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What is the correct symbol for saying "equals, but is rounded"? For example:

$$\frac{1}{17}=0.0588$$

Is not essentially correct, since it's rounded. What other equal-like symbol should I use?

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    $\begingroup$ You should use $\approx$ ($\text{\approx}$) which means "approximately." $\endgroup$ – wgrenard Jan 11 '18 at 17:42
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    $\begingroup$ I think normally authors will add a "$\ldots$" to the quantity on the right to indicate that it is exactly equal to the left side, but not exactly equal to $0.0588$, instead of weakening the sense of equality. $\endgroup$ – Erick Wong Jan 11 '18 at 17:44
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    $\begingroup$ @wgrenard : In my classes, anyone who uses "$\approx$" without specifying which approximation scheme they are using, what parameters have been used, and estimates of precision and accuracy loses points. $\endgroup$ – Eric Towers Jan 11 '18 at 17:45
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    $\begingroup$ I personally would find $\approx$ to be best. But it is perfectly okay (especially in more applied than theoretical) to simply state early on "all decimal values will be expressed within 4 significant figures". $\endgroup$ – fleablood Jan 11 '18 at 17:51
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    $\begingroup$ I personally don't like "..." for rationals as "..." indicates "and so on" which indicates either there is a pattern to continue or there is no pattern to continue. As rationals do have patterns I dislike putting "...." anywhere that isn't at a cycle break. $\endgroup$ – fleablood Jan 11 '18 at 17:55
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", which rounds to". For example "... yielding $\frac{1}{17}$, which rounds to 0.0588." This follows the old rule: "say what you mean and mean what you say".


A parable

Years ago someone gave me a fancy clock that shows the time of day as words. However, some time last week, it stopped and only shows "noon" now. There's an old adage "a stopped clock is right twice a day" (assuming a 12 hour clock). This clock is clearly correct once per day. What fraction of the day is it approximately correct? That is, for how many seconds of the day is the time "$\approx$ noon"?

Unless and until you specify what you mean by "$\approx$", the question is meaningless. For instance, with what precision is the indicated time, "noon"? Does the clock only indicate hours, also minutes, also seconds, all the way down to ... what? nanoseconds? Does it generate an additional 100 random digits, claiming absurd precision (necessarily without any promise of accuracy)? Then, when you say "$\approx$", how accurate must that value of unknown precision be? Within $\pm \frac{1}{2}$ day? $\pm \frac{1}{2}$ hour? $\pm \frac{1}{2}$ minute? $\pm \frac{1}{2}$ second? $\pm \frac{1}{2}$ nanosecond? $\pm \frac{1}{2} \times 10^{-100}$ nanosecond? What do you do with the endpoints of these accuracy intervals; are they $\approx$ or not? Do you include both endpoints, one (which one?), neither? What if you know the clock was always set an hour fast? Do you account for that systematic error in $\approx$ or not?

An approximation scheme has to specify all of these things. The function of mathematical notation is precision of expression. "$\approx$" inadequately specifies, so cannot be precise. For instance is "$100 \approx 0$" true or false? (Are we rounding to nearest millions? Where is that in the notation?)


One might claim that the same applies to my suggested phrase, "which rounds to". There are dozens of standard rounding techniques. If we restrict to computation, IEEE 754 specifies five rounding techniques. Conveniently, each of these techniques resolves all of the questions asked above. Also conveniently, the standard rounding technique is round to nearest increment with ties rounding up. If you mean some other kind of rounding, you have to say so. If, as I have suggests, you write out what you are doing to transform your exact quantity into your reported result, you have an easy place to say that you mean some other kind of rounding.

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  • $\begingroup$ You mentioned in a comment above that using the symbol $\approx$ concerns you because it leaves out information such as the level of precision and accuracy. But being in favor of using $1/17 = 0.0588\ldots $ as an alternative trades one ambiguity for another. For instance, what does $\ldots$ mean? Does it mean that the $8$s continue forever? Does it mean that $588$ repeats forever? The notation assumes that the reader understands that $\ldots$ means to continue the pattern by using long division. $\endgroup$ – wgrenard Jan 11 '18 at 17:59
  • $\begingroup$ @wgrenard : You are correct, the second paragraph does not address OP's question. (Fixed.) And to address your question: The notation says that the object on the left and the object on the right are equal; this leaves no ambiguity in the elided digits. $\endgroup$ – Eric Towers Jan 11 '18 at 18:03
  • $\begingroup$ @wgrenard : To think that the notation $\dfrac 1 {17} = 0.0588\ldots$ means something repeats forever is unreasonable. But I wouldn't write $\text{“ } = 0.0588\ldots\text{ ''} \vphantom{dfrac{\displaystyle\int}{}}$ unless I meant that that last explicit digit is actually $8$ and not rounded upward to $8,$ whereas $\text{“ } \approx 0.0588\text{ ''}$ could reasonably mean that it was rounded upward to $8. \qquad$ $\endgroup$ – Michael Hardy Jan 11 '18 at 18:08
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    $\begingroup$ Myself, I informally translate $\approx$ as "The author was content with the precision and accuracy." Thus the usefulness of such a symbol depends greatly on how the result is used, and whether those uses depend on the accuracy or not. I also find that answers your last question: if we are working on astronomical scales $100 meters \approx 0 meters$ $\endgroup$ – Cort Ammon Jan 12 '18 at 1:05
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    $\begingroup$ @EricTowers It might be dependent on your field. I'm an engineer, and in my field, I'd say easily 99% of $\approx$ convey their meaning in a sufficiently unambiguous manner. Other fields may find different results. $\endgroup$ – Cort Ammon Jan 12 '18 at 1:34
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Answer from comments:

  • Usually one writes $\dfrac 1 {17} \approx 0.0588.$

  • One can write $\dfrac 1 {17} = 0.0588\ldots,$ meaning there are further digits after the $8.$
    I would use the notation $\text{“ } =0.0588\ldots \text{ ''}$ only if that last explicit digit is $8$ and not if it's rounded upward to $8,$ whereas I would use $\text{“ } \approx 0.0588 \text{ ''}$ if it's rounded either upward or downward. (But in this case this is not an issue since $1/17 = 0.0588235\ldots$)

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  • $\begingroup$ I observe that your second (i.e., true) bullet is nonresponsive to OP's question. (I had a similar observation in my answer.) $\endgroup$ – Eric Towers Jan 11 '18 at 18:16
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    $\begingroup$ @EricTowers : Do you mean to suggest that what I said in the first bullet point is not true? $\endgroup$ – Michael Hardy Jan 11 '18 at 18:19
  • $\begingroup$ Up to the missing method, parameters, accuracy, and precision, what you said in your first bullet point may or may not be true. $\endgroup$ – Eric Towers Jan 11 '18 at 18:21
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    $\begingroup$ @EricTowers : . . . and yet my first bullet point correctly answers the question that was actually asked. Even if the symbol ought never to be used because its use in every instance is erroneous. $\endgroup$ – Michael Hardy Jan 11 '18 at 19:14
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    $\begingroup$ To me, this seems to be the correct answer. Whenever the precise way of rounding matters, it needs to be mentioned in the text, but AFAICT in this case, the OP wants to convey the numerical value of a symbolic expression, and I would just round it to a reasonable number of digits. $\endgroup$ – Toffomat Jan 12 '18 at 9:31
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I use $=$ for equaility, $\approx$ for approximately equal to, and $\doteq$ for correctly rounded to. So $$\frac{1}{17} = 0.058823529\ldots \qquad \text{and} \qquad \frac{1}{17} \approx 0.06,$$ while \begin{align*} \frac{1}{17} &\doteq 0.1 \quad (\text{1 decimal place appears so result is correctly rounded to 1 decimal place})\\ \frac{1}{17} &\doteq 0.06 \quad (\text{2 decimal places appears so result is correctly rounded to 2 decimal places})\\ \frac{1}{17} &\doteq 0.059 \quad (\text{3 decimal places appears so result is correctly rounded to 3 decimal places})\\ \frac{1}{17} &\doteq 0.0588 \quad (\text{4 decimal places appears so result is correctly rounded to 4 decimal places})\\ \frac{1}{17} &\doteq 0.05882 \,\, (\text{5 decimal places appears so result is correctly rounded to 5 decimal places})\\ \end{align*} and so on.

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    $\begingroup$ The only problem is while the first two are wide-spread usages, $\doteq $ is not nearly so well-known, so unless you are using it for an audience that you know is already familiar, you will need to explain it. $\endgroup$ – Paul Sinclair Jan 11 '18 at 21:24
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    $\begingroup$ This is true and is something I always carefully explain before first using it. $\endgroup$ – omegadot Jan 11 '18 at 21:35
  • $\begingroup$ I assumed so in your case, but I wanted to make sure others (such as the OP) were aware of this. $\endgroup$ – Paul Sinclair Jan 11 '18 at 23:56
  • $\begingroup$ What does = ... mean? $\endgroup$ – Christoffer Hammarström Jan 12 '18 at 12:02
  • $\begingroup$ The $\ldots$ means there are additional digits to come and if we were to write them all down we would have equality. $\endgroup$ – omegadot Jan 12 '18 at 20:21

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