Question

Show that f : N → N, given by

$f(x) = \begin{cases} \text{x + 1, if x is odd, \ \textless \ br /\ \textgreater \ }\\ \\ \text{x - 1, if x is even}\\ \end{cases}$
is both one-one and onto. ​

Answer 1

cheack the pic upper sode

Answer 2

For one-one :

$IF \: f(a) = f(b)$

$⇒a+1=b+1 or a−1=b−1$

$if \: both \: are \: even \: or \: odd.$

$\large\boxed{\textsf{\textbf{\blue{Note \::-}}}}$

• f(a),f(b) are not equal if one is even and other is odd, since if a is even and b is odd, a−1 is odd and b+1 is even.

$⇒a=b.$

$So, one-one \: function \\ Now, for \: onto:$

$For \: all  \: x \: ∈ \:  odd \: nos \:  f(x)  \: gives \: all \: the \: even \: nos.$

$And  \: all  \: x \: ∈ \:  even \: nos \: f(x)  \: gives \: all \: the \: odd \: nos$

$⇒range=codomain$

$So \: , onto \: function$