# Is the number 1 a unit?

In dimensionless analysis, coefficients of quantities which have the same unit for numerator and denominator are said to be dimensionless. I feel the word dimensionless is actually wrong and should be replaced by "of dimension number". For example, the Mach number is of dimension one.

Many people write, for this case:

Mach-Number | Dimension: "-" | Unit: "1"

As mentioned before, I would say 'Dimension: "1"' in this place. But what about the unit? $\text m/\text s$ divided by $\text m/\text s$ is equal to one. But is the number one a unit by definition? Or should one say that the Mach number has no unit and therefore 'Unit: "-"'?

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Related answer: physics.stackexchange.com/a/60007/17609 – Keep these mind Sep 17 '13 at 14:03
You can think of an expression such as 9.8 m/s2 as being an element of a certain group. The elements of this group can be thought of as ordered pairs $(n,u)$, where $n$ is a real number and $u$ is an element of a group representing the units. In the group that $u$ belongs to, there has to be an identity element. – Ben Crowell Sep 17 '13 at 15:48

This is analogous to the definition of an empty product in mathematics. For a finite non-empty set $S=\{s_1,\ldots,s_n\}$, the product over $S$ can be defined as $$\prod_{s\in S}s=s_1\times \cdots\times s_n.$$ For such a product you'd want disjoint unions to map into products: if $R\cap S=\emptyset$, then you want $\prod_{x\in R\cup S}x=\left(\prod_{s\in S}s\right) \times \left(\prod_{r\in R}r\right)$, but for this to make sense you want to be able to handle the empty set, and the only way to make the rules consistent is to set $$\prod_{s\in\emptyset}s=1.$$ This essentially says: if there's nothing to multiply, the result is one. (Similarly, empty sums are defined to be zero, for the same reason.) In the case in hand, you could simply say if there are no units to multiply, then you get one. As Luboš points out, this is the harmless choice as multiplying by one does not change the quantity.

Ultimately, though, this is a convention, and it is just something to get used to; it is neither 'right' nor 'wrong'. I do agree, though, that it would probably be more consistent to say that dimensionless quantities like Mach numbers should be said to have "dimension 1."

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Dear @Emilio, just a linguistic remark. The oldest dimensions were the spatial ones, measured in meters. This got generalized to other "dimensions" measuring other quantities. $1$ is simply not a dimension. As your "empty set" in your answer says, $1$ means that there is no dimension, the exponents of the elementary units are zero. That's why it would be very unwise to use "dimension 1" for dimensionless quantities. The problems would get much more severe when combined with the "dimension N operators" - dimensionless operators are surely dimension 0, not dimension 1, operators! – Luboš Motl Sep 17 '13 at 14:27
I was thinking along the lines of $[l]=\text L$, $[A]=\text L^2$, and $[v]=\text L\,\text T^{-1}$, so one could extend that to $[v/c]=\text L^0\,\text T^0=1$. – Emilio Pisanty Sep 17 '13 at 14:46
@LubošMotl Well, they can't be of dimension 0! The calculation of a dimension is \dim Q = L^\alpha T^\beta \dots. This can not become 0. – LaRiFaRi Sep 17 '13 at 15:16
@EmilioPisanty Please note, that [Q] is the unit of Q. What you wanted to write in your comment is \dim l = L, \dim A = L^2 a.s.o. – LaRiFaRi Sep 17 '13 at 15:17
Well, @Emilio, "your" convention for giving the units their name also changes when you switch to different units, doesn't it? For example, in Planck units, everything becomes "dimension 1" in your language, i.e. dimensionless in the normal physicist's terminology. As long as one understands dimensional analysis and logarithms, this is a purely terminological debate and "dimension 1" for dimensionless is very unusual among physicists while "dimension N operators" for the exponent is an important part of the modern physics terminology. – Luboš Motl Sep 20 '13 at 4:48

The number $1$ may be linguistically described as "unity". This very number is the original source of various words in the terminology, like the "unit matrix" (a matrix behaving like the number $1$).

It is a convention to write down that dimensionless quantities like the Mach number have units $1$ because the multiplication by $1$ changes nothing about the result – this is the counterpart of the multiplication by another unit like ${\rm m/s}$.

It just looks more coherent to write the unit $1$ into the tables. But verbally, one may also say that quantities with "this unit" have no units whatsoever. They are dimensionless. As long as one understands the logic, there's no problem in following these somewhat inconsistent conventions in which we sometimes say the units to be $1$ and sometimes we say that the units aren't there.

In the tables, the "unit" column means "the ratio of the full quantity and its numerical value". With this definition, the result may be calculated as $1$ without any problems. It's similar to the task to compute budget deficits as the difference of revenue and expenses. If the latter two are equal, the difference is just $0$. One may write $0$ although he could also write it as $-$ and say that the difference "doesn't exist". The numbers $0$ and $1$ play the role of the "neutral objects" for addition and multiplication, respectively.

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Thank you, actually, the word dimensionless is false by definition of the big standards. So I think, "1" should be the choice for both cases. Mach-Number, Dimension: "1"; Unit: "1". I will wait a little time, if someone has a prove. Thank you though, for an interesting answer. – LaRiFaRi Sep 17 '13 at 13:20
Hi LaRiFaRi, the word dimensionless means that the units of the quantity are $1$. There is nothing wrong about the word and the word is important. It's the same like saying that the budget is without a deficit if the deficit is zero. A zero deficit may be said to exist: the number zero exists - but equivalently, we may say that it doesn't exist because "zero" means that it is not there. For multiplication, 1 plays exactly the same role. – Luboš Motl Sep 17 '13 at 13:26

DIN EN ISO 80000-1:2012-10 on p. 26

Chapter 6.5.5: The Unit One

Translated from German by me:

The coherent SI-unit for every value of dimension number is the unit one, symbol 1. This unit is, in general, not written, if such a value is given by its number. E.g. number of turns in a coil $N = 25 \cdot 1 = 25$

Translated and summarized by me:

Special namings [I think that would be rad, degree, ...] for the unit one may be combined with SI-prefixes. The symbols % and pro-mil are part of the coherent unit one. The unit symbol 1 itself should not be combined with prefixes but then be written as powers of 10.

The standard recommends, not to use the word dimensionless or dimension 1 but "dimension number". The other definitions are "outdated". As an example, the amount of substance 5 mmol/mol is given, where the value is 5, the unit 1 and the dimension the number 0.001.

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1) Please quote carefully. Your answer (v1) seems to contain more quote than actually marked as such. Use > before each paragraph. If you want to place a comment of your own in a quote, use [ ], not ( ). 2) Is the original text freely available on the Internet? Then please provide a link. 3) What is the status of this text? Is it a proposal/a draft/ratified/accepted/etc.? – Keep these mind Sep 17 '13 at 18:16
Thanks for your tips. Corrected that. The source is not free. The original belongs to me and I am not allowed to upload that. It is a draft version of an already existing standard. The mentioned chapter is not marked as new though. (changes are discussed in the beginning of this paper) – LaRiFaRi Sep 17 '13 at 19:27

If $6$ and $7$ each have units $1$, and if this unit behaves like other units, then $6\times 7=42$ would have units $1^2$. My height is measured in metres, but half my height is in $\mathrm{m}1^{-1}$. If you multiply this by two you get back my height in metres, but if you add it to itself you get a quantity that's numerically equal to my height, but has units $\mathrm{m}1^{-1}$.

I think it would be rather difficult to make this consistent, so we have to conclude that if $1$ is a unit, it behaves rather differently from other units. And if that's so, why should we call it a unit at all?

To clarify: I do actually think it's fine to say the units of a quantity are $1$. This answer is only referring to the part of the question that says "is the number $1$ a unit by definition", which I took to mean "is it a base unit, like $\mathrm{m}$ or $\mathrm{K}$?"

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hehe, interesting post. But the last sentence "1=1/1 has no units at all" is not correct as it would have the unit 1/1 which is 1. It is always allowed to unite units together. With the typical mathematical rules. As 1^2 is 1 and 1^-1 is 1, we are still fine with the definition, that one is a unit. Mathematically seen. I understand your concerns, that's why I asked in the first place. – LaRiFaRi Sep 17 '13 at 15:11
Ok, I'll concede that if $\frac{1\:\mathrm{m}}{1\:\mathrm{m}}$ has units $1$ then $\frac{1\:\mathrm{1}}{1\:\mathrm{1}}$ should also have units $1$ -- I've updated my answer. – Nathaniel Sep 17 '13 at 15:38
The use of units in physical quantities can be formalized by making units a group, equivalent to a vector space over the complex numbers (see e.g. Wikipedia), for which the identity (or zero vector) is essential and equal to 1. If you are going to allow algebra with units (such as $3\,\text s\times 2\,\text m/\text s=6\,\text m$), then surely you need to allow such basic algebra as $1^2=1$ (and that follows from $1 x=x$ for any $x$, which is why you introduce it to begin with). – Emilio Pisanty Sep 17 '13 at 21:04
@EmilioPisanty I think everyone's agreed it's a linguistic thing, and my post was intended as a humorous way to approach that. The point is that if we treat $1$ as a base unit then it has to behave like other base units, which means not obeying such algebraic identities - the whole point was it's not going to be consistent. I would say that if dimensions are equivalent to a vector space then the base units are the basis vectors. It would be a mistake to refer to the zero vector as a basis vector, and I think it's a mistake to refer to 1 as a unit. – Nathaniel Sep 18 '13 at 1:45
@EmilioPisanty I can see that I had confounded the concept of "units" with "base unit" - I've added a clarification to my answer. – Nathaniel Sep 18 '13 at 1:50

Here's something from the top of my head:

Let us take a set $S$ of objects. For each $s$ in $S$ let us associate a field $F_s$ of real numbers with the usual binary operations and the associated unit elements for each operation (0 for additive operation and 1 for the multiplicative operation). We now have a set of fields. We denote an element of this set as $(x,s)$ where $x$ belongs to $\mathbb R$.

Let us add add some more structure to this in the form of the 'quantity multiplying' operator $D: F_s \times F_p \rightarrow F_c$, such that $D[(x,s),(y,p)]=(x \times y,s \otimes p)$ where s and p are elements of $S$.

The operation $\otimes$ involves multiplying the dimensions only and is an Abelian binary operation. $S$ thus forms a vector space over $\mathbb Q$ (integers).

For example: If you map $s=(\text{kg}^\alpha)(\text m^\beta)(\text s^\gamma)$ to the triplet of integers $(\alpha,\beta,\gamma)$, then $\otimes$ maps to addition of values in the triplet. This becomes a vector space over the rationals ($\mathbb Q$). Thus we can express elements of $S$ in terms of rational powers of the basis units, which are fundamental units.

Let the set of fields $F_s - \{(0,s)\}$ form an Abelian group with respect to $D$, with the unit element of this operation being field $F_0-\{(0,\text{unity})\}$. In the above example unity would be mapped to $(0,0,0)$.

I believe this $F_0$ encompasses all dimensionless quantities you are referring to.

EDIT: Emilio Pisanty helped me improve this answer a lot. I think I have given enough structure to generalize dimensions.

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How do you handle, though, conversions between feet and meters? – Emilio Pisanty Sep 18 '13 at 9:51
@EmilioPisanty Assuming I only included SI units, we have the relation (1,m)=(3.3,f) which establishes the dimensional equivalence of the 2. So I would not include it in the set. If I did include it, I would have to replace Fm with Ff. – dj_mummy Sep 18 '13 at 9:56
Yes, but the issue goes a bit deeper than that. Dimensional analysis is essentially doing linear algebra on a vector space over $\mathbb{Q}$ that has $\text {kg}^1$, $\text m^1$ and $\text s^1$ as a basis. This algebra is basis independent because you can take any three independent units as your basis. The complete structure must reflect this. – Emilio Pisanty Sep 18 '13 at 10:16
@EmilioPisanty Except that it is not a vector space. Multiplying of dimensions does form an Abelian group. However combining that with a real field does not work. For a simple reason: the 'vector addition' (multiplying units) does not satisfy the distributive property. It is not a linear algebra. – dj_mummy Sep 18 '13 at 10:24
It is, just not the one you are thinking of; see this WP article for details. If you map $\text{kg}^\alpha\,\text m^\beta\,\text s^\gamma$ to the triplet $(\alpha, \beta,\gamma)$, then addition is defined (multiplication of two dimensionful quantities) as well as multiplication with some rational $m/n$ (raising the quantity to the $m$th power and taking the $n$th root). With this structure the units do follow the axioms of a vector space, with scalars taken from $\mathbb{Q}$. – Emilio Pisanty Sep 18 '13 at 10:38