The validity of 0.99999... = 1
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after 2 votes the winner is...
Kaynex
Voting Style:  Open  Point System:  7 Point  
Started:  12/5/2014  Category:  Science  
Updated:  2 years ago  Status:  Post Voting Period  
Viewed:  901 times  Debate No:  66412 
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Well, here's a debate for fun. I am very much into math and this is how I roll. Anyway, the debate is a common one seen over the internet, 0.999... = 1. That is infinite 9's. I am arguing that the statement is valid. First round is acceptance.
I am debating that .99... does not equal 1. 

Thank you for accepting the debate AthiestPerson! Before I begin, I'm going to let λ = 0.99999... so it is easier to read the work. C1: They have no difference in value Subtracting λ from 1 results in zero, showing they have no difference in value. They can be used similarily in an equation. Therefore, λ = 1 C2: You cannot make it "not 1" λ to any power is still λ. This is a property that only the number 1 is capable of. a*λ is infinitessimally less than a. This is also a property that only 1 is capable of. C3: There is Mathematical proof of this Let's begin by noting that λ can be made by just adding 9's infinitely. In summation notation: λ = Σ [9 / 10^n]  {The sum goes from 1 to infinity} λ = 9 Σ [1 / 10^n] λ = 9 Σ [1 / 10]^n This is just a geometric sequence, and can be solved using: Σ a^i [from i=m to n] = [a^m  a^n] / [1  a] λ = 9[(1/10)^1  (1/10)^infinity] / [1  (1/10)] λ = 9[1/10] / [9/10] λ = 9[1/10][10/9] λ = 1
I have no clue how to use that symbol, so my representation will be ]{= .99999... First, let me ask you a question; Do you think that .3333333 is equal to 1/3? What about .666666 being equal to 2/3? Now I shall begin. Here is my proof: The function y=11/x is often used to show how the repeating decimal 0.9999... is equal to 1. When x=1, y=1; x=10, y=0.9; x=10000, y=.9999, and so on. The limit of 11/x as x approaches infinity equals 1. An assumption is often made, however, that if the limit of an expression as x approaches infinity is 1, then that expression must equal 1 when x equals infinity. Assumption: 11/x = 1 when x = infinity Subtraction: 1/x = 0 Multiplication: 1 = 0x Zero Property: 1 = 0 1 does not equal 0, therefore 11/x does not equal 1 when x = infinity. You cannot treat "infinity" like a normal number, you can only think of it in terms of limits. Some things to think about: 1. Saying ]{ = 1 is like saying 9999999999 = 10000000000. The numbers are completely different. They change completely. 2. Just because someone should not wish to write down the whole decimal out every time doesn't mean that it becomes the number one. At any time, you could make a simple symbol that represents the repeating decimal./ 

No problem that you can't use the symbol, the work was very easily read. For future reference though, you can copypaste the symbol. Just like division by 0, multiplication by infinity makes no sense and cannot be done. For example, 3x5 = 3 + 3 + 3 + 3 + 3 = 15 But 3xinfinity = 3 + 3 + 3 + 3 + 3 + 3 +... It doesn't come to a definite answer as an operation demands. Essentially, this line: "Subtraction: 1/x = 0 Multiplication: 1 = 0x" Is flawed and cannot be done. Rebuttal to 1. λ = 1 is not like saying 9999999999 = 10000000000, because this would imply a finite amount of 9's. Instead λ = 1 IS like saying 9999999999.99999... = 10000000000, and that is a statement that should be expected, since I implied that λa = a. Statement 2 is a nonissue to the claim that λ = 1.
You win 

If a woodchuck could chuck wood, he would chuck a good amount of wood. 

Yes, the woodchuck would chuck as much wood as the woodchuck could if the woodchuck could chuck wood. 
2 votes have been placed for this debate. Showing 1 through 2 records.
Vote Placed by lannan13 2 years ago
Kaynex  AtheistPerson  Tied  

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Reasons for voting decision: Concession
Vote Placed by carriead20 2 years ago
Kaynex  AtheistPerson  Tied  

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Reasons for voting decision: Con conceded.
Let _5;^W34; = L
By my definition of _5;:
L = lim[nU94;inf] (]1;[1U94;n] 9/10^n)^n
That summation is still 1, regardless if I put the ^n on it or not:
L = lim[nU94;inf] 1^n
L = 1
But this is math on paper. It has nothing to do with reality. Math is governed by a different set of rules, and I followed them to get to my answer.
As L. Ron Hubbard stated in his 1950 book on Dianetics,
Absolutes have to be considered logically unobtainable.
Albert Einstein stated that the universe itself is finite but unbounded.
"Infinity" has no actual existence.
It's Impossible to get an infinite # of 9s.
Ergo, .999 ... cannot, & does not, = 1.
As a matter of fact, if you study Quantum Physics, 1 cannot exist.