# Homework Help: Showing properties of Ordered Fields

1. Apr 18, 2013

### mliuzzolino

1. The problem statement, all variables and given/known data

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

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

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

2. Relevant equations

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

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

[a/b][1/1] = $\frac{a(1)}{b(1)}$ = [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?

2. Apr 18, 2013

### Dick

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. Apr 18, 2013

### slider142

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: Apr 18, 2013
4. Apr 18, 2013

### mliuzzolino

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. Apr 19, 2013

### Dick

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: Apr 19, 2013