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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.

Toolbox:
• 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'