The function $$f : N \to I$$ defined by $$f(x) = x - 5\left[\frac{x}{5}\right]$$, where $$N$$ is the set of natural numbers and $$[x]$$ denotes the greatest integer less than or equal to $$x$$, is:

Let $$x$$ be any natural number. Divide $$x$$ by $$5$$ using the usual division algorithm.

There exist unique integers $$q$$ and $$r$$ such that

$$x = 5q + r,$$

where $$q = \left[\dfrac{x}{5}\right]$$ and the remainder $$r$$ satisfies $$0 \le r \lt 5.$$ Explicitly, $$r \in \{0,1,2,3,4\}.$$

Substituting $$x = 5q + r$$ into the definition of the function gives

$$\begin{aligned} f(x) &= x - 5\left[\dfrac{x}{5}\right] \\ &= (5q + r) - 5q \\ &= r. \end{aligned}$$

Thus every natural number is sent to its remainder upon division by $$5,$$ and so

$$\text{Range}(f)=\{0,1,2,3,4\}.$$

To test injectivity (one-one), take two natural numbers that leave the same remainder when divided by $$5.$$ For instance, choose $$x_1 = 2$$ and $$x_2 = 7.$$ Both satisfy

$$f(2)=2,\qquad f(7)=2.$$

Since $$x_1 \ne x_2$$ but $$f(x_1)=f(x_2),$$ the function fails the injectivity criterion. Hence $$f$$ is not one-one.

To test surjectivity (onto) with respect to the stated codomain $$I$$ (the set of all integers), observe that

$$\text{Range}(f)=\{0,1,2,3,4\}\subsetneq I.$$

An integer such as $$6$$ lies in $$I$$ but never appears as an output, because every value of $$f$$ is at most $$4.$$ Therefore the function is not onto.

Because $$f$$ is neither one-one nor onto, the correct choice is the one that says “neither one-one nor onto.”

Hence, the correct answer is Option C.