Question

Show that the function f: R* $\to$** ** R* defined by f(x)= $\frac {1} {x}$ 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* ?

Answer 1

It is given that f :**** R* $\to$ R* is defined by f (x)= $\frac {1} {x}$ .
One-one: f (x)=f (y)
$\Rightarrow$ $\frac {1} {x}$ = $\frac {1} {y}$
$\Rightarrow$_ _ x=y.
∴ f is one-one.
Onto:
It is clear that for y ∈ R* , there exists x = $\frac {1} {y}$ ∈ R* ** (exists as y ≠ 0 )** such that f (x)= $\dfrac {1} { \frac {1} {y}}$=y.
∴ f __ is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function g : N $\to$ R* defined by
g (x)= $\frac {1} {x}$
we have __ g (x1)=g (x2) $\Rightarrow$ $\frac {1} {x_1}$= \frac {1} {x_2}$$\Rightarrow x1 = 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) = $\frac {1} {1.2}$

Hence, function g is one-one but not onto.