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Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Linear Algebra

1.4.2
Let F = {a + bi : a, b e Q}, where i? = – 1. Show that F is a field.

Branch of mathematics concerned with mathematical structures that are closed under operations like addition and scalar multiplication. It is the study of linear combinations, vector spaces, lines and planes, and some mappings that are used to perform linear transformations. Linear algebra also includes vectors, matrices, and linear functions. It has many applications from mathematical physics to modern algebra and coding theory.

Expert Solution

Step 1

Given $F = \{a + b i : a, b \in Q\}$ , where $i^{2} = - 1$ .

We have to show that $F$ is a field.

First of all we define addition $' +'$ and multiplication $' \cdot'$ on $F$ which is given as follows.

$\begin{matrix} (a + b i) + (c + d i) = (a + c) + (b + d) i \\ (a + b i) \cdot (c + d i) = (a c - b d) + (a b + b c) i \end{matrix}$

First we will show that $F$ forms abelian group with respect to addition $' +'$ .

(i) Let $a + b i, c + d i, e + f i \in F$

Now,

$\begin{matrix} ((a + b i) + (c + d i)) + (e + f i) = ((a + c) + (b + d) i) + (e + f i) \\ = (a + c + e) + (b + d + f) i \\ = (a + b i) + ((c + e) + (d + f) i) \\ = (a + b i) + ((c + d i) + (e + f i)) \end{matrix}$

Hence, $((a + b i) + (c + d i)) + (e + f i) = (a + b i) + ((c + d i) + (e + f i))$

Therefore, addition is associative.

(ii) $0 = 0 + 0 i \in F$

If $a + b i \in F$ then

$\begin{matrix} (a + b i) + 0 = (a + 0) + b i \\ = a + b i \\ = 0 + (a + b i) \end{matrix}$

Hence, $' 0'$ is the additive identity in $F$ .

(iii) Let $a + b i \in F$

Since, $a, b \in q \Rightarrow - a, - b \in q$

So, $- a - b i \in F$

Now,

$\begin{matrix} (a + b i) + (- a - b i) = (a - a) + (b - b) i \\ = 0 + 0 i \\ = 0 \end{matrix}$

Hence, $(- a - b i)$ is the additive inverse of $(a + b i)$ .

(iv) Let $a + b i, c + d i \in F$

Then

$\begin{matrix} (a + b i) + (c + d i) = (a + c) + (b + d) i \\ = (c + a) + (d + b) i \\ = (c + d i) + (a + b i) \end{matrix}$

Since, $(a + b i) + (c + d i) = (c + d i) + (a + b i)$

Hence, $F$ forms abelian group with respect to addition $' +'$ .

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