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 Apr 12 '16 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 Apr 13 '16 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 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.

| cite | improve this answer | |
  • 1
    $\begingroup$ What about in the case of the extended real number line? $\endgroup$ – LuuBluum Mar 21 '16 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. $\endgroup$ – Robert Israel Mar 21 '16 at 23:08
  • $\begingroup$ @Ispil: The OP is asking for a (finite-dimensional) real algebra, which the extended reals aren't. $\endgroup$ – Daniel McLaury Mar 21 '16 at 23:09
  • $\begingroup$ @DanielMcLaury, where does the OP say it should be a real algebra? $\endgroup$ – Yuriy S Apr 12 '16 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 $0$.

| cite | improve this answer | |

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 $e^{-\infty}=0$.


| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.