SOLUTION: If an infinite decimal such as .999... is to represent one specific number, it should be the limit of the sequence described by part (a), in this case 1; that is, .999...= 1. A nu

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Question 37339: If an infinite decimal such as .999... is to represent one specific number, it should be the limit of the sequence described by part (a), in this case 1; that is, .999...= 1. A number is a *limit* if the difference between it and the terms of the approximating sequence is eventually less than any distance you choose, no matter how small. What number does .4999...represent? What about 7.562999...? Why?
I have some of the answers to this problem.
.4999=0/1=0.0
1/2=0.5
2500/5001=0.49990001999
7.562999=8/1=8.0
15/2=7.5
53/7=7.57142857142857
I'm not sure this is correct nor do I understand WHY? As the questions asked. Thanks in advance for any help I can get.
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
If an infinite decimal such as .999... is to represent one specific number,
it should be the limit of the sequence described by part (a), in this case 1;
that is, .999...= 1. A number is a *limit* if the difference between it and the
terms of the approximating sequence is eventually less than any distance you
choose, no matter how small. What number does .4999...represent? 

.4999··· = .4 + .0999··· = .4 + .1(.999···) = .4 + .1(1) = .4 + .1 = .5 

What about 7.562999...? Why? 

7.562999··· = 7.562 + .000999··· = 7.562 + .001(.999···) = 7.562 + .001(1) =
7.562 + .001 = 7.563

Edwin McCravy
AnlytcPhil@aol.com
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