Show that the function

Question:

Show that the function

$f: R \rightarrow R: f(x)=\left\{\begin{array}{l}1, \text { if } x \text { is rational } \\ -1, \text { if } x \text { is irrational }\end{array}\right.$

is many - one into.

Find

(i) $f\left(\frac{1}{2}\right)$

(ii) $f(\sqrt{2})$

(iii) $f(\pi)$

(iv) $f(2+\sqrt{3})$.

Solution:

(i) $\mathrm{f}\left(\frac{1}{2}\right)$

Here, $x=1 / 2$, which is rational

$\therefore f(1 / 2)=1$

(ii) $\mathrm{f}(\sqrt{2})$

Here, $x=\sqrt{2}$, which is irrational

$\therefore f(\sqrt{2})=-1$

(iii) $f(\pi)$

Here, $x=\Pi$, which is irrational

$f(\pi)=-1$

(iv) $\mathrm{f}(2+\sqrt{3})$.

Here, $x=2+\sqrt{3}$, which is irrational

$\therefore f(2+\sqrt{3})=-1$

Ans. (i) 1 (ii) $-1$ (iii) $-1$ (iv) $-1$

 

 

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