When we create numerical models of a real-world system, we usually go through a few phases (from Abramowitz 2010):

  • a perceptual model, where we consider the relevant components of the system
  • a conceptual model, where we consider the relationships between those components
  • a mathematical model, where we come up with equations and formulae to express the conceptual model
  • a numerical model, that allows the computer to estimate the results of the mathematical model on a given spatiotemporal scale

Divide by zero errors are a pretty common cause of premature model death. Sometimes that's just due to poor programming. But if we exclude that possibility (assume a perfect programmer), is a divide by zero error an indication that the conceptual model is wrong? That is, is it reasonable to assume that the equivalent of a divide-by-zero error never happens in the real world? I guess that this question presupposes that mathematics is in some way "real", comments on the relevance of that would also be appreciated.


  • $\begingroup$ I was not the downvoter, but indeed is not clear what your question is. If your model is of the form $1/x$, every time the user input 0 you will get a divide by zero. If it makes sense that the value $x$ is zero then I would assume that there are some conceptual flaws. If not, then I think is just a bad use. $\endgroup$
    – nicoguaro
    Mar 17, 2016 at 0:23
  • $\begingroup$ @nicoguaro: Yes, I would have thought it was fairly obvious from the phrase "numerical models of a real-world system" that the variables in the model should represent something somehow "real" (even if it's an simplified empirical parameterisation of a more complex process). In other words, using your example, "is it ever reasonable to have a model that represents a real system, and that includes $1/x$ if $x$ is expected to sometimes equal zero". $\endgroup$
    – naught101
    Mar 17, 2016 at 1:00

1 Answer 1


We should keep in mind that models are just representations of a portion of reality (a narrow portion), therefore a divide by zero error or other mathematical error (negative concentration, for instance) is not necessarily a flaw in the conceptual model, only that the model is used beyond its proper range of application.

For example, let's consider a model for the growth of Saccharomyces cerevisiae, based on Monod equation at aerobic conditions (consuming glucose and oxygen). I make the approximation that there is a lot of glucose so that it is not limiting and the equations become easier.

My conceptual model would be: "biomass grows by using oxygen from the medium, besides, I know it will take other compounds such as glucose, ammonium, phosphates, and metals". The mass balance of the process in a batch reactor can be written as:

$$ \frac{dX}{dt} =Y_{X/S} \; \mu_{MAX}\frac{C_O}{K+C_O}X $$ $$ \frac{dC_O}{dt} =-\mu_{MAX}\frac{C_O}{K+C_O}X $$

However, I know that Saccharomyces cerevisiae also consumes ammonium as a source of nitrogen. If I want to follow the ammonium concentration in the process, I must write: $$ \frac{dC_{NH}}{dt} =-Y_{NH/S} \; \mu_{MAX}\frac{C_O}{K+C_O}X $$

The problem with this equation is that if ammonium is limiting, the concentration can go below zero and become negative as there is nothing in the Monod equation taking into account this limitation.

My point is that the conceptual model is not wrong, but its mathematical realisation changes depending on the range of application. This is a trivial case where the range of application is very easy to see as it depends on deciding upon which are the limiting compounds, but in other models, knowing whether the mathematical realisation follows the conceptual model can be much more complex.

To sum up, a divide by zero is not necessarily a conceptual model issue. It can be that you are applying your model where it is not supposed to hold.

  • $\begingroup$ Which could also be considered a conceptual issue - i.e. the model is incorrectly specified for the application domain. But yeah, great point, and probably relevant to my cases. Thanks! Also, great first answer! $\endgroup$
    – naught101
    Mar 16, 2016 at 23:42

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