The Instigator
SeventhProfessor
Pro (for)
Losing
0 Points
The Contender
PotBelliedGeek
Con (against)
Winning
11 Points

.999... is equal to one.

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Post Voting Period
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: 663 times Debate No: 42540
Debate Rounds (4)
Comments (3)
Votes (3)

 

SeventhProfessor

Pro

I have full BoP, .9r will represent .9 repeating to keep this debate easy to write.

First round for acceptance only.
PotBelliedGeek

Con

Fire away.
Debate Round No. 1
SeventhProfessor

Pro

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+(3-e)=2.9r
Associative property
e+3-e=2.9r
e-e=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.
PotBelliedGeek

Con

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+(3-e)=2.9r
Associative property
e+3-e=2.9r
e-e=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.
Debate Round No. 2
SeventhProfessor

Pro

SeventhProfessor forfeited this round.
PotBelliedGeek

Con

I extend all rebuttals.
Debate Round No. 3
SeventhProfessor

Pro

SeventhProfessor forfeited this round.
PotBelliedGeek

Con

I have provided logical rebuttals for all of my opponents rebuttals. My opponent has forfeited the latter rounds of this debate.
Debate Round No. 4
3 comments have been posted on this debate. Showing 1 through 3 records.
Posted by 9spaceking 3 years ago
9spaceking
wow...Pot BelliedGeek disproved the commonly-accepted math theory...
Posted by PotBelliedGeek 3 years ago
PotBelliedGeek
No prob. I was wondering tho. You don't strike me as the kind of person to do something like that.
Posted by SeventhProfessor 3 years ago
SeventhProfessor
Sorry about that, I had to leave DDO for a while.
3 votes have been placed for this debate. Showing 1 through 3 records.
Vote Placed by Beverlee 3 years ago
Beverlee
SeventhProfessorPotBelliedGeekTied
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Total points awarded:04 
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
kbub
SeventhProfessorPotBelliedGeekTied
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Total points awarded:06 
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
TheUser
SeventhProfessorPotBelliedGeekTied
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Total points awarded:01 
Reasons for voting decision: Forfeits