Since $\renewcommand{\unit}[1]{\,\mathrm{#1}} 1\unit{dm} = 10^{-1}\unit{m}$, it follows that $1\unit{dm^3} = 10^{-1} \times 10^{-1} \times 10^{-1} \unit{m^3} = 10^{-3} \unit{m^3}$.

However, in regular mathematics the following equation holds true:

$$a\,b^{3} = a\,b\,b\,b$$

By the above, the cube unit should expand as follows:

$$\mathrm{dm^3} = \mathrm{dmmm}$$

While in actual usage (as seen in the second equation) the expansion is $\mathrm{dddmmm}$, which would arise from using $\mathrm{(dm)^3}$ instead.

$$\mathrm{(dm)^3} = \mathrm{dddmmm}$$

So shortly: why aren't parentheses (commonly?) used in units?


The thing is that $\mathrm{dm}$ is a single symbol, not a combination of two symbols.

Yes, it can be understood in terms of a prefix and a base indicator, but it is still a single symbol. An analogy to the concatenation of variable is inappropriate.

Reference to an authoritative statement:

The grouping formed by a prefix symbol attached to a unit symbol constitutes a new inseparable unit symbol (forming a multiple or submultiple of the unit concerned) that can be raised to a positive or negative power and that can be combined with other unit symbols to form compound unit symbols.

Example: $\renewcommand{\unit}[1]{\,\mathrm{#1}} 2.3\unit{cm^3} = 2.3\unit{(cm)^3} = 2.3 \unit{(10^{–2}\,m)^3} = 2.3 \times 10^{–6} \unit{m^3}$

  • $\begingroup$ This answer leaves a lot of Why? type of questions unanswered. $\endgroup$ – Nit Jul 29 '14 at 17:39
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    $\begingroup$ @Nit To further clarify why it's inappropriate to consider d and m as being separate symbols, consider that you would always write dm and never md. If it really were just a case of multiplying two symbols, the latter form would be legitimate. $\endgroup$ – Kyle Jul 29 '14 at 18:00
  • $\begingroup$ @Nit: Or maybe better because the prefix does not have a meaning of it's own, so it's precedence can't be mistaken. $\endgroup$ – Jan Hudec Jul 29 '14 at 18:24
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    $\begingroup$ @Nit I've made the answer community wiki as well as sticking in GlenTheUdderboat's citation. This is just another case of a throwaway answer getting a lot of attention because the issue is accessible rather than because there is anything deep and beautiful involved. $\endgroup$ – dmckee Jul 29 '14 at 20:08
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    $\begingroup$ @JanHudec The prefix might be considered to have some meaning, especially considered in the context of dimensional analysis. For example, I may multiply by 1, or $10\:\mathrm d/1$, to cancel a "d" and get just "m". $\endgroup$ – Phil Frost Jul 30 '14 at 10:51

They're not used because it's ugly to read such texts with parentheses and it's time-consuming to write it down.

A decimeter is indeed a "product" of "deci" and a meter, so the origin is analogous to the product of two real numbers $ab$. But once we define the new derived unit ${\rm dm}$, we treat it as a single object, so it really means what you would call $({\rm dm})$.

Or if one wants to be really picky: in $ab$, the tiny space in between the two letters is a space in italics which may be interpreted as a multiplication of variables. However, the tiny space between ${\rm d}$ and ${\rm m}$ in "decimeter" is a Roman font minispace, and that isn't interpreted as a product anymore. That's why ${\rm dm}$ is interpreted as a whole.


$1\text{ dm}^3$ can be seen as a shorthand way to write $1\text{ dm}\cdot 1\text{ dm}\cdot1\text{ dm}$. The motivation behind this is that the quantity $1\text{ dm}^3$ represents the volume of a cube with each side of length $1\text{ dm}$, and the way to find the volume is to multiply the three sides: $LWH=(1\text{ dm})\cdot(1\text{ dm})\cdot(1\text{ dm})=1\text{ dm}^3$.

In other words, view "dm" as a new unit, not as a product. It might be less confusing to write $1\ (\text{dm})^3$ instead of $1\text{ dm}^3$, but that's the norm.

It may be beneficial to replace "dm" with some other measure that is represented by a single letter; this way you don't run into the issue of wanting to distribute the power. But it's probably better to get used to the norm.

As bdesham points out, it's important to note that some quantity like $4\ \text{dm}^3$ is not short for $4\ \text{dm}\cdot 4\ \text{dm}\cdot 4\ \text{dm}$, but rather $4\ \text{dm}= 4\cdot\left( 1\ \text{dm}\cdot 1\ \text{dm}\cdot 1\ \text{dm}\right)$. In other words, you have 4 cubes of side length $1\ \text{dm}$.

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    $\begingroup$ Well, $1\text{ dm}^3 = 1\text{ dm}^3 \cdot 1\text{ dm}^3 \cdot 1\text{ dm}^3$, but obviously $2\text{ dm}^3 \neq 2\text{ dm}^3 \cdot 2\text{ dm}^3 \cdot 2\text{ dm}^3$. It might be a better idea to say e.g. $4\text{ dm}^3 = 4 \cdot 1\text{ dm}^3 \cdot 1\text{ dm}^3 \cdot 1\text{ dm}^3$. $\endgroup$ – bdesham Jul 30 '14 at 18:26
  • $\begingroup$ @bdesham Excellent point. $\endgroup$ – BMS Jul 30 '14 at 20:47

In your example, ab is implied to be a multiplication - a*b; but dm is a single indivisible token.

Imagine second2 - that doesn't imply that the last letter should be squared, and it's the same with decimeters.

Mathematical notation is often ambiguous, and leaves many things implied and underspecified with the expection that the reader will fill the gaps properly - the assumption of ab = a*b is one of them.


This is just a convention and nothing more. We treat the $\mathrm{dm}$ as a single symbol. Note that there are much stranger notations of powers, such as squared sines:

$$ \sin^2(x) = \big(\sin(x)\big)^2 $$

You can not treat this formally as written - it's not a square of the sine, but the square of the value it returned.

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    $\begingroup$ It's worth noting that $\sin^2(x)$ has the perfectly sensible interpretation $\sin(\sin(x))$ — but that's not what the notation means because taking the sine of a sine is an operation that almost never comes up in a real problem. Likewise, $\mathrm{cm^3}$ is a notation that reminds you of what people actually say, "cubic centimeters"; even though it's possible to divide a cubic meter into 100 pieces can call each a "centi cubic meter," that's essentially never convenient in practice. $\endgroup$ – rob Jul 29 '14 at 20:15
  • $\begingroup$ @rob: It may also be worth noting that there are some cases where prefixes may be ambiguous, e.g. a rate of 4Kbyte/sec. may mean either 4Kbytes (4,096 bytes) per second, or it may be bytes at a rate of 4KHz (4,000Hz). $\endgroup$ – supercat Jul 29 '14 at 21:58
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    $\begingroup$ @supercat: According to the current ISO/IEC standard, $4\,{\rm kB/s}$ $=$ $4{,}000\,{\rm B/s} $ $\ne$ $4{,}096\,{\rm B/s}$ $=$ $4\,{\rm KiB/s}$. $\endgroup$ – Ilmari Karonen Jul 29 '14 at 23:37
  • $\begingroup$ Oh, not silly trig powers... When I write $f^n(x)$, I mean $f(\cdots(f(x)))$, whether $f = \sin$ or not. $\endgroup$ – michaelb958--Reinstate Monica Jul 30 '14 at 0:17
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    $\begingroup$ @supercat: The IEC binary prefixes exist precisely for that purpose. If you mean $16 \times 2^{20}$ bytes, write 16 MiB; if you mean $16 \times 10^{6}$ bytes, write 16 MB without the "i". $\endgroup$ – Ilmari Karonen Jul 30 '14 at 0:59

protected by Qmechanic Jul 30 '14 at 12:41

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