Question

Show that the function f: R → R, defined by f (x) = \(\frac{1}{x}\) is one-one 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

(i) f(x) = \(\frac{1}{x}\), If f(x1) = f(x2) ⇒ \(\frac{1}{x_{1}}\) = \(\frac{1}{x_{2}}\)

⇒ x1 = x2

Each x ∈ R, has a unique image in codomain.

⇒ f is one-one.

(ii) For each y belonging codomain, then y = \(\frac{1}{x}\) or x = \(\frac{1}{y}\), there is a unique pre-image of y.

⇒ f is onto.

(b) When domain R, is replaced by N, codomain R, remaining the same, then f : N → R,

If f(x1) = f(x2)

⇒ \(\frac{1}{n_{1}}\) = \(\frac{1}{n_{2}}\)

⇒ n1 = n2 where n1, n2 ∈ N.

⇒ f is one-one.

But for every real number belonging to codomain may not have a pre-image in N.

e.g. \(\frac{1}{\frac{2}{3}}\) = \(\frac{3}{2}\) ∉ N

∴ f is not onto.