Showing properties of Ordered Fields

But you don't have to do it all over again. You could just write the result of addition is (a, b) + (0, 1) = (a, b). That is, you don't need to show that (a, b) + (0, 1) = a/b + 0/1. This is already in your previous work. You need to show that (a, b) + (0, 1) = (a, b) and that [a/b] + [0/1] = [a/b]. That is, you need to show that the elements in the two equivalence classes are equal to each other.
  • #1
mliuzzolino
58
0

Homework Statement



Let S = {(a, b): a, b [itex]\in \mathbb{Z}[/itex] and b ≠ 0}. An equivalence relation "~" on S is defined by (a, b) ~ (c, d) iff ad = bc.

1) For any b [itex]\in \mathbb{Z} \setminus[/itex] {0}, show that [0/b] = [0/1] and [b/b] = [1/1].

2) For any a, b [itex]\in \mathbb{Z}[/itex] with b ≠ 0, show that [a/b] + [0/1] = [a/b] and [a/b][1/1] = [a/b].

Homework Equations





The Attempt at a Solution




1) First, show that [0/b] = [0/1].

For any value of b not zero, [0/1] is contained in the equivalence class when b = 1: Eb = {[0/b]: b ≠ 0}. Therefore, [0/b] = [0/1].

Next, show that [b/b] = [1/1].

For any value of b not zero, [1/1] is contained in the equivalence class when b = 1: ESUB]b[/SUB] = {[b/b]: b ≠ 0}. Therefore, [b/b] = [1/1].



2) First show that [a/b] + [0/1] = [a/b].

[a/b] + [0/1] = [itex]\frac{a*1 + 0*b}{b * 1}[/itex] = [itex]\frac{a + 0}{b}[/itex] = [a/b].

Next, show that [a/b][1/1] = [a/b].

[a/b][1/1] = [itex]\frac{a(1)}{b(1)}[/itex] = [a/b].





For 1), I'm not exactly sure that I'm doing this correctly. It seems like I should be using this equivalence class as the argument, but I'm not sure I set it up correctly.

For 2) this seems too simple. Am I missing something? Should I be adding the axioms used at each step?
 
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  • #2
mliuzzolino said:

Homework Statement



Let S = {(a, b): a, b [itex]\in \mathbb{Z}[/itex] and b ≠ 0}. An equivalence relation "~" on S is defined by (a, b) ~ (c, d) iff ad = bc.

1) For any b [itex]\in \mathbb{Z} \setminus[/itex] {0}, show that [0/b] = [0/1] and [b/b] = [1/1].

2) For any a, b [itex]\in \mathbb{Z}[/itex] with b ≠ 0, show that [a/b] + [0/1] = [a/b] and [a/b][1/1] = [a/b].

Homework Equations





The Attempt at a Solution




1) First, show that [0/b] = [0/1].

For any value of b not zero, [0/1] is contained in the equivalence class when b = 1: Eb = {[0/b]: b ≠ 0}. Therefore, [0/b] = [0/1].

Next, show that [b/b] = [1/1].

For any value of b not zero, [1/1] is contained in the equivalence class when b = 1: ESUB]b[/SUB] = {[b/b]: b ≠ 0}. Therefore, [b/b] = [1/1].



2) First show that [a/b] + [0/1] = [a/b].

[a/b] + [0/1] = [itex]\frac{a*1 + 0*b}{b * 1}[/itex] = [itex]\frac{a + 0}{b}[/itex] = [a/b].

Next, show that [a/b][1/1] = [a/b].

[a/b][1/1] = [itex]\frac{a(1)}{b(1)}[/itex] = [a/b].





For 1), I'm not exactly sure that I'm doing this correctly. It seems like I should be using this equivalence class as the argument, but I'm not sure I set it up correctly.

For 2) this seems too simple. Am I missing something? Should I be adding the axioms used at each step?

They seem to be switching notation. I think when they write [0/b] they should actually be writing [(0,b)], which is the set of all pairs (c,d) such that (c,d)~(0,b) or c*b=d*0. Similarly [(0,1)] is the set of all ordered pairs (c,d) such that (c,d)~(0,1) or c*1=d*0. So what you have to show to show [(0,b)]=[(0,1)] is that c*b=d*0 iff c*1=d*0 when b is not zero.
 
  • #3
mliuzzolino said:

Homework Statement



Let S = {(a, b): a, b [itex]\in \mathbb{Z}[/itex] and b ≠ 0}. An equivalence relation "~" on S is defined by (a, b) ~ (c, d) iff ad = bc.

1) For any b [itex]\in \mathbb{Z} \setminus[/itex] {0}, show that [0/b] = [0/1] and [b/b] = [1/1].

2) For any a, b [itex]\in \mathbb{Z}[/itex] with b ≠ 0, show that [a/b] + [0/1] = [a/b] and [a/b][1/1] = [a/b].

Homework Equations


The Attempt at a Solution

1) First, show that [0/b] = [0/1].

For any value of b not zero, [0/1] is contained in the equivalence class when b = 1: Eb = {[0/b]: b ≠ 0}. Therefore, [0/b] = [0/1].

I'm going to assume that the notation [0/b] refers to the equivalence class that contains the element (0, b), since this "/" notation is not defined in your post, but is probably what is intended. Question 1 is asking you to show that [0/b] is equal to the equivalence class [0/1] for every (each) value of b except 0. So, in other words, one of the statements that your proof must apply to is that the equivalence class [0/2] is equal to the equivalence class [0/1]. Your proof does not show this.
 
Last edited:
  • #4
I now understand that [b/b] is (b, b).

Let's say I try 1) again and try to show that [0/b] = [0/1].

This is equivalent to, show that (0, b) ~ (0, 1).

Can I do this by using the defined equivalence relation? (a,b) ~ (c,d) iff ad = bc?

Then (0, b) ~ (0, 1) would equal 0*1 = b*0. Then 0 = 0 which is true.

Does this show that [0/b] = [0/1]?



For 2) would it be something like..

[a/b] + [0/1] = [a/b]

(a, b) + (0, 1) = a/b + 0/1 = (a*1 + 0*b / b*1) = (a + 0)/b = a/b => (a, b)?

I apologize if I'm doing this entirely wrong. It just makes no sense to me.
 
  • #5
mliuzzolino said:
I now understand that [b/b] is (b, b).

Let's say I try 1) again and try to show that [0/b] = [0/1].

This is equivalent to, show that (0, b) ~ (0, 1).

Can I do this by using the defined equivalence relation? (a,b) ~ (c,d) iff ad = bc?

Then (0, b) ~ (0, 1) would equal 0*1 = b*0. Then 0 = 0 which is true.

Does this show that [0/b] = [0/1]?
For 2) would it be something like..

[a/b] + [0/1] = [a/b]

(a, b) + (0, 1) = a/b + 0/1 = (a*1 + 0*b / b*1) = (a + 0)/b = a/b => (a, b)?

I apologize if I'm doing this entirely wrong. It just makes no sense to me.

This is a pretty sloppy presentation of a problem. No wonder you are confused. They should really have given you an addition rule for ordered pairs, or at least asked you to figure one out given that you want to add them like you add fractions. So sure, you ought to define (a,b)+(c,d)=(ad+bc,bd). What you are doing is correct.
 
Last edited:

What is an ordered field?

An ordered field is a mathematical structure that combines the properties of both a field and a total ordering. This means that the elements in the field can be added, subtracted, multiplied, and divided, and also have a defined order relation such that for any two elements, one is greater than the other.

What are the properties of an ordered field?

The properties of an ordered field include closure under addition, subtraction, multiplication, and division, as well as the trichotomy property (for any two elements, one is greater than the other or they are equal) and the transitive property (if a < b and b < c, then a < c).

How do you show that a field is ordered?

To show that a field is ordered, you must first prove that it is a field by demonstrating that it satisfies the required properties of a field (such as closure, associativity, and distributivity). Then, you must show that it also satisfies the properties of a total ordering, such as reflexivity, antisymmetry, and transitivity.

What is the significance of ordered fields in mathematics?

Ordered fields are important in mathematics because they allow us to define and compare quantities in a meaningful way. They are used in many areas of mathematics, such as analysis, algebra, and geometry, and provide a foundation for more advanced structures and concepts.

Can you give an example of an ordered field?

One example of an ordered field is the set of real numbers, which satisfies all of the required properties of an ordered field. Other examples include the set of rational numbers and the set of complex numbers, both of which can also be ordered.

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