ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)

228 BRAHMAGUPTA AS AN ALGEBRAIST Let b) a (q bq ─── r₁) b (q₁ r₁q₁ ─── r₂) r₁ (q₂ r₂q₂ ─── r₃ ... ────── rₘ₋₁) rₘ₋₂ (qₘ₋₁ rₘ₋₁ qₘ₋₁ ─────── rₘ) rₘ₋₁ (qₘ rₘqₘ ───── rₘ₊₁ Then we get (when a < b, we shall have q = 0, r₁ = a) a = bq + r₁ b = r₁q₁ + r₂ r₁ = r₂q₂ + r₃ r₂ = r₃q₃ + r₄ ... ... ... rₘ₋₂ = rₘ₋₁ qₘ₋₁ + rₘ rₘ₋₁ = rₘqₘ + rₘ₊₁ Now, substituting the value of a in the given equation (1), we get by = (bq + r₁)x + c Therefore y = qx + y₁ where by₁ = r₁x + c In other words, since a = bq + r₁, on putting y = qx + y₁ (ii) the given equation (i) reduces to by₁ = r₁x + c (iii) Again, since b = r₁q₁ + r₂