Proof

(Compare this proof with the proof in Chapter 1 .) The proof is divided into two parts:

The existence of the polynomials q(X) and r(X).

The unicity.

The existence of polynomials q(X) and r(X).
We prove this by induction on the degree n of a(X). Let m be the degree of b(X). If n < m, we put q(X) = 0, r(X) = a(X). Thus we may assume that n m. In particular, a(X) 0. If a(X) is a constant (thus n = 0), then also m = 0. We may put q(X) = a/b and r = 0.

Now suppose that n > 0 and that (the induction hypothesis) the existence of polynomials q(X), r(X) has been proved for polynomials a(X) with degree at most n - 1. Consider

a(X) = a₀ + a ₁ X + ··· + a_nXⁿ ,

with a_n 0 and

b(X) = b₀ + b₁ X + ··· + b _mX ^m

with b _m 0. The degree of a(X) - (a_n/b _m )b(X)X^{n - m} is less than n. According to the induction hypothesis there are polynomials q₁ (X) and r (X) with

a(X) - ( a_n/b_m )X^{n - m}b(X) = q₁(X)b(X) + r(X), deg(r(X)) < m.

But then we have

a(X) = q(X) b(X) + r(X), deg(r (X)) < m,

where q(X) = q₁(X) + (a_n /b_m)X^{n - m}.

The unicity.
Suppose that a(X) = q(X)b(X) + r(X) and a(X) = q₁(X)b(X) + r₁(X) are two expressions as in the theorem. Subtract these two expressions:

(q(X) - q₁(X))b(X) = r₁(X) - r(X).

From this it follows that

deg(q(X) - q₁(X)) + deg(b(X)) = deg((q(X) - q₁(X))b(X))

= deg(r₁(X) - r(X))

< deg (b(X)).

Hence deg(q(X) - q₁(X)) < 0. This is only possible if q(X) - q₁(X) = 0. But then r₁(X) = r(X) as well.