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If the functions f(x) and g(x) are defin...

If the functions `f(x) and g(x)` are defined on `R -> R` such that `f(x)={0, x in` rational`
x, x in` irrational ; `g(x)={0, x in` irratinal and `x,x in` rational then `(f-g)(x)` is

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We have, `(f-g):R to R,`
`(f-g)(x)={(-x" if "x in "rational"),(x" if "x in "irrational"):}`
Clearly `(f-g)(x)` is one-one and onto.
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