If f(x) = begin{cases} x, & x in mathbb{Q} 0 | vedclass

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(D) Given $f(x) = \begin{cases} x, & x \in \mathbb{Q} \ 0, & x 
otin \mathbb{Q} \end{cases}$ and $g(x) = \begin{cases} x, & x \in \mathbb{Q} \ 0, &

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Neither one-one nor onto

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(D) Given $f(x) = \begin{cases} x, & x \in \mathbb{Q} \ 0, & x 
otin \mathbb{Q} \end{cases}$ and $g(x) = \begin{cases} x, & x \in \mathbb{Q} \ 0, & x 
otin \mathbb{Q} \end{cases}$.We need to find the function $h(x) = (f - g)(x) = f(x) - g(x)$.For any $x \in \mathbb{Q}$,$h(x) = f(x) - g(x) = x - x = 0$.For any $x 
otin \mathbb{Q}$,$h(x) = f(x) - g(x) = 0 - 0 = 0$.Thus,$h(x) = 0$ for all $x \in \mathbb{R}$.This is a constant function.$A$ constant function is neither one-one (as $h(1) = h(2) = 0$) nor onto (as the range is ${0}$,which is not equal to the codomain $\mathbb{R}$).Therefore,the function $(f - g)$ is neither one-one nor onto.

If $f(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q} \end{cases}$ and $g(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q} \end{cases}$,then the function $(f - g)$ is:

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