.999... is equal to one.
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The voting period for this debate has ended.
after 3 votes the winner is...
PotBelliedGeek
Voting Style:  Open  Point System:  7 Point  
Started:  12/17/2013  Category:  Miscellaneous  
Updated:  3 years ago  Status:  Post Voting Period  
Viewed:  611 times  Debate No:  42540 
Debate Rounds (4)
Comments (3)
Votes (3)
I have full BoP, .9r will represent .9 repeating to keep this debate easy to write. First round for acceptance only.
Fire away. 

For my first proof, I will take ,9r, and put it on both sides of an equation. .9r=.9r *10 10(.9r)=9.9r .9r 9(.9r)=9 /9 .9r=1 For my second proof, I will take an irrational number, e, and subtract its next highest integer. 3.00000... 2.71828... Looking at the digits, there is only one answer that makes sense. .28171... Now add it to e. 2.71828... +.28171...= 2.99999... As there is no end to either number, there is no term that can "bump up" any final 9 to a 10. Following this logic, the below equation ust be true. e+(3e)=2.9r Associative property e+3e=2.9r ee=0 3=2.9r 2 1=.9r My final argument is that both .9r and 1 are rational numbers, and the difference between the two must be able to be written as a rational number. While the obvious answer may be .0r1, there are two flaws with this. The first is that this is saying the one comes after infinity, which means it doesn't exist. Second, even if we did accept this logic, .0r1+.9r would be .9r1. In no way can a number be written to express 1.9r. For my first proof, I will take ,9r, and put it on both sides of an equation. .9r=.9r *10 10(.9r)=9.9r .9r 9(.9r)=9 /9 .9r=1 Here my opponent attempts to prove his assertion through algebra. He makes an error in his understanding of multiplying infinite decimals. My opponent assumes that when one multiplies said infinate decimal by ten, one simply moves the decimal place one unit to the right. Example: 9r(10)=9.9r The error here is that in this case, the number of digits after the decimal must in theory increase by a factor of ten. As the number was infinity, an increase by factor of ten remains infinity. This is paradoxical, and so as is taught in collegiate level Calc2 classes, the number of digits must first be defined before multiplying. Example: .999...(10)=9.999....0 This error in understanding infinate decimals threw off my opponents algebra. One may refute this by saying "infinate decimals can be expressed as fractions, and fractions can be multiplied easily." According to Dr. Zacheus Oguntebi of GPC, infinate decimals cannot in reality be expressed as decimals. Example: 1/3=/=3.3r it is an approximation, and cannot be expressed any other way, according to Zeno's paradox. Looking at the digits, there is only one answer that makes sense. .28171... Now add it to e. 2.71828... +.28171...= 2.99999... Here my opponent makes the error of using a calculator to discuss infinite theory. A calculator is incapable of accurately depicting and/or calculating infinite decimals, and so defaults to ten digits. This can be confirmed using any calculator. My opponents proof is faulty. As there is no end to either number, there is no term that can "bump up" any final 9 to a 10. Following this logic, the below equation ust be true. e+(3e)=2.9r Associative property e+3e=2.9r ee=0 3=2.9r 2 1=.9r My opponent bases this clause on the previous, and I have already illustrated the error therein. My final argument is that both .9r and 1 are rational numbers, and the difference between the two must be able to be written as a rational number. While the obvious answer may be .0r1, there are two flaws with this. The first is that this is saying the one comes after infinity, which means it doesn't exist. Second, even if we did accept this logic, .0r1+.9r would be .9r1. In no way can a number be written to express 1.9r Again, this can be refuted by a simple reference to Zeno's paradox. The difference between them is growing ever smaller and continues to do so as it draws infinity close to zero. This concept is the very basis of the entire college algebra 101 level. 

SeventhProfessor forfeited this round.
I extend all rebuttals. 

SeventhProfessor forfeited this round.
I have provided logical rebuttals for all of my opponents rebuttals. My opponent has forfeited the latter rounds of this debate. 
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3 comments have been posted on this debate. Showing 1 through 3 records.
Posted by 9spaceking 2 years ago
wow...Pot BelliedGeek disproved the commonlyaccepted math theory...
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Posted by PotBelliedGeek 3 years ago
No prob. I was wondering tho. You don't strike me as the kind of person to do something like that.
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Posted by SeventhProfessor 3 years ago
Sorry about that, I had to leave DDO for a while.
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3 votes have been placed for this debate. Showing 1 through 3 records.
Vote Placed by Beverlee 3 years ago
SeventhProfessor  PotBelliedGeek  Tied  

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Total points awarded:  0  4 
Reasons for voting decision: Math is not my strong suit, but Con did a good job of making this understandable. I give arguments and Conduct for the forfeiture.
Vote Placed by kbub 3 years ago
SeventhProfessor  PotBelliedGeek  Tied  

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Total points awarded:  0  6 
Reasons for voting decision: Dang con, way to go! That was a really good rebuttal. Nice job quoting doctor whoever, sources go to you for that. That was a tough debate to answer, but well done! Oh and ff. Really we'll done on Botha sides actually. Nice opening argument seventh professor.
Vote Placed by TheUser 3 years ago
SeventhProfessor  PotBelliedGeek  Tied  

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Total points awarded:  0  1 
Reasons for voting decision: Forfeits