# Is it really true 0.999... = 1 ?



## Santiago Serantes Raposo (May 10, 2012)

By their own definition, 0.999... and 1(left) are infinitesimally inferior to 1. That's the reason why I assume they equal when calculating the limit and represent 0(right) and 0(left) with an addition/substraction of 0.999...

In my opinion 1 is equal to 0.999... theoretically, but when calculating a limit it actually makes a difference in the result. Basically:
>1(left) = 1 - a/(inf) = 0.999... 
>1(right) = 1 + a/(inf)) = 2 - 0.999...
>1 = 0.99... + a/(inf) = 2 - 0.99... - a/(inf)
*Being "a" a real number; a ≠ (inf); a ≠ 0.

And you can see in the function I put as an example that it really makes a difference when calculating, so... this is why I think 1 ≠ 0.999...

P.S: I had the "great" idea to post this on 4chan /sci/ and I got banned for apparently "posting nonsense".


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## bigtex1989 (Feb 7, 2011)

yes, .99999999999999999999999 repeating does equal 1 if and only if the number of 9s is an uncountable infinity. If the number of 9s a countable infinity of less, it is a tiny bit less than 1 XD


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## Santiago Serantes Raposo (May 10, 2012)

Yes, I know that, but my doubt is concerning functions and limits. As I explain in the post and the pic.


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## Santiago Serantes Raposo (May 10, 2012)

bigtex1989 said:


> yes, .99999999999999999999999 repeating does equal 1 if and only if the number of 9s is an uncountable infinity. If the number of 9s a countable infinity of less, it is a tiny bit less than 1 XD


Yes, I know that, but my doubt is concerning functions and limits. As I explain in the post and the pic.


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## bigtex1989 (Feb 7, 2011)

I'm not exactly getting your question then. In terms of the function posted, the limit at zero from the left and right will be different, but that isn't a big deal. I don't see how that would convince you of anything. Just means that the limit at zero doesn't exist for that function which is well known.

As for your algebra, 1 = 0.99... + a/(inf) = 2 - 0.99... - a/(inf) is fine.

For the limit as x ->inf a/x -> zero. so again, your algebra shows that 1=.99....


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## bellisaurius (Jan 18, 2012)

You probably got banned because the thread pops up on a lot of forums from time to time, and people beat their heads against rocks, which wouldn't be so bad if the mathematically disenfrachised kept to themselves. However, science folks tend to also get dragged into them and when someone who knows a fact attempts to share that fact with someone who denies it vigorously, much butthurt is had on all sides. Basically, they mistook it for a troll.


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## Anonynony (Jun 24, 2012)

.999(how ever many 9's you feel like putting) is almost 1 because it will always be .0000001(or what ever) away from actually being 1. Kinda like always the brides-maid, never the bride


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## Wulfyn (May 22, 2010)

Santiago, answer this: what is 1 - 0.999... recurring?





FigureSkater said:


> .999(how ever many 9's you feel like putting) is almost 1 because it will always be .0000001(or what ever) away from actually being 1. Kinda like always the brides-maid, never the bride


If the 9's go on for infinity and never stop, will there ever be a 1 as the difference in the line of 0's?


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## Anonynony (Jun 24, 2012)

Wulfyn said:


> Santiago, answer this: what is 1 - 0.999... recurring?
> 
> 
> 
> ...


If the 9 goes on for infinity, it's .1 infinity from being 1


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## Tragic (Jan 31, 2011)

What we write down when we write a number is a representation of that number, for example:
4, IV, IIII, four
are all the same number, just written differently. They represent an abstract concept of number but they aren't the number itself. They are like a name.

Repeating decimals is a way to express a limit, nothing more, nothing less.
0.99... (recurring) is shorthand for the limit of the sequence:

0.9
0.99
0.999
0.9999
etc...

What you need to do if you want to know the value of any recurring decimal is consider the limit of the sequence. Remember it's just a shorthand and just the way of expressing a number, not the number itself. It turns out that 0.99... (recurring) can be expressed in an alternate form: 1 (or 'one' in English). It's just an equivalent way of expressing it. (If you're interested, limits are rigorously defined mathematically, the wiki article is good: Limit of a sequence - Wikipedia, the free encyclopedia)




bigtex1989 said:


> if and only if the number of 9s is an uncountable infinity.


It's a limit of a sequence. Sequences, by definition, have a countably infinite number of terms. There's no other alternative. The limit is 1.



Anyway, back to the OP's question. I'm not exactly sure what you mean, but my guess is you are interchanging limits. You can't blindly interchange limits without changing the result.

A simple example:

x(1/y) Let x and y tend to infinity.

If you let y->infinity first, then x second you get: 0
If you let x->infinity first, then y second you get: infinity

When dealing with 0.99... (recurring) it's just an expression, it's a limit of a sequence, you must take that limit first. The limit is 1. Then use 1 in your calculations. Interchanging limits will (very often) mess things up.

Hope this helps!


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## bigtex1989 (Feb 7, 2011)

I didn't think a sequence had to have a maximum of a countable infinity number of terms.....I'll go back to some books and look that up XD


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## sprinkles (Feb 7, 2010)

http://personalitycafe.com/debate-forum/76657-does-999-1-a.html <--click there


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## OldManRivers (Mar 22, 2012)

FigureSkater said:


> .999(how ever many 9's you feel like putting) is almost 1 because it will always be .0000001(or what ever) away from actually being 1. Kinda like always the brides-maid, never the bride


This is an example of needing to know and understand the mathematical concept of infinity - For the 0.999. . .to imfinity, there is not a last digit -it goes on to infinity -never ending, and it is equal to one - if there is a differencve between .999.....and one, it did not go to infinity.


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## Anonynony (Jun 24, 2012)

OldManRivers said:


> This is an example of needing to know and understand the mathematical concept of infinity - For the 0.999. . .to imfinity, there is not a last digit -it goes on to infinity -never ending, and it is equal to one - if there is a differencve between .999.....and one, it did not go to infinity.


I understand infinity. Even though it it close, .9(infinity) will never be 1. Mathematictians are just lazy. :tongue:


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## sprinkles (Feb 7, 2010)

All of this was answered in the other thread.

To redundantly repeat myself redundantly multiple times and more than once, over and over again repeatedly...

There's a reason they say it is equal to 1 because a non-zero infinitesimal is not considered a rational number. If there were a difference between 0.999... and 1, you would not be able to express it, since it will always be less than any number you come up with - always and forever less. Infinitely less.


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## OldManRivers (Mar 22, 2012)

FigureSkater said:


> I understand infinity. Even though it it close, .9(infinity) will never be 1. Mathematictians are just lazy. :tongue:


No, you do not understand infinity. I believe you are thinking in terms of a series going on and on and on - but that is incorrect. Infinity is a mathematical concept. Take a few calculus courses and you will understand. The limit cannot be less than one because that is connected to the concept of infinity, which do not understand. If it is not equal to one, infinity does not exist.
Is there a finite value for pi, the ratio of the circumference of a circle and the radius? No, pi is a proven irrational number. The value of infinite series are also proven.


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## Wulfyn (May 22, 2010)

FigureSkater said:


> If the 9 goes on for infinity, it's .1 infinity from being 1


But here you are treating infinity as a discreet number.

Can a number be 0.000 with an infinite number of 0's and then a 1? If there is ever a 1 then surely you've not had an infinite number of 0's? As soon as you consider an irrational number as rational then you can have a discrete difference. But if you were to treat 0.999 recurring as rational then the rational number you would need to use is 1 (as 1 is closer to 0.999 recurring than ANY end point of the 9's).


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## nonnaci (Sep 25, 2011)

blah, prove it by series expansion or if you want a numerical/computational proof...

Find me a real number x between .99999... and 1. If such a number exist, then in decimal notation there exist some truncated representation of x (say x to the pth digit) where truncated(x,p) > truncated(.9999..., p). Show this is a contradiction via finite representations for any p.


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## MNiS (Jan 30, 2010)

This debate again. lol

Proof that 0.999... = 1.

1/3 = 0.333...
3* 1/3 = 3* 0.333...
1. = 0.999...


That's it! There's no trick, no deception, math hasn't been proven wrong etc etc. 

Of course, if the proof doesn't convince you then there's a fundamental misunderstanding about the concept of "infinity". That however, is a user problem. Not a math problem. :wink:


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## Razare (Apr 21, 2009)

I always saw limits as a way of fancy rounding... you're getting the number that it basically is, not the number it actually is... though they may claim it's the same, but it's not! It's just indistinguishable from the same. 

But obviously I'm wrong because of the proof above ;(

Frankly, I'd argue 1/3 != .3333...

because... infinite .33333... + infinite .33333... + infinite .33333.... = infinite 0.99999....

To turn .9999.... into 1.0000, we need to add the number 0.000.....infinite....00001



But I suppose we could then say 0.0000... infinite ... 00001 is basically zero. I hate math.


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