Show that the function\(f:R→R : f(x)= \begin{cases}1,\text{ if x is

1.	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» 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

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»

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