As far as I know, the statement “$a \propto b$” is equivalent to “$a = kb$ for an arbitrary value $k$”. Is there anything wrong with the following?

$$ \begin{align} f(x) &\propto 2\,f(x) \\ 6 &\propto \pi \\ \begin{bmatrix} -5 \\ 10 \end{bmatrix} &\propto \begin{bmatrix} 1 \\ -2 \end{bmatrix} \\ A\mathbf{x} &\propto \mathbf{x} \end{align} $$

I want to use this in an informal setting and am wondering if it complete nonsense. Has this notation been (ab)used like this anywhere? Should I first clarify that “$a \propto b \Longleftrightarrow a = kb$ for an arbitrary $k$” at the top of the page?

(Other notations like “$a$ is a multiple of $b$ $\Longleftrightarrow$ $a|b$”, are too restrictive for my intended use because they sort-of imply integer multiplicity and can’t be chained together nicely, like $2(a, -a) = (2a, -2a) \propto (1, -1)$ for example.)

Thanks for your input.

Edit: I was asked to give some context: I’m just a student hand-writing his linear algebra homework. I come across myself wanting to emphasise that a vector is a scalar multiple of another in a string of equalities without having to write “where $a, a’ ∈ ℝ$” everywhere or writing words. Snippet: $$ \mathbf{q}_2 = \mathbf{v}_2 - \text{Proj}_\mathbf{q1}\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} + \frac{1}{5}\begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} \frac{6}{5} \\ -\frac{3}{5} \end{bmatrix} \propto \begin{bmatrix} \frac{2}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} \end{bmatrix} = \mathbf{u_2} $$ I realise it’s laziness, but I got curious. I thought it was pretty.

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    $\begingroup$ My understanding of $\propto$ is that we say $f\propto g$ if $f$ and $g$ are in some vector space, and $f=\lambda g$ for some scalar $\lambda$. Often the vector space will be taken to be some sort of function space. $\endgroup$ – Monstrous Moonshine May 27 '17 at 11:02
  • $\begingroup$ @MonstrousMoonshine $6 \propto \pi$ remains a valid statement with that convention for using the proportional symbol, as $\mathbb{R}$ IS a vector space over $\mathbb{R}$ and I'm not sure I like that. $\endgroup$ – Guy May 27 '17 at 11:07
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    $\begingroup$ If you edit your question to show us the context where you want this new notation perhaps someone will suggest a good alternative. Let us see several sentences. $\endgroup$ – Ethan Bolker May 27 '17 at 12:03
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    $\begingroup$ Handwritten notes are good place to be lazy and this is a clever place to reuse the symbol. In this particular example the proportionality constant does the normalization. Your notation loses that information. If you want to use it in homework you should check informally with your instructor or TA first ... $\endgroup$ – Ethan Bolker May 27 '17 at 13:16
  • $\begingroup$ I noticed that someone has downvoted on my post recently (on the same day as today). Please explain how I can improve and how I should do this. $\endgroup$ – Toby Mak May 29 '17 at 0:00

This usage of proportionality is unusual. Two "proportional" vectors are also parallel, hence the parallel symbol is more suitable for vectors $$ \begin{bmatrix} -5 \\ 10 \end{bmatrix} \parallel \begin{bmatrix} 1 \\ -2 \end{bmatrix} $$ $$ A\mathbf{x} \parallel \mathbf{x} $$ Functions are essentially vectors so $f \parallel g$ is also suitable.

  • $\begingroup$ This slipped my mind. This is a better notation. $\endgroup$ – Stefan Perko May 27 '17 at 13:02
  • $\begingroup$ I think the symbol for parallel is "//" instead of "||", i.e AB // CD. Anyway, this is a good idea because it emphasises the property that proportional vectors are parallel, which is not often shown this clearly because of notation. $\endgroup$ – Toby Mak May 27 '17 at 13:03
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    $\begingroup$ @TobyMak If you write $\parallel$, what you get is $\parallel$. While $\LaTeX$ commands are not an authorative source on the conventions of math symbols and naming (especially since there may be differences between geographic regions and mathematical fields) I think it safe to say it's not taken from nothing. $\parallel$ is also the symbol I learned in school for "is parallel to". $\endgroup$ – Arthur May 27 '17 at 13:54
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    $\begingroup$ @TobyMak I've seen $\parallel$ every time I've seen "parallel (to)" written symbolically, but I've never seen "$//$" used for the same purpose. And for what it's worth: en.wikipedia.org/wiki/Parallel_(geometry)#Symbol $\endgroup$ – M. Vinay May 27 '17 at 14:01
  • $\begingroup$ I just realised that in many competition problems, // is used because || cannot be typed. When I checked my maths textbook, it's actually ||! Sorry for the misunderstanding. $\endgroup$ – Toby Mak May 27 '17 at 23:52

To give a different viewpoint, there is actually a standard set which is the set of vectors up to scaling, and it is called $\mathbb{R}\mathrm{P}^n$ or $\mathbb{C}\mathrm{P}^n$, depending on the underlying scalar field. They are pronounced "real projective space" and "complex projective space," respectively.

Since $\mathbb{R}\mathrm{P}^n$ is defined as a set of equivalence classes, where two nonzero vectors are in the same equivalence class if and only if they are scale multiples of each other, it would be fine to use the default equivalence symbol $\sim$, so long as you define what you mean. For example, \begin{equation} \begin{pmatrix}1\\2\end{pmatrix} \sim \begin{pmatrix}2\\4\end{pmatrix}. \end{equation} (The thing I don't particularly like about $\propto$ is that it seems like a non-symmetric relation, though it is a clever use of the symbol, and if you define what you mean it is perfectly OK. In fact, you might say $v\propto w$ iff $v=cw$ for some $c\in\mathbb{R}$, and then you are allowed to say $\mathbf{0}\propto \mathbf{x}$. This is in contrast to $\mathbb{R}\mathrm{P}^n$ where we do not allow comparison with the zero vector.)


There is nothing wrong with using the notation like this since it is an actual generalization of the common usage of $\propto$. So if you find it very useful, why not? You are just giving a name to a relation (the proper definition was given by Monstrous Moonshine in the comments).

But, yes, you should most likely introduce the notation first before using it since it is not well known in this generalized version.

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    $\begingroup$ I think there is a lot wrong with this usage, and hence with this answer. The proportionality symbol has a well established meaning that does not apply in these examples. If the OP uses it this way readers will be very confused. $\endgroup$ – Ethan Bolker May 27 '17 at 11:56
  • $\begingroup$ @EthanBolker what, why?? You put the explanation at the beginning of the page and that's it; it is also completely unambigous. According to your principles you should never use notation in context it hasn't been used before. $\endgroup$ – Stefan Perko May 27 '17 at 12:11
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    $\begingroup$ No. You can and should invent notation that makes your mathematics easier to read. But you shouldn't change the meaning of well established notation "at the top of the page". You can sometimes usefully generalize, and say that's what you're doing, but in the particular examples in the question that would not help the reader. $\endgroup$ – Ethan Bolker May 27 '17 at 12:33
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    $\begingroup$ In full generality it's expanding the context so far that it's useless: why would you want to say "$17 \alpha 55.5$" ? The last two of the OP's examples would be OK with me. That's why I posted a comment asking him to provide context. I think we have to agree to disagree. $\endgroup$ – Ethan Bolker May 27 '17 at 12:43
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    $\begingroup$ Perhaps so. But I think you misunderstand me: I don't want so say $17\propto 55.5$; that's surely not the motivation to introduce this kind of notation. It's for the other examples mentoined by the OP. $\endgroup$ – Stefan Perko May 27 '17 at 12:59

Proportionality as described by Wikipedia is described as the relationship between two variables, such that “$a∝b$ is equivalent to $a=kb$" as you have described.

Therefore, none of the examples are correct examples of proportionality, since (1) compares two functions, (2) compares two numbers, (3) compares two vectors, and (4) uses only one variable.

The correct way to describes these relationships would be:

(1): $g(x) = 2f(x)$

(2): $6x = \pi$

$$ A\begin{bmatrix} -5 \\ 10 \\ \end{bmatrix} $$

$$ =\begin{bmatrix} 1 \\ -2 \\ \end{bmatrix} $$

(4): $Bx = y$

In summary, proportionality describes the relationships between two variables, so that one can be "scaled up/down" to another. In many situations it is often best to define one function in terms of another by using a constant multiplier and setting that equal to the function.

EDIT: In the context the OP has put, my opinion would still be to not use the proportionality symbol. I think that sacrificing context and quality for an abbreviation hides the true meaning of proportion, and by writing $a,a′∈ℝ$ the context for what numbers are used becomes clearer.

I think that outside of its definition of describing the relationship between two variables, an equality symbol with a constant multiplier is much more useful (it can be widely applied) and is clear-cut and well-defined.

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    $\begingroup$ I believe this is nonsense (on Wikipedia). What they really mean are functions, not variables. Or what they call a "variable" has nothing to do with what mathematicians or more specifically logicians call a "variable" (which is a meta-mathematical concept anyway). $\endgroup$ – Stefan Perko May 27 '17 at 11:42
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    $\begingroup$ @StefanPerko I disagree. Wikipedia is correct. The "variables" in that article and in this answer are easily interpreted as functions on some underlying implicit space of states of the system being studied. You don't need mathematical logic or serious meta-mathematics to make sense of them. $\endgroup$ – Ethan Bolker May 27 '17 at 12:00
  • $\begingroup$ @Stephan Perko I used a simplified version of the concept of proportionality (which is around IGCSE level). For many people like me, axioms and logic are not famliar concepts, so I don't have a good understanding of this. How would you define proportionality and what statements do you consider correct? $\endgroup$ – Toby Mak May 27 '17 at 12:04
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    $\begingroup$ @EthanBolker If it is so easy to "interpret" them as functions, then why not just say they are functions? Because they are. The way this answer is written is now is contradictory, because, when I replace "variables" by "functions", it both confirms and denies that you use "$\propto$" in connection with functions. $\endgroup$ – Stefan Perko May 27 '17 at 12:13
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    $\begingroup$ @TobyMak The wikipedia definition and your understanding both work well in this context. You needn't worry about "axioms and logic". But I would allow the proportionality symbol for the third example in the question. That's what MonstrousMoonshine's comment says. $\endgroup$ – Ethan Bolker May 27 '17 at 12:47

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