The Instigator
Shadowguynick
Pro (for)
Losing
1 Points
The Contender
kiddo
Con (against)
Winning
11 Points

0.999.... can't equal 1

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Post Voting Period
The voting period for this debate has ended.
after 3 votes the winner is...
kiddo
Voting Style: Open Point System: 7 Point
Started: 7/22/2013 Category: Science
Updated: 3 years ago Status: Post Voting Period
Viewed: 707 times Debate No: 35862
Debate Rounds (3)
Comments (5)
Votes (3)

 

Shadowguynick

Pro

Con must providde evidence as to why 0.999... Can equal
Con may start immediatly so I can refute.
Good luck.
kiddo

Con

I shall use R1 for acceptance and leave side pro to provide arguments before presenting mine.

I shall present in
R2: Arguments and rebuttals
R3: Closing statement (rebuttals if pro acts in the same manner)
Debate Round No. 1
Shadowguynick

Pro

First of I would like to state that 0.999.... is practically a good option, and should continue in use as it would mess up a fewthings were we to change it. Second off the biggest problem with 0.999... repeating should not be 1 simply because even at infinite 9s there is still infinite more ways to make it smaller. People say we can't get that 0.000..;1 to fill in the number so we should assume it's not theren However we should assume it's there because that would defeat the purpose of 0.999.... repeating because it would mean it was both 1 and infinitely less than 1. This obviously doesn't hold much weight. I await my opponent tomake his arguments. I would also like to note that this is a learning experience personally, to see if I missed any evidence concerning this issue. So bring up anything concerning this issue. Thank you.
kiddo

Con

Thank you once again Pro.
to express this mathematically
Pro arguments are inversely proportional to Con arguments
therefore, my arguments serves as rebuttals already in this case against Pro's arguments as well as a point.

Argument 1: Fraction
now lets put the equation into fractions by dividing 3
when 0.999... divides by 3, it equals to 0.333... also 1/3
when 1 divides by 3, it equals to 1/3 directly

x=0.999.... y=1
so here, x/3 = y/3
therefore, X=Y

Argument 2: Algebra
let x= 0.999...
10 x = 9.999...
10x - x = 9
9x = 9
x = 1

Argument 3: Additional mathematics geometrical progression
0.999...
Separate into: 0.9 + 0.09 + 0.009....
in this case a=0.9 r=0.1

Using sum to infinity formula
a/1-r

=0.9/ 1 - 0.1
=0.9/0.9
=1

I had fun revising my maths and add maths for tmr's paper, and still this is fun :D

http://en.wikipedia.org...
http://www.purplemath.com...;
http://www.askamathematician.com...;
Debate Round No. 2
Shadowguynick

Pro

I thank my opponent very much. His aegument was thrilling to read. And I will concede that mathematically and algebraicly the notion holds true. Bit I still don't know how it is two values at the same time. Perhaps i am just understanding how math works, butbi was always taught that multiple representations may represent one value, but one representation does't represent two.. Thank you for your insights, and I ask you once again to prove me wrong.
kiddo

Con

I once again thank my opponent for bringing up the debate, Truth be told, i think that is all the methods i have up my sleeves. Sorry ><
Truthfully it was really fun to apply what additional mathematics has taught me. Coming back from the exam, I think I may pass the paper :D Pro is a fine example of a debater with good conduct and I wish he will continue with an ever-burning flare and well mannered etiquette as he is doing now.
Debate Round No. 3
5 comments have been posted on this debate. Showing 1 through 5 records.
Posted by Shadowguynick 3 years ago
Shadowguynick
Thank you, and I see why it is considered 1. Mathematics can be pretty illogical at times.
Posted by Shadowguynick 3 years ago
Shadowguynick
Arguement*
Posted by Shadowguynick 3 years ago
Shadowguynick
I would say con held the burden of proof. But i will put forth my own arfument anway.
Posted by DT 3 years ago
DT
0.9 = 1 - (1/10)^1
0.99 = 1 - (1/10)^2
0.999 = 1 - (1/10)^3
0.9999 = 1 - (1/10)^4
0.99999 = 1 - (1/10)^5

The number 0.999999 with infinite progression of 9s can be expressed as a limit,

lim (1 - (1/10)^x) = 1, where x -> infinity

However, people forget that this answer is true only when applied with the limit theorem. All other approaches to the problem is essentially an expression of the limit theorem but re-stated in a different way.

But by definition the limit is merely the approximation of the convergence, and never really equal to the actual number it converges to.
Posted by kiddo 3 years ago
kiddo
who holds the burden of proof?
I am interested
3 votes have been placed for this debate. Showing 1 through 3 records.
Vote Placed by MisterDeku 3 years ago
MisterDeku
ShadowguynickkiddoTied
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Total points awarded:03 
Reasons for voting decision: There was definitely room for Pro to refute con's arguments. He just didn't take it. Pro conceded too quickly which is disappointing.
Vote Placed by Juan_Pablo 3 years ago
Juan_Pablo
ShadowguynickkiddoTied
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Total points awarded:05 
Reasons for voting decision: Con won the debate and held the day! Pro's interest in the topic shows an interest in wanting to understand it - which reflects well on his character. One of Con's sources explains the equality very well: http://en.wikipedia.org/wiki/0.999...#Fractions_and_long_division
Vote Placed by johnlubba 3 years ago
johnlubba
ShadowguynickkiddoTied
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Total points awarded:13 
Reasons for voting decision: Well done Con and conduct to Pro for conceding, to be honest if he never I could never tell if Con's argument was sound, but it appears to be.