1. Inspired by the topic: "Elementary Number Theory" I thought I'd make a topic with axioms.
Below Peano's axioms are written.
1. 0 (sometimes 1) is a natural number.
2.The successor of a natural number is always a natural number.
3.Two different natural numbers never have the same successor.
4. 0 is not the succesoor of any natural number.
5.The axiom of induction, if 0 has a quality P, if the successor of any natural number, n, has quality P and if n has quality P then every natural number has the quality P.

2.

3. Addition + can be defined for recursively as follows: for any natural number n, n + 1 = n* (using r* to denote the successor of the natural number r) and n + m* = (n+m)* for all natural numbers m. All the usual properties of addition follow: commutativity, associativity, etc.

Multiplication is defined thus. n1 = 1 and nm* = nm + n. Multiplication is also commutative and associative; it is moreover distributive over addition.

4. You can also define an order on . For , let for some . Then is totally ordered under and , and .

5. Originally Posted by JaneBennet
You can also define an order on . For , let for some . Then is totally ordered under and , and .
I goes beyond that even. The natural numbers are well-ordered -- every non-empty subset contains a least element.

You can in fact push the Peano axioms all the way to construct the natural numbers, integers, rational, reals and complex numbers with all of the usual algebraic operations. In addition using the order relation that you noted you can use the notion of Dedikind cuts to construct the real numbers and show that the result is topologically complete, then go on to do the same for the complex numbers. The bottom line is that everything you wanted to know about calculus and the theory of one complex variable is contained in the Peano axioms.

You can see this construction done in Landau's little book Foundations of Analysis. It is very dry, but other than that is a short and instructive read.

6. Originally Posted by DrRocket
You can in fact push the Peano axioms all the way to construct the natural numbers, integers, rational, reals and complex numbers with all of the usual algebraic operations.
Very well, let’s do it!

Define an equivalence relation ~ on by . Then ~ is an equivalence relation, and the integers are defined to be the set of all equivalence classes under ~: , where I have written to denote the equivalence class containing . (Intuitively, represents the “difference” between the natural numbers m and n.)

Now define addition , multiplication and order on as follows: , , . Yes, the two binary operations and the order relation are all well defined, as you can check for yourself.

You can also check for yourself that , with zero element and unity , satisfies all the axioms listed by Faldo_Elrith in this post: http://www.thescienceforum.com/viewtopic.php?t=8466. We have therefore constructed a well-ordered ring from the natural numbers.

Moreover, the mapping , is bijective and has the property that , , , . Hence the structure is isomorphic to a substructure of . Therefore we shall re-define our natural numbers as a subset of our integers, treating the natural number n as the integer . We will also rewrite , and more familiarly as , and (or even drop the multiplication symbol altogether). 8)

7. How do you prove that multiplication of integers is well-defined? I'm getting nowhere with it.

8. Originally Posted by JaneBennet
Originally Posted by DrRocket
You can in fact push the Peano axioms all the way to construct the natural numbers, integers, rational, reals and complex numbers with all of the usual algebraic operations.
Very well, let’s do it!

Define an equivalence relation ~ on by . Then ~ is an equivalence relation, and the integers are defined to be the set of all equivalence classes under ~: , where I have written to denote the equivalence class containing . (Intuitively, represents the “difference” between the natural numbers m and n.)

Now define addition , multiplication and order on as follows: , , . Yes, the two binary operations and the order relation are all well defined, as you can check for yourself.

You can also check for yourself that , with zero element and unity , satisfies all the axioms listed by Faldo_Elrith in this post: http://www.thescienceforum.com/viewtopic.php?t=8466. We have therefore constructed a well-ordered ring from the natural numbers.

Moreover, the mapping , is bijective and has the property that , , , . Hence the structure is ismorphic to a substructure of . Therefore we shall re-define our natural numbers as a subset of our integers, treating the integer as the natural number n. We will also rewrite , and more familiarly as , and (or even drop the multiplication symbol altogether). 8)
I'm afraid that you started out in defining your equivalence relation with already having a notion of addition of natural numbers, and in fact having the natural numbers already well-defined. The construction that starts with the Peano axioms actually goes through the process of constructing the natural numbers then defining addition of natural numbers and eventually the integers and subtraction from nothing more than the Peano axioms. It is a somewhat long process that I find a bit tedious. You can see it done in Landau's book. It is not hard and it is not very long, but it takes a lot more than one page or one frame in a forum post.

What you did is not wrong, but it assumes a lot more than just the Peano axioms.

9. Well, my process is this.
1. Start with the Peano axioms for the natural numbers. (Done by thyristor.)
2. Define addition, multiplication and order for the natural numbers first. (Done by Faldo_Elrith.)
3. Construct the integers as equivalence classes of equivalence relation of “differences” on the natural numbers.
4. Define addition, multiplication and order for the integers in terms of addition, multiplication and order for natural numbers.
5. Prove various relationships between the natural numbers and the integers.
I don’t see why anyone would want to construct the integers before defining addition, multiplication and order for natural numbers first. :? I mean, before you have your integers, wouldn’t you want to play around with your natural numbers for a little while first? :P

10. Originally Posted by JaneBennet
Well, my process is this.
1. Start with the Peano axioms for the natural numbers. (Done by thyristor.)
2. Define addition, multiplication and order for the natural numbers first. (Done by Faldo_Elrith.)
3. Construct the integers as equivalence classes of equivalence relation of “differences” on the natural numbers.
4. Define addition, multiplication and order for the integers in terms of addition, multiplication and order for natural numbers.
5. Prove various relationships between the natural numbers and the integers.
I don’t see why anyone would want to construct the integers before defining addition, multiplication and order for natural numbers first. :? I mean, before you have your integers, wouldn’t you want to play around with your natural numbers for a little while first? :P
Yes indeed. But Faldo_Elrith did not really construct or prove anything. He gave the correct definition but only waved his hands at the proof of the various properties of addition of natural numbers. It actually takes some work to get those facts from the Peano axioms and the definitions. It is not a pretty process but you can work everything out in detail (it takes Landau about 15 pages to slog through this stuff).

11. Ah, well. Faldo has obviously decided to take some results for granted. And so have I, actually. :P

But you’re right: we shouldn’t be so slipshod in our endeavours and should at least make some effort to prove some important results.

Anyway, Faldo raised this important question:

Originally Posted by Faldo_Elrith
How do you prove that multiplication of integers is well-defined? I'm getting nowhere with it.
Let’s try and do it together, shall we?

Suppose and . We want to show that .

I suppose this can be done in two stages: first show that , then show that .

Well …

Multiply by and in turn:

Add LHS of first equation to RHS of second:

And doing something similar with should yield . 8)

12. Originally Posted by Faldo_Elrith
How do you prove that multiplication of integers is well-defined? I'm getting nowhere with it.
You need to first show that multiplication of natural numbers is well-defined and does what you expect. In that light it is perhaps helpful to go back and amend your earlier post. 0 is not a natural number, but 1 is. So to start the definition of multiplication of natural numbers you need Xx1 = x and XxY*= XxY + X.

Does that help ?

13. Originally Posted by JaneBennet
And doing something similar with should yield . 8)

Multiply by and in turn:

Add LHS of first equation to RHS of second:

THANKS!!!

14. And now let’s construct the rational numbers from the integers!

NB: I will present the main details and skip the proofs of various results. These can be proved in our own time; right now, I just want to present the theory as a whole picture without being sidetracked by complicated proofs.

First, recall the defintion of zero 0 as the integer ; likewise the integer 1 was defined as . Alternatively, we may just think of 0 and 1 as the additive and multiplicative identites respectively in the ring .

Let denote the set of all nonzero integers: . Define a relation ~ on by . Then ~ is an equivalence relation, and the rational numbers are defined as the set of all equivalence classes – where I have written to denote the equivalence class containing . (Intuitively, represents the “ratio” of the integer a and the nonzero integer b.)

Now define addition , multiplication and order on as follows: , , . These definitions are well defined – proofs omitted for the moment.

These defintions make an ordered field. (Proof maybe later.)

Moreover, the mapping , is bijective and has the property that , , , . Hence the ring is isomorphic to a subring of the field . Therefore we shall re-define our integers as a subset of our rational numbers, treating the integer a as the rational number . We will also rewrite , and more familiarly as , and (or even drop the multiplication symbol altogether). 8)

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