\(f:R→R :f(x)=\begin{cases}1,\text{ if x is rational}\\ -1,\text{ if

1.	\(f:R→R :f(x)=\begin{cases}1,\text{ if x is rational}\\ -1,\text{ if x is irrational}\end{cases}\)Show that f is many-one and into.
Answer» To prove: function is many-one and into Given: \(f:R→R :f(x)=\begin{cases}1,\text{ if x is rational}\\ -1,\text{ if x is irrational}\end{cases}\) We have, f(x) = 1 when x is rational It means that all rational numbers will have same image i.e. 1 ⇒ f(2) = 1 = f (3) , As 2 and 3 are rational numbers Therefore f(x) is many-one The range of function is [{-1},{1}] but codomain is set of real numbers. Therefore f(x) is into

\(f:R→R :f(x)=\begin{cases}1,\text{ if x is rational}\\ -1,\text{ if x is irrational}\end{cases}\)Show that f is many-one and into.

Answer»

To prove: function is many-one and into

Given: \(f:R→R :f(x)=\begin{cases}1,\text{ if x is rational}\\ -1,\text{ if x is irrational}\end{cases}\)

We have,

f(x) = 1 when x is rational

It means that all rational numbers will have same image i.e. 1

⇒ f(2) = 1 = f (3) , As 2 and 3 are rational numbers

Therefore f(x) is many-one

The range of function is [{-1},{1}] but codomain is set of real numbers.

Therefore f(x) is into