I understand that significant figures is a term used for "reliably known digits". However, what I don't understand is why the 0's are not counted among these in numbers such as 0.002. Surely, if we know that the units digit is 0, and that the tenths digit is 0, and that the hundredths digit is 0, then we know these digits reliably? In other words, we know that the units digit is not 1 or 2 or 3, but 0. Thus, we know this digit reliably. Why then is it not counted as a significant figure? Why do all physics textbooks say that 0.002 only has 1 significant figure?

The "related" question is different from the one I am asking. The one there is asking about 1500 whereas my one is about 0.002, ie when the zeros come to the left of the number.


11 Answers 11


One of the logical rules for significant figures is that expressing a given number in a different order of magnitude should not make you sound like you know more or less about the number. If you start with $0.002$, we can only say that it's equal to $2\times 10^{-3}$, since you probably already appreciate the implications of adding zeros to the left of a decimal place.

Regarding the claim,

we know that the units digit is not 1 or 2 or 3

Yes, but those are extremely trivial bits of knowledge. Try saying "$002$ has three significant figures". It's obvious that there's no other constant in those places, because then we'd be dealing with a completely different number; you wouldn't call it "two". Significant figures are only a relevant thing to consider when you're debating between options which can be rounded to the same value, within reason.

  • 2
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Feb 2, 2019 at 13:45

Because significant figures measures uncertainty relative to the size of the number

Suppose you take a measurement of something and it comes out to be 0.002 meters.

You then measure something else and it comes to 345 meters.

You know that $0.002$ means $0.002 \pm 0.0005$ and $345$ means between $345 \pm 0.5 .$

The uncertainty in the numbers here are $0.0005$ and $0.5 ,$ respectively.

Notice how the difference between $345$ and $0.5$ is much greater than $0.002$ and $0.0005 .$

$345$ is $690$ times bigger than $0.5 .$ $0.002$ is only $4$ times bigger than $0.0005 .$

Thus, $345$ is a more precise relative to its size – 2 more digits precise. :)

  • $\begingroup$ If I measure out 1000m and 999m they are both +-0.5m which is almost the same uncertainty relative to the size of the number. Yet one is 1SF and one 3SF ? $\endgroup$
    – trapper
    Commented Feb 1, 2019 at 13:32
  • 15
    $\begingroup$ @trapper. That is the problem with not using scientific notation. If 1000m means 1.000km (+-0.5m) then it is 4 s.f. However, if 1000m means 1km then it is 1 s.f. and +-500m. $\endgroup$ Commented Feb 1, 2019 at 13:42
  • $\begingroup$ Hence not in any way 'relative to the size of the number' $\endgroup$
    – trapper
    Commented Feb 1, 2019 at 13:43
  • 4
    $\begingroup$ If you measure 1000m and it's accurate "to the meter" you're supposed to write it as "1000.m". The decimal point lets you know the accuracy and thus that the number is 4 significant figures. You could also write it as 1.000km as suggested by Mark $\endgroup$
    – adam_0
    Commented Feb 1, 2019 at 17:14
  • 9
    $\begingroup$ @trapper - leading zeros are never signficant, but trailing zeros may or may not be significant. That is the main reason scientific notation was invented - to provide a way of showing when trailing zeros are significant and when they are not. If you measured $1000$ m to $\pm 0.5$ m, that is $1.000\times10^{3}$, which is 4 significant figures, not 1. You may argue about how deserving it is for it have 1 more SF than $999$, but SF are only a shorthand indicator of relative uncertainty, not an exact one. $\endgroup$ Commented Feb 1, 2019 at 23:55

The notion of "significant figures" is meant to communicate how much you know about a number. A number with one sig fig means you know it to roughly one part in $10$, two sig figs mean you know it to roughly one part in $100$, and so on.

This is a useful idea, because if you multiply a number with $n$ sig figs with a number with $m$, the resulting number has $\text{min}(n, m)$. That is the point of sig figs, which is to roughly keep track of uncertainties so you know how precise your final result is.

You propose to instead define the number of significant figures to mean "the number of decimal places we are sure of the value, including anything after the decimal point". The benefit of this definition is that addition behaves nicely, but it behaves terribly under multiplication or unit changes. For example, if I convert $0.0001$ meters to $0.1$ millimeters, three significant digits disappear even though nothing about the quantity really changed. It turns out that in practice, it's much more useful to use the first notion, because keeping track of precision through multiplication or unit changes is harder than doing so for addition.

It should be kept in mind that "sig figs" is just a funny sounding word. It can in principle be defined any way you want; there is no "true" definition. The usual definition is just the more useful one.

  • $\begingroup$ So basically an ad hoc method of rounding. $\endgroup$ Commented Feb 1, 2019 at 6:38
  • 6
    $\begingroup$ @marshalcraft It's an ad hoc way of talking about confidence intervals, really. It's a shorthand (and quick) way to approximate clipping the number to "known digits" within what you're confident of. "How confident are we about this number?" "About 10" and "10±5" are identical up to precision -- one is just a little easier to work with. $\endgroup$
    – user24081
    Commented Feb 1, 2019 at 8:19
  • $\begingroup$ How are the zeros not significant thou? If I measure my desk in millimetres and it is 1000mm (+- 0.5mm) If I measured again in meters it's 1m (+- 0.5m) It's the same number but one is 1000x more accurate. $\endgroup$
    – trapper
    Commented Feb 1, 2019 at 13:08
  • $\begingroup$ @trapper Yes, the zeros in $1000$ mm are significant, because those three zeroes at the end indicate you know those digits are zero. Compare though: you could also write the very same number as $0000001000$ mm. But all those zeroes in front don't count as sig figs. $\endgroup$
    – knzhou
    Commented Feb 1, 2019 at 13:20
  • 7
    $\begingroup$ @trapper Well, the problem is that "1000" is ambiguous. It could mean you're sure of the zeroes or it could mean you aren't, depending on the person. In scientific notation we would write $1 \times 10^3$ if we're not sure and $1.000 \times 10^3$ if we are sure. If you stick with scientific notation (writing every number as a number from 1 to 10, times a power of ten), then the number of sig figs is always equal to the number of digits you write. $\endgroup$
    – knzhou
    Commented Feb 1, 2019 at 13:34

From the comments, I think part of your confusion stems from the notion that leading zeroes can't be applied in front of a decimal. This isn't accurate.

For example, in a comment you state "saying 0029 doesn't make sense because this is not a number", but it actually is a number!

You can have $29$, $0029$, $0000029$, $02.9 * 10^1$, and these are all the same number. The reason we don't typically write it the other ways, though, is because all zeroes before the first non-zero digit are not significant. Writing $0029$ instead of $29$ doesn't clarify or change anything; all it does is add more digits which don't matter, so you'll seldom see it used, but it's perfectly legal.

Likewise, the same rule applies to decimals: all zeroes before the first non-zero digit are not significant. Significant figures are used to indicate a level of precision in your number. If I gave you a measurement of 2000 feet and told you there were 4 significant figures, you'd know I measured exactly 2000 feet (with a possible decimal). But if I told you I only had 1 significant figure, you'd know my number was much more imprecise.

However, if leading zeroes in a decimal were significant, then using scientific notation would change the precision of a number. We cannot have this. In your example, $0.002$, you're arguing there'd be more than 1 significant figure, but we also know that $0.002$ = $ 2* 10^{{-3}}$. There has been no rounding; these two numbers are identical in both value and precision. We know the latter has 1 significant figure, so the first must have only 1 as well.

For your addition example, you state Tipler and Mosca say ""the result of addition or subtraction of two numbers has no significant figures beyond the last decimal place where both of the original numbers had significant figures" I think the phrasing here is very unhelpful, but this still makes sense!

$2.34 + 0.00002 = 2.34002$

$2.34$ has no significant figures beyond the hundredths place.

$0.00002$ has no significant figures beyond the hundred-thousandths place.

We can see both numbers have significant figures beyond the ones place, the tenths place, but we do not see that both have significant figures beyond the hundredths place. That means our final answer should only be that precise, so we get $2.34$ again! (Which also makes sense with our precision, as we have no idea what the hundred-thousandths place of the first number was measured to be).


The use of zeroes in a number such as $0.002$ is to enable the $2$ to go in the correct place relative to the decimal point ie in the $\frac{1}{1000}^{\rm th}$ position.

The same is true of $0.102$ but in this number you have a more significant digit (representing a larger number) which is the $1$ as it is in the $\frac{1}{10}^{\rm th}$ position.

If the number is less than one all the zeros to the left of the most significant digit are to assign the correct position relative to the decimal point to the most significant digit.

So $0.002$ is to one significant figure as the zeros are not significant they are there to place the $2$ in the correct position relative to the decimal point. and $0.102$ is to three significant figures.

Life can get complicated when the number is greater than one.
For example is $200$ to one significant figure with the zeroes putting the $2$ in the $100$'s position or to two significant figure or three significant figures.
Without further information it is impossible to say and that is why the use of scientific notation is useful.
$2 \times 10 ^2$ is to one significant figure, $2.0 \times 10 ^2$ is to two significant figures and $2.00 \times 10 ^2$ is to three significant figures.

So with your original number $0.002$ it can be written as $2 \times 10 ^{-3} $ which immediately identifies it as to one significant figure.

  • 1
    $\begingroup$ If I know that I am 29 years old, then I immediately know that I am 0.029 kilo years old, do I not? In other words, in kilo years, I know my age to 4 significant figures. In normal years, I know my age to 2 significant figures. But saying 0029 doesn't make sense because this is not a number. Please explain. $\endgroup$
    – Raghib
    Commented Jan 31, 2019 at 16:03
  • 2
    $\begingroup$ @Raghib In kilo years you know your age to two significant figures. The $0.0$ are there so that you can place the $2$ in the correct position relative to the decimal point to indicate that it represents two hundredths of a kilo year. $\endgroup$
    – Farcher
    Commented Jan 31, 2019 at 16:08
  • 2
    $\begingroup$ @Raghib, your proposed addition of 2.34 + 0.0002 necessarily requires that you add digits to the 2.34 number, such that you convert the addition problem to 2.3400 + 0.0002. This is invalid because you don't know that the first number is actually 2.3400 ... there is not enough precision to know that. The number could just as well be 2.3432, but you didn't measure the number to that precision. Hence, the addition rule tells you to stop the addition when you run out of information. $\endgroup$ Commented Jan 31, 2019 at 18:17
  • 2
    $\begingroup$ @Raghib If you find yourself having to add $2.34+.0002$ in a physics context, your analysis is not well-founded in the first place, as it's trying to account for effects that are below its ability to resolve. $\endgroup$ Commented Jan 31, 2019 at 18:57
  • 1
    $\begingroup$ @Raghib: There are formulae for working out what the error of such calculations should be, but as a general rule, for 2.34 + 0.0002 take the number of significant figures of the largest number (order-wise) and write the final answer to that number of significant figures (although in real life most sensible experiments would hardly ever present this situation). Why? Because, if this is a sensible result, the dominant value ought to correspond to the dominant process and so its accuracy is the most important contributor to the final result. $\endgroup$
    – Zorawar
    Commented Jan 31, 2019 at 20:16

How many significant digits are there in $002$?

Remember that measured quantities are not exact. The measured quantity "$2 \,\text{m}$" represents the interval $(2-1/2, 2+1/2] \,\text{m}$. Significant digits attempt to simplify working with these intervals.

Here are some variations in your example. All are measured quantities, with their significant digits indicated by the usual convention regarding decimal points.

\begin{align*} 2000 &= (2-1/2, 2+1/2] \times 10^{3} &:& \text{$1$ significant digit} \\ 2000. &= (2000-1/2, 2000+1/2] \times 10^{0} &:& \text{$4$ significant digits} \\ 2. &= (2-1/2, 2+1/2] \times 10^{0} &:& \text{$1$ significant digit} \\ 02. &= (2-1/2, 2+1/2] \times 10^{0} &:& \text{$1$ significant digit} \\ 002. &= (2-1/2, 2+1/2] \times 10^{0} &:& \text{$1$ significant digit} \\ 00000\,000002. &= (2-1/2, 2+1/2] \times 10^{0} &:& \text{$1$ significant digit} \\ 00.2 &= (2-1/2, 2+1/2] \times 10^{-1} &:& \text{$1$ significant digit} \\ 0.02 &= (2-1/2, 2+1/2] \times 10^{-2} &:& \text{$1$ significant digit} \\ 0.002 &= (2-1/2, 2+1/2] \times 10^{-3} &:& \text{$1$ significant digit} \\ 0.0020 &= (20-1/2, 20+1/2] \times 10^{-4} &:& \text{$2$ significant digits} \\ \end{align*}

Significant digits tell us how narrow is the represented interval. (Compare the last two lines, where the trailing zero has narrowed the interval by a factor of $10$.) Zeroes to the left do not reduce the size of the represented interval, so do not increase the number of significant digits.

Aside: The first two lines also show why using scientific notation for numbers is necessary to clearly indicate significant digits for some measured quantities. What if the measured quantity is $2000$ with two significant digits? There is no good place to put the decimal point. However, "$2.0 \times 10^3$" is easy to write and captures exactly this meaning.


0.002 kg equals 2 g so the 0s after point have nothing to say about it's not 1 or 2 or 3 or... 0.002 is being said as "point zero zero two" or "zero point zero zero two". It's general convention to put 0 before point to indicate there is nothing than leading 0s for example,

0.002 means 2 times 10^(-3) i.e., 0.002 = 2 x 10^(-3)

and not 0.002 equals 10.002 or 1000.02 or anything with leading non-zero digit infront of it...


Say, if you say the length of your ruler is 1m, which has one significant digit.

Few minutes later, the ruler is still 1m, but you say that it's 1000mm long -- My bad, you can't say that's it 1000mm long in physics, it's 1x103-- but you say that it's 0.001km long, which has 4 significant digits (according to your definition).

So the length of your ruler has both 1 and 4 significant digit(s) --(and maybe other infinite numbers of significant digits).

Isn't that a contradiction ?

UPDATE : To answer on your example. From your definition, say you say a book is 0.002kg, it is also equal to 0.000 002ton.

Both are the same value, but both have different significant digits (if you use your logic from your question. And we also know the zeroes in from of the 0.000 002ton is zero and it can/will have many infinite leading zeros if you want to think about it that way).

And BTW 0.002kg is likely different from 0.00200kg which has 3 significant digits in this case. And from the 0.00200kg example, if we know the last 3 "200" digits as a stated exact magnitude (to the precision of 3 significant digits), we will for sure know the magnitudes (values) of the leading zeroes, which has infinitely many leading zeroes as 000.00200kg. Therefore leading zeroes don't/can't count towards significant. Like you said, they are zeroes, we know for sure they will be infinitely many leading zeroes, (becoz) for sure we could know they don't exist. The same could not be said to trailing zeroes (which count towards significant digits), because they correspond to smaller magnitudes, we can't always know for sure their values if we try to extend the trailing digits (go further into smaller magnitudes). They are what that define the significant digits. Maybe from there someone can come down with the definition for significant digits.

Unless significant digits in a value (can) varies between different units of measurement.

Therefore the leading zero(es) does not count as significant digits here.

  • $\begingroup$ Para 2 - Just pointing out that 1m isn't 1000cm in anyone's estimation - not just physics. In fact 1m is 100cm! $\endgroup$ Commented Feb 1, 2019 at 13:17
  • $\begingroup$ Haha sorry my bad. It's been sometime since school and my brain has been kinda bricked from workin all day lol. But it's fixed now.. should be typo free.. I hope. $\endgroup$ Commented Feb 1, 2019 at 13:29

tl;dr- Leading zeros aren't significant because they trivially drop out. Still, if you prefer to keep 'em, that's fine, too; you'll just end up having a bunch of leading zeros.

And, no, this isn't something anyone should really want to do. But, the logic behind it and the consequences probably help show why.

Background: Regarding the construction of numbers

First, to define numbers:

  1. Natural numbers are defined through enumeration on a number line from $0 .$

  2. Integers as defined as natural numbers extended with decrementation (inverse enumeration) on a number line, allowing negative values.

  3. Real numbers are defined as integers with interpolation, allowing decimal values.

Conceptually, it'd be simplest if we gave each integer its own, unique symbol. But since no one wants to have to memorize arbitrarily many symbols, we tend to construct numeric identifiers through a transform$$ n ~~ \Rightarrow ~~ \sum_{i} c_i \cdot {b}^{i} \, ,$$where

  • $c_i$ is a numeric symbol selected from a limited subset of "digits" $\in \left[0,~b\right) ;$

  • $b$ is the base (and usually $10$);

then emit this construction as a string,$$ \hspace{25px} \boxed{\begin{alignat}{7} &\texttt{for}~\left(\texttt{var}~i~=~i_{\text{max}};~~~i~\ge~i_{\text{min}};~~~i\text{--}\right) \\ &\{ \\ & \hspace{2em} \texttt{Print} \left( c_i \right) ; \\ \\ & \hspace{2em} \texttt{if} \left( i ~\text{==}~ 0 \right) \\ & \hspace{2em} \{ \\ & \hspace{4em} \texttt{Print} \left( `` . " \right) ; \\ & \hspace{2em} \} \\ & \} \end{alignat}} _{~ \large{.}}$$

Background: Regarding multiplication

Since multiplication is distributive, then the product of two numbers written in the same base, $b ,$ is$$ \begin{alignat}{7} n^{\text{A}} \times n^{\text{B}} ~~ & \Rightarrow ~~ && \left( \sum_{i} c_i^{\text{A}} \cdot {b}^{i} \right) \times \left( \sum_{j} c_j^{\text{B}} \cdot {b}^{j} \right) \\[5px] & = && \sum_{i} {\sum_{j} {c_i^{\text{A}} \cdot c_j^{\text{B}} \cdot {b}^{i} \cdot {b}^{j}}} \\[5px] & = && \sum_{i} {\sum_{j} {c_i^{\text{A}} \cdot c_j^{\text{B}} \cdot {b}^{i+j}}} \, .\end{alignat} $$

Background: Regarding truncation

The above definitions are written for numbers that contain infinite information. In practice, computers (including humans) are finite (unless you find a hypercomputer), so we terminate the procedure at two ends:

  1. We declare some minimal basis, $\cdot b^{i_{\text{min}}} ,$ past which we ignore all further bases, typically under the argument that they're noisy (if from measurement/estimation) or to save on computation work (as computers do).

  2. We declare some maximal basis, $\cdot b^{i_{\text{max}}} ,$ past which we ignore all further bases. Usually, we choose to selected $i_{\text{max}}$ such that we truncate only terms in which $c_i = 0 ,$ since the zero-terms don't affect anything, anyway.

First, we note that any basis $\cdot b^k$ is affected by a noisy term if $k < i_{\text{min}} + j_{\text{max}}$ or/and $k < i_{\text{max}} + j_{\text{min}}$ – ignoring cases in which $c_i c_j \geq b ,$ which I'll mention later.

Second, we note that any basis $\cdot b^k$ is a zero-term if $k > i_{\text{max}} + j_{\text{max}} .$

Given these two constraints, we're therefore only interested in bases$$ \cdot b^k ~~ \text{such that} ~~ k \in \left[ \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)} , ~ i_{\text{max}} + j_{\text{max}} \right] \, .$$Since the number of elements in an inclusive range like this, i.e. $\left[ n_{\text{min}}, ~ n_{\text{max}}\right] ,$ is $1 + n_{\text{max}} - n_{\text{min}} ,$we're therefore interested in$$ \begin{alignat}{7} 1 + i_{\text{max}} + j_{\text{max}} - \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)} ~~ & = ~~ && 1 + \min{\left( i_{\text{max}} - i_{\text{min}} , ~ j_{\text{max}} - j_{\text{min}} \right)} \\[5px] & = && \min{\left(1+ i_{\text{max}} - i_{\text{min}} , ~1+ j_{\text{max}} - j_{\text{min}} \right)} \end{alignat} $$bases.

So, ya know how they say that, when you multiply two numbers with significant figures, the product has the lesser of the multiplicands' significant figures? That's because$$ \underbrace{1 + k_{\text{max}} - k_{\text{min}}}_{\begin{array}{c} \text{significant figures} \\ \text{in the product} \end{array}} ~~ = ~~ \min{(\underbrace{1+ i_{\text{max}} - i_{\text{min}}}_{\begin{array}{c} \text{significant figures in the} \\ \text{first multiplicand} \end{array}} , ~ \underbrace{1+ j_{\text{max}} - j_{\text{min}}}_{\begin{array}{c} \text{significant figures in the} \\ \text{second multiplicand} \end{array}} )} \, .$$This is, the product has the lesser of the number of significant figures from either multiplicand.

Except, there's one problem here: the above logic assumed that bases don't overflow. Which would be true if we were working in Base-2 (binary), but in Base-10 (decimal), we have cases in which $c_i c_j \geq b ,$ e.g. $5 \times 5 \geq 10 .$ Won't discuss that here since it's ignored by the standard rules, but I think the problem's obvious enough. That said, significant figures are meant to be an easy trick rather than used for rigorous calculations, so that they're a bit broken is kind of a given. (See also: my answer here.)

Considering leading-zeros significant

In the above derivation of significant-figure logic, we selected the rule that leading zeros are to be ignored. So, what happens if we do consider them to be significant?

Specifically, you're asking about the case in which $i_{\text{max}}$ or $j_{\text{max}}$ is less-than-zero – e.g., as in $0.01 ,$ in which $i_{\text{max}} = -2$ – and then asking why we can't consider the zeros that we still wrote, so presumably $i_{\text{max}} = 0 .$

So, let's call your alterations $i_{\text{max}}^{*}$ and $j_{\text{max}}^{*} ,$ where$$ i_{\text{max}}^{*} ~ \equiv ~\max{\left(i_\text{max}, ~ 0\right)} ~~~~ \text{and} ~~~~ j_{\text{max}}^{*} ~ \equiv ~\max{\left(j_\text{max}, ~ 0\right)} \, .$$Then, we say that we're interested in "significant" figures that include leading zeros, though we must still reference the prior notions of $i_{\text{max}}$ and $j_{\text{max}}$ because they're important to the issue of tracking the propagation of noise in the calculation.

So then, we're interested in the bases$$ \cdot b^k ~~ \text{such that} ~~ k \in \left[ \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)} , ~ i_{\text{max}}^{*} + j_{\text{max}}^{*} \right] \, ,$$which, to redo the element-count calculation, contains$$ 1 + i_{\text{max}}^{*} + j_{\text{max}}^{*} - \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)} $$members.

So, to derive the new rule for significant figures, we just need to rewrite $i_{\text{min}}$ and $j_{\text{max}}$ in terms of $i_{\text{min}}^{*}$ and $j_{\text{max}}^{*} ,$ and we're done.

So, um. If $i_{\text{max}}^{*} ~ \equiv ~\max{\left(i_\text{max}, ~ 0\right)} ,$ and we know $i_{\text{max}}^{*} ,$ then how do we calculate $i_\text{max} ?$

I mean, obviously, $i_{\text{max}}$ is either going to be equal to $i_{\text{max}}^{*}$ if $i_{\text{max}}^{*} > 0 ,$ but if $i_{\text{max}}^{*} = 0 ,$ then all we know is that $i_{\text{min}} \leq i_{\text{max}} \leq 0 .$ Our problem is that this information is lost, such that merely knowing the number of "significant figures" is insufficient to establish how many we need.

But, screw it, significant figures are a hack anyway. And, so long as we respect the bound on the least-significant basis, we can keep extra zeros if we like. Since, ya know, they don't affect anything.

So to avoid accidentally truncating leading non-zero digits, it's left to us to write the rules such that there're at least as many significant figures in the product as necessary to keep it consistent.

So, we need a number of significant figures equal to$$ \max{\left( 1 + i_{\text{max}}^{*} + j_{\text{max}}^{*} - \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)} ~~~~\forall \begin{array}{l} i_{\text{max}} \in \left[i_{\text{min}}, ~ \max{\left(0, ~ i_{\text{max}}^* \right)}\right] \\ j_{\text{max}} \in \left[j_{\text{min}}, ~ \max{\left(0, ~ j_{\text{max}}^* \right)}\right] \end{array} \right)} \, ,$$which reduces to$$ 1 + i_{\text{max}}^{*} - i_{\text{min}} + j_{\text{max}}^{*} - j_{\text{min}} \, ,$$or$$ \underbrace{1 + i_{\text{max}}^{*} - i_{\text{min}}}_{\begin{array}{c} \text{significant figures in the} \\ \text{first multiplicand} \end{array}} + \underbrace{1 + j_{\text{max}}^{*} - j_{\text{min}}}_{\begin{array}{c} \text{significant figures in the} \\ \text{second multiplicand} \end{array}} - 1 \, .$$In other words, the new rule is that we need to retain a number of significant figures equal to the sum of the significant figures of the multiplicands, minus one, where the new "significant" digits, if any, are leading zeros.

Which is a rule you can have, but it then requires that you recount the significant figures of the product afterward before doing further operations, as this logic's non-conservative.

Problem: Number of significant figures grows

To avoid improper truncation, we had to keep at least as many leading zeros as to ensure that nothing was dropped. If these calculations are repeated, the bloat of leading zeros may continue to grow. (Which I haven't actually worked out; typing this all out took longer than I originally estimated, and, honestly, I'm bored. =P)

But, since the premise of this derivation is that we don't mind doing trivial work, being why we rejected dropping terms $c_i c_j b^{i+j} = 0$ from the calculation, that's presumably not an issue for someone who'd want to use this logic.


Long story short, you can consider leading zeros "significant" if you want, then maintain a bunch of leading zeros in front of numbers to maintain that logic.

It doesn't really mean anything, as it's basically just keeping extra zeros, but it's a mathematically consistent calculation approach one could take if they were so inclined.


You are conflating precision with order of magnitude.

Significant figures are a (approximate) measure of precision, in turn a rough specification of the implied error bars in a number. In the case of the value 0.002 the implication is that the error bars are less than plus/minus 0.0005, so that one can accurately state that the number is closer to 0.002 than to either 0.001 or 0.003. As such the significant figures, ie precision, are a fundamental attribute of the specified value when expressed as a percentage of the value.

Order of magnitude however is dependent on the units in which a value is specified. As such it is not a fundamental attribute of the measurement being expressed, but rather of the units in which the measurement is expressed.

So whether I state that a value is 1 Kelvin or 1 * 10^3 milliKelvin, or 1 * 10^6 microKelvin, or 0.001 kiloKelvin, the significant figures remain 1 while the magnitude tacks the units in which the value is expressed.

In the good old days of slide rules the distinction was easier to track as the slide rule itself only provided precision - one always needed to track order of magnitude manually, whether in one's head or on a separate piece of paper. Electronic calculators and spreadsheets are wonderful things, but they steepen the learning curve for some concepts.


The term "relative" was used in other answers. If you are working with an object that is 1 meter in length, then 0.002 meters can be significant. However the "significant digits" of a number is relative to itself. 0.002m can be written 2mm or 2000 micrometers. What is common is the number "2" surrounded by, effectively, placeholders that help you relate the "2" to the units of meters, millimeters or micrometers respectively. Nothing has changed about the accuracy of the number. Our shorthand for this discussion is "significant figures"


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.