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Comments

Catalyst

I'm sorry, sir, but how exactly do you define "Troll"? It seems to me that you're just saying a troll is anyone in opposition to you.

The definition of an Internet troll is some one who is attempting to elicit hateful or angry comments or actions (in other words, to provoke).

Chimaera posted his beliefs, which were contrary to your own. He was not attempting to provoke you to anger, and probably wasn't even trying to debate, as he left fairly quickly. His actions were nowhere near that of a troll. And, by drawing it out, YOU sir, are the troll, trying to entice his anger.

If I were you, I'd advise letting the matter drop with him. You seem to be defending this far too aggressively. The man let it drop, why do you feel the need to pressure him?

Oh, wait... that's right... he made an ANALOGY.

Let me explain for those too faint of mind or common sense. An analogy is when you compare one thing to another using terms such as "is" or "like".

You are pursuing the fact that he compared arguing with you to arguing with an Atheist over God's existence. The fact that you seem to have pressured him when he had clearly wanted to end it suggests that you took offense to this in some way. His post did not in any way suggest you held this as a religious belief, but that you instead were so held to this thing that it was nearly impossible to convince you otherwise, in the same way it is hard to convince an Atheist of any proof of God.

Please, Sir, let it die with Chimaera. You are trying to drag things out in a way which does not need to be done. He's the bigger man in this case, because he actually knew when to stop fighting because he knew it was done. You sir, are a troll trying to drag out the fight. I'm deeply saddened by the actions of one who seems fairly intelligent.

PAStheLoD

I've commented on the original post, but here Daniel said almost the same thing as I did.

0.999... = 1 is ture
0.999... IS (identical to) 1 isn't true :]

julian

I find it funny that no math professors have weighed in on this. It's just a bunch of amateur enthusiasts arguing over their heads because someone found a cute trick with algebraic semantics ... lots of noise by so called "expects", much like the rest of the web. My reasoning is that I haven't seen a single proof of this that goes beyond first year college calculus. That's a dead giveaway when arguments about number theory come up.

gaudentius

This is waaayyy too funny! Wake up people!!! He's gonna have a full belly when this is over. He's catching a lot of fish . . .

IQ70

I just had to add in something here that everyone else seems to not understand or ignore.

Mathematically speaking (read, not in the real sense), the number 0.0000.....000001 is ignored.

So on one end, 0.99....99 is stretched to 1, while on the other end 0.00000....01 is stretched to 0 on the number line.

This problem occurs only when a fraction is being written as a decimal number. If you take a circle and cut it into 1/3rd and 2/3rd, you do not have a small part missing.

This whole "problem" is nothing but an example of rounding off at the infinite decimal place to be able to correctly represent fractions in the decimal system. It is written to make up for a deficiency.

In reality, 0.9999..99 is not 1, but in mathematics it is rounded off to 1 because 0.99999....999 never truely completely represents 2/3.

QED

Edward Flick

Lets get things correct shall we? Let ^ represent the overline (i.e.: 0.9^ = 0.9 with a line over the 9). We all agree that: 1=3/3=1/3+2/3=0.3^+0.6^ right? What about the following: 1=0.3^+0.6^=(0.3+1/30) + (0.6+2/30) = (0.33+1/300) + (0.66+2/300), correct? Then would it not be correct to assume that to say that 0.x^ is not actually equal to 0.xxxx... but is rather an approximation, and the true equality can only be written as a fraction or the sum of a decimal and its fractional remainder. No matter how many decimal places this is worked out to there will always be the fractional remainder, and to ignore it would be bad math. The problem in this case is the assumption that 0.3^=0.3333... which is an untrue and inconsistent axiom.

Edward Flick

Lets get things correct shall we? Let ^ represent the overline (i.e.: 0.9^ = 0.9 with a line over the 9). We all agree that: 1=3/3=1/3+2/3=0.3^+0.6^ right? What about the following: 1=0.3^+0.6^=(0.3+1/30) + (0.6+2/30) = (0.33+1/300) + (0.66+2/300), correct? Then would it not be correct to assume that to say that 0.x^ is not actually equal to 0.xxxx... but is rather an approximation, and the true equality can only be written as a fraction or the sum of a decimal and its fractional remainder. No matter how many decimal places this is worked out to there will always be the fractional remainder, and to ignore it would be bad math. The problem in this case is the assumption that 0.3^=0.3333... which is an untrue and inconsistent axiom.

foo foo foo

Hey, I just figured out why polymath is such an idiot: read his religion links. God nuts have no problem saying "It is this way because I say so!"

Pedantic

ADAM DRAKE:

>> If |r| > proof, then -1 > maybe I'm
>> misunderstanding?

Your step (3) concludes that the geometric series converges because |r| a, which is blatantly false.

Pedantic

Apparently half my comment vaporized. Never mind then.

< 1 is not strong enough, you have to say -1 < |r| < 1 and r != 0, because if |r| had been 0 it would've converge to 9/10, but your proof suggest it does.

Polymath

Ok, jokes up: I was just kidding! It was part of a study to see web reaction to trolls. Thanks for participating, everyone.

jks

If .999|9 == 1, then shouldn't they be interchangable in other equations? If that isn't true, then it seems like the statement ".999|9 == 1" is nothing more then a conceptual argument which, while interesting, has to bearing on actual mathmatics.

That is to say, if .999|9 == 1, then shouldn't .999|9 - x == 1 - x?

Obviously that isn't true:
.999|9 - .666|6 = .333|3
while:
1 - .666|6 = .444|4

unless, of course, you want to now argue that .333|3 == .444|4.

Seems that the problem here is really about arguing about the accuracy of "imperfect" representations of numerical ratios. 1/3 == .333|3, but a number repeating to infinity only has any real meaning as an approximation, that is, only as .333... with x number of '3's repeated. Eventually you round in order to give it any meaning.

So, I would say that the statement ".999 repeating and 1 are the same" is an interesting one, but inherently useless and downright incorrect in applied mathmatics in that if you perform the exact same operations on 1 and on .999|9 you end up with differing results.

Pedantic

>> Ok, jokes up: I was just kidding! It was part of a study to see web reaction to trolls. Thanks for participating, everyone.

Sweet. So how much did you pay people all those people to passionately say things like 9x = 8 so 90x = 88?

You paid those people, right? Right?

Pedantic

Obviously that isn't true:
.999|9 - .666|6 = .333|3
while:
1 - .666|6 = .444|4

Okay, let's go slow. 0.66 is clearly less than 0.666, right?

And 1 - 0.666 is 0.334

Next!

Keith

You're initial argument is a little wrong, I believe.
when you multiply your x by 10, and then subtract x
e.g.:

10x = 9.99999...
-x = 0.999999...
-----------------
9x = 8.99999...999991

This is not rational to think of because you say the there's an infite number of nine's, unfortunately, there's always exactly one more 9 in the 0.99999... than there is in the 9.9999...

I think Georg Cantor is rolling in his grave right now.

Bob

In your original 'proof' you used the example of:

10x = 9.999999...

but for it to be a 'proof' the result should hold true for

nx = n*9.999999...

So can you please show those as well?

3x=2.999999999...7
-x=0.999999999...9
------------------
2x=1.999999999...8
x!=1
x=0.999999999...9

However if you tell me that I cannot assume to know what the last digit of the infinite sequence of nines is, then what is 3*0.9999...?

klm

Next you'll probably be trying to tell us that the cardinality of the real set is equal to the cardinality of natural numbers!

Robert

.9999999... does not equal one and I'll show you how:

Lets say you take a test that had 100 multiple choice questions. And let's say you get 1 wrong. Your score will be 99%, right? Now, 99% = .99 since 99/100 = .99. This isn't an approximation. Now, let's say you have a test with 10,000 questions, and you get 1 wrong. Your score will be 99.99% or .9999 if you divide by 100. Now, let's say you took a test with 1,000,000,000 (1 billion) questions and you got 1 wrong. Your score would be 99.9999999% or 0.999999999 if you divide by 100. You still don't have 1 (100/100).

In all these cases, no matter how many trailing 9's you had behind the decimal point, you'll never get 100%, or 1 if you divide by 100, simply because you got 1 question wrong. Even if you were to continuously add questions to the end of a test that only had one question wrong from now to the end of time, you would never, ever, ever reach 100%. You'd get infinitely close, but you'll never reach it.

Here's my proof that .999999... will never equal 1:

99/100 = .99
999/1,000 = .999
999,999/1,000,000 = .999999
9,999,999/10,000,000 = .9999999
99,999,999/100,000,000 = .99999999
999,999,999/1,000,000,000 = .999999999
999,999,999,999/1,000,000,000,000 = .999999999999
999,999,999,999,999/1,000,000,000,000,000 = .999999999999999
and on and on into infinity...

So you see, as long as you have 1 less than the total possible, you'll never reach 1 even if you were to add infinite questions to the end of the test.

If you can disprove this, then please do so.

bunk

1 - .6|6 = .4|4
.9|9 - .6|6 - .3|3

1 != .9|9

bunk

ignore the above post, i get it ;)

Phord

Sorry, Bob. There is no "last digit" in the product (3*0.9999...).

However, perhaps this helps: the nth digit after the decimal is 9 for all values of n.
Ergo, 3*0.9999... = 2.9999...

Davis

Hey polymath, now that you've gotten the attention of the math trolls, it would be a great time for a discussion of Cantor diagonalization!

As for the .999...=1 refuters -- why exactly do you seem to think you understand math better than the mathematicians?

The Pancake

What, then, is the following solution...

x = .999...

while(true) {
x = x^.999...
}

Wouldn't x approach 0?

Bob

@Phord - so we just ignore the very last digit in the sequence?

And so suddenly 3*9 != 27? Somebody please tell me where the 7 goes and what happens to the carry of the 2?

There is no way 3*0.9999... simply equals 2.99999...

ib

It's amazing how many people openly defy what is widely accepted mathematical fact.

When I first heard about it in elementary school (about 20 years ago) it came as a shock to me too.

I had always considered 0.999 repeating to be the "next number down" from 1. But then I thought about it hard. If that's true, then that would mean there's a quantifiable non-zero distance between 1 and 0.999->
But since every non-zero quantity can be divided, the entire premise topples over.


The only way you'll ever disprove this is if you can provide a value for x such that x <> 0 and x = 1 - 0.999->

I'll save you some time. You can't.

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