Linear algebra - Field equations

In summary: You might also need to mention the fact that the field axioms imply the cancellation property.For iii), you can just say that multiplication is associative and commutative in a field (since it is a commutative ring with unity) and then use the fact that ab^-1a'b'^-1 = aab^-1b'^-1 = (ab)(a'b')^-1, so the result follows.For iv), you can use the fact that (ab^-1)(a'b'^-1) = (aa')(b^-1b'^-1) = (aa')(bb')^-1, and then use part 1) to show that this is equal to (ab')(a'b)^-1
  • #1
karnten07
213
0

Homework Statement



Let F be a field. For any a,b [tex]\in[/tex] F, b[tex]\neq[/tex]0, we write a/b for ab^-1. Prove the following statements for any a,a' [tex]\in[/tex]F and b,b' [tex]\in[/tex] F\{0}:

i.) a/b = a'/b' if and only if ab' = a'b.

ii.) a/b + a'/b' = (ab' + a'b)/bb'

iii.) (a/b)(a'/b') = (aa')/(bb')

iv.) ((a/b)/(a'/b')) = (ab')/(a'b) if in addition a' [tex]\neq[/tex] 0

Homework Equations





The Attempt at a Solution



Okay I am getting really confused in these questions because i don't know when it means to divide by and when it is showing that a/b = ab^-1

i.) want to show a/b = a'/b' if and only if ab' = a'b.

from ab' = a'b a = a'b/b' and a' = ab'/b

So substituting in values of a and a'

a/b = (a'b/b')/(ab'/b)

Carrying out division:

(a'ba')/(b'ab') and because a'b = b'a, substitute this in and cancel like terms:

(ab'a'/ab'b') = a'/b'

therefore a'/b' = a/b

is this even right because its just rearranging stuff?
 
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  • #2
Yeah, i can't be doing it right, it must mean i need to write in the inverses. Thoughts?
 
  • #3
So for i.)

a/b = a'/b' becomes a(b)^-1 = a'(b')^-1

So rearranging a'b = b'a gives a = a'b/b' and b = b'a/a', substitute this in:

(a'b/b') (b'a/a')^-1

Substitute in b' = a'b/a into the non inverse bracket:

(a'ba/a'b)(b'a/a')^-1

Then by the multiplicative inverse property of a field, elements and their inverses become the identity element, which is 1 in this case (do i need to show its 1?):

So, (a'b/b)(b')^-1

the b's cancel to give a'(b')^-1 which is written as a'/b'

So a/b = a'/b' as asked

Is this the right idea? Any tips to improve my setting out and explanations as i go along as i don't want to miss anything? Thanks in advance
 
  • #4
Should i write that i can cancel the b's by existence of multiplicative inverse as bx1/b = 1?
 
  • #5
I think i have done number ii.) by substituing in values for a and b, then using the inverse to rearrange into the desired form.
 
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  • #6
I think you're really overthinking things. All of those just require you to rewrite x/y as xy^-1 then use the properties of field multiplication.
 
  • #7
Mystic998 said:
I think you're really overthinking things. All of those just require you to rewrite x/y as xy^-1 then use the properties of field multiplication.

Yea i thought so, so in my third post do you think that i have the right idea in my proof?
 
  • #8
Yeah, you'd just rewrite it as [itex]ab^{-1}a'b'^{-1}[/itex], then rearrange using commutativity of multiplication.

Edit: Or you could use part 1 to show equality. In this case, it might be a better idea, because the question seems to be alluding to the kind of stuff you have to do to show that the field of fractions of an integral domain is actually a well-defined field.
 
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  • #9
Mystic998 said:
Yeah, you'd just rewrite it as [itex]ab^{-1}a'b'^{-1}[/itex], then rearrange using commutativity of multiplication.

Edit: Or you could use part 1 to show equality. In this case, it might be a better idea, because the question seems to be alluding to the kind of stuff you have to do to show that the field of fractions of an integral domain is actually a well-defined field.

How do i show equality? What do i do differently than what i have done already?
 
  • #10
For part iii.) can i do it like this:


(a/b)(a'/b') = (aa')/(bb')

(ab^-1)(a'(b')^-1) for LHS then say that it can be rearranged to right hand side by law of associativity? Feels like one of those questions that has a suspiciously simple solution.
 
  • #11
Commutativity and associativity, yes. At least, that's the way it seems.
 

Related to Linear algebra - Field equations

1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations and their representations in vector spaces. It involves the use of matrices, vectors, and linear transformations to solve problems in various fields such as physics, engineering, and computer science.

2. What are field equations in linear algebra?

Field equations in linear algebra refer to the set of rules and operations that define a field, which is a mathematical structure used to represent numbers. Field equations are essential in linear algebra as they provide a foundation for solving equations and manipulating vectors and matrices.

3. How are field equations used in real-world applications?

Field equations in linear algebra are used in various real-world applications such as computer graphics, data analysis, and machine learning. They are also essential in physics to describe the behavior of physical systems, such as electromagnetic fields and quantum mechanics.

4. What are the basic operations in linear algebra?

The basic operations in linear algebra include addition, subtraction, multiplication, and division of vectors and matrices. These operations are used to solve equations, find the inverse of matrices, and perform transformations on vectors.

5. How does linear algebra relate to other branches of mathematics?

Linear algebra is closely related to other branches of mathematics such as calculus, differential equations, and abstract algebra. It provides a foundation for these subjects and is used extensively in their applications. For example, linear algebra is used in solving systems of differential equations and in finding the critical points of multivariable functions in calculus.

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