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Give examples of two functions \(f : N \to Z \) and \(g: Z \to Z\) such that \(g\;o\;f\) is injective but \(g\) is not injective.

(Hint : Consider f(x) = x and g(x) = |x|).    

1 Answer

  • 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 two functions $f:A \to B $ and $g:B \to C$, then composition of $f$ and $g$, $gof:A \to C$ by $ gof (x)=g(f(x))\;for\; all \;x \in A$
  • $=>x=y$
Lets define a function $f:N \to Z$ as $f(x)=x$ and $ g:Z \to Z$ as $g(x)=|x|$
$\textbf {Step 1: Example function such that g is not injective}$
We observe that for $-1 \in Z$ $g(-1)=|-1|=1$ and since $g(1)=|1|=1$, we can see that since $-1 \neq 1$, $g$ is not injective.
$\textbf {Step 2: Testing if gof is injective}$
Now, $gof(x)=g(fx)=g(x)=|x|$ and $gof(y)=g(f(y))=g(y)=|y|$
$\Rightarrow$ $|x|=|y|$, and since $x,y \in N$, both are positive. Therefore $x=y$.
Hence, $gof$ is injective.


answered Feb 27, 2013 by meena.p
edited Mar 20, 2013 by balaji.thirumalai

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