Mathematics in Ancient and Medieval India c7.s5.42

c7.s5.42

If there be (n-1) number of partial quotients in the Table II, then : $u/v = (q ₂a + cp_n-1)/(q ₂b + cq_n-1)$ Āryabhaṭa I knew that : $p_n · v - q_n · u = -c [n = odd] (3)$ For p_n · v - q_n · u $= p_n (q₂b + cq_n-1) - q_n (q ₂a + cp_n-1)$ $= c (p_nq_n-1 - q_n p_n-1) [ p_n = a, q_n = b]$ $= -c [n = odd]$