If the functions $$f(x)$$ and $$g(x)$$ are defined on R→R such that $$f(x)=0$$,x∈ rational x,x∈ irrational and $$g(x)=0$$,x∈ irrational x,x∈ rational then (f$$−$$g)(x) is

Question

If the functions $$f(x)$$ and $$g(x)$$ are defined on R→R such  that $$f(x)=0$$,x∈ rational x,x∈ irrational  and $$g(x)=0$$,x∈ irrational x,x∈ rational  then (f$$−$$g)(x) is

Infinity Learn · Accepted Answer

We are given thatf:R→R,$$f(x)=0$$,x∈ rational x,x∈ irrational g:R→R,$$g(x)=0$$,x∈ irrational x,x∈ rational ∴ (f$$−$$g):R→R such that (f$$−$$g)(x)=$$−x, if x∈ rational x, if x∈ irrational  Since $$(f$$−$$g):R→R for any x, there is only one value of  $$($$ $$f(x)$$−$$g(x))$$ whether x is rational or irrational. Moreover,  as x∈R,$$f(x)$$−$$g(x)$$ also belongs to R . Therefore, (f$$−$$g) is $$one-one$$ and onto.

$If the functions f (x) and g (x) are defined on R \to R such that f (x) = \{\begin{cases} 0, x \in rational \\ x, x \in irrational \end{cases}$
$and g (x) = \{\begin{matrix} 0, x \in irrational \\ x, x \in rational \end{matrix} then (f - g) (x) is$

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If the functions f(x) and g(x) are defined on R→R such that f(x)=0,x∈ rational x,x∈ irrational and g(x)=0,x∈ irrational x,x∈ rational then (f−g)(x) is

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$If the functions f (x) and g (x) are defined on R \to R such that f (x) = \{\begin{cases} 0, x \in rational \\ x, x \in irrational \end{cases}$
$and g (x) = \{\begin{matrix} 0, x \in irrational \\ x, x \in rational \end{matrix} then (f - g) (x) is$