YES.

And your analysis of finite decimals:

0.9.....9
`---´
n ∈ N¹

...will never show that 0.(9) ≠ 1. Shows precisely the opposite.

if you consider the sequence s[k] : {0.9, 0.99, 0.999, 0.9999, ... }
for all Δ>0 exists an 'N' such that |s[k] - 1| < Δ whenever k>N.
Therefore

And now try to guess what 0.(9) means? 1.

This is the question in this thread, answered thousands of times, thousands of different ways. Everyone knows 0.9 < 1, 0.99 < 1, etc. But not everyone seems to understand why 0.(9) = 1. And someone keeps refusing the concepts of limit and infinite series.

Last edited by kylekatarn (2007-02-10 01:54:43)

Anthony.R.Brown · 2007-02-10 01:58:16

To kylekatarn

Quote:

"But not everyone seems to understand why 0.(9) = 1."

All you have to do is show how the last? if possible .9 disappears!! and becomes = to 1 ?

kylekatarn · 2007-02-10 02:02:27

There is no "last 9". That's why we use limits to understand what really means 0.(9).

I bet you didn't look to the definition of limit yet. : (

Last edited by kylekatarn (2007-02-10 02:11:05)

luca-deltodesco · 2007-02-10 02:02:51

why would it dissapear? youre making no sense.

and either way, how can you show how the last one dissapears, as youve said, there is no end to the nines, there is no last 9, because that means theres an end to them, but there isnt, there is a start, but no end, no last 9.

Anthony.R.Brown · 2007-02-12 02:18:04

----------------------------------------------------------------------------------------------------------------
INFINITE 0.9 <> 1 PROOF : FORMULA : By,Anthony.R.Brown,12/02/07
----------------------------------------------------------------------------------------------------------------

A = 1 " Single Start Value For 1 "

B = 0.9 " Single Start Value For 0.9 "

C ( A/B ) x B " Infinite 0.9 Value "

D ( A - C ) " Infinite < 1 Value "

C <> ( C + D ) " INFINITE 0.9 <> 1 "

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

Maelwys · 2007-02-12 02:45:04

Anthony.R.Brown wrote:

----------------------------------------------------------------------------------------------------------------
INFINITE 0.9 <> 1 PROOF : FORMULA : By,Anthony.R.Brown,12/02/07
----------------------------------------------------------------------------------------------------------------
A = 1 " Single Start Value For 1 "
B = 0.9 " Single Start Value For 0.9 "
C ( A/B ) x B " Infinite 0.9 Value "
D ( A - C ) " Infinite < 1 Value "
C <> ( C + D ) " INFINITE 0.9 <> 1 "
----------------------------------------------------------------------------------------------------------------

If:
D = A - C
Then:
C + D = C + A - C
C + D = A

If:
C = A/B * B
Then:
C = A

If:
C + D = A and C = A;
Then:
C + D = C

Therefor:
Since A = 1;
Then: C = 1 and D = 0

kylekatarn · 2007-02-12 08:13:22

Anthony.R.Brown wrote:

----------------------------------------------------------------------------------------------------------------
INFINITE 0.9 <> 1 PROOF : FORMULA : By,Anthony.R.Brown,12/02/07
----------------------------------------------------------------------------------------------------------------
A = 1 " Single Start Value For 1 "
B = 0.9 " Single Start Value For 0.9 "
C ( A/B ) x B " Infinite 0.9 Value "
D ( A - C ) " Infinite < 1 Value "
C <> ( C + D ) " INFINITE 0.9 <> 1 "
----------------------------------------------------------------------------------------------------------------

OK, but be happy!
0.(9) equals 1... and the world doesn't end if you refuse it.

Last edited by kylekatarn (2007-02-12 08:14:54)

Anthony.R.Brown · 2007-02-15 01:57:11

To Maelwys

Quote:

(1) If: D = A - C
(2) Then: C + D = C + A - C
(3) C + D = A
(4) If: C = A/B * B
(5) Then:C = A
(6) If: C + D = A and C = A;
(7) Then: C + D = C
(8) Therefor: Since A = 1;
(9) Then: C = 1 and D = 0

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

A.R.B

(1) " Yes D = 0.001...etc
(2) " Yes C + D = 1 " = " Yes C + A - C = 1 "
(3) " Yes C + D = A " equals 1
(4) " Yes C = A/B x B " equals the same as 0.999...etc
(5) " Wrong! Then:C = A " C = 0.999...etc A = 1
(6) " Yes C + D = A " equals 1
" Wrong! and C = A " C = 0.999...etc A = 1
(7) " Wrong! Then: C + D = C " C + D = 1 and C = 0.999...etc
(8) " Yes Therefor: Since A = 1 "
(9) " Wrong! Then: C = 1 and D = 0 " C = 0.999...etc and D will always = ( A - C )
0.001...etc < 1

Maelwys · 2007-02-15 02:27:13

Anthony.R.Brown wrote:

(1) " Yes D = 0.001...etc
(2) " Yes C + D = 1 " = " Yes C + A - C = 1 "
(3) " Yes C + D = A " equals 1
(4) " Yes C = A/B x B " equals the same as 0.999...etc
(5) " Wrong! Then:C = A " C = 0.999...etc A = 1
(6) " Yes C + D = A " equals 1
" Wrong! and C = A " C = 0.999...etc A = 1
(7) " Wrong! Then: C + D = C " C + D = 1 and C = 0.999...etc
(8) " Yes Therefor: Since A = 1 "
(9) " Wrong! Then: C = 1 and D = 0 " C = 0.999...etc and D will always = ( A - C )
0.001...etc < 1

Can you explain why 5, 6, 7, and 9 are wrong? Without using numbers, I mean. I'm just looking at the letter equations, and solving them without substituting in the values yet. For example, line 5 is simply solving for line 4 (C = A/B x B), which you agree with. Using basic rules of multiplication, A/B x B = A, there's no other thing that it COULD equal. If I said A = 4 and B = 2, would you argue that 4/2 x 2 DOESN'T equal 4? Or if A = 5 and B = 7, would you argue that 5/7 * 7 doesn't equal 5? A/B x B will always equal A (anything divided by a number and then multiplied by the same number equals the original number, it's basic logic). So if A/B x B = A, then C = A, hence line 5. The rest is just subsitution of C = A into the other formulas, that follows out to show A = C = 1, and D = 0.

Anthony.R.Brown · 2007-02-17 01:36:46

To Maelwys

First it's important to look at the Calculated Facts in my,

--------------------------------------------------------------------------
INFINITE 0.9 <> 1 PROOF : FORMULA : 12/02/07
--------------------------------------------------------------------------

( f1 ) A " Will always be > than B " " Because A - B = 0.1...etc "

( f2 ) A " Will always be > than C " " Because A - C = 0.001...etc "

( f3 ) A " Will always be > than D " " Because A - D = 0.999...etc "

( f4 ) A " Will always only be = to A and ( C + D ) " " Because A = A and A = ( C + D ) "

I only answered the way you wrote it down! thinking you understood what was happening in the Calculations! that's why it's important to stick with the original Formula including the brackets ()

Quote: " 4 Using basic rules of multiplication, A/B x B = A, "

A.R.B

C ( A/B ) x B " Infinite 0.9 Value "

C = ( 1 / 0.9 ) = ( 1.111...etc ) x ( 0 .9 ) = 0.999...etc

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

Quote : ( 5 ) " Wrong Because of ( f4 ) above "

Quote : ( 6 ) " Wrong Because of ( f4 ) above "

Quote : ( 7 ) " Wrong Because of ( f4 ) above "

Quote : ( 9 ) " Wrong Because of ( f4 ) and ( f3 ) above "

Ricky · 2007-02-17 04:43:40

ARB, for any real numbers A and B, A/B x B = A.

Are you saying that they are not real numbers?

Maelwys · 2007-02-17 05:54:29

Anthony.R.Brown wrote:

( f1 ) A " Will always be > than B " " Because A - B = 0.1...etc "
( f2 ) A " Will always be > than C " " Because A - C = 0.001...etc "
( f3 ) A " Will always be > than D " " Because A - D = 0.999...etc "
( f4 ) A " Will always only be = to A and ( C + D ) " " Because A = A and A = ( C + D ) "
I only answered the way you wrote it down! thinking you understood what was happening in the Calculations! that's why it's important to stick with the original Formula including the brackets ()
Quote: " 4 Using basic rules of multiplication, A/B x B = A, "
A.R.B
C ( A/B ) x B " Infinite 0.9 Value "
C = ( 1 / 0.9 ) = ( 1.111...etc ) x ( 0 .9 ) = 0.999...etc
---------------------------------------------------------------------------------------------------------------
Quote : ( 5 ) " Wrong Because of ( f4 ) above "
Quote : ( 6 ) " Wrong Because of ( f4 ) above "
Quote : ( 7 ) " Wrong Because of ( f4 ) above "
Quote : ( 9 ) " Wrong Because of ( f4 ) and ( f3 ) above "

I agree that C = 0.999..., the way that you wrote it. However, You also have to agree that (A/B) x B = A. Ignoring the values that you've assigned to A and B, even with the brackets there, it's basic math. If I ask you what (A/F) x F is, without knowing the value of F, the answer is still A. By definition, division and multiplication are the inverses of each other, and order of operations is irrelevant (so (A/B) x B = (AxB)/B = A either way).

The other problem with the above is that you're starting with an assumption, and then trying to prove it using your initial assumption. You're starting with the assumption that C < A, and then using that proof (f2, f3) to prove (f4) that A cannot equal C. But A can equal C for cases where D = 0, and since you're starting with A and B and solving for C and D, you can't automatically assume that C < A and D > 0, you have to assume nothing about them, and THEN prove that it's true. So ignoring your initial assumptions, we can see that A/B x B = A, therefor C = A, therefor since A = C + D, D = 0, and all the rest of the formulae fit into place.

Anthony.R.Brown · 2007-02-19 02:25:06

To Maelwys

Quote:

"I agree that C = 0.999..., the way that you wrote it. However, You also have to agree that (A/B) x B = A "

A.R.B

It is impossible for (A/B) x B = A because there is an Infinite Difference!

from my Formula C = (A/B) x B = 0.999....etc

A - C will always have an Infinite Difference! 0.001...etc

The rest of what you have put forward is Answered in my Formula!

The Infinite Difference's can never be made up without the use of the Algebra + sign.

Maelwys · 2007-02-19 03:23:04

Anthony.R.Brown wrote:

To Maelwys
Quote:
"I agree that C = 0.999..., the way that you wrote it. However, You also have to agree that (A/B) x B = A "
A.R.B
It is impossible for (A/B) x B = A because there is an Infinite Difference!
from my Formula C = (A/B) x B = 0.999....etc
A - C will always have an Infinite Difference! 0.001...etc
The rest of what you have put forward is Answered in my Formula!
The Infinite Difference's can never be made up without the use of the Algebra + sign.

You're still suffering the same problem. You're starting with the assumption that C < A, and then using that assumption to justify why C < A. If you want to make a proof, you have to start without any assumptions, and then get to the answer that you need. When the algebraic proof shows that A = C, you can't discount that because you want C to be less than A, and then say that it's proof that C is less than A. (A/B) x B = A, every time. I don't care if A is 43 and B is 762, (A/B) x B still equals A. Your argument is basically the same as if I said "X=2, Y=1, X-1=Y, but because I want to prove that 1=2, X=Y, so -1=0 must also be true, and since -1=0, if I add 2 to both sides I get 1=2, which proves that it!". The logic doesn't work if you start with your assumption, and then try to prove it using your conclusion as proof of itself.

Anthony.R.Brown · 2007-02-21 00:43:23

Let's have a look at why there will always be an Infinite Difference between A and C and why the Difference Remains the same from the start Onwards in my Formula!
And why the Algebra Sign + (Difference Value) is the Only way to make up the Difference's!

A = 1 " Single Start Value For 1 "
B = 0.9 " Single Start Value For 0.9 "
C ( A/B ) x B " Infinite 0.9 Value "
D ( A - C ) " Infinite < 1 Value "
C <> ( C + D ) " INFINITE 0.9 <> 1 "

I am going to give three examples below from the start onwards! and then the Infinite example!

Example (1) ex1A ( A - C) " Stage one/Decimal place " = ( 1 ) -( 0.9 ) = ex1D ( 0.1 )

Example (2) ex2A ( A - C) " Stage Two/Decimal place's " = ( 1 ) -( 0.99 ) = ex2D ( 0.01 )

Example (3) ex3A ( A - C) " Stage Three/Decimal place's " = ( 1 ) -( 0.999 ) = ex3D (0.001)

In the above Examples! for A to be able to Equal 1 again using the Algebra Sign + (Difference Value) the Anwers are shown Below!

Example (1) ex1A + ex1D = 1

Example (2) ex2A + ex2D = 1

Example (3) ex3A + ex3D = 1

Now below I will give the Infinite Example!

Example (I) exIA ( A - C) " Infinite Decimal place's " = ( 1 ) -( Infinite 0.9 ) = exID ( Infinite
0.1 )

Example (I) exIA + exID = 1

In the above Infinite Example! the amount of .9's and .1 (Difference) will always be the Same Length as each other from the Start onwards!
And more Important the Acutual Value Difference will always be the same!!
What I mean by this is shown Below!

0.999999999999999999999999999999999999999999999999999999999999999..........
0.000000000000000000000000000000000000000000000000000000000000001..........

For the above Example we can remove the .9's and the .0's After Stage one/Decimal place!

Because they are the same lenghth! and show the Difference is Permanent!

What we end up with is the Same Two Values! as from the Start Onwards!!

0.9
0.1

The two Final Calculations Below are the Same!!

Example (1) ex1A + ex1D = 1
Example (I) exIA + exID = 1

Last edited by Anthony.R.Brown (2007-02-22 00:15:31)

Maelwys · 2007-02-21 02:32:35

Anthony.R.Brown wrote:

C <> ( C + D )

What is the mathmatical basis for this assertion? This is only true if D <> 0, and you can't assume that D <> 0 when you're trying, in essance, to prove that D <> 0 (and the rest of us are proving that D = 0 through various methods). You need a proof for D <> 0 before you can use it as fact to determine the rest of your proofs.

Anthony.R.Brown wrote:

In the above Infinite Example! the amount of .9's and .1 (Difference) will always be the Same Length as each other from the Start onwards!
And more Important the Acutual Value Difference will always be the same!!
What I mean by this is shown Below!
0.999999999999999999999999999999999999999999999999999999999999999..........
0.000000000000000000000000000000000000000000000000000000000000001..........

The problem is that if they're the same length, and it's an infinite length, that means that there's an infinite number of 0s. And 0.000... (with an infinite number of 0s) = 0, so D = 0, so the above assertions are false. You cannot possibly have an infinitely long string of 0s with a 1 on the end, it's not possible within the realms of mathmatics. If you continue to assert that it is possible within your definition of math, I'll have to assume that you're inventing your own version of math where 1 > 1, and then I can't argue against you because I don't understand your personal "math" system.

Anthony.R.Brown wrote:

For the above Example we can remove the .9's and the .0's After the Decimal Point!
Because they are the same lenghth! and show the Difference is Permanent!
What we end up with is the Same Two Values! as from the Start Onwards!!
0.9
0.1

What? So 0.999... = 0.9? Does that mean that 0.5555 = 0.5? And 0.0004 = 0.4? And 0.9991 = 0.1 (because you can remove the 9s, since they're the same). I'm confused how you came up with that one!

mathsyperson · 2007-02-21 05:32:05

Me and one of my friends somehow started talking about this a few days ago.

I said that 0.999... = 1, and told him about the proof that involves taking x away from 10x to give 9, meaning x is 9/9 which is 1. He didn't believe me, and said that that couldn't be true on the grounds that 1 is a natural number. If 0.999... = 1, then 0.999... should be a natural number as well. It clearly isn't though, because its decimal form has things other than a bunch of zeroes after the point.

That isn't enough to convince me that 0.999... <> 1, but at the same time, I'm not entirely sure why that argument isn't valid.

Thoughts?

Dross · 2007-02-21 06:32:09

mathsyperson wrote:

Me and one of my friends somehow started talking about this a few days ago.
I said that 0.999... = 1, and told him about the proof that involves taking x away from 10x to give 9, meaning x is 9/9 which is 1. He didn't believe me, and said that that couldn't be true on the grounds that 1 is a natural number. If 0.999... = 1, then 0.999... should be a natural number as well. It clearly isn't though, because its decimal form has things other than a bunch of zeroes after the point.
That isn't enough to convince me that 0.999... <> 1, but at the same time, I'm not entirely sure why that argument isn't valid.
Thoughts?

In the context of what your friend is talking about, 0.9999... is not a number. And neither is 1 - they are just funny little symbols that represent numbers. Just because the funny little symbol that looks like 0.9999... or 0.9999999999999.......... or looks like a circle and then a dot and a bunch of the same numbers and then some more dots, does not change what it represents, and that is the number that is the smallest positive integer - however you write it.

Zhylliolom · 2007-02-21 06:34:57

mathsyperson wrote:

Me and one of my friends somehow started talking about this a few days ago.
I said that 0.999... = 1, and told him about the proof that involves taking x away from 10x to give 9, meaning x is 9/9 which is 1. He didn't believe me, and said that that couldn't be true on the grounds that 1 is a natural number. If 0.999... = 1, then 0.999... should be a natural number as well. It clearly isn't though, because its decimal form has things other than a bunch of zeroes after the point.
That isn't enough to convince me that 0.999... <> 1, but at the same time, I'm not entirely sure why that argument isn't valid.
Thoughts?

But it is a natural number. Since 0.999... = 1, it is a positive integer and thus a natural number. Just because it doesn't look like it is doesn't mean it isn't. One big reason people have trouble accepting the fact that 0.999... = 1 is that they just don't look the same. But in reality it is just the same as saying 5(3n + 6)/(15n + 30) = 1, they clearly don't look the same, but you'd never say that they aren't equal.

Theorem. Assume x ≥ 0. Then for every integer n ≥ 1 there is a finite decimal r[sub]n[/sub] = a[sub]0[/sub].a[sub]1[/sub]a[sub]2[/sub]...a[sub]n[/sub] such that r[sub]n[/sub] < x ≤ r[sub]n[/sub] + 1/10[sup]n[/sup].

Proof. Let S be the set of all nonnegative integers less than x. Then S is nonempty, since 0 ∈ S, and S is bounded above by x. Then by the Completeness Axiom of the real numbers, S has a supremum, which we denote a[sub]0[/sub] = sup S. a[sub]0[/sub] ∈ S, so that a[sub]0[/sub] is a nonnegative integer. a[sub]0[/sub] is then the greatest integer in x minus one, [x] - 1. Clearly, a[sub]0[/sub] < x ≤ a[sub]0[/sub] + 1. Now let a[sub]1[/sub] = [10x - 10a[sub]0[/sub]] - 1, the greatest integer in 10x - 10a[sub]0[/sub] minus one. Since 0 < 10x - 10a[sub]0[/sub] = 10(x - a[sub]0[/sub]) ≤ 10, we have that 0 ≤ a[sub]1[/sub] ≤ 9 and a[sub]1[/sub] < 10x - 10a[sub]0[/sub] ≤ a[sub]1[/sub] + 1. Then a[sub]1[/sub] is the largest integer satisfying a[sub]0[/sub] + a[sub]1[/sub]/10 < x ≤ a[sub]0[/sub] + (a[sub]1[/sub] + 1)/10. More generally, having chosen a[sub]1[/sub],..., a[sub]n-1[/sub] with 0 ≤ a[sub]i[/sub] ≤ 9, let a[sub]n[/sub] be the greatest integer satisfying a[sub]0[/sub] + a[sub]1[/sub]/10 + ... + a[sub]n[/sub]/10[sup]n[/sup] < x ≤ a[sub]0[/sub] + a[sub]1[/sub]/10 + ... + (a[sub]n[/sub] + 1)/10[sup]n[/sup]. Then 0 ≤ a[sub]n[/sub] ≤ 9 and we have r[sub]n[/sub] < x ≤ r[sub]n[/sub] + 1/10[sup]n[/sup], where r[sub]n[/sub] = a[sub]0[/sub].a[sub]1[/sub]a[sub]2[/sub]...a[sub]n[/sub]. This completes the proof.

It is easy to see from this theorem that we can define an infinite decimal representation of x. For x = 1, we see that a[sub]0[/sub] = 0, and a[sub]n[/sub] = 9 for all n > 1. Then 1 = 0.999....

Last edited by Zhylliolom (2007-02-23 05:49:47)

George,Y · 2007-02-21 15:42:00

Zhylliolom wrote:

mathsyperson wrote:
Me and one of my friends somehow started talking about this a few days ago.
I said that 0.999... = 1, and told him about the proof that involves taking x away from 10x to give 9, meaning x is 9/9 which is 1. He didn't believe me, and said that that couldn't be true on the grounds that 1 is a natural number. If 0.999... = 1, then 0.999... should be a natural number as well. It clearly isn't though, because its decimal form has things other than a bunch of zeroes after the point.
That isn't enough to convince me that 0.999... <> 1, but at the same time, I'm not entirely sure why that argument isn't valid.
Thoughts?
But it is a natural number. Since 0.999... = 1, it is a positive integer and thus a natural number. Just because it doesn't look like it is doesn't mean it isn't. One big reason people have trouble accepting the fact that 0.999... = 1 is that they just don't look the same. But in reality it is just the same as saying 5(3n + 6)/(15n + 30) = 1, they clearly don't look the same, but you'd never say that they aren't equal.
Theorem. Assume x ≥ 0. Then for every integer n ≥ 1 there is a finite decimal r[sub]n[/sub] = a[sub]0[/sub].a[sub]1[/sub]a[sub]2[/sub]...a[sub]n[/sub] such that r[sub]n[/sub] < x ≤ r[sub]n[/sub] + 1/10[sup]n[/sup].
Proof. Let S be the set of all nonnegative integers less than x. Then S is nonempty, since 0 ∈ S, and S is bounded above by x. Then by the Completeness Axiom of the real numbers, S has a supremum, which we denote a[sub]0[/sub] = sup S. a[sub]0[/sub] ∈ S, so that a[sub]0[/sub] is a nonnegative integer. a[sub]0[/sub] is then the greatest integer in x minus one, [x] - 1. Clearly, a[sub]0[/sub] ≤ x < a[sub]0[/sub] + 1. Now let a[sub]1[/sub] = [10x - 10a[sub]0[/sub]] - 1, the greatest integer in 10x - 10a[sub]0[/sub] minus one. Since 0 < 10x - 10a[sub]0[/sub] = 10(x - a[sub]0[/sub]) ≤ 10, we have that 0 ≤ a[sub]1[/sub] ≤ 9 and a[sub]1[/sub] < 10x - 10a[sub]0[/sub] ≤ a[sub]1[/sub] + 1. Then a[sub]1[/sub] is the largest integer satisfying a[sub]0[/sub] + a[sub]1[/sub]/10 < x ≤ a[sub]0[/sub] + (a[sub]1[/sub] + 1)/10. More generally, having chosen a[sub]1[/sub],..., a[sub]n-1[/sub] with 0 ≤ a[sub]i[/sub] ≤ 9, let a[sub]n[/sub] be the greatest integer satisfying a[sub]0[/sub] + a[sub]1[/sub]/10 + ... + a[sub]n[/sub]/10[sup]n[/sup] < x ≤ a[sub]0[/sub] + a[sub]1[/sub]/10 + ... + (a[sub]n[/sub] + 1)/10[sup]n[/sup]. Then 0 ≤ a[sub]n[/sub] ≤ 9 and we have r[sub]n[/sub] < x ≤ r[sub]n[/sub] + 1/10[sup]n[/sup], where r[sub]n[/sub] = a[sub]0[/sub].a[sub]1[/sub]a[sub]2[/sub]...a[sub]n[/sub]. This completes the proof.
It is easy to see from this theorem that we can define an infinite decimal representation of x. For x = 1, we see that a[sub]0[/sub] = 0, and a[sub]n[/sub] = 9 for all n > 1. Then 1 = 0.999....

Great argument worth debating against...

First, I agree with you that one thing may have different forms. But before you equate one form to another, one thing shall be resolved- that is, is the form VALID?

Well, you may say that if I want it to be valid, or some famous mathematicians, the maths society, or the book say it is valid, then it is valid. However, this is far from a rational answer. The infinite 9's, as I have proved earlier, has an inherent logic flaw or delima that prevents it from standing. I may repeat it again(post 107), that is, when the 9's turn from representing something finite small to something infinite small?

However, it is difficult for people to forget what they have accepted. Then another problem emerges. Is the "proof" a FAIR one?

We know some games are designed so unfair that one side is sure to lose. Such games include:
4 sticks, two players. Each one should take away one or two sticks at a time and take turns, the one who takes away the last sticks (literally the last one or the last two) wins. If the two players are clever enough the first player (who takes one stick first) definately wins. But does the first player Really wins? In fact Not, because the rule of the games FAVORs him/her.

By examing the rule of the common "proof" Zhylliolom has provided and much earlier Ricky has provided, one can easily find taking turns and setting turn sequences are an essential part of it.

Last edited by George,Y (2007-02-21 16:13:39)

Ricky · 2007-02-21 17:56:55

I'm not quite sure how we can communicate, George. I provided a definition for decimal expansion as a limit of an infinite series, and you didn't accept this. Since we can't even agree with what a decimal expansion even is, how can we talk about the decimal expansion of 0.999...?

Perhaps Zhylliolom and others feel otherwise, but I don't believe any further discussion can be progressive until the inconsistencies of how we define things can be resolved.

George,Y · 2007-02-21 22:21:28

So literally you mean Post 108 and Post 125??

All you did were embeding infinite digits(a self-contradictory concept) into your defination of r-somes and claiming you need it. Do you call that valid? Can a 1/10[sup]∞[/sup] or r[sub]∞[/sub] exist?

Hiding a problem doesn't solve it.

George,Y · 2007-02-21 22:27:45

Hiding it into a defination does not either.

Anthony.R.Brown · 2007-02-22 00:22:27

from A.R.B

Below is a correction in my Post Above 23:43:23

" For the above Example we can remove the .9's and the .0's After Stage one/Decimal place! "

Anthony.R.Brown · 2007-02-22 00:40:26

There are some problems that have to be solved and shown to be solved before the Argument Infinite 0.9 = 1 can be considered an Argument!
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(1) Explain how it is possible for a Number that starts with a lower value than another Number! is or becomes equal to the other Number! without using an Algebra sign.

(2) Explain how it is possible for a Number that starts with a Decimal Point! losses the Decimal Point! without using an Algebra sign.

(3) Give a Maximum Possible Value for the Amount of Infinite 0.9's. ( there has to be a Value ? ) otherwise the .9's will Continue!

Math Is Fun Forum

#351 2007-02-10 01:47:03

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

#352 2007-02-10 01:58:16

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

#353 2007-02-10 02:02:27

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

#354 2007-02-10 02:02:51

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

#355 2007-02-12 02:18:04

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

#356 2007-02-12 02:45:04

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

#357 2007-02-12 08:13:22

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

#358 2007-02-15 01:57:11

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

#359 2007-02-15 02:27:13

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

#360 2007-02-17 01:36:46

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

#361 2007-02-17 04:43:40

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

#362 2007-02-17 05:54:29

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

#363 2007-02-19 02:25:06

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

#364 2007-02-19 03:23:04

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

#365 2007-02-21 00:43:23

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

#366 2007-02-21 02:32:35

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

#367 2007-02-21 05:32:05

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

#368 2007-02-21 06:32:09

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

#369 2007-02-21 06:34:57

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

#370 2007-02-21 15:42:00

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

#371 2007-02-21 17:56:55

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

#372 2007-02-21 22:21:28

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

#373 2007-02-21 22:27:45

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

#374 2007-02-22 00:22:27

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

#375 2007-02-22 00:40:26

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

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