Exercise · ZW-POLYNOMIAL-DIVISION-ALGORITHM-009 · ★★★ Advanced

In the uniqueness proof, if \(f=gq_1+r_1=gq_2+r_2\) with \(q_1\ne q_2\), what contradiction appears?

Choose one option.

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  • B. \(g(q_1-q_2)=r_2-r_1\) would have degree \(\ge\deg g\) and also \(<\deg g\)

Explanation

Subtracting gives \(g(q_1-q_2)=r_2-r_1\). If \(q_1\ne q_2\), the left side has degree at least \(\deg g\) (a field has no zero divisors), but the right side has degree less than \(\deg g\).

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Review the concept: Polynomial division algorithm

Given polynomials \(f\) and \(g\neq 0\) over a field \(F\), there exist a unique quotient \(q\) and a unique remainder \(r\) satisfying the equality below, with \(r\) zero or of degree less than that of \(g\).

\[f=gq+r,\qquad r=0\ \text{ or }\ \deg r<\deg g\]

More practice problems: polynomial division algorithm

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