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