I saw this link and had a problem with the first proof on the accepted answer, namely:
$\newcommand{\R}{\mathbb{R}}$ $\newcommand{\raw}{\rightarrow}$ $\newcommand{\N}{\mathbb{N}}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\Raw}{\Rightarrow}$
Suppose that $f:\R\raw\R$ and define a function $g:\R\raw\R$ by $g(x)=1/f(x)$. Prove that $g$ has no roots.
I found this hard to make sense of. For certain functions, for example $f(x) = x^2$ it is obvious that $1/f(x)$ will have no roots, however I became confused when considering the case $f(x) = 1/x^2$, because then $g(x) = x^2$ which has a repeating root at $(0,0)$. Can anyone explain to me where I am making a mistake?
Best Answer
As stated, there is no answer to the proposed claim because the domain of any such $g$ must be a subset of $\Bbb R-\{0\}$ for it to be defined. Thus if $g$ is defined, the range of the function $f$ must be a subset of $\Bbb R - \{0\}$.
To prove that $g$ has no roots, we can proceed as follows.
If we write $y = f(x)$, then $g(x) = 1/f(x) = 1/y$. Now we claim that $1/y \ne 0$. If this were so, then we could multiply by $y^2$ on both sides of our equation to obtain $0 = y^2\cdot 0 = y^2\cdot 1/y = y$. However, this says that $y = 0$, which is impossible because $y$ belongs to the range of $f$. We conclude that $g$ can have no roots.
Also, you apparently have some misunderstanding of the definition of a function. For instance, the map $x\mapsto 1/x^2$ is not a function $\Bbb R\to \Bbb R$, since it is not defined for $x = 0$.