I was thinking about the likelihood ratio test and it seemed to me that both positive and negative argument should be preferred instead of neither positive nor negative when we define the sign of zero.

Let me explain what led me to this conclusion. Assume we have apples and pears in a box and they are transmitted from a transmitter to the receiver one by one via an unfriendly transporter. As a results an apple might be undergone a heavy damage during the travel and might seem like a pear or vice verse, i.e., a pear might look like an apple.

our aim at the receiver side is to decide if an apple was sent or a pear. To do this we have a deformaton profile both for an apple and a pear. Those are the probability density functions $f_{apple}$ and $f_{pear}$. Whenever we receive a deformed fruit, say $y$, which is known to us previously that it is either apple or pear, we give a decision based on our models, namely $f_{apple}$ or $f_{pear}$ as follows

$$l(y)=\frac{f_{apple}(y)}{f_{pear}(y)} \stackrel{\mbox{apple}}{\underset{\mbox{pear}}{\gtrless}} 1$$

In words, if the probability that $y$ comes from our model $f_{apple}$ is greater than the probability that it comes from $f_{pear}$, we decide for $f_{apple}$, else for $f_{pear}$.

When we take the $\log$ of both sides we get

$$\log l(y)=\log \frac{f_{apple}(y)}{f_{pear}(y)} \stackrel{\mbox{apple}}{\underset{\mbox{pear}}{\gtrless}} 0$$

Now we are interested in what happens if $f_{apple}(y)=f_{pear}(y)$? definitely we can not produce any useful decision. But we can say fifty-fifty, namely $50\%$ it is apple or pear.

In this problem decision for an apple means a sign of positive and a decision of a pear means a sign of negative. When we are at point $0$, we see that the decision involves $50\%$ apple and $50\%$ pear, as a result one would not expect that $y$ is a banana. It is either an apple or a pear. As a result at point $0$ we have both an apple and a pear with a $0.5$ probability. This suggests that $0$ is both positive and negative which is also in agreement with the use of signed zeros..

My question is: is there any example which could indicate that the pair neither-nor is more suitable for zero? or are there more examples supporting my ideas above?

Thanks for reading this post and for any nice comment, even for not nice ones.

  • $\begingroup$ If $0$ is both positive and negative, how do you define $\operatorname{sgn}0$? $\endgroup$ – yo' Nov 18 '13 at 16:41
  • $\begingroup$ @tohecz I cannot define but $0$ can still be both positive and negative. I mean simultaneously positive and negative, not either of them at a time. $\endgroup$ – Seyhmus Güngören Nov 18 '13 at 16:46
  • $\begingroup$ I mean examples like this: "The derivative of $x\mapsto |x|$ is $+1$ for positive $x$ and $-1$ for negative $x$. $\endgroup$ – yo' Nov 18 '13 at 16:49
  • $\begingroup$ @tohecz i see your point. Thanks for the comment. $\endgroup$ – Seyhmus Güngören Nov 19 '13 at 15:21

Zero has no "sign". We have the positive numbers, $x>0$, the negative numbers, $x<0$, and the zero. That is, zero is a breaking point, so to speak.

To dwell into more details, suppose you have a field $F$, that is, a set $F$ with operations of addition and multiplication, a $1$ and a $0$, $1\neq 0$, and suppose moreover we have a subset $P$ of that field, called the "positive numbers" (or positive cone) with the following properties:

$(1)$ Whenever $x,y\in P$, $xy,x+y\in P$,

$(2)$ For any $x\in F$; either $x\in P$, or $-x\in P$, and

$(3)$ $0\notin P$.

We ask, moreover, that $F=\{0\}\sqcup P\sqcup P'$ where $P'=-P=\{-x;x\in P\}$.

Then we can define an order in $F$ by stating $x>y$ iff $x-y\in P$. We thus say $x>0$ when $x\in P$, that is $x$ is "positive". Since $0$ is not in $P$ (the positive numbers) and neither is in $P'$ (the negative numbers), we cannot speak of $0$ "having a sign."

  • $\begingroup$ so you think neither-nor is more suitable? $\endgroup$ – Seyhmus Güngören Dec 23 '13 at 21:33
  • $\begingroup$ @SeyhmusGüngören I haven't read your question when I answered this, merely the title. I would say zero is nor positive nor negative, i.e. it has no "sign." $\endgroup$ – Pedro Tamaroff Dec 23 '13 at 21:35

In some situations, we might use the number 0 as a shorthand for a very small quantity $x$ appoaching to zero. In this case, the size of $x$ may not be so relevant, but its sign may be. For example, $1/x \approx +\infty$ for $x$ small positive and $1/x \approx -\infty$ for $x$ small negative.

In these cases, it's helpful to distinguish between positive zero and negative zero. (IEEE floating-point math makes this distinction). We might say $1/+0 = \infty$ and $1/-0 = -\infty$.


At one time, the Bourbaki school considered 0 as both positive and negative but this is not a common position today. Today, it is more common to regard it as neither positive nor negative.

I found an interesting footnote in this book: The Number System by H.A.Thurston at the bottom of page 15.


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.