# Number system with $e^x = 0$ for some $x$

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?

• 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 – Yuriy S Apr 12 '16 at 13:44
• @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. – cpiegore Apr 13 '16 at 20:53
• 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 – Yuriy S Apr 13 '16 at 21:00

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.
• 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. – Robert Israel Mar 21 '16 at 23:08
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 $0$.
In this system, $-\infty$ is a number and $e^{-\infty}=0$.