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Decimal approximations

  1. Jul 21, 2008 #1
    I was reading about decimal approximations in one of my math books and I like his explanation of why 5 and over we round up and so on (as it's closer to rounding down). I even understood the explanation of how to calculate what the error could be given a series of decimal numbers that have been approximated.

    However, this paragraph confused me:

    "...Thus, by the method of article 34, 23/24 = .95833+. Expressed to four decimal places the real value of this fraction lies between .9583 and .9584; .9583 is .00003+ less than the true value, and .9584 is .00006+ greater. Therefore, .9583 is nearer the correct value and is said to be correct to four decimal places. Similarly, .958 is correct to three places and .96 to two."

    I understand .95833 - .00003 = .9583. But shouldn't .9584 be .00007+ greater than .95833?
     
  2. jcsd
  3. Jul 22, 2008 #2

    Integral

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    I am not sure I understand his use of the + in this notation. Let me guess.

    .9584 - .958333... = .0000666...

    I think .00006+ discribes this number better then .00007+ clearly it is LESS then .00007 and greater then .00006.
     
  4. Jul 22, 2008 #3

    HallsofIvy

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    If you had 0.95833 exactly then 0.9584 would be exactly 0.00007 greater. But 23/24= 0.95833333... where the "3" keeps repeating. The "+" in 0.00006+" means "0.00006 plus more terms after that (in this case the "3333..."). The difference is a little less than 0.00007, again because of that continuing "3333...".
     
  5. Jul 22, 2008 #4
    Thanks guys. There are usually few errors in his books so I didn't think it would be an error. The explanation above makes sense, the plus sign indicates that more figures are to be added, so .9583 is .00003+ less than the true value, which is the closest approximation at that level, and likewise for the other number, the .00006+, which is closer than .00007 because of the addition of the plus sign.
     
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