0.9999... equals 1
Vote Here
The voting period for this debate does not end.
Voting Style:  Open  Point System:  7 Point  
Started:  3/27/2010  Category:  Miscellaneous  
Updated:  7 years ago  Status:  Voting Period  
Viewed:  1,790 times  Debate No:  11563 
Debate Rounds (3)
Comments (10)
Votes (2)
The parameters of this debate are as follows:
1. By E(i=a,b) x_i, I mean the summation over a set of numbers. (1) 2. By inf, I mean infinity. 3. By 0.999..., I mean a decimal number with an infinite number of nines, represented by E(i=1,inf) 9/(10^i) 4. By "A is equal to B", I mean that A is the same number, regardless of its possible decimal expressions, as B, such that they represent the same quantity. The proposition to be proved by the Pro is "0.999... can be shown to be equal to 1.0". (1) http://mathworld.wolfram.com...
Mathmatically and logically .9999... does NOT equal one. First of all lets look at math. Lets pretend there is a space and this space is infinite. 1 represents the distance out 5 you will be traveling across this infinite expanse. Same with .999... . Because this space is infinite the fact that .999 goes on infinitely is irrelevant because it will end up infinitely away from 1 in this imaginary space. Only a distance of 1 can land at the same place as 1. Also , logically, the only thing equal to 1 is 1. No matter how close something is to it 1 will be the only thing 1 is equal to. Finally I'll bring up a metaphor. If there is a pot in greece from old times (this represents one.) no matter how close we get to making a replica of this vase (.999...) it will NEVER be the original vase. So no, 1 is not equal to anything but itself (and please dont bring up something stupid like '21 = 1'. 

leafy forfeited this round.
The pro has forfeited the round. This means two things. First of all, it means that my arguments still stand. Second, I have no alternative than to believe that the opponent agrees with what I say because of his inability to refute my arguments. I hope my opponent decides to say something and I look forward to his response. Thank you. 

I did forfeit the round, but not voluntarily. I had forgotten about the debate  Con's inferences to my motives for inadvertantly forfeiting show his lack of robust logic.
While there are many arguments that 0.999... equals 1, I will not be presenting less rigorous arguments already debated at DDO. I will here present the argument from the limit definition. Firstly, a formal definition of the expression 0.999... is needed. What do we mean when we say 0.999...? When we refer to 0.999..., the elipses following the nines mean an infinite number of nines are following the zero. This could be represented in summation form as follows: E(i=1,inf) 9/(10^i) This is exactly what we mean when we refer to 0.999...: We mean nine tenths (0.9) plus nine hundredths (0.09) plus nine thousandths (0.009), and so on  the 9/(10^i) means that the i is the current index of summation  the place we are in the decimal chain. Observe that for the first three summation terms I enumerated (0.9, 0.09, 0.009) correspond to the first three summation terms (9/10, 9/(10^2) = 9/100, 9/(10^3) = 9/1000). Hopefully this demonstration is adequate to show that this summation expression is the same concept we refer to when we say 0.999... Next, we must consider the convergence of this sum. Sums can sometimes converge on a single value. There are many tests used in mathematics to determine this (**1**). Let's consider one test: the ratio test. If the limit as k approaches infinity of the (k+1)th number in the series divided by the kth number in the series is less than 1, then the series converges. If the limit is more than one or does not exist, then the series diverges. If the limit is equal to 1, then the test is inconclusive (**2**). To apply this to the series, we must find algebraic formulas for the kth number and the (k+1)th number. The first number is 0.9, or 9/10^1. Note it corresponds to the summation notation earlier: E(i=1,inf) 9/(10^i). The ith term is when you substitute in i in the notation. The kth term would then be 9/10^k, and the (k+1)th term would be 9/10^(k+1). By substituting in the formula for the ratio test, the limit is equal to (9/(10^(k+1))) / (9/(10^k)) (9/(10^(k+1))) * ((10^k)/9) 10^k / 10^(k+1) 1/10 The limit is equal to one tenth. Thus, by the ratio test, the series converges. To find what the series converges on, we can actually solve it since it is an example of a geometric series, or a series with a constant ratio (1/10) between each term (**3**). For a sum from k=0 to infinity of a * r^k, the solution is a/(1r). In our example, a is 9/10, r is 1/10. Thus, the solution is a / (1  r). 9/10 / (1  1/10) is 9/10 divided by 9/10. Thus, the sum is equal to 1. I look forward to Con's response to this proof. (1) http://mathworld.wolfram.com... (2) http://mathworld.wolfram.com... (3) http://en.wikipedia.org...
Alright. The Aff has tried to prove that .99999... = 1 by showing a math equation, which will finally result in .9999.... = 1. This math WILL in fact result in the Aff being true, but what he doesn't understand is that our math is not advanced enough to calculate infinity. Using infinity in our math can lead to just about ANYTHING in the world I want it to. I will give and example here. (inf)*1 = 3000*(inf) First let's work it our to see if it's true. (inf)*1 = 3000*(inf) (inf)*1 = (inf) (inf)*3000 = (inf) (inf) = (inf) So the expression: (inf)*1 = 3000*(inf) is obviously true. Not let's take that expression and work it in a different way. (inf)*1 = 3000*(inf) (inf)*1 = 3000*(inf)   (inf) (inf) (Divide infinity by both sides) 1 = 3000 So using our mathematics with infinite terms, it is obvious that one is equal to threethousand. This, of course, is illogical and therefore we must abdicate our use of these mathematics and resort to logic (or the logical mathematics I presented earlier). As I proved earlier it is illogical for .9999.... to EVER equal 1. One is equal to one, and no matter how close anything EVER gets to it, there will never be another 1. It is because my opponent missed 1/3 of the match and because I refuted his arguments using a combination of logical math and pure logic that you should vote Negative. Thank you. 
2 votes have been placed for this debate. Showing 1 through 2 records.
Vote Placed by Nails 7 years ago
leafy  Mr_Jack_Nixon  Tied  

Agreed with before the debate:      0 points  
Agreed with after the debate:      0 points  
Who had better conduct:      1 point  
Had better spelling and grammar:      1 point  
Made more convincing arguments:      3 points  
Used the most reliable sources:      2 points  
Total points awarded:  0  4 
Vote Placed by Mr_Jack_Nixon 7 years ago
leafy  Mr_Jack_Nixon  Tied  

Agreed with before the debate:      0 points  
Agreed with after the debate:      0 points  
Who had better conduct:      1 point  
Had better spelling and grammar:      1 point  
Made more convincing arguments:      3 points  
Used the most reliable sources:      2 points  
Total points awarded:  2  4 
1/3 * 3 = 1
1/3 = .333(3)
.333(3) * 3 = .999(9)
one third(1/3) multiplied by three(3) equals both .999(9) and 1, so they must be equal. I gave PRO before and after.
That said, PRO's arguments (especially in light of CON's rebuttal) don't sufficiently prove the resolution true to me, so I give CON arguments. CON also gets conduct because PRO forfeited.
NOTE: This is why you meet your deadlines. That last rebuttal would definitely have made a difference. Starting your arguments in the last round puts you at a strategic disadvantage since you can't contest your opponent's argument or better explain your own.
lim as x goes to infinity of 1/x is zero, because the function gets very close  arbitrarily close  to 0 as x increases without bound.
tl;dr take a calculus course.
Try the Negative 0 debate.