It is well known that $e^x \ne 0$ for all $x \in \mathbb{R}$ as well as $x \in \mathbb{C}$. Upon reading this article and doing a bit of research I have found that this also applies to the quaternions $\mathbb{H}$, the octonions $\mathbb{O}$ as well as the space of $m$ by $n$ matrices with real or complex entries.

My question is whether there is ANY number system at all for which $e^x = 0$ for some $x$, that is, $\log 0$ is defined and has a finite value. Preferably the example should be finite-dimensional and should not be constructed by arbitrarily assigning a value to $\log 0$, such as $\log 0 := 42.$

Additionally, for the purposes of this question, none of the usual properties of arithmetic or the exponential function are assumed true, though I suppose this makes my question somewhat meaningless.

Edit: I am intrigued at Yuriy S's idea of defining $e^x = 0$ for all $x$. My question now is what is the most "well behaved" algebra we can come up with if $e^x$ is required to be identically zero?

  • $\begingroup$ Your last paragraph says it all. If you want to get rid of some essential properties of the exponent, then tell us which properties you want to keep. Otherwise, I'll just define my own function $e^x=0$ for all $x$ and be done with it $\endgroup$
    – Yuriy S
    Commented Apr 12, 2016 at 13:44
  • $\begingroup$ @YuriyS What I want is to assign a finite value to $\log 0$ while also obeying as many "standard properties" as possible. I am sorry if this comes across as vague and unhelpful. $\endgroup$
    – cpiegore
    Commented Apr 13, 2016 at 20:53
  • $\begingroup$ cpiegore, see this: $$\log(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\cdots$$ Thus, we can (not rigorously) write $$\log(0)=-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\cdots$$ which is the usual Harmonic series. They diverge, but there are ways to assign a finite sum for them: see here $\endgroup$
    – Yuriy S
    Commented Apr 13, 2016 at 21:00

3 Answers 3


If you want the properties $$e^{x+y}=e^xe^y\quad\hbox{for all $x,y$}$$ and $$e^0=1$$ to remain true, then we have $$e^xe^{-x}=1\quad\hbox{for all $x$}$$ and so $e^x$ can never be zero.

  • 1
    $\begingroup$ What about in the case of the extended real number line? $\endgroup$
    – LuuBluum
    Commented Mar 21, 2016 at 23:00
  • 8
    $\begingroup$ Note that $e^{x+y} = e^x e^y$ is not true in non-commutative algebras such as the quaternions, although $e^x e^{-x} = 1$ is true. $\endgroup$ Commented Mar 21, 2016 at 23:08
  • $\begingroup$ @Ispil: The OP is asking for a (finite-dimensional) real algebra, which the extended reals aren't. $\endgroup$ Commented Mar 21, 2016 at 23:09
  • $\begingroup$ @DanielMcLaury, where does the OP say it should be a real algebra? $\endgroup$
    – Yuriy S
    Commented Apr 12, 2016 at 17:13

Adding to what David said, even if you give up on the homomorphism part of the exponential, the series definition (which holds in any Banach algebra, which includes the number systems) make so that $e^X e^{-X}=1$, so $e^X$ can't be $0$.


The floating-point number system (for a given number of bits) is a finite subset of the extended real number line. It relaxes various algebraic identities so that it can remain closed under as many operations and inputs as possible. As you put it, the "usual properties of arithmetic or the exponential function" are not true in this system, although they are approximately true for the most part.

In this system, $-\infty$ is a number and $e^{-\infty}=0$.



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