Show that `f: N to N` given by
`f(x)={(x+1,"if x is odd"),(x-1,"if x is even"):}`
is both one-one and onto.

Explanation

for one-one
CASE1: When` x_1,x_2`,are odd natural number
`f(x_1)=f(x_2)` `Rightarrow``x_1+1=x_2+1 forall x_1,x_2 in N
so x_1=x_2`
i.e. one -one .
Case II: When` x_1,x_2`are even natural number
`f(x_1)=f(x_2)rightarrow x_1-1=x_2_1`
`Rightarrow``x_1=x_2`
i.e. f is one-one
Case(iii): When` x_1` is odd and `x_2` is even natural number
.`f(x_1)=f(x_2)``Rightarrow``x_1+1=x_2-1`
`Rightarrow``x_2-x_2=2`which is never possible as the difference of odd and even number number is always add number
.Hence in this case `f(x_1)nef(x_2)`
i,e.` f `is one-one
Case (iv): When `x_1` is even and `x_2`is odd natural number similar as case III, We can prove `f` is one-one
For onto:
`f(x)=x+1` if x is odd
`=x-1`if x is even
For every even number `y` of codomain exists odd
number `y-1`in domain and for every add number y
of codomain exists even number `y+1` in domain.
i.e. f is onto function
Hence f is one-one onto function