Question

Check the injectivity and surjectivity of the function $f:N\to N$ given by $f\left(x\right)=x^2$

Answer 1

$f\left(x\right)=x^2$

To check the injectivity:

$f\left(x_1\right)=x_1^2$

$f\left(x_2\right)=x_2^2$

Putting $f\left(x_1\right)=f\left(x_1\right)$

$\Rightarrow x_1^2=x_2^2$

$\Rightarrow x_1=x_2$ or $x_1=-x_2$

Since $x_1$ and $x_2$ are natural numbers, they are always positive.

Hence,$x_1=x_2$

hence, it is one-one(injective)

$f\left(x\right)=x^2$

Let $f\left(x\right)=y$ such that $y\in N$

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

Put $y=2$

$\Rightarrow x=\sqrt{2}=1.41$

Since $x$ is not a natural number.

Given function $f$ is not onto.

So, $f$ is not onto(not surjective)