sqrt(MC - XI) = sqrt(IX) + sqrt(CM)This is rather elegant in Roman numerals! Is it merley a coincidence, or is something deeper going on here?
For example, does the equation hold in other bases? For this question to make sense, the values I, X, C, and M should be interpreted as 1, b, b2 and b3, where b is the base. The proposed identity is then:
sqrt((b3 + b2) - (b + 1)) = sqrt(b - 1) + sqrt(b3 - b2)This certainly doesn't look to be generally true, but in fact, it is!
LHS = sqrt(b2 * (b + 1) - (b + 1)) = sqrt((b + 1) * (b2 - 1)) = sqrt((b + 1) * (b + 1) * (b - 1)) = (b + 1) * sqrt(b - 1) = b * sqrt(b - 1) + sqrt(b - 1) = sqrt(b3 - b2) + sqrt(b - 1) = RHSSurprisingly, the identity holds for all bases. Of course, each expression is integral only when b is one more than a square.