# Why the function 1/x does not increase between (-infinity,0) and (0,+infinity)?

I know that the domain is: $(-\infty,0) \cup (0,\infty)$, or all Reals except Zero.

But if a take the "nearest negative number to zero" and then, the "nearest positive number to zero", my function will hugely increase (Y will approach +infinity).

And my domain have all numbers that approach Zero in both sides, except Zero.

So, why the function 1/x does not increase between $(-\infty,0)$ and $(0,\infty)$? (Sorry for any english error)

• The function does increase as you have pointed out, but it is never an increasing function on any interval. – Matt Samuel May 21 '17 at 22:28
• A graph would probably clear things up. As you approach $0$ from the left, $1/x \to -\infty$ by the way. It approaches positive infinity as you tend to 0 from the right. – Nap D. Lover May 21 '17 at 22:29
• There is no nearest negative or positive number to $0$ by the way, as I'm sure you know, and reasoning along those lines is bound to lead you to trouble. – Matt Samuel May 21 '17 at 22:29
• It sounds like you're trying to make sense of a claim made by someone else. If so, could you please quote that claim? – Chris Culter May 21 '17 at 22:49

## 2 Answers

The question is how you define increasing function ? Suppose we define increasing function as one for which $a,b \in \text{dom}(f)$ with $a\leq b$ implies $f(a)\leq f(b)$. Then, for $f(x) = 1/x$ for $x \in \mathbb{R} - \{ 0 \} = (-\infty,0) \cup (0, \infty)$ we certainly cannot claim $f$ is increasing. Consider, $-1<1$ and $-1,1 \in \text{dom}(f)$ yet $f(-1) = -1 \nleq 1 = f(1)$.

On the other hand, we can also define $f$ to be increasing on $U \subseteq \text{dom}(f)$ if whenever $a,b \in U$ and $a \leq b$ we obtain $f(a) \leq f(b)$. In this terminology, a function is increasing iff it is increasing on its domain.

I think where students get confused is in calculus where we talk about derivatives and their connection to increase and decrease of a function. That local idea is really played out on a small neighborhood of the point. If we discard the role of the subset in the discussion of increase and decrease then it leads to the sort of confusion that is seen in your question.

To avoid pathological things from a geometric perspective, focus your attention to intervals. An interval is a connected subset of real numbers.

• +1 for "I think where students get confused is...where we talk about derivatives....[The] idea is really played out on a small neighborhood of the point" and giving corresponding def'ns. – Jacob Manaker May 22 '17 at 0:12

There are two phenomena occurring here. This first is that, in an informal sense, the region where $x\mapsto\frac{1}{x}$ is increasing is also where it is infinite, so we traditionally exclude that from the domain. The second is that this region has all been squashed together, so that $x\mapsto\frac{1}{x}$ has no time to increase.

To see this, consider $f:\mathbb{R}\setminus\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\to\tilde{\mathbb{R}}$; $$f(x)=\begin{cases} \frac{1}{x+\frac{\pi}{2}} & x\leq-\frac{\pi}{2} \\ \frac{1}{x-\frac{\pi}{2}} & \frac{\pi}{2}\leq x \end{cases}$$ (I'm using $\tilde{\mathbb{R}}$ to denote the extended reals, so that $f\left(\pm\frac{\pi}{2}\right)=\pm\infty$.)   Both $\tilde{\mathbb{R}}$ and $\mathbb{R}\setminus\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ are still ordered sets under the standard ordering inherited from $\mathbb{R}$, so it makes sense to ask whether $f$ is increasing.   And, lo-and-behold, this function does act the way you want it to: it decreases on $(-\infty,-\frac{\pi}{2}]\cup[\frac{\pi}{2},\infty)$ and increases on $\left\{-\frac{\pi}{2},\frac{\pi}{2}\right\}=[-\frac{\pi}{2},\frac{\pi}{2}]\cap\left(\mathbb{R}\setminus\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right)$.

(Note that I've done two things here:

1. I've passed to the extended reals, and

2. I've shifted apart the two "pieces" of $x\mapsto\frac{1}{x}$.)

What is the problem with (1.) above?   At two points, $f$ is infinite.   Now, $x\mapsto\frac{1}{x}$ is normally not considered on the extended reals (for a reason I will get to below), so we obtain a closer analogy by taking $f$ on everywhere where it is finite.   The region wherein $f$ is finite is, of course, $\left(\mathbb{R}\setminus\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right)\setminus\left\{-\frac{\pi}{2},\frac{\pi}{2}\right\}$.   But then we've created a problem: $f$ increases on the very small (two-point!) region $\left\{-\frac{\pi}{2},\frac{\pi}{2}\right\}$, and by cutting out the infinite points, we also just lost the section where $f$ is increasing.   So one (trivial) reason why $x\mapsto\frac{1}{x}$ does not increase between your two regions is that you have purposefully cut out the regions on where it is increasing, because those point(s) are infinite.

Now, what is the problem with (2.)?   But what if we left in the infinite parts, and pulled the two pieces together?   Then $f(0)=-\infty$, because the point in the "left" part with rightmost abcissa has ordinate $-\infty$.   But we also have $f(0)=\infty$, because the point in the "right" branch with leftmost abcissa has ordinate $\infty$.   If $f$ is to be a function (even one into $\tilde{\mathbb{R}}$), $f$ needs to have one value at $0$, not two! We can't have both, can we? Actually, we can, iff $-\infty=\infty$, and that definition is one convention for infinities used to extend $x\mapsto\frac{1}{x}$ to $0$.   If we do that, then we can't order the extended real line, because then \begin{align*} -1&<0<1<\cdots\\ &<1000<\cdots\\ &<\infty=-\infty<\cdots\\ &<-1000<\cdots\\ &<-1\end{align*} So the only way to join our two pieces of $x\mapsto\frac{1}{x}$ involves losing the notion of "increasing" we wanted in the first place.