Show that the function f: R → R : f(x) = {1, if x is rational -1, if x is irrational is many- one into.

Question

Show that the function

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

is many - one into.

Find

(i) \(f(\frac{1}{2})\)

(ii) \(f(\sqrt{2})\)

(iii) \(f(\pi)\)

(iv) \(f(2+\sqrt{3}).\)

Answer 1

Answer is (i) 1 (ii) - 1 (iii) - 1 (iv) - 1

 (i) \(f(\frac{1}{2})\)

Here, x = 1/2,which is rational

∴f(1/2) = 1

 (ii) \(f(\sqrt{2})\)

Here, x = √2,which is irrational

∴f(√2) = - 1

 (iii) \(f(\pi)\)

Here, x = ∏, which is irrational

\(f(\pi)\)= -1

 (iv) \(f(2+\sqrt{3}).\)

Here, x = 2 + √3, which is irrational

∴f(2 + √3) = - 1