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 * ?

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Solution

The given function f: R * → R * is defined by f( x )= 1 x .

Assume, f( x )=f( y ) for some x,y∈ R * .

⇒ 1 x = 1 y ⇒x=y

So, f is one-one.

For every y∈ R * , there exists x= 1 y ∈ R * , such that,

f( x )= 1 1 y =y

So, f is onto.

Hence, the function f is one-one and onto.

Now, consider function g:N→ R * defined by g( x )= 1 x .

Assume, g( x 1 )=g( x 2 ), for some x 1 , x 2 ∈N.

⇒ 1 x 1 = 1 x 2 ⇒ x 1 = x 2

So, g is one-one.

For 3.6∈ R * , there does not exist any x in N such that, g( x )= 1 3.6 .

So, g is not onto.

Thus, the function g:N→ R * defined by g( x )= 1 x is one-one but not onto.

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