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 September 30th, 2016, 12:43 PM #11 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 830 Thanks: 67 The difference between .5000... and .4999... approaches 0 as n approaches infinity, but it never equals 0. So .49999999....... never equals .50000000..... But, you say, if .499999... doesn't equal .500000... in the limit as n approaches infinity, there must be a number between them. NO. because in the limit as n approaches infinity they are real numbers and the real numbers are complete: there is no real number between them.
 September 30th, 2016, 02:35 PM #12 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,354 Thanks: 2085 Math Focus: Mainly analysis and algebra In which case they are the same real number because for any two real numbers $a\lt b$ there exists a third real number $\frac{a+b}{2}$ and $a \le \frac{a+b}{2} \le b$. What you are saying about the elements of the sequence is obvious. But you have now contradicted your original claim that all decimals represent different real numbers because you have said that the infinite decimals 0.4999... and 0.5000... are equal. Last edited by v8archie; September 30th, 2016 at 02:42 PM.
 October 1st, 2016, 05:43 AM #13 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 830 Thanks: 67 $\displaystyle \lim_{n\rightarrow \infty}$1/n=0 but 1/n never equals 0.
 October 1st, 2016, 05:59 AM #14 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,354 Thanks: 2085 Math Focus: Mainly analysis and algebra ${1 \over n} \ne 0$ for finite $n$, correct. But that has nothing to do with infinite decimals.
October 1st, 2016, 09:05 PM   #15
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Quote:
 Originally Posted by v8archie But you have now contradicted your original claim that all decimals represent different real numbers...
"Decimal representation is unique" is the title, not a claim made in the original post; "all decimals represent different real numbers" wasn't claimed at all.

 October 2nd, 2016, 02:54 AM #16 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,354 Thanks: 2085 Math Focus: Mainly analysis and algebra Those two phrases are equal. The interpretation that "each decimal representation represents exactly one real number" has nothing whatever to do with the content of Zylo's posts and is anyway a trivial consequence of the definition of the reals as a limits of Cauchy sequences. Last edited by v8archie; October 2nd, 2016 at 03:09 AM.
 October 2nd, 2016, 03:38 AM #17 Global Moderator   Joined: Dec 2006 Posts: 16,223 Thanks: 1150 The phrases aren't equal - in the original post, it's plainly stated that the decimal representations referred to have n decimal places (which makes them unique), whereas you used the second phrase in a context that suggests that infinite decimal representations are included.
 October 2nd, 2016, 04:37 AM #18 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,354 Thanks: 2085 Math Focus: Mainly analysis and algebra Do you not think that there is a clear implication that this extends to "infinity". With that, the statement is one of complete banality isn't it?
 October 2nd, 2016, 04:51 AM #19 Global Moderator   Joined: Dec 2006 Posts: 16,223 Thanks: 1150 It was indicated by zylo in a later post that uniqueness is maintained as n approaches infinity, which hints that infinite decimal representations would be considered next, but zylo never got that far. Unfortunately, the uniqueness of the finite representations doesn't imply uniqueness for infinite representations.
October 3rd, 2016, 01:34 AM   #20
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Quote:
 Originally Posted by zylo If I measure 1/8 seven times in succession with a ruler accurate to 3 decimal places, the result is not accurate to 3 decimal places.
Eh? If you measure the length of something with a ruler with a precision of 0.001 m, then the measurement error is at best $\displaystyle \pm 0.001$.

Quote:
 7x.125 = .875, But if .125 is really .1254, then 7x.125 = .878
Yes, that's why we have propagation of errors in quadrature, which allows us to calculate the error on derived values. In your case, if the error on the original value is $\displaystyle \pm 0.001$, then:

$\displaystyle 7 \times (0.125 \pm 0.001) = 0.875 \pm 0.007$

Because the error just scales with the multiplicative factor. There is a general technique for any sequence of operations.

This hasn't really got anything to do with the decimal number system.

Quote:
 Had rules for rounding off decimal arithmetic explained in chemistry. Never understood or used them. An engineer (ok, me) calculates .12x.3135 as .03762, but call it .0376 because that's the closest you can measure with a micrometer (.124x.3135=.038874).
I think you are trying to say that there if there is some sort of error in fundamental arithmetic then the rest of mathematics would be wrong. Thankfully this is not the case!

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