Show that the function f : R. → R. defined by 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.?
It is given that R. → R. by one - one
f(x) = f(y) x = y
f is one- one onto
It is clear that , there exists
(Exiasts as y
0) such that f(x) =
= y
f is onto
Thus the given function (f) one-one and onto
Now, consider function g : N R defined by g(x)=
We have g (x1) = g(x2)
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) =
hence function g is one-one but not onto.