Show that if f1 and f2 are one-one maps from R to R, then the product

Solution:

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

Proving $f_{1}$ is one-one:

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

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

$\Rightarrow x=y$

Proving $f_{2}$ is one-one:

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

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

$\Rightarrow x=y$

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

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

Given:

$\left(f_{1} \times f_{2}\right)(x)=f_{1}(x) \times f_{2}(x)=x \times x=x^{2}$

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

$\left(f_{1} \times f_{2}\right)(x)=\left(f_{1} \times f_{2}\right)(y)$

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

$\Rightarrow x=\pm y$

So, $\left(f_{1} \times f_{2}\right)$ is not one-one.