Can Infinity Change the Value of .9 Repeating?

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Discussion Overview

The discussion centers around the mathematical concept of the repeating decimal 0.9 repeating (0.999...) and its equivalence to the number 1. Participants explore various viewpoints regarding the implications of infinity in this context, including measurement limits, definitions, and the nature of non-terminating decimals.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants argue that 0.9 repeating is defined as the limit of the sequence 0.9, 0.99, 0.999, etc., and therefore is exactly equal to 1.
  • Others propose that the concept of having an infinite number of zeros followed by a 1 could create a distinction, although this idea is challenged by participants who emphasize the nature of infinity.
  • One participant suggests that the approximation of 0.999... to 1 is due to measurement limits, while another counters that this is not about approximation but rather an exact equality.
  • There are claims that different representations of numbers, such as 1.000... and 0.999..., are equivalent, but some participants question the validity of these representations.
  • A participant expresses skepticism about the validity of the arguments presented, suggesting that some statements may not be serious or are misinterpretations of mathematical principles.

Areas of Agreement / Disagreement

Participants generally disagree on the interpretation of 0.9 repeating and its relationship to the number 1. Multiple competing views remain regarding the implications of infinity and the nature of decimal representations.

Contextual Notes

The discussion includes various assumptions about the nature of infinity, the definitions of limits, and the implications of non-terminating decimals, which are not fully resolved.

Justin Horne
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Hi. I'm a freshman in High school, and my algebra teacher gave me a problem. She said that .9 repeating = 1 Now apart from all the ways this didn't make sense to me, I thought of a way that might make it not work. If you were to put something like .0 [infinity of 0's] 1, wouldn't that be able to be added to .9 repeating, making 1? Correct me if I am wrong here, please. Thanks,
Justin.
 
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There is a limit to how much we can measure. In fixing this limit we also fix our approximations. When there are infinites of nine we say that it is very close to one and hence approximate it to 1. We could approximate 0.999999999999999999999999999999999999999999999... to 0.9, but 0.9999999999999999999999999999999999999999... is more closer to 1 than 0.9. Similarly, 0.0000000000000000000000000000000000000000001 is more closer to 0 than 1. Thus it is basically because there is a limit to our measurements and approximations that we approximate to te closest digit possible.

Sridhar
 
Originally posted by sridhar_n
we approximate to te closest digit possible.
Sorry, this is not correct. [itex]0.\overline{9}[/itex] is not approximately 1, it is absolutely 1. Remember, I'm not talking about a lot of nines, I'm talking about an infinite number of nines. That's a whole different bucket o' spaghetti.

- Warren
 
I know it is hard to accept but even teachers are right ocassionally! The problem with you idea is that you can't have an "infinite number of 0s" and then put a 1 on the end. With an infinite number of 0s there is NO end.

A non-terminating decimal such as 0.99999... is DEFINED as the limit of the sequence .9, .99, .999, .9999, etc. and that limit IS 1.0.
 
There is a limit to how much we can measure. In fixing this limit we also fix our approximations. When there are infinites of nine we say that it is very close to one and hence approximate it to 1. We could approximate 0.999999999999999999999999999999999999999999999... to 0.9, but 0.9999999999999999999999999999999999999999... is more closer to 1 than 0.9. Similarly, 0.0000000000000000000000000000000000000000001 is more closer to 0 than 1. Thus it is basically because there is a limit to our measurements and approximations that we approximate to te closest digit possible.

Sridhar

I really have to object to this. When we say that 0.999... is 1, we are not talking about "measurement" and we are not talking about approximations.

0.99999... is not close to 1 it is 1: exactly equal to 1. As I said in another post, 0.99999... is defined as the limit of the sequence 0.9, 0.99, 0.999, ... and it is easy to show that that limit is 1.0.
 
When 1.000... and 0.999... are two representations of the same number then:

1.00... = 0.999...

0.100... = 0.0999...

0.0100... = 0.00999...

0.00100... = 0.000999...

0.000100... = 0.0000999...

0.0000100... = 0.00000999...

Therefore we can write:

0.100... + 0.0100... = 0.0999... + 0.00999...

0.0100... + 0.00100... = 0.00999... + 0.000999...

But this is not true because:

0.1100... not= 0.0999... + 0.00999... = 0.10999...8

0.01100... not= 0.00999... + 0.000999... = 0.010999...8

and so on ...


Digit 8 is not the last digit but the limit digit or the unreachable digit of 0.999...

Therefore 1.000... is not the limit of 0.999...
 
Last edited:
I sincerely hope that that was your idea of a joke.
 
Check out the other thread with the identical post to convince yourself that it wasn't a joke... :frown:
 
  • #10
[moderator hat]

This thread should be locked, since it just parallels the other one.

[/moderator hat]

- Warren
 

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