Show that the function f : R∗ → R∗ defined by f(x) = 1/x is one-one and onto, where R∗ is the set of all non-zero real numbers. Is the result true, if the domain R∗ is replaced by N with co-domain being same as R∗ ?

Class 12, NCERT Chapter 1,  Exercise 1.2, Q1

It is given that f: R∗ →R∗  is defined by f(x) =1/x
For one – one:
Let x, y ∈ R∗ 
such that f(x) = f(y)
⇒1/x =1/y
⇒ x = y
Therefore,  f is one–one.
For onto:
For y ∈ R∗,

there exists x =1/y ∈ R∗                      [as y ≠ 0] 

such that
⇒f(x) = 1/ [1/y]= y
Therefore, f is onto.
Thus, the given function f is one-one and onto.
Now, consider function g:N→R∗  defined by g(x) =1/x
We have,
⇒g(x) = g(y)
⇒1/x=1/y
⇒x = y

Therefore, g is one–one.
Further, it is clear that g is not onto as for 1.2 ∈ R∗, 

there does not exit any x in N
such that g(x) =1/1.2