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

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

Toolbox:
• A function f:A?B where for every x1,x2?X,f(x1)=f(x2)?x1=x2 is called a one-one or injective function.
• ,A functionf:X?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?Y, there exists an element x in X such that f(x)=y.
Step1: Injective or One-One function:
Let $f:R \to R$ and
$f(x)=x^2$
let x1 =-1 and x2=1 be two elements in R
$f(-1)=f(1)\; since\; (-1)^2 =(1)^2$
but $-1 \neq 1$
$f : R\to R \;given\; by\; f(x)\; = x^2$ is not injective as f(x1) =f(x2) does not imply x1=x2
Step 2: Surjective or On-to function:
Let $f:R \to R$ and
Let us consider the element -2 , for $-2 \in R$ there does not exists $x \in R$ such that $f(x)=x^2=-2$
$\sqrt {-2} \notin Z$
$f : R\to R \;given\; by\; f(x)\; = x^2$ is not surjective as for every $y\in R$ there does not exist $x\in R$ such that f(x)= y
Solution:Hence$f : R\to R \;given\; by\; f(x)\; = x^2$ is neither injective nor surjective
edited Jan 24, 2014