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Abstract Algebra: An Introduction

Found in: Page 94

Book edition 3rd

Author(s) Thomas W Hungerford, David Leep

Pages 608 pages

ISBN 9781111569624

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Short Answer

If $f (x), g (x) \in R [x]$ and $f (x) + g (x) \neq 0_{R}$ , show that
$d e g [f (x) + g (x)] \leq m a x \{d e g f (x), d e g [g (x)]\}$

It is proved that $d e g [f (x) + g (x)] \leq m a x \{d e g f (x), d e g [g (x)]\}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Polynomial Arithmetic:

If any given function R[x] is a ring, then the commutative, associative, and distributive laws hold such that the function f(x)+g(x) exists.

Step 2: Polynomial Operations

Let the given functions be:

$f (x) = \sum_{i = 0}^{m} a_{i} x^{i} . . . . . . . \{m = d e g f (x)\} g (x) = \sum_{j = 0}^{m} b_{j} x^{j} . . . . . . . \{n = d e g g (x)\}$

Now, assuming:

$i > m \Rightarrow a_{i} = 0 j > n \Rightarrow b_{j} = 0$

We have,

$\begin{array}{rcl} f (x) + g (x) & = & \{\sum_{i = 0}^{m} a_{i} x^{i}\} + \{\sum_{j = 0}^{m} b_{j} x^{j}\} \\ = & \sum_{i = 0}^{m a x (m, n)} (a_{i} + b_{i}) x^{i} \end{array}$

This implies that the degree of above expression will lie within:

$0 \leq i \leq m a x (m, n) . . . . . . . \{a_{i} + b_{i} \neq 0\}$

.

This results: $d e g [f (x) + g (x)] \leq m a x \{d e g [f (x)], d e g [g (x)]\}$

Hence proved, $d e g [f (x) + g (x)] \leq m a x \{d e g [f (x)], d e g [g (x)]\}$ .

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