Mathematics in Ancient and Medieval India c7.s5.33

c7.s5.33

Now α₁ = a₃ t + q₁ {ll} = a_3 ( (r_2\ q_1 - c)/r_3 ) + q_1 & \because t = (r_2\ q_1 - c)/r_3 / = (q_1\ (\ r_2\ a_3 + r_3\ ) - a_3\ c)/r_3 / = (q_1\ r_1 - a_3\ c)/r_3 & ( p. 196 ) v = a₂ α₁ + t {ll} = a_2\ {\ (q_1\ r_1 - a_3\ c)/r_3\ } + (r_2\ q_1 - c)/r_3 / = (q_1\ (a_2 r_1 + r_2) - c\ (a_2 a_3 + 1))/r_3 / = (q_1 b - c\ (a_2 a_3 + 1))/r_3 & ( p. 196 ) u = a₁ v + α₁ {l} = a_1\ {\ (q_1\ b - c\ (a_2\ a_3 + 1))/r_3\ } + (q_1\ r_1 - a_3\ c)/r_3 / = (q_1\ (a_1\ b + r_1) - c\ {a_1\ (a_2\ a_3 + 1) + a_3\ })/r_3 / = (q_1\ a - c\ {a_1\ (a_2\ a_3 + 1) + a_3})/r_3 {ll} u/v & = (q_1\ a - c\ {a_1\ (a_2\ a_3 + 1) + a_3})/(q_1\ b - c\ (a_2\ a_3 + 1)) / & = (q_1\ a - cp_3)/(q_1\ b - cq_3) If there be (n-1) number of partial quotients in the Table I, then u/v = (q₁ a - p_n-1)/(q₁ b - q_n-1) The last n th convergent is always a/b hence p_n/q_n = a/b.