#### Question

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**?

#### Solution

It is given that* f*: **R**_{*} → **R**_{*} is defined by `f(x) = 1/x`

One-one:

f(x) = f(y)

`=> 1/x = 1/y`

=> x = y

∴*f* is one-one.

Onto:

It is clear that for *y*∈** R**_{*}, there exists ` x= 1/y in R ("Exists as y" != 0)`such that

`f(x) = 1/((1/y)) = y`

∴*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_1) = g(x_2) => 1/x_1 = 1/x_2 => x_1 = x_2`

∴*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`

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