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Home  >>  CBSE XII  >>  Math  >>  Relations and Functions
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True or False: Let A={0,1} and N be the set of natural numbers.Then the mapping $f:N \rightarrow A$ defined by f(2n-1)=0,f(2n)=1,$\quad n\in N$,is onto.

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  • A function $f: N \to A \qquad A=\{0,1\}$ is onto if the exists every element y an element in N such that $f(x)=y$
$f:N \to A$ defined by
 
$f(2x-1)=0$
 
$f(2x)=1 \qquad n \in N$
 
Let y=0
 
f then there exist $x=(2x-1) \in N$ such that $f(x)=0 =>f(2n-1)=0$
 
Also let y =1
 
then there exists $x=2x \in N$ such that $f(x)=1 =>f(2x)=1$
 
Therefore for every element $y \in A$ there exists $f(x)=y$
 
f is onto
 
The given statment is 'True'

 

answered Mar 6, 2013 by meena.p
 

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