Prove that a function $g$ has no roots

Solution 1:

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 $x = 0$.

Solution 2:

Your issue is a misunderstanding of the notation $f:\mathbb{R}\to \mathbb{R}$.

In modern math, the notation $f:A \to B\;$means that $f$ is function whose domain is exactly equal to $A$, and whose range (or image) is a subset of $B$. In particular, if we have $f:A \to B$, then $f(x)$ is defined for all $x \in A$.

Thus, the notation $f:\mathbb{R}\to \mathbb{R}$ means that $f$ is a real-valued function such that $f(x)$ is defined, for all $x \in \mathbb{R}$.

Now suppose $f:\mathbb{R}\to \mathbb{R}$. If $f$ has a real root, $r$ say, there is no function $g:\mathbb{R}\to \mathbb{R}$ such that $g(x) = {\large{\frac{1}{f(x)}}}$, else $g(r)$ would be undefined. Based on that observation, the problem itself needs to be reworded so as to assert that $g(x)$ exists. So let's assume that the intended problem was this:

• Prove that if $f:\mathbb{R}\to \mathbb{R}$ and $g:\mathbb{R}\to \mathbb{R}$ are such that $g(x) = {\large{\frac{1}{f(x)}}}$, for all $x \in \mathbb{R}$, then $f,g$ have no real roots.

The proof is easy . . .

Let $r \in \mathbb{R}$.

If $r$ is a root of $f$, then $$g(r) = \frac{1}{f(r)} = \frac{1}{0}$$ which is undefined, contrary to the assumption that $g:\mathbb{R}\to \mathbb{R}$.

Hence $r$ is not a root of $f$.

If $r$ is a root of $g$, then \begin{align*} &g(r)=0\\[4pt] \implies\;&\frac{1}{f(r)}=0\\[4pt] \implies\;&f(r)\left(\frac{1}{f(r)}\right)=f(r)(0)\\[4pt] \implies\;&\require{cancel} \cancel{f(r)} \left( {\small{\frac{1}{\cancel{f(r)}}}} \right)=f(r)(0)\qquad\text{[since $f(r) \ne 0$]}\\[4pt] \implies\;&1=0\\[4pt] \end{align*} contradiction.

Hence $r$ is not a root of $g$.

It follows that $f,g$ have no real roots.

Solution 3:

As $g(x):=1/f(x),$ we can write $f(x)=1/g(x)$, which would cause $f$ to be undefined at the roots of $g$.