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Let's take what I believe to be the typical representation employed for notion of direct proportionality: y = k * x

My understanding is the 'k' represents a constant while 'y' and 'x' represent mathematical objects that have a multiplication operation defined ( here denoted by *). But what if the the set of mathematical objects that you are dealing with does not have an identity element defined under its multiplication operation. I believe that if this is the case, then there also exist no inverse operation for multiplication under the set. With no inverse operation (like Division, denoted as '/'), it seems that we are unable to rewrite the typical form for direct proportionality as y/x = k.

So with all that said, is it still possible to have a proportional relationship between the mathematical objects 'x' and 'y' with no inverse operation defined for multiplication under its set? It seems like the notion of ratios and proportional relationships can still persist even when you are unable to represent them in fractional like forms, but given that mathematics is not a strong point of mine, I wanted to reach out to the community.

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Yes, it is possible to have that kind of relations without the presence of an inverse element. Take for instance the case of matrices and consider the following:

$$A=\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} =\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1& 0 \\ 1 & 0 \end{pmatrix} = B \cdot C$$

But B and C are singular, so you can't consider their inverse. Here you don't make use of the unitary element of matrices either.

As you said, what you can't do is to write this as $A \cdot C^{-1} = B$ or $B^{-1} \cdot A = C$

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  • $\begingroup$ Hi @Adolfo, thanks for responding. I just got finished grilling for about 6 hours so I have not had a chance to research your answer. I say research because it has been about 20 years since I have dealt with matrices and so I need to take a little time to familiarize myself with some of the concepts you have discussed. But I wanted to thank you for taking this to respond. $\endgroup$ – Alonzo Archer Sep 7 '15 at 2:34

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