.99999999999999999999999999999999999999

[szw]

Yes or no, does .999999... exactly equal 1?

I can't edit my poll but its supposed to be .999999repeating, ignore the ~

[skandalz]

i'm gonna go w/ no.... 1 = 1, what's this for?

[toyboy88]

yes :eek:

[DUBBED]

No it doesn't.

[buh_buh]

1=1. no other number does.

[skandalouz_08]

It does not equal 1, but for mathematical equations we make it equal 1 rather than write out tons of .9's and make a simplified solution

[FiveFreshFish]

Mathematically, no.

For most other intents and purposes, yes.

[khtm]

This is really like:

(9/10) + (9/100) + (9/1000) + ....

which is the sum of (9/10^n) where n goes from 1 -> infinity

the limit of this equation = 1

So the answer is yes. And yeah, I'm a math geek.

[Civic_Stylez403]

rounded, yes

[BerserkerCatSplat]

Does it equal 1 exactly? No.

However, as the number of digits approaches infinity, the limit is 1 (as khtm stated). So, it depends on the context of the question and the number of significant digits involved.

Why do you ask?

[szw]

Just saw it on another forum, wanted to see the responses. i'm too lazy to wait for more votes so i'll just post a proof here.

x=0.999r
10x=9.999r
10x-x = 9.999r - 0.999r
9x = 9.0
x = 1

[Rocky]

same as this:

(1/3) + (2/3) = 3/3 = 1

but as a decimal, it only works out to .9999999999

Equals is equals, it doesn't matter, by definition nothing can be equal to something without being exactly the same. So as far as I'm concerned, any amount of 9's after a decimal does NOT equal 1 :P

[kaput]

.

[gpomp]

x=0.999r
10x=9.999r
10x-x = 9.999r - 0.999r
9x = 9.0
x = 1

so you subtract x from one side and 0.999r from the other???

then you should get 9x = 9r and x = r

unless i'm doing my math wrong

[D. Dub]

yes everything is relative

[wildrice]

Originally posted by gpomp


so you subtract x from one side and 0.999r from the other???

then you should get 9x = 9r and x = r

unless i'm doing my math wrong

no, cuz x=0.999r
10x-x = 9.999r - x is the same thing as 10x-x = 9.999r - 0.999r

[iceburns288]

Originally posted by szw
Just saw it on another forum, wanted to see the responses. i'm too lazy to wait for more votes so i'll just post a proof here.

x=0.999r
10x=9.999r
10x-x = 9.999r - 0.999r
9x = 9.0
x = 1

What the hell is this proof:nut: . r just randomly disappeared and you've got variables flyin around like nobody's business.

You're asking to prove if one number equals another number. It's impossible. Wanna ask me how 5=2? No? Then why would you ask this?

[gpomp]

Originally posted by wildrice


no, cuz x=0.999r
10x-x = 9.999r - x is the same thing as 10x-x = 9.999r - 0.999r
ok so 10x-x gives you 9x and 9.999r - 0.999r gives you 9r. you get 9x = 9r....

[szw]

9.9999
- .9999
_____
=9.0000

[szw]

Originally posted by iceburns288


What the hell is this proof:nut: . r just randomly disappeared and you've got variables flyin around like nobody's business.


America.

[Hakkola]

Originally posted by iceburns288


What the hell is this proof:nut: . r just randomly disappeared and you've got variables flyin around like nobody's business.


:rofl: :clap: That's gold. I think he's right though, just didn't show all his work.

The way I did it I ended up with x-r = r-x
2x=2r
x=r
x and r can both be expressed as 1/1

Because then 1/1 = 1/1
1 = 1

Math has always been by far my worst subject though.

[pinoyhero]

Originally posted by khtm
This is really like:

(9/10) + (9/100) + (9/1000) + ....

which is the sum of (9/10^n) where n goes from 1 -> infinity

the limit of this equation = 1

So the answer is yes. And yeah, I'm a math geek.

:werd:

[natejj]

REminds me of the math problem made from Ancient greek....

A turtle and a man line up to race. The turtle starts 20 feet in front of the man, so lets say the man starts at A, and the turtle starts at B. They both start moving. By the time the man is at B, the turtle is now at a new point C, because they are both constantly in motion. By the time the man is at C, the turtle is now at a new point, D. By the time th man reaches D, the turtle is now at E......How is it possible the man passes the turtle, because everytime the man reaches where the turtle was, he will be at a new spot!

IT HURTS MY HEAD!!!!!!

[habsfan]

Originally posted by szw
Just saw it on another forum, wanted to see the responses. i'm too lazy to wait for more votes so i'll just post a proof here.

x=0.999r
10x=9.999r
10x-x = 9.999r - 0.999r
9x = 9.0
x = 1

so you agree that you can just plug in values whenever you feel like it? i.e. from:

10x=9.999 repeated
to:
10x-x=9.999r-0.999r

and then, you divide out the x, when you know the value of x in the last step

9x=9.0<- right here, you say x=.999 repeated, so that means you should do
9(.9999999....)=9, which is not true.

so long story short
for a mathematician, 1 does not equal 0.999999999r
for a engineer/technician, 0.999999r= close enough

[Hakkola]

Originally posted by natejj
REminds me of the math problem made from Ancient greek....
IT HURTS MY HEAD!!!!!!


LOL, that's easy, it doesn't say points a,b,c,d,e,f etc... are all equal distance apart.

Man
A........B.....C...D..E.FC
Turtle
..start.B.....C...D..E.FC

Edit: Damn vBulletin It doesn't show the spaces bettween letters like it should.

[szw]

Originally posted by habsfan


so you agree that you can just plug in values whenever you feel like it? i.e. from:

Sorry, I don't understand what you are trying to say. Are you saying there is an error?

Or are you wondering how you can subtract the same value from both sides of an equation?

[habsfan]

no im not saying you cant subtract a same value from both sides of an equation, you surely can, but on one side you are using the value, and on the other you are using x, and then incorrectly manipulating the equation to get the desired result. even so, we can agree that it is ok to subtract the same value from both sides, my second point still stands that you are subtracting x's when you know the value for x.

[Hakkola]

Originally posted by habsfan
my second point still stands that you are subtracting x's when you know the value for x.

Doesn't matter as long as you do it to both sides, that's what I did, which gave me x-r = r-x, you end up with the same answer, that is, if my math is correct.

[habsfan]

im not sure if i'm reading what you mean wrong Hakk, but in his equation "r" means repeated, not a value. and also x-r doesnt equal r-x if "r" actually has a value.

[zhulander]

Originally posted by szw
Just saw it on another forum, wanted to see the responses. i'm too lazy to wait for more votes so i'll just post a proof here.

x=0.999r
10x=9.999r
10x-x = 9.999r - 0.999r
9x = 9.0
x = 1

x=0.999r
10x = 9.99r
10x-x = 9.99r - 0.999r


10 x 0.999r should be 9.99r and not 9.999r because it only uses 3 significant digits. :dunno:

[szw]

habsfan, I'm still unsure as to why you feel a mistake has been made. I think its pretty simply put. We are not incorrectly manipulating anything.

[sabad66]

r just means repeating...its not a friggin variable

[szw]

Originally posted by zhulander


x=0.999r
10x = 9.99r
10x-x = 9.99r - 0.999r


10 x 0.999r should be 9.99r and not 9.999r because it only uses 3 significant digits. :dunno:

Sorry, r isn't a value. r simply means repeating.

so 9,99r is the same as 9.999r or 9.9999999999999999r

I only wrote r because I dont know how to type 9bar

[habsfan]

because, you've already stated that x=0.999r, so therefore it should be plugged into the equation, and not solved for. to put it really simply, you're basically stating that x=0.999r, and then halfway through saying x has no value, lets solve for x

[jaysas_63]

that proof is pure bull shit......if you want to prove that please be my guest and use a viable method such as induction....."Left hand equals right hand" isn't used much past grade 4 lol

[szw]

Originally posted by jaysas_63
that proof is pure bull shit......if you want to prove that please be my guest and use a viable method such as induction.....&quot;Left hand equals right hand&quot; isn't used much past grade 4 lol

Are you saying LHS=RHS isn't valid then?

[habsfan]

LHS=RHS is quite valid, i'm taking a couple math classes in college right now(mechanical engineering) and LHS=RHS still shows up from time to time. so it'd vaild, if done right, unlike here.

[FatboyTheHungry]

Originally posted by khtm
This is really like:

(9/10) + (9/100) + (9/1000) + ....

which is the sum of (9/10^n) where n goes from 1 -&gt; infinity

the limit of this equation = 1

So the answer is yes. And yeah, I'm a math geek.

The LIMIT as n approaches infinity is 1 but never equal to 1.

I say no. And I'm a math geek too.

[szw]

Originally posted by habsfan
LHS=RHS is quite valid, i'm taking a couple math classes in college right now(mechanical engineering) and LHS=RHS still shows up from time to time. so it'd vaild, if done right, unlike here.

Please ask one of your profs to look at it. There is nothing wrong with any of the steps in the proof.

[habsfan]

dude, read what i said above

you state that x=0.999999999999....

and then you solve for x

you have already stated a value for x

get it?

[szw]

The other proof posted has been ignored also.

1/3 == 0.333333333
2/3 == 0.666666666
+ +
3/3 == 1 == 0.999999999999999

[szw]

Originally posted by habsfan
dude, read what i said above

you state that x=0.999999999999....

and then you solve for x

you have already stated a value for x

get it?

This is a proof. it doesn't matter.

You start with an expression that you know to be true. You manipulate it so that you still know it is true. You end with a true equation with 1 variable and 1 unknown.

The fact that they are equal shows that, mathematically, .99999r is the same as 1.

[habsfan]

i do agree with you that the way your proof was done, it comes out valid, if it wasnt for the fact that x is already defined.

EDIT: i completely see what you're saying, starting with an known value, keeping it known, and then coming out with the answer. im looking at it from the point of view that if you have a known value, it should be plugged into the equation. the way you are leaving it is technically valid, so technically i agree that this proof is justified. i just dont feel this is the best, and ultimately, correct way to go about solving this equation.

[szw]

Originally posted by habsfan
i do agree with you that the way your proof was done, it comes out valid, if it wasnt for the fact that x is already defined.

I will try to take a different approach to the same proof.

x=0.999r
10x=9.999r
10x-x = 9.999r - 0.999r
9x = 9.0
x = 1

Ok, so we stop at line 4 ignoring line 5 for now. Do you think line 4 is wrong? If so, where did we go wrong?

According to you, we must have made a mistake somewhere. where is it?

Anyway according to your edit you see what I'm saying so I'll shut up now. I'm sure it could have been explained easier but i'm not that good at explainations.

[sabad66]

k

[habsfan]

it's right if you look at it how you are looking at it(which i see and agree with). the whole time i've been looking at it as x=0.999r being a given value, and then the next line being the start of the actual equation. in that case, then the way i was going would be right because at line 4 it would then be 9(.999r)=9 which doesnt end up as true.

so to summarize i was looking at the problem differently, and saw that by plugging in known value, it's wrong

when looking at it purely from the manipulation point of view, which is how you wanted it to be looked at, it checks out and i have no objection.

[KLCC]

ask something rhetoric like, chicken and the egg, which one came first?

[BerserkerCatSplat]

Now, what happens if you put the equation on a conveyor belt moving backwards... ;)

[Team_Mclaren]

Originally posted by habsfan


for a mathematician, 1 does not equal 0.999999999r
for a engineer/technician, 0.999999r= close enough

are u sure?

for a mathematician, as N approaches infinity in an equation such as (9/10^n), the limit is stated as 1.

in another common example: 1/n as N approaches infinty is = 0 but 1/N never "equals" to 0 in an equation, but it is stated that 0 as an answer.

arr... im a math geek too...


Originally posted by sabad66
k

i dont know what you are proving...:nut:

[KLCC]

p.s. SZW, your equation is kinda flawed....
let me reword the question and the proof:

Assuming X = 0.999999999(repeat) is true

10X = 9.9999999999999(repeat) has to be true

10X - X = 9.999999999999(R) - 0.999999999999(R)

9 X = 9.0

X = 1

which conflict the fact that X = 0.9999999(repeat) in the first place, therefore. 0.99999999999(r) is not equal to 1

*it should've been....didn't realized until I redo the math*
9 X = 8.99999999999999999(R)....1
then you will see, X never turns out to be 1

QED

**************************************
using the same example....

X=0.9
10X = 9
10X-X = 8.1
9X=8.1
X=0.9

x=0.999
10X=9.99
10X-X = 8.991
9X=8.991
X= 0.999

x=0.999999999
10X=9.99999999
10X-X = 8.999999991
9X=8.999999991
X= 0.999999999

x=0.999999999..............
10X=9.99999999.............
10X-X = 8.99999999..............1
9X=8.99999999...............1
X= 0.999999999..................

[szw]

that is exactly the reason that 0.999999999 does equal 1, they are the same.

[mo_money2supe]

Here's another approach to this trivial case:

If 1/9 = 0.11111...
2/9 = 0.22222...
3/9 = 0.33333...
and so on...

then 9/9 = 0.99999... based on the definition in the pattern.

However, as far as we understand fractions in mathematics, 9/9 should = 1 as well.

In other words, 0.99999... = 1 !!!


Of course, that's just my way of trying to prove this. But in my personal understanding of this case, I agree with what was aforementioned in previous posts, 0.99999... only approaches 1, but is never EXACTLY equal to 1.

[Goblin]

Originally posted by szw
Just saw it on another forum, wanted to see the responses. i'm too lazy to wait for more votes so i'll just post a proof here.

x=0.999r
10x=9.999r
10x-x = 9.999r - 0.999r
9x = 9.0
x = 1

You can do all that math, but you asked, does .9999 equal exactly one, and the answer is no.

What you just did was round it up in a complex way.


Oh, btw, put that equation on a graphing calculator, it will come close to, like guy above me said, but it will NEVER touch the X axis (or y, however you type it in.)

[Hakkola]

Originally posted by habsfan
im not sure if i'm reading what you mean wrong Hakk, but in his equation &quot;r&quot; means repeated, not a value. and also x-r doesnt equal r-x if &quot;r&quot; actually has a value.

:banghead: LOL, good call.

[szw]

Originally posted by Goblin


What you just did was round it up in a complex way.


Oh, btw, put that equation on a graphing calculator, it will come close to, like guy above me said, but it will NEVER touch the X axis (or y, however you type it in.)

No, I didn't. Where does the proof 'approximate'?

And the calculator is not an argument, you can't infinitely repeat 9's on your calculator.

Another way to look at it, there is no possible number that falls between 1 and .9r

[KLCC]

X=0.9
10X = 9
10X-X = 8.1
9X=8.1
X=0.9

x=0.999
10X=9.99
10X-X = 8.991
9X=8.991
X= 0.999

x=0.999999999
10X=9.99999999
10X-X = 8.999999991
9X=8.999999991
X= 0.999999999

x=0.999999999..............
10X=9.99999999.............
10X-X = 8.99999999..............1
9X=8.99999999...............1
X= 0.999999999..................

X doesn't equal to 1.

[Aleks]

Do you guys ever get laid? :confused: :rofl:

[Si_FlyGuy]

Originally posted by Aleks
Do you guys ever get laid? :confused: :rofl:

They do if you approximate holding hands as getting laid.:nut: :rofl: :rofl:

[iceburns288]

Originally posted by szw
The other proof posted has been ignored also.

1/3 == 0.333333333
2/3 == 0.666666666
+ +
3/3 == 1 == 0.999999999999999
No, we only see 1/3 as .3333 because we have no way to actually state its true value. 1/3 is a ratio, not a number, and you just completely cocked that up by trying to make it one. It's a ratio, not a number.



Originally posted by BerserkerCatSplat
Now, what happens if you put the equation on a conveyor belt moving backwards... ;)
LOL :rofl:


Originally posted by Si_FlyGuy
They do if you approximate holding hands as getting laid.:nut: :rofl: :rofl:
Holding their own hands, that is...

[wildrice]

Originally posted by szw
Just saw it on another forum, wanted to see the responses. i'm too lazy to wait for more votes so i'll just post a proof here.

x=0.999r
10x=9.999r
10x-x = 9.999r - 0.999r
9x = 9.0
x = 1

This is actually wrong.

9.999r-0.999r = 9r and not 9.0
the r's don't subtract each other.

[habsfan]

^the r stands for repeated, not a value

[khtm]

Originally posted by FatboyTheHungry


The LIMIT as n approaches infinity is 1 but never equal to 1.

I say no. And I'm a math geek too.
No, the LIMIT of that sum I posted above does = 1.

It breaks down to .9 / (9/10) = 1.

I guess all the math I took during my 5 years of engineering might be wrong, though? ;)

[three.eighteen.]

Originally posted by szw




Another way to look at it, there is no possible number that falls between 1 and .9r

that still does not make 1 = .9r

besides .9 is different from .99, which is also different than .999 etc etc, .9r can never be expressed and if it is (as a flaw in our numbering system i guess) you can always add one more .xxxxxxx9 to make it closer to 1

[finboy]

this thread got boring after the first post, lemme see if i can help you guys out...































http://www.wrc-wallpaper.de/wall/Carmen/Carmen%20Electra%2013.jpg

[szw]

Originally posted by three.eighteen.


that still does not make 1 = .9r

besides .9 is different from .99, which is also different than .999 etc etc, .9r can never be expressed and if it is (as a flaw in our numbering system i guess) you can always add one more .xxxxxxx9 to make it closer to 1

Actually it does. Between any two DIFFERENT real numbers there will be an infinite amount of numbers between them. There are none in this case, and so they are defined as the same number.

[szw]

Originally posted by KLCC

x=0.999999999..............
10X=9.99999999.............
10X-X = 8.99999999..............1
9X=8.99999999...............1
X= 0.999999999..................

This is wrong because infinity is infinity.

Where do you define infinity-1?

[KLCC]

basic algebra

X = 0.9
10X = 9.0 (NOTICE shifting the digit?)

10X - X = 9.0 - 0.9
9X = 8.1

so same RULE applies to something that's infinite

X = 0.999~
10X = 9.999~ (with one of the infinite digits over to the left)

you substract the two X, you will end up with some
8.99999~1

and you do the DIVISION, X will equal to 0.9999~

This thread isn't gonna yield some science breaking result...

so Here is another question for ya:

HONDA, NISSAN or TOYOTA which one has a faster car???:D
:goflames:

[szw]

Originally posted by KLCC

you substract the two X, you will end up with some
8.99999~1

This is still wrong but if you want to just post pics of hot chicks instead I won't argue. :goflames:

[iceburns288]

My math teacher (friggin genius... I dunno how he does it) says the only reason I don't get it is that my understanding of infinity is incomplete:confused: .

So szw, you're right, but you got there a completely wrong way;)

[THREE40SEVEN]

WOW...
well.. I'd love to add to the discussion, but I'm gonna go shave my balls.

[max_boost]

Originally posted by szw
The other proof posted has been ignored also.

1/3 == 0.333333333
2/3 == 0.666666666
+ +
3/3 == 1 == 0.999999999999999

SOLD. I believe you man. :drama: :rofl:

[KLCC]

if 0.9999r is truly equals to 1
counter prove this :dunno:

****************************************
"I wish to walk to the shop. However, in order to do so, I must first walk 9/10 of the way there. And after that, I must walk 9/10 of the remaining distance. And after that, I must again walk 9/10 of the remaining distance. And so on. I will get closer and closer to the shop, but I will never reach it.""

Let us see how you move to the target::

1st step (9/10)* 10 = 9 miles (total 9 miles) (remaining 1 miles)
2nd step (9/10)* 1 = 0.9 miles (total 9.9 miles) (remaining 0.1 miles)
3rd step (9/10)* 0.1 = 0.09 miles (total 9.99 miles) (remaining 0.01 miles)
4th step (9/10)* 0.01 = 0.009 miles (total 9.999 miles) (remaining 0.001 miles)


Though you have come as close as possible to the target, but there is ""ALWAYS"" SOME DISTANCE REMAINING THAT YOU ""HAVE"" TO COVER. So if you honesty walk the precise distance you are supposed to move, it is indeed true that, YOUL WILL NEVER REACH THE SHOP.

If you want to prove my statement is wrong then you have to prove that the remaining distance vanishes at some point! Tell me when it happens.

You will not be able to tell 'when', becuase it tends to infinity. You will have to keep moving, FOREVER. So 0.9999' is not equal to 1. :thumbsup:

****************************************

p.p.s. I will post a picture of some chick if you can find ONE counter case to this problem

[DC2]

wow there is quite a stirrup about this.

KLCC i like his explaination so simple. very very simpo...

fatboythehungry, did you use your graphing calculator for that!?

EDIT: where the hell did his post go?
EDIT: nevermind it was being edited

[FatboyTheHungry]

Originally posted by khtm

No, the LIMIT of that sum I posted above does = 1.

It breaks down to .9 / (9/10) = 1.

I guess all the math I took during my 5 years of engineering might be wrong, though? ;)

I guess it's your AMAT vs my 5 years of PMAT and CompSci...
For practical purposes, yes it is close enough. But from a purely mathematical theory standpoint, the limit is 1, but is never reached.

As n approaches infinity,
the SUM of (9/10) + (9/10^2) + (9/10^3) + ... + (9/10^n) = 1

But you never truly get to the nth degree, because then you have to account for (n+1), (n+2), (n+3)...(n+m)
where m is some arbitrary number infinitely larger than n, and then you account for m+1, etc.

KLCC's example above is pretty clear.

[FatboyTheHungry]

Originally posted by DC2
wow there is quite a stirrup about this.

KLCC i like his explaination so simple. very very simpo...

fatboythehungry, did you use your graphing calculator for that!?

EDIT: where the hell did his post go?

Sorry, made an edit for this. no, I don't use a calculator. We weren't allowed to use them when I was in Uni. ;)

[szw]

Originally posted by KLCC


Dude. Because of the concept of infinity your example does not apply.

In a purely mathematical point of view, .9r==1.

Do you believe that 1/3 = .3r? Yes it is not exact if you round it at any point, but the concept of the bar over the 3 means that 1/3 equals exactly the value of .3r.

Its the same thing.

If we were using base 5 then 0.4r==5

EDIT: i'm sorry I posted that proof. There are MANY proofs for this, I posted the one that contains only basic algebra because it should have been easy for everyone to follow.

BTW didn't any of you learn this in school? They taught this to us in gr 10 I remember, but I forgot about it again until recently reading about this topic.

[szw]

my last post in here (maybe)

I understand how people can look at these proofs and think &quot;it smells fishy&quot;, but they are mathematically sound. Questioning it is definately the smart thing to do, as you shouldn't believe everything that is shown to you that is supposed to be clear. I do agree that the proofs are not good because they do not explain why they are equal numbers, they simply show that they are.

If you look up the completeness property for real numbers, there will always be a number inbetween two numbers. But by definition, there is no such number that exists. Ultimately, this is why they are the same number.

They are written differently due to simply a weird artifact of the way we write out numbers. To put it another way, .9r is just like writing out 1^2 or 2/2 or 2-1 or any other way of writing 1.

[szw]

Originally posted by habsfan
dude, read what i said above

you state that x=0.999999999999....

and then you solve for x

you have already stated a value for x

get it?

FINE i'll keep posting.

YES WE SET A VALUE for x....however we manipulate the equation so that we KNOW IT IS STILL CORRECT. And the only way it can still be correct, is if X is ALSO equal to 1. Therefore...x=1 as well as the original value, therefore they ARE the same value.

The reason you are pointing out is the exact reason that they are equal.

For you guys that love wiki here is one of their pages.
http://en.wikipedia.org/wiki/Proof_that_0.999..._equals_1


edit: I didn't think this thread would end up like this, I thought it would be a bunch of people saying no then most people going OHhhh yeah! (like I did when I read this on another forum). Then we'd all laugh about it and post some porn and the thread would get moved to ask leo.

[KLCC]

Read the discussion, you will see that not everyone is convinced with the article. Wiki isn't the de facto of truth.
For all practical calculations, I can treat endless nines as one. But, I am unwilling to say with absolute certainty that 0.99999(Repeat) EQUALS to one.... :angel:

now where is the porn???:guns:

[mikey_mo_love]

Originally posted by BerserkerCatSplat
Now, what happens if you put the equation on a conveyor belt moving backwards... ;)

:rofl: :rofl: :rofl:

[Si_FlyGuy]

let 0.99999r = self-love
let 1x = getting laid.

1x = 0.9999r ????

obviously LHS does not equal RHS.:rofl:

[khtm]

Originally posted by FatboyTheHungry


I guess it's your AMAT vs my 5 years of PMAT and CompSci...
For practical purposes, yes it is close enough. But from a purely mathematical theory standpoint, the limit is 1, but is never reached.

As n approaches infinity,
the SUM of (9/10) + (9/10^2) + (9/10^3) + ... + (9/10^n) = 1

But you never truly get to the nth degree, because then you have to account for (n+1), (n+2), (n+3)...(n+m)
where m is some arbitrary number infinitely larger than n, and then you account for m+1, etc.

KLCC's example above is pretty clear.
Sigh...why even bring up CompSci math. You guys take about half of what engineers have to take. CompSci is a freakin' joke. I don't know what AMAT or PMAT is...I didn't go to U of C.

Anyways, I just googled around and found something to back up what I was saying:

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

Seems like a lot of mathematicians would disagree with all the people that said "NO".

I guess the easiest way to explain this is that .99999... ONLY doesn't equal 1 once you STOP EXPANDING IT. But that never happens as it continues on forever.

Or: 1.00000000...
- 0.99999999...
----------------------
0.00000000...

[KLCC]

I guess I found case to back up my thoughts as well, so does that makes us stalemate? :D


For example, the open set (0,1) includes the number 0.9999ֹ, i.e. 1-e for some infinitesimal. This number is NOT 1, as 0.9999 recurring never equals 1, but only tends to 1

Reference
1)http://www.users.globalnet.co.uk/~perry/maths/nonstandardinterval/nonstandardinterval.htm

2)http://www.askmehelpdesk.com/forum/showthread.php?t=8574

[Carfanman]

1/3 does not definitly .333333333r. 1/3 times three equals one. .333333333r times 3 equals .999999999r. So 1/3 is only .333333r if .9999999r=1, not the other way around. In other words, saying that 1/3=.333333333r does not prove anything, because whether or not that is true depends on the of the original argument, it doesnt prove it. I explained that poorly but oh well.


And the way I see it there is no answer because .99999r isn't an actual number, its just a theory or an idea. It doesn't exist because it has no limit, and it can't exist anywhere other than in theory. You can't have .99999r of an object, because thats not a real number.
Therefore this is a moot point.




Originally posted by Si_FlyGuy
let 0.99999r = self-love
let 1x = getting laid.

1x = 0.9999r ????

obviously LHS does not equal RHS.:rofl:

Not necessarily...

First we need to define getting laid.
The generally accepted definition of laid is engaging in sex.
There are many different kinds of sex.
There is vaginal.
There is anal.
There is oral.
And there is also hand sex, also known as a hand job.
The generally accepted definition of a hand job is having your nether regions stroked by a hand. That definition doesn't specify who does it. So any hand of any person including the owner of the stick that is being stroked will be considered giving a hand job.
Therefor self love is a hand job, which is hand sex, which is sex, which is getting laid.
So self-love IS equal to getting laid.
:rofl: :rofl: :rofl:

[Team_Mclaren]

Originally posted by Carfanman


Not necessarily...

First we need to define getting laid.
The generally accepted definition of laid is engaging in sex.
There are many different kinds of sex.
There is vaginal.
There is anal.
There is oral.
And there is also hand sex, also known as a hand job.
The generally accepted definition of a hand job is having your nether regions stroked by a hand. That definition doesn't specify who does it. So any hand of any person including the owner of the stick that is being stroked will be considered giving a hand job.
Therefor self love is a hand job, which is hand sex, which is sex, which is getting laid.
So self-love IS equal to getting laid.
:rofl: :rofl: :rofl:

:rolleyes:

funny how some people can't contribute anything intelligent in a thoughtful argument.

[FatboyTheHungry]

Originally posted by khtm

Sigh...why even bring up CompSci math. You guys take about half of what engineers have to take. CompSci is a freakin' joke. I don't know what AMAT or PMAT is...I didn't go to U of C.


Apparently you are also unaware that the CPSC dept is house inside the department of mathematics at the U of C.
AMAT = Applied Math
PMAT = Pure Math

I'm sure the 75% failure rate in CPSC qualifies it as a joke then...

Apparently you also missed the part where I said:


Originally posted by fatboythehungry

As n approaches infinity,
the SUM of (9/10) + (9/10^2) + (9/10^3) + ... + (9/10^n) = 1


but never gets there BECAUSE of the fact that it is infinitely expanding.

[Carfanman]

Originally posted by Team_Mclaren


:rolleyes:

funny how some people can't contribute anything intelligent in a thoughtful argument.


You mean like the ten other people who posted trivial BS but noone said anything?


At least my first two paragraphs were relevant, and the last part was a joke, although technically accurate.

[szw]

Originally posted by Carfanman
1/3 does not definitly .333333333r. 1/3 times three equals one. .333333333r times 3 equals .999999999r. So 1/3 is only .333333r if .9999999r=1, not the other way around. In other words, saying that 1/3=.333333333r does not prove anything, because whether or not that is true depends on the of the original argument, it doesnt prove it. I explained that poorly but oh well.



This is so wrong it scares me.

[szw]

Originally posted by KLCC
I guess I found case to back up my thoughts as well, so does that makes us stalemate? :D



Reference
1)http://www.users.globalnet.co.uk/~perry/maths/nonstandardinterval/nonstandardinterval.htm

2)http://www.askmehelpdesk.com/forum/showthread.php?t=8574

Um...no
Non-standard analysis uses hyper-real numbers to explain the presence of infinitesimal numbers, i.e. numbers which are less than 1/n for all n, but however never 0.
It is not the same argument.

[Super_Geo]

0.9999999999999repeating = 1.

The algebric explinations beforehand are accurate. However, if you want a more philisophical explination, here it is:

Two numbers are not equal only if there is a number between them. So, for example... 3 is not equal to 2 because there's 2.5.... which is so obvoius that it's trivial.

This changes for 0.99999999 repeating. There is no number larger than 0.99999repeating that is not 1. In other words: there is no number that is between 0.99999repeating and 1 (think about it...). The only number bigger than 0.9999999repeating is 1.000000000001 (add as many zeros as you want before the 001), the condition of which is that it is >1. Therefore, this simply proves that 0.999999repeating = 1.

The equation solution before is accurate though, it explains it perfectly.

[Super_Geo]

Carfanman wrote

1/3 does not definitly .333333333r. 1/3 times three equals one. .333333333r times 3 equals .999999999r. So 1/3 is only .333333r if .9999999r=1, not the other way around. In other words, saying that 1/3=.333333333r does not prove anything, because whether or not that is true depends on the of the original argument, it doesnt prove it. I explained that poorly but oh well.


And the way I see it there is no answer because .99999r isn't an actual number, its just a theory or an idea. It doesn't exist because it has no limit, and it can't exist anywhere other than in theory. You can't have .99999r of an object, because thats not a real number.
Therefore this is a moot point.

This is why you should go to school. You have no idea what you're talking about. I'm not even going to bother explaining to you how off the mark you are. Please, go to school and educate yourself.

[alloroc]

Originally posted by Team_Mclaren


:rolleyes:

funny how some people can't contribute anything intelligent in a thoughtful argument.

Arguments aren't thoughtful.

Debates and discussions are thougthful.


Does your car make 299.999999r horsepower?

Or.

Do you say 300!

If you were having intercourse with 2 chicks at the same time ( switching from one to the other ) and finished up in girl A) and not girl B).

The next day would you say you had intercourse with one chick or two!

anyone who says ... .9999999r does not equal 1 is lying <G>

I say .99999999r does equal 1

and it does not equal 1.00000r.

Come on folks wake up and smell the sig figs.


Oh ya and get laid :)

[alloroc]

Originally posted by KLCC
ask something rhetoric like, chicken and the egg, which one came first?

The egg.

There were eggs before chickens.

[Carfanman]

Originally posted by Super_Geo
This is why you should go to school. You have no idea what you're talking about. I'm not even going to bother explaining to you how off the mark you are. Please, go to school and educate yourself.

I'm not wrong.


.333r and 1/3 are only equal if .999r=1 because .333r*3=.999 and 1/3*3=1. Since we haven't decided whether .999r and 1 are equal, we can't decide whether .333 and 1/3 are equal.


.999r can't exist in reality. Explain to me how it is possible to have an object that is .999r complete. It's infinite. You can't have an infinite object. so .999r is only and idea.

[bliggityblo]

x=0.999r
10x=9.999r
10x-x = 9.999r - 0.999r
9x = 9.0
x = 1

when you subract x (which equals .99999999repeating) from the ten shouldnt you get

9.00000000000000000000000(repeating forever)

but somewhere theres got to be a one so it will round up to ten

so it would be

9.000000000000000...1x = 9.0

so then you cant make them equal.

maybe?

i dunno?

[Goblin]

2 does not equal 1

3 does not equal 1

5 does not equal 1

.8 does not equal 1

.9 does not qual 1

.9999999 does not equal 1

1.000000000000000000000000000001 does not equal 1


god its not that hard!

[szw]

Originally posted by Carfanman


I'm not wrong.

Short answer: Yes, you are wrong.

Long answer: Ok it'll still be short. Pick up any math textbook grades 3 and up and you will see that you are wrong.

Anyone who has seen any of the proofs and still disagrees are basically saying they don't believe in algebra.

And Carfanman, where did we ever say that something has to physically exist in order to think about it? Does "1" even physically exist? Is it real? What does "1" look like? "1" is an idea also, as is our entire number system.

[Hakkola]

Well, I think the fact that only 30% of beyond's community (smart as it may be), chose that 0.9999r is the same as 1 helps reaffirm my belief that it is.

Most people don't have the mathamatical knowledge to have the answer to this kind of question.

On brain buster questions, whatever option the majority picks is usually wrong, that's why the question becomes popular.

[szw]

Originally posted by Hakkola

Most people don't have the mathamatical knowledge to have the answer to this kind of question.


Yeah, it's not really something you derive yourself. It's something someone teaches you and you learn how to understand it.

Really, it doesn't have much bearing on anything, it is simply something people like to argue about that mathematics people (not me!) know are true.

BTW on previous forums I've seen this on the spread is usually 60/40 +-5

[Hakkola]

I'd like to see how you came up with that number. ;)

Are you actually stating that on most boards the answer has been about 60% for yes? That would really surprise me.