how to prove the division algorithm for polynomials using induction

Everly

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11-01-2025

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The division algorithm for polynomials is a fundamental concept in algebra. It states that for any polynomials f(x) and g(x), where g(x) is not the zero polynomial, there exist unique polynomials q(x) (the quotient) and r(x) (the remainder) such that:

f(x) = g(x)q(x) + r(x)

where the degree of r(x) is less than the degree of g(x). This article will demonstrate how to prove this algorithm using mathematical induction on the degree of f(x).

Understanding the Base Case (Degree of f(x) = 0)

Before embarking on the inductive proof, we need to establish the base case. When the degree of f(x) is 0, f(x) is a constant. Let's represent this constant as 'a'.

If the degree of g(x) is also 0 (g(x) = b, where b is a constant and b ≠ 0), then:

q(x) = a/b and r(x) = 0. This satisfies the division algorithm.

If the degree of g(x) is greater than 0, then:

q(x) = 0 and r(x) = a. Again, this satisfies the condition that the degree of r(x) (which is 0) is less than the degree of g(x).

Therefore, the base case holds true.

The Inductive Step: Assuming Truth for Degree k

Our inductive hypothesis assumes that the division algorithm is true for all polynomials f(x) with degree k. That is, for any polynomial f(x) of degree k and any non-zero polynomial g(x), there exist unique polynomials q(x) and r(x) such that:

f(x) = g(x)q(x) + r(x)

with deg(r(x)) < deg(g(x)).

Proving the Algorithm for Degree k+1

Now, we need to prove that the algorithm holds for polynomials f(x) of degree k+1. Let f(x) be a polynomial of degree k+1, and let g(x) be a non-zero polynomial. Let the leading term of f(x) be a_k+1x^k+1 and the leading term of g(x) be b_mx^m, where m is the degree of g(x).

We can construct a new polynomial f'(x) as follows:

f'(x) = f(x) - (a_k+1/b_m)x^k+1-mg(x)

Notice that the leading term of (a_k+1/b_m)x^k+1-mg(x) cancels out the leading term of f(x). This means the degree of f'(x) is at most k.

Since the degree of f'(x) is at most k, we can apply our inductive hypothesis. There exist unique polynomials q'(x) and r(x) such that:

f'(x) = g(x)q'(x) + r(x)

with deg(r(x)) < deg(g(x)).

Substituting the definition of f'(x), we get:

f(x) - (a_k+1/b_m)x^k+1-mg(x) = g(x)q'(x) + r(x)

Rearranging this equation, we obtain:

f(x) = g(x)[(a_k+1/b_m)x^k+1-m + q'(x)] + r(x)

Let q(x) = (a_k+1/b_m)x^k+1-m + q'(x). Then:

f(x) = g(x)q(x) + r(x)

This proves that the division algorithm holds for polynomials of degree k+1.

Conclusion

By the principle of mathematical induction, we have proven the division algorithm for polynomials. We started with a base case (degree 0) and showed that if the algorithm holds for degree k, it also holds for degree k+1. This completes the proof. The uniqueness of q(x) and r(x) can be proven separately, but this inductive proof establishes their existence.