Question

Show that the function $f:R\to R,$ defined as$f\left(x\right)=x^2$, is neither one-one nor onto.

Answer 1

Solve for one-one:
Given: $f:R\to R$
$f\left(x\right)=x^2$
For, one-one
$f\left(x_1\right)=(x_1{)}^2$
$f\left(x_2\right)=(x_2{)}^2$
Now, $f\left(x_1\right)=f\left(x_2\right)$
$x_1^2=x_2^2$
$x_1=x_2$ or $x_1=-x_2$
i.e. For same image, elements are different.
Hence, it is not one-one.

Solve for onto.
$f:R\to R$
$f\left(x\right)=x^2$
Let $f\left(x\right)=y$, such that $y\in R$
$x^2=y$
$x=\pm \sqrt{y}$
Clearly, $y$ is a real number, so it can be negative also.
Which is not possible as root of negative number is not real.
Hence,$x$ is not real.
So, $f$ is not onto.
Hence, $f$ is neither one-one nor onto.