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Would the Four Colour Problem Solution! still Stand in the Universe! with things Like Black Holes! where an Area' s Dimension is in More than one Place at the Same time!!
A.R.B
To Maelin
Quote:
"Sorry, are you suggesting that 0.999... (infinite digits long) is equal to 0.9 (one digit long)? Or that 0.999 (three digits long) is equal to 0.9 (one digit long)?"
It is According to The Infinite Difference Remaining the Same! No Matter How Long they
Both are!
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 "
In my Formula Above D will always have a Value! so that Only ( C + D ) or A can = 1
To Maelin
I do agree with you! Infinite 0.9 Starts as 0.9 and so Remains 0.9 forever! no matter How long it is!!
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!
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! "
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
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.
To Maelwys
First it's important to look at the Calculated Facts in my,
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INFINITE 0.9 <> 1 PROOF : FORMULA : 12/02/07
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( 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
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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 "
[ By posting this theory here, you willingly allow others to discuss the point in a civilized manner. Moderator action has been taken against those who are not civilized by Ricky and Zach. ]
INFINITE 0.9 <> 1 PROOF By,Anthony.R.Brown,15/01/07.
The Variables below A & B are Both Single Start Values.
A = 1
B = 0.9
Below Proves A is Infinitely > B And Gives The Variable C This Value.
C = A / B = 1.1111111111......
Below is How Infinite 0.9 is Calculated.
0.9 x C = 0.9999999999 ......
Below Proves Infinite 0.9 is <>1 By Using The Original Variables A & B.
A x C = 1.1111111111 ......
B x C = 0.9999999999 ......
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The Variables below A & B are Both Single Start Values.
A = 1 "This has to be the Start Value for 1"
B = 0.9 "This has to be the Start Value for Infinite 0.9"
C = A / B = 1.1111111111......
A x C = 1.1111111111 ......"This is the same Calculation to Make a Number Infinite! as with B"
B x C = 0.9999999999 ......"This is now Infinite 0.9"
The two Values for A and B are different! because Infinite 0.9 <> 1
If the Result for A did Equal B then Infinite 0.9 does = 1
Even If! by some Magic B does become 1 it wil still be Infinite 0.11111..... Less than A
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A starts as = 1
B starts as = 0.9
A x C results in a number > 1
B x C results in a number < 1
The point is A is 1
and B never = 1
If we are going to Infinitely multiply a number so that is becomes 1 "if possible"
Then a number that is already 1 must always be Greater!!
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A starts as = 1
B starts as = 0.9
A x C results in a number > 1
B x C results in a number < 1
The point is A is 1
and B never = 1
If we are going to Infinitely multiply a number so that it becomes 1 "if possible"
Then a number that is already 1 must always be Greater
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A x C = 1.11111...
B x C = 0.99999....
It Only takes One Calculation for A x C to = 1.1111.....Because A is! and A eqauls 1
It takes B x C to = 0.9999999..... Which is One Calculation Because B is < 1 to start with!
If B was Now to Some how now Reach! or Equal 1 by Multiplication! this would only then = 1
This would be would still make B 0.1 < A because B never started as 1 or = 1 the same as A
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LET'S MAKE THIS AS CLEAR AS POSSIBLE! USING NEW VARIABLES.
N1 = 0.9 This is a Single Number Value
N2 = Infinite 0.9 This is an Infinite Number!
N3 = 1 This is a Single Number Value
N1..............( Here between N1 and N3 there are no other Numbers! Because N1 is the Single last Number!).............N3.
N2..............( Here between N2 and N3 there are an Infinite Amount of Numbers! Because N2 is an
Endless stream of Numbers!).............N3.
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If a Number has to be Infinitely Multiplied! To Try and Reach a Target Value! Then the Target Value! Must be Greater than the Number being Infinitely Multiplied! Reason Being! If the Target Value was also Infinitely Multiplied! Then the Target Value would again be Greater!
Number Infinitely Multiplied! = 0.9 Target Value! = 1
[Anthony.R.Brown 20/01/07
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I would now like to Add another Variable to this Example! N4
This is for anyone who feels it's not possible to place another Number between N2 and N3.
N4 = The Maximum possible Number of .9's
Anyone who feels its not possible to place another Number between N2 and N3, all they have to do to prove how they feel! Is giving a Value for N4.
So now we have!.....
N2..............( Here between N2 and N3 there are an Infinite Amount of Numbers! Because N2 is an
Endless stream of Numbers!).............N3.
N2..............( How Ever! If..N4...is proved! Then the above is not True!).............N3.
N2..............( But as it Stands!..N4...Can have + .9's i.e More .9's Between! Because N2 is Infinite!).............N3.
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LET'S MAKE THIS EVEN MORE CLEARER!
N1 = 0.9 This is a Single Number Value
N2 = Infinite 0.9 This is an Infinite Number!
N3 = 1 This is a Single Number Value
N4 = The Maximum possible Number of .9's ?
N1..............( Here between N1 and N3 there are no other Numbers! Because N1 is the Last Single Number! Before ).............N3.
N2..............( Here between N2 and N3 there are an Infinite Amount of Numbers! Because N2 is an Endless stream of Numbers! Before ).............N3.
Below if Anyone can Give! and Prove! a Value for N4 Then the Above will be False! and not True! for then another .9 will not be able to be Inserted Between N2..and..N3.
N2..............(.N4.)...........N3.
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N1 = 0.9 This is a Single Number Value And Yes! .9 is The Last Digit! Because its not an
Infinite Number the same as N2
N4 is the Variable to Represent the .9,s which is Between N2 and N3
Because N4 is an Infinite number then Any amount of .9,s can be Inserted to Represent N4
Unless! " Anyone can Give! and Prove! a Value
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N1 = 0.9 This is a Single Number Value
and is the last Single Value then 0.2,0.3,0.4,0.5,0.6,0.7,0.8, Then N1 and Then 1
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Below is Needed because No One! is prepared to put forward an Example,in a line by line way! the same as my Proof! to explain what they are trying to say!
The main problems are not saying at what stage A & B are at! i.e Single Start Values! or Calculated x C.
I am going to Add some new Variables so everyone knows what is Happening!
Sometimes new Variables have to be Add to explain things in an easier way,a good example is the N2 = Infinite 0.9999..."is it possible to insert anymore Numbers Between! " N3 = 1
as soon as I introduced the N4 Variable = The Maximum possible Number of .9's ?
Everyone was stumped! Because as they know as soon as anyone gives a Value for N4,I will answer with N2..............(.N4 + "as many more .9s I want to Insert!.)...........N3.
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INFINITE 0.9 <> 1 PROOF By,Anthony.R.Brown,29/01/07.
The Variables below A1 & B1 are Both Single Start Values.
A1 = 1 "This has to be the Start Value for 1"
B1 = 0.9 "This has to be the Start Value for Infinite 0.9"
C = A1 / B1 = 1.1111111111...."There is only one Value & Variable for C"
A2 = A1 x C = 1.1111111111 ......"This is the same single Calculation as B2"
B2 = B1 x C = 0.9999999999 ......"This is now Infinite 0.9 and a single Calculation"
If B1 or B2 was at anytime to = A1 Then either would end up with the same Result as A2,but because it will take more than one single Calculation! for B1 or B2 to = A1
Then its Impossible from the Start for Infinite 0.9 to ever = 1 in a single Calculation.
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INFINITE 0.9 <> 1 PROOF By,Anthony.R.Brown,02/02/07.
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( 1.1 ) x 0.9 = 0.99 " One Decimal Place = " 0.1 < 1
( 1.11 ) x 0.9 = 0.999 " Two Decimal Place's = " 0.1 < 1
( 1.111 ) x 0.9 = 0.9999 " Three Decimal Place's = " 0.1 < 1
( 1 / 0.9 ) x 0.9 = 0.9..... " Infinite Decimal Place's = " 0.1 < 1 "
The 0.1 Difference Above is Permanent! Because it is an Infinite Difference!!
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To Back up my Proof above! I would like to Make an Immortal Quote:
" IF TWO NUMBERS START WITH A DIFFERENCE IN THEIR VALUES! AND BOTH ARE MULTIPLIED BY THE SAME AMOUNT! ONE VALUE WILL ALWAYS BE GREATER THAN THE OTHER!! "
----------------------------------------------------------------------------------------------------------------INFINITE 0.9 <> 1 PROOF : CONCLUSION By,Anthony.R.Brown,06/02/07.
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After all the Thoughts! and Arguments! the Solution is really Quite Simple!
V1 = 0.9999999999...." = The 0.9 Value from the Start Onwards! "
V2 = 0.0000000001...." = The < 1 Value from the Start Onwards! "
(a) The Above will always be True! as Calculated in the Equation Below
( V1 + V2 ) = 1
(b) There will always be Two Values from the Start Onwards
(c) Neither of the Individual Values will ever Equal 1
(d) There will always be an Infinite Difference ( V1 <> V2
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I thought I made it Clear!
V1 = 0.99999999999999999999999999999999..Infinite.0.9...." = The 0.9 Value from the Start Onwards! "
V2 = 0.00000000000000000000000000000000..Infinite.0.1...." = The < 1 Value from the Start Onwards! "
V1 = Infinite 0.9
V2 = The Infinite Difference
No Matter how Long V1 is! V2 will always be the same Length!
Because V1 starts as 0.9 and Because V2 starts as 0.1 The Difference will always be the same!
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INFINITE 0.9 <> 1 PROOF : TRUTH TABLE : By,Anthony.R.Brown,07/02/07.
----------------------------------------------------------------------------------------------------------------Below V1 = The Infinite 0.9 Sequence. : Below V2 = The < 1 Infinite Difference Sequence.
V1 = ( 0.9 ) + V2 = ( 0.1 ) = 1
V1 = ( 0.99 ) + V2 = ( 0.01 ) = 1
V1 = ( 0.999 ) + V2 = ( 0.001 ) = 1
V1 = ( 0.9999 ) + V2 = ( 0.0001 ) = 1
V1 = ( 0.99999 ) + V2 = ( 0.00001 ) = 1
V1 = ( 0.999999 ) + V2 = ( 0.000001 ) = 1
V1 = ( 0.9999999 ) + V2 = ( 0.0000001 ) = 1
V1 = ( 0.99999999 ) + V2 = ( 0.00000001 ) = 1
V1 = ( 0.999999999 ) + V2 = ( 0.000000001 ) = 1
V1 = ( 0.9999999999 ) + V2 = ( 0.0000000001 ) = 1
The Table above clearly shows no matter how long V1 is! it will always need the algebra + symbol for V1 to equal 1 ( V1 + V2 ) = 1
This proves Infinite 0.9 <> 1 because if ( V1 + V2 ) = 1 is allowed! it will be a contradiction to the term Infinite 0.9.
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1 - 0.9 = 0.1 " Notice this is a number! with a 1 on the end " " If it's not? then funny if you add " 0.1 to 0.9 you get 1
1 - 0.99 = 0.01 " Notice this is a number! with a 1 on the end " " If it's not? then funny! if you add " 0.01 to 0.99 you get 1
1 - 0.999 = 0.001 " Notice this is a number! with a 1 on the end " " If it's not? then funny! if you add " 0.001 to 0.999 you get 1
The Above is Only a Small sample from the Start Onwards! But I can Assure you! is Infinitely possible!
Because 1 - ( Another Number < 1 ) Will always have an Infinite Difference!!!
P.s the 1 "at the End!" and the 9 "at the End!" is a term to decribe! at the End of an Infinite Sequence! for as far as we have Calculated! and can see on paper!
We all know it's not an actual End!! but in the Above three examples,we can see them at the End Three Times!!
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INFINITE 0.9 <> 1 PROOF : THE INFINITE 10 % PERCENT DIFFERENCE 03/03/07
by Anthony.R.Brown.
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One of the most Accurate and sound ways to prove Infinite 0.9 <> 1 is to give the Infinite Percent Difference Values from the Start Onwards!
The example below is for the first Ten Decimal Place's,which clearly shows the count increasing as more and more Infinite .9's are looked at! But remain a constant Difference of 10% in relation to how far the example has traveled.
Count = ...... ( 1 ).....( 2 )....( 3 )...( 4 )....( 5 )...( 6 )....( 7 )....( 8 )....( 9 ).....( 10 )
Infinite 10 %
Difference =..(10%).(10%).(10%).(10%).(10%).(10%).(10%).(10%).(10%)...(10%)
Infinite 0.9 = ( 0.9 ) ( 1.8 ) ( 2.7 ) ( 3.6 ) ( 4.5 ) ( 5.4 ) ( 6.3 ) ( 7.2 ) ( 8.1 ) ( 9 )
................... ( x 1 ) ( x 2 ) ( x 3 ) ( x 4 ) ( x 5 ) ( x 6 ) ( x 7 ) ( x 8 ) ( x 9 ) ( x 10 )
Infinite 0.9 Onwards! there will always be a 10 % Difference making it Impossible for Infinite 0.9 to ever = 1
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INFINITE 0.9 <> 1 PROOF : FORMULA : By,Anthony.R.Brown,12/02/07
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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 "
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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
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INFINITE 0.9 <> 1 PROOF : FORMULA : By,Anthony.R.Brown,12/02/07
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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 "
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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 ?
To
krassi_holmz
I just heard that the Genius "Michael Faraday (1791-1867)"
Never hardly used any Math!
A.R.B
Quotes:
"Well, if we have an infinite number of nines after the decimal place, you cannot do this, since the 9's never end."
"So the answer to your question is an infinite number of 9's after the decimal place."
So now we all agree! there is no End to the .9's
To Maelwys
Quote:
"I agree with your first three statements. All of those numbers do add together to get 1. Your fourth statement however ("The Above is Only a Small sample from the Start Onwards! But I can Assure you! is Infinitely possible!") is a fallacy. Your pattern holds for every number of a finite series only."
If my pattern only holds for every number of a finite series only ?
Can you Give me the ( Maximum Possible finite series for 0.9 ) onwards!
and Can you Give me the ( Maximum Possible finite series for 0.1 ) onwards!
How Sad!!
1 x 0 = 1
0 x 0 = 0
I need more of a Challenge!!
A.R.B
1 - 0.9 = 0.1 " Notice this is a number! with a 1 on the end " " If it's not? then funny if you add " 0.1 to 0.9 you get 1
1 - 0.99 = 0.01 " Notice this is a number! with a 1 on the end " " If it's not? then funny! if you add " 0.01 to 0.99 you get 1
1 - 0.999 = 0.001 " Notice this is a number! with a 1 on the end " " If it's not? then funny! if you add " 0.001 to 0.999 you get 1
The Above is Only a Small sample from the Start Onwards! But I can Assure you! is Infinitely possible!
Because 1 - ( Another Number < 1 ) Will always have an Infinite Difference!!!
P.s the 1 "at the End!" and the 9 "at the End!" is a term to decribe! at the End of an Infinite Sequence! for as far as we have Calculated! and can see on paper!
We all know it's not an actual End!! but in the Above three examples,we can see them at the End Three Times!!
To Dross!
Do Something Constructive and answer the Question Below!
Instead of trying to make your own self centred opinion about everything!
I waste so much time Paging past the Rubbish you Post on this Forum!
All we want from you is Answers Only! to do with Math! Keep your opinions to your self!!
From another post of mine! can you give a Value for N4 = " Maximum possible Infinite 0.9's "
Because I can always give ( N4 ) + another
To George,Y
Wonder full Stuff! Now I think the Penny is Starting to Drop! with the Non - Believers!
I think we will have to set up a post for the!
" Past Proved Wrong! Math is Fun Forum Members! "
A.R.B
" IF TWO NUMBERS START WITH A DIFFERENCE IN THEIR VALUES! AND BOTH ARE MULTIPLIED BY THE SAME AMOUNT! ONE VALUE WILL ALWAYS BE GREATER THAN THE OTHER!! "
P.s I do need to add! to the above Quote: refers to Numbers Only! and No Algebra Symbols!!
A.R.B
Any Challenges on this!
To
kylekatarn & Dross
Don't worry we all detect your Humorous Nervousness! it's always a sign when confronting the Truth!
A.R.B
To kylekatarn
From another post of mine! can you give a Value for N4 = " Maximum possible Infinite 0.9's "
Because I can always give ( N4 ) + another .9
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INFINITE 0.9 <> 1 PROOF : TRUTH TABLE : By,Anthony.R.Brown,07/02/07.
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Below V1 = The Infinite 0.9 Sequence. : Below V2 = The < 1 Infinite Difference Sequence.
V1 = ( 0.9 ) + V2 = ( 0.1 ) = 1
V1 = ( 0.99 ) + V2 = ( 0.01 ) = 1
V1 = ( 0.999 ) + V2 = ( 0.001 ) = 1
V1 = ( 0.9999 ) + V2 = ( 0.0001 ) = 1
V1 = ( 0.99999 ) + V2 = ( 0.00001 ) = 1
V1 = ( 0.999999 ) + V2 = ( 0.000001 ) = 1
V1 = ( 0.9999999 ) + V2 = ( 0.0000001 ) = 1
V1 = ( 0.99999999 ) + V2 = ( 0.00000001 ) = 1
V1 = ( 0.999999999 ) + V2 = ( 0.000000001 ) = 1
V1 = ( 0.9999999999 ) + V2 = ( 0.0000000001 ) = 1
The Table above clearly shows no matter how long V1 is! it will always need the algebra + symbol for V1 to equal 1 ( V1 + V2 ) = 1
This proves Infinite 0.9 <> 1 because if ( V1 + V2 ) = 1 is allowed! it will be a contradiction to the term Infinite 0.9.
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To Maelwys
Quote 00.43.22
"
The main one is your definition of V2. You can't have a number that contains and infinite number of 0s, followed by a 1. Infinite means "never ending". That means that as long as you can put another number at the "end" of the string of numbers, it MUST be another zero "
A.R.B "There is No zero at the end of 0.999999......"
You are Missing the point of what I am saying!
We Cannot actually write down all the 0.9999's
All we know is that it starts 0.1 < 1 There is no reason possible! why at some stage Later!
The missing 0.1 is Add to 0.9
Maelwys
Quote 00.43.22
" Therefor, there is no possible number that could be added to 0.999... to give 1. And THAT, is why 0.999... equals 1, because if there's no possible number that you could add to it to achieve 1, then there's no difference between the two, which means that they're the same number."
A.R.B
We are not trying to find a Number that can be Add to 0.999... to give 1. we already know 0.999.. will always have it missing! and that's fine!
As I have already said " I am Quite Happy Knowing there is a Number Infinitely Smaller than 1
A Quote: from the Burnt! Toaster!
" Why don't we just forfeit... it would be so much easier..."
A.R.B
Thank you Toast! The First sensible Post I have ever seen you put forward!!