The Instigator
MissLenaElan
Pro (for)
Tied
0 Points
The Contender
alric8
Con (against)
Tied
0 Points

One Point Nine Repeating is Equal to Two

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Post Voting Period
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Voting Style: Open Point System: 7 Point
Started: 4/18/2015 Category: Education
Updated: 2 years ago Status: Post Voting Period
Viewed: 533 times Debate No: 73731
Debate Rounds (2)
Comments (3)
Votes (0)

 

MissLenaElan

Pro

I have had heated arguments with several math teachers discussing whether or not 1.99999...=2. I believe it is, and will be arguing that point. I wish the best of luck to my opponent.
alric8

Con

Its recurring - not repeating.

1.9r is infinitesimally less than 2. I will use the following statements to prove it.

2 - 1.9r = n

N must be higher than 0 as proved by this equation.

n + 2 = n + 1.9r

If I was to subtract 1.9r from both sides I end up with an infinitesimal number. as supposed to 2.

1.9rs final digit is 9 and therefore it is an odd number. 2 is an even number. (if you don't rebut to that you then are a bit of a fail)

The factors of 4 are 1,4 and 2,2. 1.9r,1.9r don't give you the same answer. It is simply impractical to be precise.

Here is my most important argument though: Multiply 1.9r by ten and then it reduces in size one decimal place that means it is not infinitesimal and therefore it is definitely not the same. On the other hand 10(1+1) = 10(0.5+1.5) and therefore 10 X 2. For any two concepts to be the same you should be able to multiply them by 10 and get the same answer as proved.
Debate Round No. 1
MissLenaElan

Pro

Thank you for your acceptation of this debate! I've never heard of some of those arguments before, but I'll do my best to offer up a solid rebuttal and arguments of my own.

"2 - 1.9r = n

N must be higher than 0 as proved by this equation.

n + 2 = n + 1.9r

If I was to subtract 1.9r from both sides I end up with an infinitesimal number. as supposed to 2."

n+2=n+1.9r if considered true only proves my point. When n is subtracted from both sides, you're left with 2=1.9r. However, if the equation is true, your statement "N must be higher than 0" must be false and n=0 because when you do subtract 1.9r from both sides, you're left with 2n=n, which is only possible if n=0.

"1.9rs final digit is 9 and therefore it is an odd number. 2 is an even number. (if you don't rebut to that you then are a bit of a fail)"

Your very first statement is false. Assuming you are correct and 1.9r is less than 2, 1.9r would not be an integer. Parities (aka whether or not a number is even or odd) only apply to integers so 1.9r is neither if not equal to two.

"The factors of 4 are 1,4 and 2,2. 1.9r,1.9r don't give you the same answer. It is simply impractical to be precise."

The point I am arguing is that 1.9r=2, therefor anything divisible by two is divisible by 1.9r (see my arguments below).

"Here is my most important argument though: Multiply 1.9r by ten and then it reduces in size one decimal place that means it is not infinitesimal and therefore it is definitely not the same. On the other hand 10(1+1) = 10(0.5+1.5) and therefore 10 X 2. For any two concepts to be the same you should be able to multiply them by 10 and get the same answer as proved."

Let x=1.9r

Then, 10x=19.9r

So, 10x-x=19.9r-1.9=18.000...

So, 9x=18, which implies n=2.

My argument applies infinitely. Meaning if a is any integer, a.9r=a+1. When you substitute our aforementioned x for a, 10x=19.9r, 10x=20.

There are infinite numbers in existence, and yet there is no real number between 1.9r and 2. This is impossible, unless my previous statement and main argument is true and a.9r=a+1.

My favorite point has to do with devision. 1/3=0.3r. This should mean 0.3r*3=1, however different calculators give you different results, and some say 0.3r=0.9r. I agree with both statements, as they're both mathematically correct. When you add one to each side, you're left with either (0.3r*3)+1=2 or (0.3r*3)+1=1.9r. Because of the law (if a=b and a=c, b=c), that means 2=1.9r.

Thank you for the debate, and I look forward to reading your rebuttals and closing statement.
alric8

Con

Yes, well what is 2-x if x is infinitesimal? 1.9r, it cant be less.

Therefore I end up with this equation: x + n = n.

But since 2 - x = 1.9r, not 2, that equation is untrue.
This is pretty hard to argue, I would have loved to hear what your maths teachers said but anyway.
Debate Round No. 2
3 comments have been posted on this debate. Showing 1 through 3 records.
Posted by bluesteel 2 years ago
bluesteel
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>Reported vote: FuzzyCatPotato // Moderator action: REMOVED<

3 points to Pro (arguments). Reasons for voting decision: Mathematical proof

[*Reason for removal*] This is a vote bomb. It doesn't offer any sort of coherent explanation for why Con's argument was wrong or unpersuasive.
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Posted by vi_spex 2 years ago
vi_spex
find me 1,9 random object
Posted by MissLenaElan 2 years ago
MissLenaElan
Another way to see this is that there are many ways to represent a number.
For example, 2 can be written as 2, sqrt(4), 1 + 1, 4/2, 16/8,
20000/10000, and so on. It just so happens that one representation of 2
is 1.99999.... no more or less intuitive than any other representation.
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