$f(x) = \left\{ \begin{array}{r} x,\ if\ x\ is\ rational \ 0,\ if\ x\ is\ irrational \end{array} ight.\$ and $f(x) = \left\{ \begin{array}{r} 0,\ if\ x\ is\ rational \ x,\ if\ x\ is\ irrational\ \end{array} ight.\$. Then, $f - g$ is

Question

$f(x) = \left\{ \begin{array}{r} x,\ if\ x\ is\ rational \ 0,\ if\ x\ is\ irrational \end{array} ight.\$ and $f(x) = \left\{ \begin{array}{r} 0,\ if\ x\ is\ rational \ x,\ if\ x\ is\ irrational\ \end{array} ight.\$. Then, $f - g$ is

Dr. Abubacker · Accepted Answer

(d)
Let $\phi(x) = f(x) - g(x)$
$= \left\{ \begin{array}{r}
x,\ x \in \mathcal{Q} \
 - x,\ x 
otin \mathcal{Q}
\end{array} ight.\$
For one-one
Take any straight line parallel to x-axis which will intersect $\phi(x)$ only at one point.
$\Rightarrow \phi(x)$ is one-one.
Foe onto
As, $\phi(x) = \left\{ \begin{array}{r}
x,\ x \in \mathcal{Q} \
 - x,\ x 
otin \mathcal{Q}
\end{array} ight.\$, which shows
$y = x$ and $y = - x$ for irrational values $\Rightarrow y 
otin$ real numbers.
$	herefore$ Range=Codomain
$\Rightarrow \ \ \ \phi(x)$ is onto.
Thus, $f - g$ is one-one and onto.

Q. $f(x) = \left\{ \begin{array}{r} x,\ if\ x\ is\ rational \\ 0,\ if\ x\ is\ irrational \end{array} \right.\$ and $f(x) = \left\{ \begin{array}{r} 0,\ if\ x\ is\ rational \\ x,\ if\ x\ is\ irrational\ \end{array} \right.\$. Then, $f - g$ is

Explanation

Reviewed by Dr. Abubacker

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