Mathematics in Ancient and Medieval India c5.s14.9

c5.s14.9

By following this, √41 is found as √41 = 6 + 5/12 - ((5/12)²)/(2(6 + 5/12)). This formula is also used to calculate the values of √41, √105, √487, √889, √339009. The rule is attributed to Greek Heron (200 A.D.) and is restated by al-Ḥaṣṣarār (c. 1175 A.D.?) and medieval Arabian algebraists.¹