Suppose f1 and f2 are non-zero one-one functions from R to R.
Question:

Suppose $f_{1}$ and $f_{2}$ are non-zero one-one functions from $R$ to $R$. Is $\frac{f_{1}}{f_{2}}$ necessarily one-one? Justify your answer. Here, $\frac{f_{1}}{f_{2}}: R \rightarrow R$ is given by $\left(\frac{f_{1}}{f_{2}}\right)(x)=\frac{f_{1}(x)}{f_{2}(x)}$ for all $x \in R$.

Solution:

We know that $f_{1}: R \rightarrow R$, given by $f_{1}(x)=x^{3}$ and $f_{2}(x)=x$ are one-one.

Injectivity of $f_{1}$ :

Let x and y be two elements in the domain R, such that

$f_{1}(x)=f_{2}(y)$

$\Rightarrow x^{3}=y$

$\Rightarrow x=\sqrt[3]{y} \in R$

So, $f_{1}$ is one-one.

Injectivity of $f_{2}$

Let x and y be two elements in the domain R, such that

$f_{2}(x)=f_{2}(y)$

$\Rightarrow x=y$

$\Rightarrow x \in R$

So, $f_{2}$ is one-one.

Proving $\frac{f_{1}}{f_{2}}$ is not one-one:

Given that $\frac{f_{1}}{f_{2}}(x)=\frac{f_{1}(x)}{f_{2}(x)}=\frac{x^{3}}{x}=x^{2}$

Let x and y be two elements in the domain R, such that

$\frac{f_{1}}{f_{2}}(x)=\frac{f_{1}}{f_{2}}(y)$

$\Rightarrow x^{2}=y^{2}$

$\Rightarrow x=\pm y$

So, $\frac{f_{1}}{f_{2}}$ is not one-one.