Is $1/x$ strictly decreasing?

Question

Is $1/x$ a strictly decreasing function over all $\mathbb{R}$? The question arises because when we differentiate$1/x$ we get $-1/x^2$ (all calculations exclude $0$), which is always negative and thus indicates strictly decreasing. But if we look at the graph then we see an anomaly at $0$. So, what should be the correct answer?

Answer 1

To answer your question, we need to find a definition of "strictly decreasing" that is precise enough to settle any disputes. Here I paraphrase Spivak's Calculus:

A function $f$ is [strictly*] decreasing on an interval if $f(y)<f(x)$ for all $x$ and $y$ in the interval with $y>x$. (We often say simply that $f$ is decreasing, in which case the interval is understood to be the domain of $f$.)

Immediately we run into trouble because $1/x$ is undefined when $x=0$, meaning that the domain of the function $f(x)=1/x$ is $\Bbb{R}\setminus\{0\}$, which is not an interval.** So according to Spivak, the "monotonicity" of $f$ itself is undefined. A similarly restrictive definition is given in Baby Rudin.

However, in Tao's Analysis I, the definition of "strictly decreasing" is stated more carefully:

Let $X$ be a subset of $\Bbb{R}$, and let $f:X\mapsto\Bbb{R}$ be a function. We say that $f$ is strictly decreasing iff $f(y)<f(x)$ whenever $x,y\in X$ and $y>x$.

According to this definition, $f$ is not strictly decreasing. Take $x=-1$ and $y=1$, for instance. We can however say that $f$ is strictly decreasing on each connected part of its domain, which is likely what people mean when they say $f$ is strictly decreasing.

The good news is that both of the above definitions agree that $f$ is strictly decreasing on $(-\infty,0)$, and on $(0,\infty)$. Here is the proof that $0<x<y\implies 1/y<1/x$ given in Baby Rudin, based upon the field axioms:

If $y>0$ and $v\le0$, then $yv\le0$. But $y \cdot 1/y = 1>0$. Hence, $1/y>0$. Likewise, $1/x>0$. If we multiply both sides of the inequality $x<y$ by the positive quantity $(1/x)(1/y)$, we obtain $1/y<1/x$.

The reader is invited to fill in the details of how to prove $f$ is strictly decreasing on $(-\infty,0)$.

---

*Some authors use "strictly decreasing" to mean $y>x \implies f(y)<f(x)$ and "decreasing" to mean $y>x\implies f(y)\color{red}{\le}f(x)$. Others, including Spivak, use "decreasing" and "non-increasing" respectively.

**Can we agree on the definition of an interval, please?

Answer 2

It is for every value in the domain. Notice, $0$ is not in the domain of neither $f(x)=\frac{1}{x}$ nor $f'(x)= \frac{1}{x^2}$. Looking at the plot of $f(x)$, we see on the negative side of the domain, the function is always decreasing, and on the positive side, the function is always decreasing. The function is neither increasing nor decreasing at $x=0$ as it is not even defined there. The function nor its derivative 'see' this value, they simply see the other values where, as noted, they see the function decreasing.

Answer 3

We can define: $f$ is strictly decreasng on a set $A$, where $A$ is a subset of the domain of $f$. In your case, $1/x$ is strictly decreasing on $(0,+\infty)$ or any subset of it; and $1/x$ is strictly decreasing on $(-\infty,0)$ or any subset of that. But $1/x$ is not strictly decreasing on the set $\{-2,7\}$, for example.

The criterion with $f'(x) < 0$ can be used to prove $f$ is strictly decreasing on an interval. But it cannot be used in the $\{-2,7\}$ example.