Problem 1
Question
Let \(a(x)\) and \(b(x)\) be polynomials of positive degree. By the division algorithm, we may divide \(a(x)\) by \(b(x)\) $$ a(x)=b(x) q_{1}(x)+r_{1}(x) $$ Prove that every common divisor of \(a(x)\) and \(b(x)\) is a common divisor of \(b(x)\) and \(r_{1}(x)\). It follows from part 1 that the ged of \(a(x)\) and \(b(x)\) is the same as the ged of \(b(x)\) and \(r_{1}(x)\). This procedure can now be repeated on \(b(x)\) and \(r_{1}(x)\); divide \(b(x)\) by \(r_{1}(x)\) : Next, $$ \begin{aligned} r_{1}(x) &=r_{2}(x) q_{3}(x)+r_{3}(x) \\ & \vdots \end{aligned} $$ Finally, $$ r_{n-1}(x)=r_{n}(x) q_{n+1}(x)+0 $$ In other words, we continue to divide each remainder by the succeeding remainder. Since the remainders continually decrease in degree, there must ultimately be a zero remainder. But we have seen that $$ \operatorname{gcd}[a(x), b(x)]=\operatorname{gcd}\left[b(x), r_{1}(x)\right]=\cdots=\operatorname{gcd}\left[r_{n-1}(x), r_{n}(x)\right] $$ Since \(r_{n}(x)\) is a divisor of \(r_{n-1}(x)\), it must be the ged of \(r_{n}(x)\) and \(r_{n-1}(x)\). Thus, $$ r_{n}(x)=\operatorname{gcd}[a(x), b(x)] $$ This method is called the euclidean algorithm for finding the gcd.
Step-by-Step Solution
VerifiedKey Concepts
Division Algorithm
- \(a(x) = b(x) q_1(x) + r_1(x)\)
- where \(0 \leq \deg(r_1(x)) < \deg(b(x))\)
Greatest Common Divisor (GCD)
- If \(d(x)\) is the GCD, that means both \(a(x)\) and \(b(x)\) can be expressed as multiples of \(d(x)\).
- This ensures that \(d(x)\) is common to both, verifying that there are no larger polynomials that meet these conditions.
Common Divisors in Polynomials
- For any two polynomials \(f(x)\) and \(g(x)\) if \(d(x)\) divides both \(f(x)\) and \(g(x)\), then \(d(x)\) divides \(Af(x) + Bg(x)\) for any polynomials \(A(x)\) and \(B(x)\).
Polynomial Division Process
- Begin by dividing \(a(x)\) by \(b(x)\) to get a remainder \(r_1(x)\).
- Then, divide \(b(x)\) by \(r_1(x)\) to get \(r_2(x)\), and continue this process.