One-to-One Function:
A function f(x) is said to be one-to-one (or injective) if every element in the domain of f(x) is mapped to a distinct element in the range. In other words, no two elements in the domain are mapped to the same element in the range.

To determine if f(x) = (x^2 - 1)/(ax) is a one-to-one function, we need to investigate the conditions under which two different values of x yield the same value of f(x).

Setting up the Equation:
Let's assume that there are two values of x, x1 and x2, such that x1 > m and x2 > m, and f(x1) = f(x2). We can set up the equation as follows:

(x1^2 - 1)/(ax1) = (x2^2 - 1)/(ax2)

Simplifying the Equation:
To simplify the equation, we can cross-multiply:

x1^2 - 1 = x2^2 - 1

Since the constant term -1 is the same on both sides of the equation, we can cancel it out:

x1^2 = x2^2

Taking the square root of both sides, we get:

x1 = x2 or x1 = -x2

So, we have two possible cases:

1. x1 = x2: This implies that the two values of x are the same, contradicting our assumption that x1 and x2 are different. Therefore, this case is not possible.

2. x1 = -x2: This implies that the two values of x are negatives of each other. However, since we assumed x1 and x2 to be greater than m, this case is also not possible.

Conclusion:
Since neither case is possible, we can conclude that f(x) = (x^2 - 1)/(ax) is a one-to-one function for x > m.

Finding m:
To find the value of m, we need to determine the smallest value of x for which the function becomes one-to-one. From the above analysis, we can see that the function is always one-to-one for x > m, regardless of the value of m.

Therefore, m can be any real number, and the function will still be one-to-one as long as x is greater than m.