1=.999999...?

[Wonderer]

there have apparently been circulations of a "mathematical" proof that goes as follows

x=.999[bar] the bar signals recurring numbers.

10x=9.99[bar]

10x -(x)= 9.99[bar] - (.999[bar])

9x=9

x=1

at first you get kind of dumb founded that this seems to be valid... upon further examination this is what i come up with.

when you multiply .999[bar] by ten, those ten "infinitesimal differences from ten" are all represented by a single infinitesimal. what i mean by this is that when you multiply the .999[bar] there is a loss pf precision because the infinitesimal difference from .999[bar] to 1 can always be made more infinitesimal. meaning a 9 can always replace where there would have been an 81.

heres a mental image that does the trick.

imagine a cube that is .999[bar] inches in height. next imagine a wall with inch markers on it from 1 to 10 starting from the ground up.

take the .999[bar] cube and place it next to the 1 inch mark. it is below the 1 line but impossibly close to notice.

take another .999[bar] cube and place it on top of the existing one... logically the difference would be twice as great from the 2 inch mark than the first block is from the 1 inch mark.

add a third and the difference would be even greater.

when you delve into the nature of these differences the only way to describe them is as infinitesimals.

1 divided by infinity...

the problem is that ten infintesimals are represented as one infinitesimal when a number ends like .999[bar]

an obvious misrepresentation... thoughts?

[Anthem]

It's an informal proof, but it still works. You need infinite series to do the formal proof...convergence is a weird, weird thing. Here's another informal proof:

Again, some examples can help make this clear:
_
0.1 = 1/9
__
0.12 = 12/99
___
0.123 = 123/999

and so on.

So, here's something to consider: What fraction corresponds to
_
0.9 = ?

It has to be something over 9, right?
_
0.9 = ? / 9

The _only_ thing it could possibly be is
_
0.9 = 9 / 9

right? But that's the same as 1.

Let's hope colin doesn't come in here

you don't even know whats maths is
mathematicans cant even dont what .999 is not 1
let me be your accountantill take your money
haha
stuipd mathmeticans

[trahne]

haha
stuipd mathmeticans

[Yorick]

an obvious misrepresentation... thoughts?

The way the decimal system works and unintuitive infitesmals contribute to the misunderstanding that 0.999[bar]=/=1. In fact, 1 does equal 0.999.

http://mathforum.org/dr.math/faq/faq.0.9999.html

I couldn't believe it either the first time I heard about it.

[Wonderer]

LOL this is nonsense... it does not equal one...

the reason you guys think .999[bar] equals 1 is because you cannot measure the difference...

what fraction correspondds to .9?

9/10...

infantesimals cannot be quantified.! lol i still don't get why every body thinks that .999[bar]=1 it's ludacris.

.1 is 1/10 ... not 1/9... i dont get what is gained by that example...

i think i've already demonstrated how and why infintesimals don't exist or cannot be quantified which is physical proof that .999[bar] dosen't equal one.

frankly i doubt that .999[bar] exists. show me a fraction using real numbers that creates a .999[bar].

and for the record .333[bar] + .333[bar] +.333[bar] = 1

perhaps i don't understand where the argument is coming from. i have already disgarded the 10x-x example and shown why it is futile with numbers and {bar]'s. is there something im missing here?

[Anthem]

It very much does equal one. The fraction that equals .999 bar is 9/9. Think about it: 1/9 = .111111... 4/9 = .4444444444...

What does 9/9 equal? 1, obviously, but also .999999...

you just said .3333333... equals 1 when added together three times. What's 3 times 3? 9. 1/3 * 3 = .9999... = 1

Infintesimals can be quantified; that's the weirdness of math. If they don't exist, neither does .999...

You could look at the formal proof using infinite series if you don't believe the simple math tricks. It's not that much harder.

Here's yet another way of looking at it. Let's say you wanted to go the other way the same distance .999... is away from 1. Where the hell would you put the .0000...0001? At the last possible point: infinity. But we can never get there. So you're stuck with this 1 that you can never tack onto the 1 after the decimal place.

[Maytacera]

The way I always saw it broken down was like this.

1 / 3= .3Repeated

3 x .3Repeated= .9Repeated.

It would "logically" follow that .9Repeated=1.

However, when we apply these numbers to real things, we see that the theory does not pan out. For instance, when dividing a pie in to three pieces, I am left with 3/3s. Together, they equal a whole.

[Laurentius]

I have actually asked my teacher to explain it to me and it looked rather simple when he calculated it and I can see in the theory how it can work but I can’t see it in reality. 1 – 0.999… = 0.000…1. If any body have a proof from reality that 1=0.999... then please show me.

You guys might want to read about Georg Ferdinand Ludwig Philipp Cantor he is famous for dealing with this sort of mathematic problems. According to some people it is the solution of Zenon’s paradox.

[Wonderer]

what would happen if you added up .333333[bar] and .3333[bar]... YOU CANNOT...

thats the simple truth. they talk about .33[bar] +.33[bar] +.33[bar] =.999[bar] but thats just not the case.

what .333[bar] means is that the 3's go on forever... you cannot possibly add them up because you could never finish writing them all down.

what we are left with is a number that can only be as precise as how far into it you care to calculate...

that is the difference between 1/3 and .333[bar]... though both mean the same thing, the 1/3 is the most accurate because when you try to manipulate .333[bar] you are left with greater than infinitesimal remainders.

.999[bar] can only be represented Presley as follows.

1 -(1/infinity)= .999[bar]

here are some questions we can now ask... what happens when we subtract (1/infinity) from 1? what happens when we add (1/infinity) to 1?

basically you cannot assuming the infinitesimal could always have been smaller... but if you could, it certainly wouldn't still be 1...

now when you take 1 third, it is not 1 infinitesimal away from anything noticable... it is just .333's forever.. however, if it were only composed of 3's than it wouldn't truly be 1 third.

then you add another third. and it is composed of nothing but 6's, though again if it were only 6's it wouldn't truly be 2 thirds.

this "not true third" thing must be brought into light. how can you expect your "proof" to be accurate if the nature of .333[bar] won't allow it.

.333[bar] is lacking in absolute precision to begin with.. therein lies the fault and the missing infinitesimal between .999[bar] and one.

not trying to offend anyone but i think it is easy to be convinced by such a math theory that .999[bar]=1 if you have a weak imagination... especially pertaining to "infinity"

[Yorick]

Sorry mates, pretty much every professional mathematician on earth is of the opinion you're wrong. .999... does equal 1. Look it up. Its been explained bazillions of times all over the internet.

Don't feel bad, everyone falls for this one.

[ladyjane]

Sorry mates, pretty much every professional mathematician on earth is of the opinion .999... does equal 1

just show mathematicians are stupid conservatives as colin leslie dean says
their error is so simple

.99999999.. [R]
IS NOT A SUM
yes .99999999 is 9/10 +9/100 etc
but that is not a real sum
if you sum all that then it converges to 1
but .999999 [r] is not a convergent number as it just goes on for ever it converges not to 1

when you say
1= .999999 [r]

ie
1= the number .999999 [R] sums to over infinty
you could also say
x= the number .33333333 [R] sums to over infinty
then

x would be bigger than .333333.[R]
then 3 * x would be bigger than 3 -which is a contradiction

you are misconstring a number for a convergent sum
PROOF
1/3 = .33333333 [R]
WHICH IS
3/10+ 3/100 etc
so what does that sum converge to
it might converge to a number bigger than .3333333.. [R]

THEN WHEN YOU ADD THREE OF THEM TOGETHER YOU GET A NUMBER BIGGER THAN 3-WHICH YOU DONT


just another example proving colin leslie deans point that maths ends in meaninglessness

[Anthem]

Yup, I knew he'd be in here soon enough. Hold on guys and gals, let me take care of this...

cOLIN lESLIE iDIOT. ATTENTION. YOU DO NOT UNDERSTAND CONVERGENCE AND YOU NEVER WILL BECAUSE YOU REFUSE TO TRY. STOP ATTEMPTING TO REFUTE ALL OF MATHEMATICS WHEN YOU UNDERSTAND ONLY THE MOST BASE ATTRIBUTES THAT AN ELEMENTARY SCHOOL STUDENT UNDERSTANDS. THAT IS ALL.

You have no idea what you're talking about, and I suspect you never will. Do us all a favor and sew your fingers to your lips.

[ladyjane]

You have no idea what you're talking about, and I suspect you never will. Do us all a favor and sew your fingers to your lips.

if .999999 [R] converges to 1
then
what does .33333 [R] converge to
add three of them together and you get a bigger number than 1- a contradiction

so what number does .33333 [R] converge to

[Wonderer]

still no one has rebuked my cube example... i don't see how simpler it could be to explain it...

.333[bar]=1/3

1/3+1/3+1/3=3/3=1

QED

.333[bar]+.333[bar]+.333[bar]=1

i don't care what the majority of Internet mathematicians think... all scientists once had to say the earth was the center of the universe or they would be severely punished... what's worse more than half of than half of them believed it...

it's no surprise that things like this which can easily baffle a weak mind have been adopted by Internet mathematicians (politicians)...

until someone can make a response to my cube example my beliefs will remain unswayed

[Yorick]

so what number does .33333 [R] converge to

It converges to 1/3.

[Yorick]

until someone can make a response to my cube example my beliefs will remain unswayed

imagine a cube that is .999[bar] inches in height.

Already at this point, we disagree. I am of the opinion that the cube is exactly 1 inch in height so the rest of the example fails. The rest of the example is based on the assumption that the cube is *not* 1 inch in height which is what it attempts to show so it shows nothing.

you cannot possibly add them up because you could never finish writing them all down.

You don't *actually* write them down. You don't need to. Think about it this way: We have the two infinite series of 3's. We place them next to each other and add each 3 to its counterpart.

0.333333333333333...
0.333333333333333...+
-------------------
0.666666666666666...

Note that even though we don't write all the 3's and 6's we can very well imagine them

i don't care what the majority of Internet mathematicians think... all scientists once had to say the earth was the center of the universe or they would be severely punished... what's worse more than half of than half of them believed it...

While it is true that mathematicians don't know everything, they know a LOT more than you about math and you have done little to convince me otherwise.

[ladyjane]

333[bar]=1/3

1/3+1/3+1/3=3/3=1

QED

.333[bar]+.333[bar]+.333[bar]=1

sorry
.333[bar]+.333[bar]+.333[bar]=.999999[bar]

[ladyjane]

It converges to 1/3.

sorry 1/3 = .33333333 [R]
but what does .333333R] converge to

add three of them together and your number is bigger than 1

[Pandora]

http://en.wikipedia.org/wiki/Proof_that ... ._equals_1

[ladyjane]

the proof is rubbish
as i said

.99999999.. [R]
IS NOT A SUM
yes .99999999 is 9/10 +9/100 etc
but that is not a real sum
if you sum all that then it converges to 1
but .999999 [r] is not a convergent number as it just goes on for ever it converges not to 1

when you say
1= .999999 [r]

ie
1= the number .999999 [R] sums to over infinty
you could also say
x= the number .33333333 [R] sums to over infinty
then

x would be bigger than .333333.[R]
then 3 * x would be bigger than 3 -which is a contradiction

you are misconstring a number for a convergent sum
PROOF
1/3 = .33333333 [R]
WHICH IS
3/10+ 3/100 etc
so what does that sum converge to
it might converge to a number bigger than .3333333.. [R]

THEN WHEN YOU ADD THREE OF THEM TOGETHER YOU GET A NUMBER BIGGER THAN 3-WHICH YOU DONT


just another example proving colin leslie deans point that maths ends in meaninglessness

[Anthem]

I reported you for not putting any effort into that last post.

You're wrong. If you can't put enough thought into understanding why 1 = .999... then I'm very sorry.

colin's just upset because if .999... converges to a number (which it does) you could 'point to it on a number line' and that would throw off her whole life's philosophy.

[gib]

The problem here is in the way we write these quantities out. Writing '1' is a way of expressing the quantity 1 exactly and instantly. Writing '.999[bar]' is a way of expressing the same quantity but by way of approaching that quantity over time - you do get there in the end though. The way we imagine this is as though it were a process - that is, we imagine 9s being constantly added to '.99999...' until we get to the end. But of course, we never reach the end since that would take an eternity. But the fact of the matter is, we are always approaching 1, no matter how long it takes. The problem, though, is in that, like all limits in mathematics, it takes an eternity to actually get there. We think, therefore, that it CAN'T be reached. But this is only true for a finite amount of time (which brings us to a finite decimal expansion - which indeed is less than 1). The fact of the matter is, however, .999[bar] is infinitely long. Therefore, as it stands, it has already reached the limit of 1, and therefore already equals 1.

[Wonderer]

heres an example....

say we want to travel 4 miles east in an electric car.

the problem is that each time you can only travel half the distance that you just traveled.

the first distance you travel is 2 miles, the next is 1 then .5, .25, .125... and so on.

you will for eternity try to travel a total of 4 miles but it will never happen.

the sequence of numbers converges to 4, but still will never equal four.

similarily, if .999[bar] really exists than it is not converging to 1, it is converging to 1-(1/infinity)

i have already demonstrated why infintesimals cannot be quantified, and if .999]bar] exists (the 9's can go on forever) than an infintesimal must exist.

once again i have shown why infintesimals are not quantified in decimal values...

don't just say im wrong... prove me wrong...

don't be a slave to symbols and numbers, use reason to prove me wrong, not the same old proof i have already objected to

[Wonderer]

]Already at this point, we disagree. I am of the opinion that the cube is exactly 1 inch in height so the rest of the example fails. The rest of the example is based on the assumption that the cube is *not* 1 inch in height which is what it attempts to show so it shows nothing.

no your denial proves nothing.

this example was made to show what actually happens when you multiply .999[bar]. i didnt use the cube example to prove anything... there was other explanations for that.

but in the same way that the cube example doesnt prove ur theory wrong ur theory dosen't prove my cube example wrong.

if.999[bar] =1 than it wouldnt start out with a .9...

i keep concluding that you people are all slightly confused and are as a result appealing to the majority

[Wonderer]

i doubt you guys can even prove that .999[bar] exists.. in the practical world it doesn't...

the only way to represent it is like this [1-(1/infinity)]

and we don't know if infinity really exists