SRL Archive - Discussions & Debates

0.9999…so on=1 how? you may ask, it's simple algebra :p
x=0.999999..
10x=9.999999…
10x-x=9x
9.999999…-0.999999…=9
9x=9
x=1 anyone agree?

yeh i agree :confused:

on the last line u put x=1 in maths if x is unknown then it is counted as 1
edit : Question : y the explination?

You are incorrect.
0.9999… Is an undefined number, it isn't used in algebra.
x=0.999999.. x cannot equal an undefined number, it's like dividing by zero!
10x=9.999999…
10x-x=9x
9.999999…-0.999999…=9 You've fucked up your equation here, you've just got rid of the x which isn't allowed and makes the equation invalid.
9x=9 IS NOT EQUAL TO NINE! It is equal to 8.999…
x=1 anyone agree? NO!

My advice? Go back to school and learn BASIC maths!

EDIT: Pureblood, x is never equal to 1 if it is unknown.

EDIT EDIT: You may be thinking about if it's to the power of zero then it is 1.

erm…sorry? say that in english? algebra isn't just a defined list of numbers :P and if 9.9999..-0.999…=8.99… then the world is about to explode.

What's hard to understand about that? You've proven x=0.999… as per your original statement. Also, refer to my previous statement telling you to go back to school, undefined numbers cannot be put into an equation, as the name suggest, they are undefined!


EDIT:

if 9.9999..-0.999…=8.99… then the world is about to explode.

Why would it explode?

idk, plus the fact i still am at school, and once again please prove me wrong on this, how is my theory mathematically incorrect? and 10x-x=9x? how is that wrong? i mean really "learn basic algebra"
edit:undefined? are you thick? it doesn't have to be all letters in algebra, seriously think before you post

Well then pay attention in school, if you're going to round 8.999…to 9
then you have to round 9.999 to 10, which basically means you're going back to 1=1

I don't know what year of school you're in but i'll try to "dumb" it down for you,

0.999… ad infinitum

is not a real number. Why you ask? Simple, do 1 x 0.999… ad infinitum in your head. You can't can you? Neither can a PC or any other computing device as 0.999… ad infinitum doesn't have an exact value, it goes on for ever and ever.

So you cannot put this into an equation, it's a divide by zero error.

Theres no simpler way to do it than this! What does an apple look like? What does 0.999… ad infinitum of an apple look like? Nothing as you cannot possible get anything "material" to reach infinity and that's what you need to know, that you cannot use an infinite value in maths.

EDIT:

edit:undefined? are you thick? it doesn't have to be all letters in algebra, seriously think before you post

What are you talking about? If it was all letters it would be unsolvable! I never said it needed to be all numbers, learn what an undefined number is before you call someone "thick". You need to think before you post or atleast learn maths.

sandos's equation is correct.

.9 recurring = 1.


Kiloch: You're saying that every recurring decimal doesn't have a value.

.999… ad infintum is not 1, any recurring value does not have a real value, unless put into a real number (2/3 for example) An image taken off of google does not make it correct, try actually learning and studying during maths to remember the basic fact that you cannot

USE AN UNDEFINED NUMBER IN AN EQUATION

You can write it down but it does not mean it is correct. Take Division of zero for example.

0 x X = 0

therefore

0/0 = x

It cannot be used in an equation. It is not mathematically sound.

Try reading the image off of google, and see that it is correct. 0 CAN be used in an equation. 0 x 5 + 3 = 3.

0 Can be used in an equation, 0 can't be divided by as it is an undefined number, as are all numbers that go on ad infinitum. What i'm saying is that it doesn't have a real value, it has an undefined value so a correct assumption would be 0.999 ~=~ 1, not that it exactly equal to one.

You aren't very open-minded, are you? Seeing as you're so "mathematically correct", it's apparent that you haven't taken any complex mathematics classes, such as calculus or trigonometry. .9 recurring = 1. Ask your teacher.

@ kiloch nice fight but i think u have lost
@ the rest of the people who was against kiloch a tip for u in life…
There is no point fighting against people who won't listen except for the pleasure at the end of it all.

Oh no! Trigonometry is complex! Calculus is complex also! Oh no! Damn! Too bad I passed those subjects…

Go sit down and reread your Calculus textbook, this is Algebra, not calculus, they are two different parts of maths, true they can relate to each other but you cannot use calculus in a purely algebraical equation if you arn't integrating or differentiating.

F(x) = 0.9999999
F'(x) = 0

and intergrating

= 0.999…ad infinitum x

They don't have anything to do with the equation.

And what has trigonometry got to do with this? I'm not sure about your school curriculum but we were taught that at 14.

EDIT: Pureblood, how have I lost? That the popular consensus says i'm wrong does not mean I am wrong, your argument goes both ways, i'm not listening to you and your not listening to me, so which one is correct? The majority? There is no winner and no point in being educationally mis-guided by the popular choice.

A bit ago, there was an entire thread of people proving the equation correct.

You still haven't proven to me that it's wrong.

You mean

it doesn't have a real value, it has an undefined value so a correct assumption would be 0.999 ~=~ 1, not that it exactly equal to one.

0.999… ad infinitum is taken as one as it goes on forever, with no end to it, it is easier just to round it off to one and say "job done".

So, 0.999… is ~=~ 1 as we cannot express it as it's own value, but it is still not exactly equal to one.

.9 recurring can NOT be rounded to 1.

You are incorrect.

There is no possible way to go from .9 recurring to 1.

Therefore, that is another way that the two are equivalent.

So you are saying that 0.999… ad infinitum cannot be rounded to one? Oh lets see,

0.99999999999999999… ad infinitum rounded to 1 d.p
1.0

well wasn't that easy?

How did you round it?

It's impossible to add .01 to it, or any other number, to make it to 1.

You made an assumed round, though you actually did nothing to prove how you rounded it.

2.9 can be rounded to 3 by adding .1, but 2.9 recurring cannot be rounded to 3 by adding any number, because it would overflow and return a number greater than 3.

lets call this a draw???

It's not a draw.

It's been mathematically proven that .9 recurring = 1.

Google it.

I'm done.

EDIT: Eh i'm getting tired of this now, If you can link me to a reliable source saying it is correct, then I will take back all of my previous assertions.

No, it's possible to subtract it.

Any number subtracted by itself is 0.

First, .999… is exactly equal to 1.

http://en.wikipedia.org/wiki/.999

They have lots of proof on that page. And don't you dare say "it isn't valid" because it's wikipedia.

Second, was this thread inspired by the fact that I put this in my sig the other day?

nah i was just thiking of some random stuff in maths the other day and ran it by my teacher, so i thought i'd post it :P, and thanks TViYH i was looking for a website with that on it :P.

I linked Kiloch to that link via PM earlier.

From: Kiloch
Oh no! Trigonometry is complex! Calculus is complex also! Oh no! Damn! Too bad I passed those subjects…

Go sit down and reread your Calculus textbook, this is Algebra, not calculus, they are two different parts of maths, true they can relate to each other but you cannot use calculus in a purely algebraical equation if you arn't integrating or differentiating.

F(x) = 0.9999999
F'(x) = 0

and intergrating

= 0.999…ad infinitum x

They don't have anything to do with the equation.

And what has trigonometry got to do with this? I'm not sure about your school curriculum but we were taught that at 14.

EDIT: Pureblood, how have I lost? That the popular consensus says i'm wrong does not mean I am wrong, your argument goes both ways, i'm not listening to you and your not listening to me, so which one is correct? The majority? There is no winner and no point in being educationally mis-guided by the popular choice.

Your arrogant and wrong. How is .9999… Undefined? is .333… undefined? So isn't .99999….. = .33333…. * .333333…..? Correct?

Tell me, how did you pass calculus when you where obviously asleep through at least a week of the class. Every calc teacher I know of introduces the fact that .9999…. = 1 when talking about infinity and its implication in calculus (actually, my pre-calculus teacher brought this one up)

How can you say that calculus and algebra are different parts of maths? They are the same thing! Where you asleep during the delta-epsilon proofs, or when you took derivatives the long way? How about when you where calculating limits, what did you use to do that? Magic? Gee, no, you used algebra… Calculus is so tightly interwoven with algebra that only an idiot would claim that you can have Calculus without having algebra.

The reason calculus is brought up because central to the theme of calculus is the idea of what happens when dealing with infinity on a general basis.

You are wrong, because you can't prove yourself right and other have and can continue to prove you wrong using many simple algebraic equations.

1/3 = .333…..

2/3 = .666…..

3/3 = .999….

1/3+2/3= 3/3, or .999….

which = 1?

No matter how many nines you stick onto the end of 0.999 etc., it will never be 1. If it was 1, it would be written as 1, not 0.999999…Just because it is impossible to make that number 1, it does not mean to say that it is 1.

From: Ejjman1
1/3 = .333…..

2/3 = .666…..

3/3 = .999….

1/3+2/3= 3/3, or .999….

which = 1?

I think that pretty much sums it up. :p

1/3 + 2/3 = .999999…
1/3 + 2/3 = 3/3 (without changing the fractions to decimals)

by the transitive property..

.99999… = 3/3
.99999… = 1

:O

From: R1ch
No matter how many nines you stick onto the end of 0.999 etc., it will never be 1. If it was 1, it would be written as 1, not 0.999999…Just because it is impossible to make that number 1, it does not mean to say that it is 1.

you are just closed minded. The math says that it should be right. And i asked my teacher, he tried it, and then got all distressed because it was 1 :p.

Lim(.999[…]) = 1
Lim(1) = 1

It's a constant, so isn't that good enough?

Where I disagree is:
9.999999…-0.999999…=9

Since you don't know the decimal place of either number to any precision, you can't know the decimal precision of the answer either. Maybe I'm wrong, but that's just the way I see it.

A G E N T: Any number subtracted by itself = 0.

I am not going to waste my time, trying to teach math to you guys who don't believe. I asked my calc III teacher last time this came up, just make double-triple sure, and she said yes. Also e-mailed my highschool math teacher (PhD), he also said yes. If you don't believe it, ask your math teacher, if you still don't believe it, gtfo, we don't care, you can have you ignorance.

I retract all previous statements made by myself, 0.999… ad infinitum does in fact equal 1.

From: TViYH
A G E N T: Any number subtracted by itself = 0.

Thanks… :P
What I was trying to get at is that: are you sure that .999…. = .999…..

From: A G E N T
Thanks… :P
What I was trying to get at is that: are you sure that .999…. = .999…..

Uh, yes. Any number is equal to itself.

No it isn't

You've proven yourself wrong, 1= 0.999… so in effect it could be a diffrent number.

From: Kiloch
No it isn't

You've proven yourself wrong, 1= 0.999… so in effect it could be a diffrent number.

How? The thing is, they're not actually different numbers. They're just different ways of writing the same number.

From: Kiloch
No it isn't

You've proven yourself wrong, 1= 0.999… so in effect it could be a diffrent number.

Oh and I thought you understood. 1 = 0.999…. = 1 = 0.999…. How does that equate to 1 = 2 or 1 = 3? Just because 0.999…. = 1, it doesn't mean that .999… can equal some other number (unless that number is just another way of writing 1. For example cos 0 = 1, so .999… = cos 0).

From: boberman
Oh and I thought you understood. 1 = 0.999…. = 1 = 0.999…. How does that equate to 1 = 2 or 1 = 3? Just because 0.999…. = 1, it doesn't mean that .999… can equal some other number (unless that number is just another way of writing 1. For example cos 0 = 1, so .999… = cos 0).

.999 is a number

cos 0 is an equation.

From: Ejjman1
.999 is a number

cos 0 is an equation.

cos 0 is not an equation because it has a fixed value that will never change no matter what. Equations imply some variability to the expression (cos 0 is an expression, but so is .999…). For example, are you going to argue that sqr(1) is an equation? After all the square root operator is being called. how about 3/3 would that be an equation because the division operator is being called?

cos 0 is not an equation.

Even if it was, it wouldn't invalidate my argument in any way, shape or form because cos 0 always equates to 1, no matter what.

From: senrath
Uh, yes. Any number is equal to itself.

You've misunderstood me again. .999… is an undefined value (AFAIK). How can you equate two undefined values to each other?

EDIT: At any rate, this one makes way less sense to me: e^(pi*sqrt -1) = -1.

From: A G E N T
You've misunderstood me again. .999… is an undefined value (AFAIK). How can you equate two undefined values to each other?

EDIT: At any rate, this one makes way less sense to me: e^(pi*sqrt -1) = -1.

.999… is not undefined. Is .333… Undefined? no. You seem to misunderstand what the term "Undefined" means. .999.. is very well defined, we know exactly how it behaves on to infinity. (just like .333… is very well defined)

.999… behaves the same where ever it is placed and however it is used, just like 1 behaves the same in every situation it is placed in.

Nope.

If anyone isn't getting it, think about 2 things:

1) A number can have more than one representation.
2) If you can't deal with numbers going on forever, because it isn't what you're used to, look at the 1/3 proof posted by Ejjman

we use that repeating-decimal-to-fraction process to figure this out, no? so who's to say that that isnt wrong and that .999…. <> 1?…just wondering

From: Dan Cardin
we use that repeating-decimal-to-fraction process to figure this out, no? so who's to say that that isnt wrong and that .999…. <> 1?…just wondering

That was ONE of the methods used to prove that .999… = 1, however it isn't the ONLY method used to prove it. There are several proofs using different methods of proving it true. Take a look at the wikipedia link. They have like 5 or six valid proofs of .999… = 1.

Look, I've been thinking about this issue for a whole damn day straight now.

No matter how you put it, the difference between between 0.99.. and 1 will always remain as the 1^-..infinity (..0001). As long as there is some difference, the numbers aren't identical, and hence, 0.99.. is not the same as 1.

Nope.

Difference between 0.99 and 1 is 0.01. Difference between 0.999 and 1 is 0.001. Difference between 0.9999 and 1 is 0.0001. With each 9 you add, you add a 0 before you get to the 1. So if the 9s never stop, you never stop adding 0s, and you never get to the 1.

From: EvilChicken!
Look, I've been thinking about this issue for a whole damn day straight now.

No matter how you put it, the difference between between 0.99.. and 1 will always remain as the 1^-..infinity (..0001). As long as there is some difference, the numbers aren't identical, and hence, 0.99.. is not the same as 1.

If you have an infinite number of digits, How do you tack a number on the end? You can't The 0s never end so the 1 never exists at the end. If the difference is infinitely small, it is non-existent.

I +1 boberman's response.

If the 0's are infinite, there is no end. How can you put something on the end of something that doesn't end?

From: TViYH
I +1 boberman's response.

*cough* 9 minutes *cough* :p

Boberman, so by your logic any number with continuing digits = a different number simply because the .01 wouldn't have a chance to be placed is physical terms?

"If you have an infinite number of digits, How do you tack a number on the end? You can't The 0s never end so the 1 never exists at the end. If the difference is infinitely small, it is non-existent."

This logic goes against your algebra.
****_
So .89, by your logic must = .9 because that last digit would never be "tracked."

Algebra disproves this because

x = .8999…
10x = 8.999…
8.9 - x = 8.099…1, not .9 as with what happens with .99…


I am not here to disprove that .999… = 1. I simply saw a flaw in bober's logic, which implied that any repeating 9 = something larger.

From: A G E N T
Where I disagree is:
9.999999…-0.999999…=9

Since you don't know the decimal place of either number to any precision, you can't know the decimal precision of the answer either. Maybe I'm wrong, but that's just the way I see it.

Good point, I'd love to see an answer to this one.

Brown: We're talking about .9, not .89.

A G E N T: To my knowledge, any number (positive) subtracted by itself is 0.

From: Brown
Boberman, so by your logic any number with continuing digits = a different number simply because the .01 wouldn't have a chance to be placed is physical terms?

"If you have an infinite number of digits, How do you tack a number on the end? You can't The 0s never end so the 1 never exists at the end. If the difference is infinitely small, it is non-existent."

This logic goes against your algebra.
****_
So .89, by your logic must = .9 because that last digit would never be "tracked."

Algebra disproves this because

x = .8999…
10x = 8.999…
8.9 - x = 8.099…1, not .9 as with what happens with .99…


I am not here to disprove that .999… = 1. I simply saw a flaw in bober's logic, which implied that any repeating 9 = something larger.

Algebra confirms it, if you do correct algebra, if you use incorrect algebra then you can say whatever you want, though it doesn't prove or disprove anything.

First off, 8.9 - .89999 != 8.09991, it is equal to 8.00001. Therefore, 8.9 - .899…. is equal to, surprise 8. because that .8999.. is equivalent to saying .9.

Next time, check your algebra before saying it disproves anything. You can't say that there is a number at the end of an infinite sequence, by saying that you are saying your sequence isn't infinite. (infinite BY DEFINITION means never ending). A 1 after and infinite number of zeros doesn't exist because those zeros don't end. Heck, we might as well put pi, e, or ln(pi^e) in there, it doesn't matter because they don't exist.

Infinity is less of a number and more of a concept (a very strange concept the deeper you go into the rabbits hole, Look up cantor and his theories on infinity). If you don't have a cursory understanding on what infinity is and implies, then yes this won't make sense to you. However, you can trust me and every other person with a PHd in Mathematics that .999…. is in fact = 1.

10/3 = 3.33..
10/30 = .33..
.33.. x 3 = .99..
therefore..
3*10/30 = .99
30/30 = .99..
1/1 = .99..
1 = 0.99..

figured that mehord out my self :D

0.999… = 1 and 0.999… <> 1?

0.9 recurring can be proved by mathematically logical steps to be equal to 1

this has been known since the time of ancient greeks.
i dont even know why we're debating about this, its not anyones opinion, this is a mathematical fact, theres no room for debate or interpreting.


the only time I've ever debated against maths was satirically saying that pi = 3, which was only to point out the flawed attidute of religion.

pi equals three when I need to draw a sinus graph ^_^.