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 September 30th, 2016, 07:21 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,082 Thanks: 87 Decimal representation is unique .99..... and 1.00..... to n decimal places are not the same no matter what n is, but their difference approaches zero as n approaches infinity, just as the difference between any two n-place decimals in consecutive order does as n approaches infinity.
 September 30th, 2016, 07:29 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,878 Thanks: 2240 Math Focus: Mainly analysis and algebra If they are different, you will be able to give a real number that lies between them. There's no need for weasel words, just give the counter-example in decimal form. As usual, you don't understand the difference between potential infinity and actual infinity. The length of an infinite decimal does not approach infinity (which in a literal sense is a meaningless phrase, it has a strict definition which does not involve "infinity" at all), it is infinite. Last edited by v8archie; September 30th, 2016 at 07:47 AM.
September 30th, 2016, 07:58 AM   #3
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Quote:
 Originally Posted by zylo .99..... and 1.00..... to n decimal places are not the same no matter what n is, but their difference approaches zero as n approaches infinity.
The statement above is correct. However, in the decimal number system, it is defined that

$\displaystyle 0.\dot{n} \equiv 0.nnnnnnnn...$

where $\displaystyle n \in \{0,1,2,3,4,5,6,7,8,9\}$ and the ellipsis (...) denotes that an infinite string of digits follows. This recurring operator exists in order to allow the decimal number system to represent adequately all rational numbers.

Therefore

$\displaystyle 0.\dot{9}$

is a valid decimal number and this decimal number is equivalent to 1. Other examples of non-unique representations are $\displaystyle 0.\dot{0} = 0$ and $\displaystyle 1.\dot{0} = 1$. Therefore, the statement that "the decimal representation is unique" is incorrect.

If you wish to craft a decimal number system that is unique, then you can simply discard the recurring operator and force trailing zeroes to be removed, making all strings have a finite length, but then your new system cannot represent all rational numbers.

Last edited by Benit13; September 30th, 2016 at 08:08 AM. Reason: Typos and missing details... need another coffee I think...

 September 30th, 2016, 08:16 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,878 Thanks: 2240 Math Focus: Mainly analysis and algebra Moreover, if you decide that all decimal representations are unique, you then have that not all real numbers have a decimal representation. This then leads to the obvious problem of showing that numbers having no (infinite) decimal representation are the limit of a convergent sequence of rationals. There is a sister problem here, that the convergent sequence of rationals $$\frac{9}{10},\frac{99}{100}, \frac{999}{1000},\ldots$$ Now converges to both $0.\dot9$ and $1$, which makes it divergent.
 September 30th, 2016, 08:19 AM #5 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,082 Thanks: 87 The decimal system to n places is an approximation. It neither gives 1/3 precisely nor pi. But the n-place decimal representation of 1/3 is unique. To 3 places, It is not .332 and it is not .334, it is .333. It becomes increasingly precise as n approaches infinity, while always remaining unique.
 September 30th, 2016, 09:11 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,878 Thanks: 2240 Math Focus: Mainly analysis and algebra So what? If 0.999... is not equal to 1, you are insisting that the limit of the sequence $\{1-\frac1{10^n}\}$ is not equal to 1 and thus that the limit of the sequence $\{\frac1{10^n}\}$ is non-zero. One likely effect of this is that 0.333... no longer approximates $\frac13$. You are asserting the existence of a positive number smaller than every positive decimal. Essentially, you are creating the hyperreal numbers while trying to avoid distinguishing between the infinitesimal hyperreal numbers and the reals. This leads to some real problems, not the least of which is that the definition of the limit must change, but that changes the definition of the reals themselves. It appears that all consistency is lost. Thanks from Joppy
 September 30th, 2016, 09:20 AM #7 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,082 Thanks: 87 The subject is not accuracy of decimal arithmetic (see title). That's a whole other story. 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. 7x.125 = .875, But if .125 is really .1254, then 7x.125 = .878 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).
 September 30th, 2016, 09:32 AM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,878 Thanks: 2240 Math Focus: Mainly analysis and algebra Who mentioned accuracy? Apart from your talk of approximating $\frac13$. The point is that if real numbers exist that are smaller than all positive decimals, then 0.333... no longer approximates because there are now real numbers between the limit of the sequence $\frac3{10},\frac{33}{100},\frac{333}{1000},\ldots$ and $\frac13$ (as far as the concept of a limit, and hence real numbers, make any sense under those conditions).
 September 30th, 2016, 09:43 AM #9 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,082 Thanks: 87 Straw Man
 September 30th, 2016, 09:56 AM #10 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,878 Thanks: 2240 Math Focus: Mainly analysis and algebra No. It's a case of following your premise. The fact that you don't like the conclusion because it contradicts your wishes isn't my fault.

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