# Check the injectivity and surjectivity of the following functions: $$f : Z\to Z \;given\; by\; f(x)\; = x^2$$

A) $f$ is injective and surjective B) $f$ is injective only C) $f$ is surjective only D) $f$ is neither injective nor surjective     Note: This is part 2 of a 5  part question, split as 5 separate questions here.

Toolbox:
• A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given $f:Z \to Z$ defined by $f(x)=x^2$
Step1: Injective or One-One function:
let us consider two elements$f(-1)=f(1)\; since\; (-1)^2 =(1)^2$
but $-1 \neq 1$
$f:Z \to Z$ defined by $f(x)=x^2$ is not injective
Step 2: Surjective or On-to function:
Also $-2 \in Z$ but there does not exist $x \in z$
Such that $f(x)=x^2=-2$
$\sqrt {-2} \notin Z$
$f:Z \to Z$ defined by $f(x)=x^2$ is not surjective
Solution: Hence  $f:Z \to Z$ defined by $f(x)=x^2$ is nether injective nor surjective

edited Mar 20, 2013 by meena.p