I posted this question earlier today on the Mathematics site (https://math.stackexchange.com/q/3988907/96384), but was advised it would be better here.

I had a heated argument with someone online who claimed to be a school mathematics teacher of many years standing. The question which spurred this discussion was something along the lines of:

"A horseman was travelling from (location A) along a path through a forest to (location B) during the American War of Independence. The journey was of 22 miles. How far was it in kilometres?"

To my mind, the answer is trivially obtained by multiplying 22 by 1.6 to get 35.2 km, which can be rounded appropriately to 35 km.

I was roundly scolded by this ancient mathematics teacher for a) not using the official conversion factor of 1.60934 km per mile and b) not reporting the correct value as 35.405598 km.

Now I have serious difficulties with this analysis. My argument is: this is a man riding on horseback through a forest in a pre-industrial age. It would be impractical and impossible to measure such a distance to any greater precision than (at best) to the nearest 20 metres or so, even in this day and age. Yet the answer demanded was accurate to the nearest millimetre.

But when I argued this, I was told that it was not my business to round the numbers. I was to perform the conversion task given the numbers I was quoted, and report the result for the person asking the question to decide how accurately the numbers are to be interpreted.

Is that the way of things in school? As a trained engineer, my attitude is that it is part of the purview of anybody studying mathematics to be able to estimate and report appropriate limits of accuracy, otherwise you get laughably ridiculous results like this one.

I confess I have never had a good relationship with teachers, apart from my A-level physics teacher whom I adored, so I expect I will be given a hard time over my inability to understand the basics of what I have failed to learn during the course of the above.

  • 17
    $\begingroup$ This isn't substantial enough to leave as an answer, but basically you're correct for the exact reasons you stated. And although nobody likes to be scolded by a teacher, especially when the teacher is wrong, that rarely (never?) happens here as long as questions are in good faith like yours is. I'm sorry about your previous encounters with teachers; maybe we can make up for them a bit here. $\endgroup$ – Thierry Jan 17 at 18:25
  • 14
    $\begingroup$ I've seen numerous instances of this teacher's error, converting to a new unit of measure and producing an absurd level of accuracy. I was taught at a very young age that normal human body temperature is 98.6 degrees Fahrenheit. More recently, 100.4 degrees is the official border for suspicion of covid19. These temperatures happen to result from converting Celsius temperatures of 37 and 38 degrees, respectively. (See also my answer at matheducators.stackexchange.com/questions/1572 for a similar example.) $\endgroup$ – Andreas Blass Jan 17 at 18:53
  • 8
    $\begingroup$ ancient mathematics teacher --- Apparently not too ancient, as this would have been ridiculously silly until calculators arrived (in my school this was 1975, when two or three students had one). I looked at several books I have from within 20 years of this, and most didn't even have English-metric conversions. Two that did, Dolciani's Modern Algebra. Structure and Method. Book 1 (1973 edition) and Lankford/Clark's Basic Ideas of Mathematics (1953), only gave the approximations 1 km = 0.6 mile (Dolciani, p. 577) and 1 km = 5/8 mile (L/K, p. 497) $\endgroup$ – Dave L Renfro Jan 17 at 21:05
  • 8
    $\begingroup$ Related: Brian Kernighan (of K&R C book) gave a guest lecture to Harvard's CS 50 course a decade back that was almost entirely making fun of innumeracy errors, including this over-precision class of misunderstanding: youtu.be/kw9KwjJCJH8?t=2170 $\endgroup$ – Daniel R. Collins Jan 18 at 2:49
  • 10
    $\begingroup$ Because of the coastline paradox, it's not really clear that that level of precision is even meaningful. en.wikipedia.org/wiki/Coastline_paradox $\endgroup$ – Ryan_L Jan 18 at 3:26

11 Answers 11


You're right. The random, anonymous person you met online is not competent. This is basic mathematical literacy, as taught in every freshman chemistry and physics class.

  • 23
    $\begingroup$ Your answer may have shown the source of the problem. I was taught the principle "don't produce significant digits out of thin air" in a physics class, but never in a mathematics class. If that experience is widespread, people may have picked up the idea that this principle is limited to applications classes, whereas in mathematics classes we should "be precise" even if the precision is garbage. $\endgroup$ – Andreas Blass Jan 18 at 16:31
  • 7
    $\begingroup$ +1 for the abruptness of the answer $\endgroup$ – David Z Jan 18 at 19:34
  • 1
    $\begingroup$ @AndreasBlass's point could extend to thinking about spurious precision in the input $\endgroup$ – Chris H Jan 19 at 15:52
  • 3
    $\begingroup$ Outside of SE and some other isolated pockets, I'd suggest the first sentence of this answer is near universal. $\endgroup$ – Lamar Latrell Jan 20 at 4:51
  • 1
    $\begingroup$ @AndreasBlass: Yes, I'm sure you're right. I think the underlying problem is that to get this kind of thing right, you need strong number sense, which by far the majority of K-6 teachers lack. The following is a question that the majority even of my college-level science majors can't do without extensive help: If a real number is rounded to the nearest integer, what is the maximum rounding error? You'll also see this lack of number sense in the desire to teach and learn heuristics as if they were absolute rules, or the belief that such things are class rules set by each teacher. $\endgroup$ – Ben Crowell Jan 20 at 14:43

The product of two numbers should be given with as many significant digits as the least precise of the numbers multiplied (see https://www.nku.edu/~intsci/sci110/worksheets/rules_for_significant_figures.html). 1.60934 km/mile has six significant digits (or, if a mile is defined to be an exact number of km, then the conversion factor has an infinite number of significant digits). 22 miles has two significant figures. We take the smaller of these two, which is two significant figures from the 22 miles. This means that rounding to 35 km is correct.

It is a good idea to use, during one's work, at least one significant digit more than the final quantity needed, so it would have been good practice to use the conversion factor of 1.61 if this were a test, but for a casual online conversation, 1.6 is fine.

The importance of getting significant figures correctly pales in comparison of basic decency. Even if this person had been correct, scolding you would not be. If you believe that someone is in error, you should express that view politely. It appears that this person's civility may have atrophied from having a captive audience with such a power differential that they have been able to dispense with basic politeness.

  • 12
    $\begingroup$ What dismays me most is the thought of those generations of students whose experience of mathematics will have been irretrievably compromised. "Follow these rules! Don't argue! You are not allowed the privilege of even expecting it to make any sense!" $\endgroup$ – Prime Mover Jan 18 at 7:19
  • 2
    $\begingroup$ The one struggle I still have with significant digits - convert 20 miles. Use 1.6, and get 32. Will you choose to keep significant digits to 1, or treat the zero of 20 as significant? (I realize, scientific notation takes care of this, indicating whether the zero was significant or not.) $\endgroup$ – JTP - Apologise to Monica Jan 18 at 12:59
  • 7
    $\begingroup$ @JTP-ApologisetoMonica This is a problem of all numbers whose SF are fewer than the number of figures on the left of the DP. Are the zeroes actually significant figures or are they merely placeholders? In such a situation it is up to the person communicating the number to specify to how many SF "20" is reported to -- either 1 or 2. This issue crops up over and over again. $\endgroup$ – Prime Mover Jan 18 at 13:26
  • 1
    $\begingroup$ @JTP-ApologisetoMonica You would have to take the context into consideration, e.g. if there are other, similar, measurements to 2 S.F. in the same document, use that. Or you could write "about 32 km" to indicate that the amount is not necessarily as accurate as written. $\endgroup$ – Andrew Morton Jan 18 at 13:26
  • 2
    $\begingroup$ If they wanted an answer of 35.405598 km they would have had to specify the rider rode 22.000000 miles. $\endgroup$ – Schwern Jan 20 at 23:10

Here's a joke I like to tell when people could use a reminder about precision vs accuracy:

A tour guide at Giza was explaining how the Pyramids were 4507 years old. Someone in the crowd asked: "That's oddly specific. How do we know this?"

"Well. I was told they were 4500 years old when I started working here 7 years ago."

I'm not sure the grumpy teacher you mentioned would be amused, though.

  • 3
    $\begingroup$ Heh! I had that in mind when I was pondering on this while driving around earlier: "This hominid skull is one hundred thousand and seventeen years old. And five months and two days." $\endgroup$ – Prime Mover Jan 18 at 23:31
  • 3
    $\begingroup$ What a great joke! Good one! $\endgroup$ – Fattie Jan 19 at 18:07

Just to play the devil's teacher's advocate here: one can make a point that rounding should be generally avoided but measurement uncertainty instead be expressed explicitly. Specifically, rounding errors should always be much smaller than measurement errors. Now, if you have a figure of 22 miles, I'd interpret this as $(22\pm0.5)\mathrm{mi} = (35.4\pm0.8)\mathrm{km}$. I specified one more digit, but not only did I represent the center value better (which in your rounding adds a whopping 50% error), I also captured that the inaccuracy of that result is even bigger than simply $35\:\mathrm{km}$ would suggest. In particular, $36\:\mathrm{km}$ is also within the range!

How many digits to write out is then uncritical; in physics convention is to write two non-significant digits in both the value and uncertainty figure. One is usually enough, but when completely omitting non-significant digits you do introduce excessive extra error. If the numbers are just stored in a computer, you should typically keep all the digits of the binary number representation – with double precision that means you keep a rather absurd 16 decimals! It doesn't really increase the precision, but it also doesn't really cost anything or suggest too high precision (because uncertainty is stored separately), and it makes sure that rounding really will have no contribution to the error of the final result.

  • 3
    $\begingroup$ Where does this convention of two non-significant digits come from? I'm not sure I've ever heard of it, so I would certainly disagree with saying that it's the convention in physics. $\endgroup$ – David Z Jan 18 at 19:39
  • 2
    $\begingroup$ @DavidZ it's not a universal convention, but it does seem to be used by most big-scale experimental physics projects nowadays. It is also the form in which NIST lists physical constants, e.g. the vacuum impedance is $376.730 313 668(57)\: \Omega$. $\endgroup$ – leftaroundabout Jan 18 at 20:19
  • 2
    $\begingroup$ Comments are not for extended discussion; I've moved the conversation that was attached to this answer to chat. $\endgroup$ – Chris Cunningham Jan 19 at 20:16
  • 1
    $\begingroup$ Or nitpick if "International Mile" or "US Statue Miles" was used? those have different conversion factors and for civil war era the quoted 1.60934 would be wrong if survey miles are used, those would be rounded to 1.60935. So the whole thing is kind of pointless if you have no agreement on measurment precision and time. $\endgroup$ – schlenk Jan 20 at 0:07
  • 2
    $\begingroup$ @EricDuminil Actually there is somewhat infamous contract during colonization of Africa, where german negotiators wrote miles in a contract and the local chieftain only knew british miles, but the negotiators insisted they meant german miles (7532,5 meters). Quite a significant difference and led to the Hereo and Nama genocide later. So mind your miles or people might die. $\endgroup$ – schlenk Jan 20 at 22:17

When a tutoring student asks me about rounding, I tell them that absent specific instructions from a teacher, common sense should apply.

For a conversion, 22 miles isn’t 22.0000 miles, there’s the assumption it’s been rounded. You can’t convert and find yourself with 6 digits of accuracy beyond the decimal. As you note, there’s a number of digits that result to be the nearest meter, millimeter, etc. which is absurd. Before GPS, I’d give directions accurate to 1/10 mile, as that’s what a car odometer reflects. Even that was often called a bit obsessive.

My home scale gives me my weight to .1 lbs. Would it really be of value to have an extra digit of accuracy?

A person’s height? The nearest inch will do.

The one thing I warn about - don’t round while doing interim steps. This is a sure way to find that the final result may be off by enough to be graded as wrong. This issue commonly presents itself with trig functions which ask for a triangle side to the nearest 1/100. Rounding should be done as the final step.

  • $\begingroup$ There is more than an assumption that the figure has been rounded. There is a context into which the question has been placed. I think I understand what the main problem here is now: people seem to assume that the context is there merely to provide a pretty little story to keep the unmotivated students on board. No, the context is there to provide a scenario to be analysed. $\endgroup$ – Prime Mover Jan 20 at 9:25
  • 1
    $\begingroup$ Didn't my examples offer that context? We can go off on multiple tangents here, from the observation that not all math problems offer the context required, to the fact that ultimately, perhaps unfortunately, I often ask a (tutoring) student "Do you want to be right, or do you want the credit for your answer?" You are not going to change that person. But you do have 10 answers here that are in your favor. I don't see one that sides with that teacher. $\endgroup$ – JTP - Apologise to Monica Jan 20 at 11:23
  • $\begingroup$ I was not arguing against you, but I was specifically elaborating your statement "there's the assumption it's been rounded." Oh, and I'm not sure where I saw it now, but there are some answers here which appear to suggest that what you do in a maths class (i.e. calculate the numbers exactly) is different from what you do in a physics etc. class, because "maths is pure" or some such. And someone did say that the context in a word problem is just there to make it interesting and engage the students. Utter piffle, of course. $\endgroup$ – Prime Mover Jan 20 at 12:06
  • $\begingroup$ I agree, 100%. Good talk. $\endgroup$ – JTP - Apologise to Monica Jan 20 at 12:12

When I was in school, I once got an answer marked as error for having too many digits. IIRC it was in trigonometry and I had just written down as many digits as the calculator displayed. (I was able to discuss it away, but was told to avoid unreasonable amounts of digits in the future)

That was in the 1990s in continental Europe, but I think it is still good enough for s counterexample: Not all teachers are like that.


IF the horse ride were 22.00000000 miles then the other person would be right.

Else if it were 22 miles then you should round the answer to zero decimal places.

Some people are illogically pedantic without any rational reason for what they promulgate.

  • 5
    $\begingroup$ It's not decimal places you should have in mind, but significant figures, surely? $\endgroup$ – Prime Mover Jan 18 at 17:37
  • $\begingroup$ I presumed that the numbers given were to sig figs limit $\endgroup$ – practical man Jan 20 at 3:56
  • $\begingroup$ @practicalman You would still round to sig digs though, not decimal places. Your number has 10 sig digs total, two left and 8 right of the radix. Your wording implies that your result should always have 8 digits right of the radix. For example: 22.00000000 miles = 35200.00000000 meters, when it should be 35200.00000 meters (assuming 1 mile = 1600 meters with infinite precision for simplicity). $\endgroup$ – DKNguyen Jan 20 at 14:04

You're both right, depending on the domain of discourse and the rules of engagement.

In pure math, the traditional expectation is that the numbers given are exact unless stated otherwise, and answers are also to be exact unless stated otherwise. So when the mathematician read "22 miles," he's using a tradition that means "exactly 22 miles."

But in the physical sciences, all measurements are understood to be inexact and approximations and rounding are either "allowed" or "expected" (depending on the logical rigor applied).

In this case, the correct answer boils down to a question of semantics and assumptions. What about this question:

If a man traveled 22 miles, how far did he travel in kilometers?

How would you answer that? The "If" complicates things. Some would say that it turns the question into a hypothetical that ignores the physical difficulties in measuring exactly 22 miles and turns it into a "given." It's not a stretch to read the original question as a hypothetical, even without an explicit "If" at the beginning.

Some traditions say that integers are always expected to be exact and that the question should have used "approximately 22 miles," "22.0," or a bar on the last significant digit to show it's a real number instead of an integer.

Even in the physical sciences, scenarios used for pedagogical purposes are sometimes idealized in order to remove confounding factors that might distract from the main point being taught. I don't think we know enough about the source of this question to know about what assumptions or simplifications are being made.

You may argue that the use of "a man riding on horseback through a forest in a pre-industrial age" implies a real situation and an actual, inexact measurement. A counterargument is that the use of abstract identifiers "A" and "B" to to designate the starting and stopping point suggest an idealized situation.

I would agree this is a good question for Mathematics Educators Stack Exchange. It emphasizes that in the classroom (as well as life) it's important to be explicit about assumptions and expectations and to lay out the ground rules.

Adding a summary, based on comments, that tries to be more direct:

  1. Use of significant digits only applies to inexact numbers such as measurements.
  2. In a problem like "Convert 22 miles to kilometers," there is no reason to think 22 miles is a measurement. Rather, it is a "given": Something that is to be assumed or taken for granted for the sake of the problem.
  3. I think this question boils down to this: In the original question, is "22 miles" to be taken as a given or a measurement?

I don't think we can tell. (At least not without more context about where the question came from and why it was asked.) The original question could merely be "Convert 22 miles to kilometers," dressed up in a story to make it engaging or interesting.

My reading of some of the comments suggests a point of view of "If the problem resembles a real-world situation, then it must be interpreted as a real-world situation." Or more succinctly: If 22 miles could be interpreted as a measurement, then it must be interpreted as a measurement. Or that by phrasing the question in a historical, real-world context, that somehow forces the measurement interpretation. I don't follow that. It ignores the way real-world people write, talk, and teach.

  • 3
    $\begingroup$ The subtext here, if mathematics and the sciences are taught this way is: "This is mathematics, here within this walled garden. The other side of the wall is physics and chemistry and messy stuff like that, and we don't have anything in common with them. We use numbers precisely, and those smelly hairy apes over there use (shudder) approximations." Oh, and the reason for using "A" and "B" is because I could not remember the actual place names. It is implicit that this is a real-world scenario being modelled mathematically ... $\endgroup$ – Prime Mover Jan 18 at 23:34
  • 2
    $\begingroup$ ... and so the expectation is that the student reads all of the question and puts the entire situation into context. You cannot honestly say: "We're mathematicians and so we scoff at the real world because we work with ideals." You are given the real world situation and it is an important part of mathematics to be able to translate accurately and appropriate the full context of a "word problem" into the correct mathematical model. TL;DR: This is not a pure mathematics problem. It is at base an a exercise in mathematical modelling. $\endgroup$ – Prime Mover Jan 18 at 23:38
  • 3
    $\begingroup$ It seems like a very odd choice to treat real-world, continuous measurements as integers, since it effectively implies infinite precision of the measurement. Integers seem appropriate for counting problems (i.e. there are exactly 22 apples in the basket), but not at all for this type of measurement problem (it's not even possible to measure something as exactly 22 miles away with infinite precision). $\endgroup$ – Nuclear Hoagie Jan 19 at 14:31
  • 3
    $\begingroup$ "It ignores the way real-world people write, talk, and teach." Does it? This lack of explanation and succinctness of this statement implies that you think it is an given. It is very unclear to me that this is actually the case. You have some explaining to do because if anything, the way people write, talk, and teach goes completely in the opposite direction in that the assumption is lack of precision in the absence of explicit information, rather than infinite precision. $\endgroup$ – DKNguyen Jan 19 at 20:34
  • 2
    $\begingroup$ @DKNguyen. I regret that I gave that impression. I don't think we have enough information to know what interpretation works best. If my post seems to advocate for the "given" interpretation, I think it's because most of the posts that I read seemed to imply felt it was "obviously" a measurement (my phrasing, not theirs.) So I did want to make a case that were was an alternate interpretation. I appreciate your feedback. $\endgroup$ – Syntax Junkie Jan 19 at 21:00

It depends

I agree with just about everyone that the answer is 35, or perhaps 35.4 (a number I like better, see below). An answer of 35.405598 km is precise to the millimeter. I've ridden horses; they don't work in millimeters.

Update: For what it's worth, after all this discussion, I think that the right number is "about 35 and a half" (not 35.5) kilometers. Thirty five and half has about the same uncertainty as "22 miles" (maybe even more), and is within "horseshoes and hand-grenades" of the exact answer of "just about 35.4 exactly".

As you acknowledge, the intermediate answer you came up with (using an approximate conversion factor) of 35.2 km is wrong; 35 km is a correct answer, but 35.2 km is just plain wrong. It makes sense to consider that a distance of "22 miles" is likely more precise than "something between 21.5 and 22.5" which is what considering 22 as having only two significant figures means. It's more like 22.0 miles (i.e., between 21.95 and 22.05 (which gives you an uncertainty of about 500 feet (about 160 m)).

But, when you multiply 22.0 by 1.6, then your answer should definitely only have 2 significant figures (not because of the 22, but because of the 1.6). You can tell that your 3 significant figure result is off, the "completely precise" number is off by 0.2 km (200 m) from your figure. Horses are more accurate than hundreds of meters.

What you want to do working with numbers is to get an understanding of both the precision and the accuracy of the measurement. Saying something is about 22 miles, give or take 500 feet makes 22.0 about the right number to use.

When doing a conversion, it's always best to use the most precise number you have for all intermediate work, and only round back to the correct number of significant figures at the end of the calculation. When doing distance calculations, I always use the fact that one inch is exactly 2.54 centimeters (i.e. 2.54000000000, as many zeros as you want). If I've got a calculator (or a slide-rule) handy, I'd do this:

22 miles * 5280 ft/mile * 12 in/ft * 2.54 cm/in / 100 cm/m / 1000 m/km
= 35.405568 km

Note that that number is off by 30 millimeters from what you quote. My number is correct. Also note that I carried the units through the calculation. That way, I can do some dimensional analysis and see that I get an answer in km, and that it's what I expect: (miles * (ft/mile) * (in/ft) * (cm/in) / (cm/m) / (m/km) works out to km).

I'd look at that number and say "yes, it's 35.4 km." Also note that all those intermediate conversion constants are exact (the number of inches in a foot is exactly 12 - so you can treat 12 like 2.54, it has as many zeros as you want).

But then again

Way back when I was a student, I had a math prof who'd get upset at us engineers for saying the answer is about 35.4km. He's say that two numbers can be equal, but "about equal" or "approximately equal" have no mathematical meaning. Then he'd point out that it would be pretty easy to figure out that one was about equal to zero - and at the point, everything breaks.

So, if you are in a math class and the teacher says "The relationship between miles and kilometers is 1.609344 km/mile, how many kilometers are there in 22 miles?", then the answer is 35.405568 km, not 35.4 km.

Note the absence of the horse in this phrasing of the question.

  • 1
    $\begingroup$ I never gave $35.2$ as an answer. It was an interim calculation based on the necessities of the question. $22$ miles is only ever going to be a guessification. You don't need a conversion factor of anything better than $1.6$ km per mile in such circumstances. It's the same when I drive across Europe. I like to know how many miles and how many km left. I only need an approximate number. So I multiply by 8 and divide by 5 (or vice versa) and I can do it in my head. $\endgroup$ – Prime Mover Jan 20 at 6:00
  • 1
    $\begingroup$ @DKNguyen Applied mathematics uses units. And unit conversion was part and parcel of the mathematics curriculum throughout the whole length of my high school, and even earlier, although early education did not put "names" on the classes so much, it was all just "lessons". $\endgroup$ – Prime Mover Jan 20 at 7:28
  • 4
    $\begingroup$ @DKNguyen. Q: "If you're truly a math class, why is the professor using units at all?" a: to demonstrate that units need to be treated as algebraic quantities than can be manipulated as any other variable; (2) To demonstrate that having an answer with sensible units can help a student check their work. Q: "Or real world examples of things?" A: To test the student's ability to extract the relevant details from a body of text and convert that into mathematical symbolism. $\endgroup$ – Syntax Junkie Jan 20 at 17:25
  • 1
    $\begingroup$ There are contexts (such as land measurements in the United States) where 1 inch is not exactly 2.54 cm. civilgeo.com/blog/when-a-foot-isnt-really-a-foot $\endgroup$ – Jasper Jan 21 at 1:08
  • 1
    $\begingroup$ ... It's very common, often a requirement, that math courses include applications with real-world units. See any math textbook in a course that engineers might take. Pedagogically most people find that making work concrete helps students get traction. Moreover, how else can students practice manipulating units before entering their applied/engineering courses? My dept.'s liberal-arts math course is mostly about teaching precisely that topic. $\endgroup$ – Daniel R. Collins Jan 21 at 1:45

It depends on the level of the class.

I would expect someone who has a recent undergraduate degree in mathematics to have experienced significant figures at at least some point in their life, either in high school or in college. I would also expect common sense to kick in and say that the level of accuracy proposed is unreasonable.

But it is reasonable to dodge the topic when teaching arithmetic, algebra, etc., because the students usually have a hard enough time as it is. Sometimes you can arrange for numbers that come out evenly anyway, but if you're stuck trying to teach an awkward conversion (miles to km) or if the task is to teach something about decimals or fractions then you may be unable to avoid it.

For example, "Alex had five cookies and split them evenly with Blake. How many did each of them get?" Two and a half, and we aren't going to quibble over how precisely half of a cookie was achieved.

If your students are advanced enough to be working with more precise numbers (and, presumably, starting to question what level of accuracy is acceptable) then the best way to dodge it is to simply specify what rounding you want in the question: "Answer to x decimal places." That way you can specify the correct precision without the students having to understand how to calculate what the correct precision should be.

That's much simpler for the student to understand than the official way, which is according to NIST:

The precision of your conversion should be based on relative error. If error isn't specified, then you can infer it from the number of digits in the values given. Use a conversion factor with equal or more precision to that to preform the calculation. Then you round the result to produce a relative error that is of the order of the original.

$22$ miles implies an error range of plus or minus $\frac{1}{2}$ mile which is $2.\overline{27}\%$.

Using a conversion factor of $1.61$ kilometers per mile (which has better than $2.\overline{27}\%$ accuracy, note that $1.6$ is not accurate enough)... $22*1.61=35.42$ km. We could also use $1.609$ or any more precise conversion, it does not matter because we will be rounding. (For example, in this case, $22*1.609 = 35.398$ km.)

Now we round... $40$ km would have a relative error $5\div40\approx12.5\%$ which is too much, $35.4$ would have a relative error of $0.05\div35.4\approx0.141%$ which is too little. $35$ km has a relative error of $0.5\div35\approx 1.43\%$ which is just right. Note that we get the same (rounded) answer regardless of how much precision we used in the conversion factor, as long as the conversion factor met a minimum level precision.

Question: Why do we assume plus or minus one-half mile? Wouldn't a distance of 22 miles be measured more accurately than that?

Answer: No. If anything it is likely to be much worse. (Disclaimer, I'm not doing sigfigs in this section, I just can't be bothered.)

In American revolutionary war era from the New York Public Library, Thomas Jefferson measuring exactly 22 miles would have actually gone over 22.3 miles (and he was a bit obsessive about measurements):

Before he left on the trip, Jefferson bought from a Philadelphia watchmaker an odometer that counted the revolution’s of his carriage’s wheel. He had measured distance based “on the belief that the wheel of the Phaeton [his carriage] made exactly 360. revoln’s in a mile.” After the trip, though, he re-measured circumference of the wheel and found that it made only 354.95 revolutions in a mile. So for every seventy-one miles Jefferson thought he traveled, he had actually traveled seventy-two.

But I use my car odometer, not a carriage! It's much more reliable! ...nope. From motus.com, if your car odometer says 22 miles then it could be anywhere between 21.12 to 22.88 miles:

Surprisingly, there is no federal law that regulates odometer accuracy. The Society of Automotive Engineers set guidelines that allow for a margin of error of plus or minus four percent.

Actually I use GPS, that's very accurate! ....nope again. GPS has a margin of error on every position measurement made, plus error from the distance between measurements. Essentially your path is like a coastline and the GPS can suffer from the coastline paradox. From singletracks.com (with pictures and a good explanation: In a very, very bad case (steep trail, lots of switchbacks) your GPS may report 22 miles when you've actually gone 44 miles! Holy guacamole.

[...]GPS reports the full loop is right at 1 mile long. In fact, everyone else who rides this trail gets roughly the same measurement. But the trail always “feels” much longer than that.

Recently I started riding with a wheel-based cyclocomputer, which I calibrated and verified during one of our track tests. Measuring this particular trail with the cyclocomputer reveals the trail is actually closers to 2 miles long, meaning our GPS units are underestimating distance by half!

I'm not going to find sources for inaccuracy of distance calculated by counting steps, the time it took to travel, etc. It's pretty obvious that no one (and no horse) actually moves at that even of a pace for 22 miles.

So accepting 22 miles on a trail as being between 21.5 and 22.5 is actually pretty generous. Better to just call it a day and say it was "some distance".

  • $\begingroup$ Exactly. However, when working with familiar integers in conversation, it's not the same as working from a specification. If someone says "22 miles", I'm assuming that the uncertainty is less than +/- 0.5 mile. That would be "about 22 miles". Perhaps my estimate of +/- 500 feet is too narrow, but I suspect that it's not that far off. The real number may be "35 and a half" (not 35.5), there's a lot of slop in that wording. Question, if someone says "10 miles", how much uncertainty is there in that number - is it 1 sig fig, or 2, or, 1.5? I believe that "22 miles" has less slop than "10 miles" $\endgroup$ – Flydog57 Jan 20 at 23:56
  • $\begingroup$ @Flydog57 No. I added into my answer at the bottom, because it was much too long for a comment, but the short answer is that everyday tools for measuring distances traveled are really inaccurate. $\endgroup$ – user3067860 Jan 21 at 13:12

I would say that based on the words of the question, the answer is 35.

It is not a distance or measurement of 22 miles between points A & B. It is a journey between locations A & B.

Next time you see the guy who scolded you, ask him how one should answer if asked, “What is the numerical value of pi minus e?”


Your Answer

