
#19
Oct2011, 01:38 PM

P: 799

What you are calling values, the rest of us call numbers. If your objections go beyond these syntactic substitutions, please advise. Otherwise, use standard terminology to avoid confusion. 



#20
Oct2011, 01:58 PM

P: 623





#21
Oct2011, 02:06 PM

P: 15,325

If this were commerce forum where discussions of PINs were common and there were some ambiguity in terminology, yes, you'd be asked to use that industry's terminology. 



#22
Oct2011, 02:39 PM

P: 623

My copy of Chambers Dictionary of Science and Technology says;
Can we clarify what/whose definitions we're using here, then? 



#23
Oct2011, 03:12 PM

P: 12

In my mind comparing 0.9999... and 1 is apples to oranges: one is real, one isn't. The sum mention above approaches 1 as the number of terms approaches infinity, yes. But you won't EVER have that infinite term, so they won't ever be equal as long as both numbers are real. Once you hit that infinite term, 0.99... becomes nonreal. What solidified my issues was this proof 1/3 = 0.3333... 3(1/3) = 0.99999.... 3/3 = 1 



#24
Oct2011, 03:16 PM

Mentor
P: 16,548

0.9999... is a real number and it is equal to 1. 0.9999... is as real as 1.00000.... There is no difference between recurring 9's and recurring 0's. 



#25
Oct2011, 03:30 PM

P: 12

0 is not a member of the set of values in the sequence 9/10^n for n[1,Inf). This sequence is always changing; the derivative is also nonzero. Therefore its sum will never settle at any number, let alone 1. The sum will approach 1, get closer and closer, but never get there, unless you are at Infinity, which doesn't exist. I don't understand the break down in this logic. 



#26
Oct2011, 03:35 PM

Mentor
P: 16,548

[tex]0.9,~0.99,~0.999,...[/tex] will never settle. It will always come closer and closer to 1. The point is that the very definition of 0.9999.... is a limit, namely the limit of the series [tex]\sum_{n=1}^{\infty}{\frac{9}{10^n}}[/tex] this is the definition of 0.9999.... The above series will approach 1, and the limit of the above series will equal one. Thus 0.9999..., being defined as a limit, will equal 1. 



#27
Oct2011, 03:41 PM

P: 12

I get it. 0.9999... is just short hand for a limit. It isn't a real number in the traditional sense.




#28
Oct2011, 03:48 PM

Mentor
P: 16,548

We call a real number L the limit of [itex](x_n)_n[/itex] if [tex]\forall \varepsilon>0:\exists n_0:\forall n>n_0:x_nL<\varepsilon[/tex] So you see that only real numbers can be limits, by definition. 0.9999... is simply notation for the real number defined as the limit of the series [tex]\sum_{n=1}^{\infty}{\frac{9}{10^n}}[/tex] This series has a limit which is a real number. We denote this real number by 0.9999... But 1 is also a limit, thus 1=0.99999... 



#29
Oct2011, 04:05 PM

Mentor
P: 20,967

For example, we can write the number 4 in a variety of ways: 4, IV (in Roman numerals), 100 (in binary). These are different numerals that all represent the same number. In the same way, 0.999... and 1 are different representations of the same number. I can't believe we're having this same discussion again! 



#30
Oct2011, 04:11 PM

P: 12





#31
Oct2011, 04:17 PM

Mentor
P: 16,548

Limits are real numbers. 



#32
Oct2011, 04:20 PM

P: 12





#33
Oct2011, 05:06 PM

P: 66

There is also the infinitesimal approach.
There is a hyperreal number [itex]\epsilon[/itex] that is smaller than the smallest real number, so we can define the following: [itex] 1  \epsilon = .999...[/itex]. This implies that [itex]1[/itex] and [itex]1  \epsilon[/itex] (.999...) are different numbers in the hyperreal numbering system. 



#34
Oct2011, 05:11 PM

Mentor
P: 16,548





#35
Oct2011, 06:12 PM

Mentor
P: 14,434

This topic has come up so often here that we have an FAQ that addresses this concept: http://www.physicsforums.com/showthread.php?t=507001.
Thread closed. 


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