Does 0.9999(repeating) = 1?

Yes 705 54.95%
No 578 45.05%
Voters: 1283. You may not vote on this poll

Does 0.9999(repeating) = 1?

First Riot Post
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idDobie

Senior Member

01-23-2013

Quote:
Originally Posted by Kaolla View Post
thank you for proving my point. limit = approximation
Limits are not approximations. They are different in different contexts but for most mathematics it is something like every value of something past some point being some small tolerance away from another value(the limit).

Approximations are not exact - limits are exact. A limit isn't a range of acceptable values it is ONE value. An approximation is a value that is close enough to service the needs of whatever application you're using them for.


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Ryedan

Visual Designer

01-23-2013
25 of 32 Riot Posts

http://www.math.umt.edu/tmme/vol7no1...e1_pp.3_30.pdf

I like this explanation


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Adumu

Senior Member

01-23-2013

Quote:
Originally Posted by WhattayaBrian View Post
They are equal within the standard definition of Complex Numbers. But don't take my word for it.

[turley]TINfzxSnnIE[/turley]

[turley]wsOXvQn3JuE[/turley]
I really like this response.


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Kaolla

Senior Member

01-23-2013

Quote:
Originally Posted by Noobgrenade View Post
A repeating decimal is not a process
nope. it is exactly a process

1/3 is one divided by three is .333 which is the result of you processing that fraction.

if at any point you STOP the process, you do not have an equivalency. if the process is running, fine.


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Kaolla

Senior Member

01-23-2013

Quote:
Originally Posted by idDobie View Post
A limit isn't a range of acceptable values it is ONE value
it's one value approximately equal to the real value


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Kaolla

Senior Member

01-23-2013

Quote:
Originally Posted by Imogen Poots View Post
Infinity is not a real number (only part of the extended real number line); 0.999... infinitely repeating is a real number.
wrong, both are processes


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Noobgrenade

Senior Member

01-23-2013

Quote:
Originally Posted by Kaolla View Post
nope. it is exactly a process

1/3 is one divided by three is .333 which is the result of you processing that fraction.

if at any point you STOP the process, you do not have an equivalency. if the process is running, fine.
Nope, it is exactly not a process.

1/3 and .333... are the same number, with different representations.
If, at any point, you start a process to determine the position of the last 3, you are misunderstanding what .333... represents.
While you may invoke a process to determine the representation of a number in a different system, this process isn't a fundamental part of the number itself.


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Imogen Poots

Senior Member

01-23-2013

Quote:
Originally Posted by Kaolla View Post
i've said it before, what's with people taking the limit of an infinite function and assuming they have anything other than an approximation?

hi i'm going to LIMIT my function to 10 decimal places!

.999999999 = 1

wow this statement is FALSE
LOL. I see your problem now. You don't have have the first clue what limit people are talking about. ROFLMAO.

In response to the bolded part: Oh, I don't know, it probably has something to do with the fact that they, in fact, are more than an approximation. I would point you to the displacement/speed functions again, but given what you have just said above, I don't think you have the foggiest idea what I am talking about.


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Aliath

Senior Member

01-23-2013

leave ot please


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Kaolla

Senior Member

01-23-2013

Quote:
Originally Posted by Noobgrenade View Post
Nope, it is exactly not a process.

1/3 and .333... are the same number, with different representations.
If, at any point, you start a process to determine the position of the last 3, you are misunderstanding what .333... represents.
you can't create the "number" .333 without resorting to a process