Is 1 = 0.9999......

[zayened]

why is this thread still going?

 

[ndee]

1=1, 0.999999=0.999999, is that so hard to understand?




This thread receives the most stupid thread EVAR award.

 

[Darien]

Originally posted by: zayened
why is this thread still going?

Beats me. People want it to hit 1,000 posts?

 

[compudog]

Originally posted by: ndee
1=1, 0.999999=0.999999, is that so hard to understand? This thread receives the most stupid thread EVAR award.

After reading through all 19 pages of this thread, I couldn't agree more.

<---- 1=1 Fanboy

 

[luvya]

Why do we play into Bleed's hand? Let's all stop it here! He merely wants to make this thread over 1000 posts! Let's stop and piss him off! I wish mod could lock this thread!

 

[Darien]

Originally posted by: compudog

Originally posted by: ndee
1=1, 0.999999=0.999999, is that so hard to understand? This thread receives the most stupid thread EVAR award.

After reading through all 19 pages of this thread, I couldn't agree more.

<---- 1=1 Fanboy

No one will argue if those 9s terminate. But they don't


lim x-> infinity of 1 - 1/x = 1

 

[Nemesis77]

Maybe I'm an idiot or something, but if 0.9999999 = 1 why would they call it 0.9999999 and not 1?

 

[ndee]

so if everything has endless numbers after the comma, it just gets upped to the next bigger number? Like 0.3999999999999... would get 0.4?

 

[elkinm]

Originally posted by: ndee
so if everything has endless numbers after the comma, it just gets upped to the next bigger number? Like 0.3999999999999... would get 0.4?

Exactly, now you're getting it. Although .388888888..... will not eaqual .39 simply because .3899 is greater then .38888888....... and .3899 is less then .39. But what number is less then .34 but greater then .3999999999999...... I don't think you will find that number and the only case were you cannot find a real number that is less then another real number and greater then a diferent number will of the form were you need to find a number between .34 and .34 which is because .34 = .34

The same holds for 1. If we cannot find a number between 1 and 1 then 1 = 1, now that was hard. I also don't know of a number between 1 and .9999999....... so I guess 1 = .9999999........ strange isn't it. Please prove me wrong, find an x such that 1 < x < .999999......

There was a statement that we don't know what .9999999..... realy is so we can't compare 1 and .99999999....... Yet we can know what the square root of 2 is. Square root of 2 is irrational meaning that it cannot be written in the form of a fraction in lowest terms. And although not an official definition, it means that there will never be an end to the decimal representation of square root of 2 and most importantly there will never be a pattern to the representation of square root of 2. If there is a pattern then theoreticaly if you create a system running a simple infinite loop, then you could output the entire number in an infinite amount of time. And if the loop results in an accurate number then the number much be rational.
Now take a program, print "." then set up an infinite loop to print "9" right after the last character. See it run and after an infinite amount of time you will see .9999........ But if that is true then .9999...... must be rational and can be written as a fraction, but what fraction. I can only think of the fraction of 1/9 + 8/9 = 9/9 as mentioned earlier. But 9/9 = 1


I sometimse like to say that the difference between 1 and .999999..... is .0000000........repeating1 but that in itself is not correct. Since it repeats infinite amount of times and infinity cannot be compared to a number then it would be true, but if we are saying that 1 is not .999999...... and .00000000..........1 is not 0 then we cannot say that infinity = 2 times infinity or even infinty + 1 = infinity. If we say that that 1 after the zeros will ever come then infinity must be treated as an exact term such that only infinity = infinity. Threfore if I take my term .0000000......1 and .99999999......... and add them then I would simply get .9999999999........1 as we have both the 0 and the nine repeating infinetly many times and one after that we add the final 1. So what we need to do is remove one of the 0s from the .0000000..........1 term or basicaly multiply it by ten to get to a .00000000..........*10 1 term. and then you could say if infinity ever stopped then you could provide an exact number to add to .999999........ repeating to equal to 1.
But as you have noticed that this entire argument is irrational and has gaps on to many levels and it also is the only argument that could ever proove that .999999...... is not equal to 1 so you either agree with my ramblings above or aggree that 1 = .9999999........ or you could ingore both options and simply agree that 0 = 1 = 2 = 3 = 4 = 5 = 6 = ..........

 

[Kyteland]

Thread about fallacies

This was posted today. It seems to me that some of the people arguing in this thread need to carefully read this. I won't mention names.


Oh and ... Bump

 

[ElFenix]

only a few more posts....
.9999 still != 1, btw

 

[Kyteland]

Burden Of Proof:

the claim that whatever has not yet been proved false must be true (or vice versa). Essentially the arguer claims that he should win by default if his opponent can't make a strong enough case.

There may be three problems here. First, the arguer claims priority - but why is it him who wins by default? Second, he is impatient with ambiguity, and wants a final answer right away. And third, "absence of evidence is not evidence of absence."

 

[bleeb]

Originally posted by: Kyteland

Burden Of Proof:

the claim that whatever has not yet been proved false must be true (or vice versa). Essentially the arguer claims that he should win by default if his opponent can't make a strong enough case.

There may be three problems here. First, the arguer claims priority - but why is it him who wins by default? Second, he is impatient with ambiguity, and wants a final answer right away. And third, "absence of evidence is not evidence of absence."

0.9999...!=1

 

[Kyteland]

Originally posted by: bleeb

0.9999...!=1

Argument By Repetition (Argument Ad Nauseam):

if you say something often enough, some people will begin to believe it. There are some net.kooks who keeping reposting the same articles to Usenet, presumably in hopes it will have that effect.

 

[Kev]

Is 1 = 0.9999......



Shouldn't it read "Does 1 = 0.9999......


By the way, this is a really dumb thread. Which is why I'm posting in it!

 

[ElFenix]

Originally posted by: maladroit
Is 1 = 0.9999......



Shouldn't it read "Does 1 = 0.9999......


By the way, this is a really dumb thread. Which is why I'm posting in it!

this thread was dumb enough without your post, thanks

 

[Kyteland]

Originally posted by: maladroit
Is 1 = 0.9999......



Shouldn't it read "Does 1 = 0.9999......


By the way, this is a really dumb thread. Which is why I'm posting in it!

That depends on how you mentally interpret "="

if "=" is equivalent to "equal" then "Does 1 equal 0.9999..." makes sense

if "=" is equivalent to "equal to" then "Is 1 equal to 0.9999..." makes sense

But since this is my thread (Yes that's right. I own it.) we are using the second method.

 

[MadRat]

Originally posted by: Darien

No one will argue if those 9s terminate. But they don't

lim x-> infinity of 1 - 1/x = 1

But if they reach to infinite then they can be decribed only to the limit of infinity. Since you cannot represent it in a fraction without the use of the term infinity then it must not exist on the Real Number line. There will always be a remainder of 1 possible to compliment every nine, even at the infinite ninth. If you can represent it in a fraction using infinite then the boundary for the 9's is infinity and it still falls short of 1. No matter how you try to argue it, 1<>.999...

 

[bleeb]

Next thing you know, people will start saying that F != ma.

 

[zCypher]

Originally posted by: bleeb
Next thing you know, people will start saying that F != ma.

 

[VBboy]

This thread must die

 

[Stojakapimp]

I think the most interesting proof i've seen so far is where someone divided 1 by 1. Instead of going into 1, he said it went in 0 times with a remainder of 1, so then he had to divie 1 into 10, which he then said went in 9 times with remainder 1, and so on. So if you kept on carrying out the long division, you would end up with .9999 with all these remainder 1's to infinite. But I also like the like the 1/9 + 8/9 things.

so yes, 1 = 0.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...

 

[MadRat]

So if you use inaccurate math then its okay. Thank you for your contributions.

 

[ElFenix]

Originally posted by: Stojakapimp
I think the most interesting proof i've seen so far is where someone divided 1 by 1. Instead of going into 1, he said it went in 0 times with a remainder of 1, so then he had to divie 1 into 10, which he then said went in 9 times with remainder 1, and so on. So if you kept on carrying out the long division, you would end up with .9999 with all these remainder 1's to infinite. But I also like the like the 1/9 + 8/9 things.

so yes, 1 = 0.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...

i don't think thats a real proof.

though i still think madrat is arguing a different system.

 

[silverpig]

Originally posted by: ElFenix

Originally posted by: Stojakapimp
I think the most interesting proof i've seen so far is where someone divided 1 by 1. Instead of going into 1, he said it went in 0 times with a remainder of 1, so then he had to divie 1 into 10, which he then said went in 9 times with remainder 1, and so on. So if you kept on carrying out the long division, you would end up with .9999 with all these remainder 1's to infinite. But I also like the like the 1/9 + 8/9 things.

so yes, 1 = 0.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...

i don't think thats a real proof.

though i still think madrat is arguing a different system.

<--- Long division guy

It definitely shows that 0.9r = 1, but there are more rigorous proofs out there. It was just another offering.