1. ## arithmetics relations

On which axioms <, > relations is based? (Bigger than, smaller than)
Can I exchange the relations by =?

3. ## Re: arithmetics relations

A "relation" between sets X and Y is a collection of ordered pairs, the first member of each pair from X the second from Y. A "relation" on a set X is a collection of ordered pairs, both members from X. An "order relation" on set X is a relation on X that is "transitive"- that is, that if (x, y) and (y, z) are in the relation then so is (x, z).

Writing that relation as "<", that is, that (x, y) is in the relation is represented by "x< y", then the condition is that "if x< y and y< z then x< z". That is the only "axiom" required for an order relation.

4. ## Re: arithmetics relations

I think that what you describe is a definition on group theory axioms.
Can this definition base on arithmetic (the arithmetic theory itself)?

5. ## Re: arithmetics relations

Originally Posted by policer
On which axioms <, > relations is based? (Bigger than, smaller than)
Can I exchange the relations by =?
One approach to this is the following.
Axiom: In the field of real numbers there exists a set $\mathbb{P}\subset\mathbb{R}$ having the properties
i) for each $x\in\mathbb{R}$ exactly one of $\bf{x=0,~x\in\mathbb{P},\text{ or }-x\in\mathbb{P}}$ is true.
ii) the set $\mathbb{P}$ is closed with respect to addition and multiplication.

Now using that axiom we define: $\large\boxed{a<b~~\text{ if and only if }~(b-a)\in\mathbb{P}}$

6. ## Re: arithmetics relations

Moved to the Advanced Algebra Forum.

-Dan

7. ## Re: arithmetics relations

Originally Posted by Plato
Now using that axiom we define: $\large\boxed{a<b~~\text{ if and only if }~(b-a)\in\mathbb{P}}$
EDIT
This should read: If $\large{a~\&~b\text{ are two numbers }\boxed{a<b~~\text{ if and only if }~(b-a)\in\mathbb{P}}}$

8. ## Re: arithmetics relations

Originally Posted by policer
On which axioms <, > relations is based? (Bigger than, smaller than)
Can I exchange the relations by =?
In response to your second question, can you turn the strict ordering relation into an equivalence relation? Short answer: no. The reason is because the strict ordering relation is not reflexive or symmetric. It is transitive, though. If you were to say that it were both reflexive and symmetric, then it would, indeed, be an equivalence relation, but it would not look anything like the strict order relations that you are used to.

For example, let us augment the < relation to make it both reflexive and symmetric.

We have $<^*$ where for any two numbers $x,y$ we have $x<^* y$ if and only if $x<y$ or $y<x$ or $x=y$.

But, then we wind up with all numbers being equivalent.