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Another significant digit question

  1. Feb 15, 2005 #1

    tony873004

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    I believe the rule for multiplying is you round off your answer to the least number of significant digits contained in your factors.

    5 * 5 = 25
    5 has 1 significant digit, therefore
    5 * 5 = 30

    But, 5 can be used to express any value between 4.5 and 5.5.
    Multiplying the 2 extremes, values just under the midpoint, and the midpoint:


    4.5 * 4.5 = 20.25
    4.9 * 4.9 = 24.01
    5.0 * 5.0 = 25
    5.5 * 5.5 = 30.25

    So in the original problem, rounding 5 * 5 to 30 would fail me if the 5's represented 4.9's, or any combination where they did not multiply to at least 25.
     
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  3. Feb 15, 2005 #2

    dextercioby

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    What do you mean by rounding 5*5 to 30????????5*5 is DEFINED AS 25.There's no rounding there..

    Daniel.
     
  4. Feb 15, 2005 #3

    tony873004

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    5.0 * 5.0 = 25
    In 5 * 5 = 25, the factors only contain 1 significant digit, where the answer contains 2 significant digits. This violates the rule of rounding to significant digits, as I understand the rule.
     
  5. Feb 15, 2005 #4

    dextercioby

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    Apply your so-called "rule" to 16 times 16.

    Daniel.
     
  6. Feb 15, 2005 #5

    tony873004

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    16 * 16 = 260 (rounded to significant digits)

    It's not my rule. The beginning of my physics book contains a chapter on significant figures. And according to them, this is how its done. There was one homework problem where I got an answer of 14.37 m/s, but given in the problem one of the values was 5km, which is only 1 significant digit.

    When I looked in the back of the book, it said the answer was 10 m/s. Either I did the problem wrong, or they rounded to significant digits. The answer is restricted to 1 significant digit since 5km only has 1 significant digit. 14.37 rounds to 10.
     
  7. Feb 15, 2005 #6

    dextercioby

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    It means that your book is totally idiotic.Zero's before the dot are to be considered significative digits.E.g.1000 has 4 significative digits.

    14.37 with 2 sign digits is 14.

    Daniel.
     
  8. Feb 15, 2005 #7

    tony873004

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    But 14.37 with 1 sign digit is 10.

    And the answer is limited to 1 significant digit because 5km is only 1 significant digit.
     
  9. Feb 15, 2005 #8

    dextercioby

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    No.10 has 2 significative digits...

    Daniel.
     
  10. Feb 15, 2005 #9

    dextercioby

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    If you want to,14.56749724 with 1 significative digit is 1.

    Daniel.
     
  11. Feb 15, 2005 #10

    tony873004

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    Not according to my book and the two following websites on significant figures. 10 only has one significant digit. You'd have to put a decimal point after it if you wanted to represent it with 2 significant figures.

    http://www.fordhamprep.com/gcurran/sho/sho/lessons/lesson23.htm
    http://academic.umf.maine.edu/~magri/tools/SigFigsAndRounding.html#SigFigRules

    The 2nd link has a significant figure calculator link at the bottom of the page. Putting 5 x 5 (it likes x instead of * for multiplication), and 25 for your answer makes it say "wrong... it's 30". Same for the 16 * 16 problem.
     
  12. Feb 15, 2005 #11

    Doc Al

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    FYI: Writing a number as "100" is ambiguous as to the number of sig figs intended. Could be 1, 2, or 3. If it's a measured result and sig figs matter (as opposed to a pure number) then you'd better write it explicitly (1.00 x 10^2) or state the number of sig figs. Don't try to be a mind reader. Often, 100 would be considered as having 1 sig fig.
     
  13. Feb 15, 2005 #12

    BobG

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    Tony's right. If you want to be anally technical about it, which the book will be, at least during the significant digit chapter, 5 * 5 = 30, since you can only use one significant digit.

    Of course, this is the only part of the book where they're going to follow that rule to such an extreme. The real goal is to get the student to use a little common sense. Thanks to calculators, if a right triangle that has one leg about 3.2 cm and one leg about 5.6 cm, the student will say the hypotenuse is exactly 6.44980619864 cm long. It would be interesting to see the student try to physically verify that.

    Surprisingly, you even see an embarrassing number of the professional commercial sector use the same flawed logic. You fill a tanker truck to some line on the guage, say 200 gallons. If the liquid is just a little bit below the line, the customer might complain about being short changed, so the liquid probably at least reaches the 200 gallon line. Still, 200 gallons is recorded as the 'official' amount of liquid pumped into the tanker truck. Then, using the chemical composition of the liquid pumped into the tanker, the company's costs are computed to around 12 decimal places, hoping such a detailed cost accounting will make the theoretical numbers match the real world numbers. Of course they won't.

    By the way, we didn't have chapters on significant digits way back when we all used slide rules.
     
  14. Feb 15, 2005 #13

    tony873004

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    My book uses significant figures through the entire book. And this book is 1200 pages long. Luckily, my professor said he won't enforce significant digits on homework or tests. But I'm also taking an Astronomy class, and the professor said she would enforce significant figures through the whole course. So I'm desperately trying to learn them, and they're not that easy.

    I still can't figure out why 5 * 5 = 30 . Because 5 * 5 (one sig digit each) could represent 4.9 * 4.9 which = 24.01 which would have to round down to 20 for 1 significant digit, making my 30 answer useless.
     
  15. Feb 15, 2005 #14

    Doc Al

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    Realize that using significant figures is just an approximate way of handling errors. There are more rigorous approaches that can be used. (Specifying error distributions, for one.)
     
  16. Feb 15, 2005 #15

    BobG

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    Your original numbers were 5 (plus or minus .5). Your final answer is 30 (plus or minus 5). Being right in the middle shows the problem that crops up with enforcing this to such an extreme measure. It also shows why using scientific notation is such a good idea if you want to avoid any ambiguity in what you mean.

    By the way, there's a difference between multiplying 5 amps times 5 ohms to get approximately 25 volts (or 30 volts rounded off to 1 significant digit) and placing five 5 ohm resistors in series. In the second, the 5 is a constant is considered to be accurate to an infinite number of decimal places, so your resistance would be 25 ohms plus or minus 2.5 ohms instead of rounding off to 30.
     
  17. Feb 15, 2005 #16

    cepheid

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    What difference does it make what the measured value of 5 could be in reality? The whole point of this exercise in significant figures is to take into account the fact that your measuring devices have limited precision, and that affects the accuracy of your value (how close it is to the "true" value). (Note, the calibration of the instrument and many other factors could also affect the accuracy of the value, regardless of precision. It is possible to have a very precise but inaccurate measurement. But I digress). So yeah, I think the reason why they stress the sig fig stuff so much is to drum it into students' heads that it is not possible to take an exact measurement in science, and how you evaluate the accuracy of your experimental results is crucially important.

    So, if you were measuring a length in cm on a metric ruler that only had markings to the nearest whole number of centimetres, and you got a value pretty damn close to 5 cm then you would report your answer to two significant figures: One would be certain: the 5. The second would be uncertain (estimated). In general, the precision of your intrument is considered to be to about half the value of the smallest divisions, ie. 0.5 cm. That's just a rule of thumb, and it depends on how well you think you can estimate where in between the two nearest markings your length lies. In stating that precision, you are saying that the value could be anywhere between 4.5 and 5.5 cm, but the precision of your ruler (or lack thereof) does not allow you to determine it any more closely. You would therefore report your measurement as 5.0 cm +/- 0.5 cm. I hope that you can see how the precision of the instrument determines both the certainty of the measurement (ie the number of significant figures), and the uncertainty (the error range). If you were measuring the area of a 5 x 5 square with this ruler, you'd report the area to be 25 cm^2

    I hope also that you can see that if you have a very precise instrument and your measurement is taken to many sig. figs, then (assuming you did the measuring properly and all other things being equal), then the chance of that measurment being more accurate is greater (ie your measurement has greater certainty. More sig figs = more certainty). What you can see from what I've told you is that although for the purposes of the exercise, what you have done is correct, measured values expressed to only one sig. fig would probably be pretty rare in reality. What would be the use of taking such an imprecise measurement? Furthermore, if your ruler were precise only to the nearest centimetre, what would the spacing of the markings be? :confused: (every 2 cm, I *guess*).

    One final point. If the value you are dealing with is an *exact* value, not a measured one (for example if you're really just multiplying by 5 in your calculation), then you are completely certain about that value. You can express this by saying it has infinite sig figs (after all, you could write it as 5.0000000...) if you wanted to. So it doesn't affect the product when you're considering which factor had the least number of sig figs. Obviously you express the product of two exact values exactly. 5 X 5 = 25.

    All of this is just what I got out of it when I was taught sig figs, so if anyone thinks I have some misconceptions here, let me know.
     
    Last edited: Feb 15, 2005
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