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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.

1 Answer

  • 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
$f(2x)=1 \qquad n \in N$
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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