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I am looking for an intuitive answer that will explain me why there are only two significant figures in say the number 1500.

Also definition from wikipedia:

The significant figures of a number are those digits that carry meaning contributing to its precision

So are we not considering last two zeros as meaningful ? Why like that ?

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    $\begingroup$ Counting significant figures is not a good way to describe precision. There could be two significant figures in 1500 or there could only be one and you wouldn't know it. The precision of a measurement has to be specified in addition to the actual value. A better but still crude way of doing this are e.g. by adding braces: 150(0), which indicates that the last decimal is uncertain but the second to last is not. Even better would be to give an error interval, e.g. 1500[+3,-7] for a case with an asymmetric uncertainty or 1500[+-4] for the symmetric case. $\endgroup$ – CuriousOne May 20 '15 at 8:49
  • $\begingroup$ You are confusing the number and the measurement. If 1500 is the number of pounds of fertilizer in a pile the precision of the measurement could be 6 digits or not even one. $\endgroup$ – Hot Licks May 20 '15 at 17:53
  • $\begingroup$ More on significant figures. $\endgroup$ – Qmechanic Apr 17 '16 at 14:14
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This depends on the context. If you have 1500 of something and you counted them yourself and you are sure you have precisely 1500, then all four figures are significant. On the contrary, if you're guessing that you have 1500, implying a certainty of order 100, then only the leading two figures are significant. In scientific notation, one would write the former as $1.500\times 10^3$ and the latter as $1.5\times 10^3$. Explicitly stating the trailing zeros indicates that those figures are indeed significant.

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    $\begingroup$ You are still cutting yourself short with this method. If you have made multiple measurements and the main source of error is statistical in nature rather than systematic, then one can recover a higher precision by averaging, but only if we aren't introducing rounding errors in addition to the actual measurement error. Rounding a value to its individual precision introduces a new and completely unnecessary error source which has both a statistical error and, in unfortunate cases, also a systematic bias. $\endgroup$ – CuriousOne May 20 '15 at 8:59
  • $\begingroup$ Sure, I wasn't implying at all that one should use this degree of rounding internally. You are right, those errors are unnecessary. $\endgroup$ – Jonas May 21 '15 at 9:16
  • $\begingroup$ Also, depending on the discipline, I totally support your comment that one should give a proper error bar on measurements. $\endgroup$ – Jonas May 21 '15 at 9:18
  • $\begingroup$ It's a tricky business. I like to err on the side of giving too many significant digits and a proper error estimate. That, at least, is honest in the sense that people looking at my data can see what I saw. In the past that was frowned upon, but in my opinion for no good reasons. The game, of course, has changed. Today one can buy true 6-7 digit voltmeters with short term stable 8 digit readout, which makes certain measurements rather high resolution and independent of e.g. human readout bias. Frequency measurements can be even higher resolution, but the statistics is even more tricky. $\endgroup$ – CuriousOne May 21 '15 at 17:03
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I teach high school in the United States. I want to preface with that, because conventions common in one context are not necessarily universal.

That being said, it's pretty standard teaching practice here (at least in every class I've ever taken, taught, or known a colleague to teach) to assume that trailing zeroes are not significant unless otherwise noted. So, in the example of $1500$, I automatically assume that measurement to have two significant figures unless noted otherwise. There are multiple ways that you could note a different amount of significant digits:

  1. Use Scientific Notation- This is probably the best way, in general, to indicate the number of digits that are significant, because any digit in your multiplier prior to the power of ten is significant. For example $1.5 \times 10^3$ has two significant digits, because $1.5$ has two significant digits. If I wanted to indicate that all four digits are significant, I would instead write $1.500 \times 10^3$ where intentionally writing the extra zeros after the decimal shows that they are significant. I could easily repeat the process for three significant digits, using only one trailing zero after the decimal. However, depending on the order of magnitude of your measurement, it may feel silly or cumbersome to use scientific notation so you could...

  2. Use other markers to indicate significance- At least around here, it's pretty common to end your number with a decimal point to indicate significance on all digits. For example, if I wanted to explicitly indicate significance on all four digits in your example, I could write it as $1500.$ with a decimal point after the final zero. Because the decimal is not needed for any other reason, it's only there to indicate significance of the zero's to the left. Well what if the first zero is significant but the second zero is not, for a total of three significant digits? Then I would indicate the last significant zero with a bar either above or below the zero. So either $15\bar00$ or $15\underline00$ could represent three significant digits. Like I said above $1500$ with no other marks augmenting it is typically assumed to only have two significant figures.

Now, I fully recognize that at the research level, there are far better ways to indicate precision and relative uncertainty. However, this strikes me as a high-school level question, so there's your high-school level answer.

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    $\begingroup$ Terminating with a decimal point to indicate significance doesn't work if the number is at the end of a sentence. $\endgroup$ – David Richerby May 20 '15 at 16:12
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    $\begingroup$ I think the best part of your answer is contained in the last two sentences. I admit that I don't know of a way to teach the subtleties of measurement errors properly at the high school level, but it would be great, if we could. $\endgroup$ – CuriousOne May 21 '15 at 1:09
  • $\begingroup$ @DavidRicherby those situations would be fairly rare. Normally in physics your measurements have units, so if for example you were measuring distance, you would say $1500. m$ or if you were measuring force you would say $1500. N$. $\endgroup$ – Sean May 21 '15 at 15:42
  • $\begingroup$ I think this answer is spot on. And good to see a high school teacher on this site! I am sure there are (many) others and I just haven't noticed...? $\endgroup$ – Floris May 21 '15 at 16:07
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In physics, all numbers are imprecise. Significant figures are conceptually the digits that it is meaningful to include in the reported result.

The concept

When you read a gauge, the reading error is half the smallest interval on its scale - you basically take the value of the tick nearest to the hand's position (digital gauges do this for you).

So, if the reading is 1503 (the scale step is 1), it basically means that it's actually 1503±0.5.

  • 150 are precise
  • 3 is precise, too (in the true value, it can be 3 or 2 but when rounded to 4 digits, it will always be 3)

So, it's meaningful to only include the 4 digits into the result. Including further ranks would be meaningless since they can be anything in the true value - they wouldn't actually carry any additional information.

On the other hand, if a result (of a calculation, probably) is something like 1534.35056±3.50584 (the relative error is 2.3%)

  • the leading 153 are precise
  • 4 is somewhat precise (in the true value, it can be anything from 0 to 7)
  • the following figures are spurious (can be anything in the true value)

Since 4 is rather imprecise, it's reasonable to include the next digit as well: 1534±4 gives 2.6% error (worse than it actually is) while 1534.4±3.5 gives the correct 2.3%.

Conclusion

So,

  • the concept (and thus the number) of significant figures only has meaning if the precision is given or implied.
  • when the precision is implied

    • it's typically implied the error is ~0.5 of the last significant rank (i.e. all the digits given are implied to be strictly precise). When it's not so, the error must be specified explicitly. This implication mostly stands for numbers written in scientific notation (n.nnnE[-]n). (let's call it the "scientific implication")
    • For integer numbers written in plain (especially large round ones and especially in "inaccurate" domains like statistics), it's as usually implied that only the leading non-zero digits are "precise" while the zeros are just the padding for readability (in these domains, all numbers are often padded to thousands/millions/etc). (let's call it the "layman implication")

Application

In addition to the general cases given above, there are a few rules made out of pure convenience:

  • leading zeros are never needed: 0.000045 is much more readable as 4.5e-5.
  • trailing zeros are not needed unless precision is implied and they're the "precise" digits: 12.40 implies precision 0.01 (error 0.005) while 12.4 implies precision 0.1 (error 0.05). In 1500000, if only two zeros are actually precise, it should be written as 1.500e6. It's because of these implications that it's so important to watch the trailing zeros - to avoid giving a false implication.
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    $\begingroup$ Nitpick: In a digital instrument where there is no doubt the displayed digits are correct, the discretization error is $1/\sqrt{12}$ times the scale step, not $1/2$, if you're going for the standard deviation. $\endgroup$ – user10851 May 21 '15 at 5:09
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You are completely right, it is a confusing case you have. The number $1500$ does have 4 significant figures as it is.

But, you are told it has only 2. This is strictly speaking not correct. But it is just a shorthand way of writing $1.5 \times 10^{3}$ (or $15 \times 10^{2}$). It is an easier way to write a number with not so high an order of magnetude.

When you see $1501$ you know it is 4 significant figures. But when you see something like $1500$ you actually don't know if it is just this "shorthand" notation, as I will call it. You must know the context to be sure. If you do not know, you can only assume 4 significant figures.

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There's a lot of context that goes into deciding which digits are significant. I like to use the "anger management method" for deciding which digits are significant.

Suppose you are shopping for a fancy television. You see an advertisement that there's one you like on sale for \$1369.99. You know that your area has 10% sales tax, so you predict that when you buy the television your total will be about \$1500. You buy the television, and your actual total is \$1506.98. Are you mad about that extra seven dollars? Probably not — in which case the trailing zeros in your estimate were not significant.

The next week, the water main in your front yard fails and your yard fills with water. Your plumber digs a trench in your yard and replaces the pipe, and you write him a check for \$1500. When your bank statement arrives you notice that the bank actually debited \$1506.98 when it honored the check. Are you mad about that extra seven dollars? Goddamn right you're mad! You'll probably double-check the copy of the check to make sure you wrote it correctly and it wasn't modified on its way to the bank, and you probably won't use that plumber again. In that case the trailing zeros were significant.

When in doubt in the physical sciences, you write the uncertainty explicitly: $1500\pm100$ or $1500\pm10$ or $1500\pm1$ or $1500.0\pm0.1$.

Supposedly when the height of Mt. Everest was first measured the altitude came to $29\,000\pm1\rm\,ft$. The surveyors thought that if they reported the height as $29\,000\rm\,ft$ then people would assume their measurement was sloppy, so the last digit was fudged and the altitude was reported as $29\,002\rm\,ft$.

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