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

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 the 1st part of a 5 part question, which is split as 5 separate questions here.

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  • A function $f: A \rightarrow B$ 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 $f : N\to N\; given\; by\; f(x)\; = x^2$
Let $x$ and $y$ be two elements in $Z$.
For an injective or one-one function, $f(x) = f(y)$
$ \Rightarrow x^2 = y^2$$ \Rightarrow x = y.$
Therefore $f:Z \rightarrow Z$ defined by $f(x)=x^2$ is one-one or injective.
For an on-to function, for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
$ \Rightarrow$ For every $y \in Z$ there must exist $ x$ such that $ f(x)=x^2= y$.
However, we see that for $y=2$, there is no $x \in Z$ such that $f(x) = x^2 = 2$.
Therefore $f:Z \rightarrow Z$ defined by $f(x)=x^2$ is not onto or surjective.
Solution: $f:Z \rightarrow Z$ defined by $f(x)=x^2$ is only one-one or injective but not onto or surjective.
answered Mar 14, 2013 by meena.p
edited Mar 17, 2013 by balaji.thirumalai

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