The arithmetics of natural numbers

 The standard way of introducing the arithmetics of natural numbers is via Peano axioms -- it is a very ascetic way. However, I'd like a comfortable way that will lead to the number theory much faster.

Many people have an idea of natural numbers as a set  N = {1 2 ...}  together with operations  +  and  *  performed on natural numbers, i.e. on elements  1 2 ...  of  N.  In this wlog this idea will get a precise meaning.

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Asscio (N +)
----------------

We start with a set  N, and with an associative operation binary (2-argument) operation  +.  Being an operation means that  a+b  belongs to  N  for all  a b  belonging to  N.

Thus, the pair

                           (N +)

forms an ascio -- it's just a very simple kind of algebra. Operation  +  is called  addition.

Remark 0:  The standard terminology for  ascio  is semigroup.

By definition, associativity of  +  in  N  means:

(Axiom 0)  associativity:     (a+b)+c = a+(b+c),         for all elements  a b c  of  N.

We also assume the following axioms:

(Axiom 1)  commutativity:   a+b = b+a,   for all  a b  belonging to  N;

(Axiom 2)  uniqueness:    if  a+b = a+c  then  b=c,  for all  a b c  belonging to  N;

(Axiom 3)  natural progress:  a+b =/= a  for all  a b  belonging to  N.

Remark 1   Thanks to associativity of  +  (Axiom 0) we can write for instance  a+b+c  with the understanding  a+b+c := (a+b)+c = a+(b+c).

========================

Inequalities
---------------


The binary inequality relation < in N is defined as follows:

Definition 0:  For arbitrary  a b  in  N,

              a < b  <=:=>  there exists  x in N  such that  a+x = b

Theorem 0:  (Transitivity) For arbitrary  a b c  in  N,

                      (a < b   &   b < c)   ==>   a < c

Proof:  Let  a b c  in  N  be such that  a < b  &  b < c.  Then there exist  x y  in  N  such that

                     a + x = b       &.       b + y = c

Then by the associativity Axiom 0:

                    a + (x+y) = (a+x) + y = b+y = c

i.e. there exists  z  in  N,  namely  z := x+y,  such that  a+z = c.  End of Proof

Theorem 1:  For arbitrary  a b  in  N,

                              never (a < b   &   b < a)

Proof:  By contradiction, let arbitrary  a b  belonging to  N  be such that  a < b  and  b < a. Then there would be  x y  in  N  such that

                                 a+x = b   &   b = a+y

Then, by associativity:

                     a + (x+y)  =  (a+x) + y  =  b+y = a

in a contradiction to the natural progress Axiom 3 (where Axiom's  b  is substituted by  x+y).

Remark 2:  In general, the inequality relations are called orders. In view of Theorem 1, the inequality relation  <  in  N  is said to be a linear order.

========================

Monoid (N  *  1)
---------------------

Set  N  admits more than just the addition operation +, there is also the multiplication operation  *  in  N,  it satisfies some new axioms:

(Axiom 4)  associativity:     (a*b)*c = a*(b*c),         for all elements  a b c  of  N.

This means that  (N *)  is another ascio. It is commutative, and it has the uniqueness property:

(Axiom 5)  commutativity:   a*b = b*a,   for all  a b  belonging to  N;

(Axiom 6)  uniqueness:    if  a*b = a*c  then  b=c,  for all  a b c  belonging to  N.

However, we deal here with more than ascio  (N *),  we work here with monoid  (N * 1), and with more axioms:

(Axiom 7)  unit:  1 is an element of  N -- it is called the unit element;

(Axiom 8) neutrality:    1 * a = a   for every  a  belonging to  N.

Remark 3   Thanks to associativity of  *  (Axiom 4) we can write for instance  a*b*c  with the understanding  a*b*c := (a*b)*c = a*(b*c).


==========================================================

Natural arithmetics  (N + * 1) -- part 1
-------------------------------------------------

The addition  +  and the multiplication  *  together with  1  need to be connected in order to obtain full natural arithmetics  (N + * 1).  This requires more axioms:

(Axiom 9) distributivity:    (a+b)*c  =  (a*c) + (b*c)     for all  a b c  belonging to  N.

Actually, we always assume that multiplication  *  binds stronger than addition  +  hence for instance

                  (a*c) + (b*c)  =  a*c + b*c

On the other hand,  natural numbers  (a+b)*c  and   a + b*c  are almost always different;  of course

                 a + b*c  =  a + (b*c)

but that, as a rule, is different from  (a+b)*c. We see that the distributivity equation (from Axiom 9) can be written simply as:

                       (a+b)*c  =  a*c + b*c

---

Natural number  1  is given as one of the initial notions of the natural arithmetics. Then we define more specific natural numbers as:

    2 := 1+1        3 := 2+1          4 := 3+1          etc.

Then while we already know that  1*n=n (Axiom 8), we easily prove that:

                  2*n = n+n

Indeed,  2*n = (1+1)*n = 1*n + 1*n = n+2. Great!  Then we show that:

         3*n = n+n+n               4*n = n+n+n+n              etc.

=====================================================

Exponentiation  ^  and powers  a^b
----------------------------------------------

Binary operation  ^  in  N  satisfies two axioms:

(Axiom 10)    a^1 = a    for every  a  in  N

(Axiom 11)    a^(b+c) = (a^b)*(a^c)   for all  a b c  belonging to  N

Now we obtain the algebraic identities known from high school (or even elementary school) such as:

                                    (a+b)^2 = a^2 + 2*a*b + b^2

Indeed,
                (a+b)^2  =  (a+b)^(1+1)  =  ((a+b)^1)*((a+b)^1)  =  (a+b)*(a+b)  =

                a*(a+b) + b*(a+b)  =  a*a + a*b + b*a + b*b  =  a^2 + 2*a*b + b^2

Great!

     

We also obtain implications such as:

                                    a+b = c  ==>  a*(b+c) + b*2  =  c^2

Indeed,

              a*(b+c) + b^2  =  a*(a+2*b) + b^2  =  a^2 +2*a*b + b^2  =  (a+b)^2  =  c^2

Great!

=======================

Induction -- arithmetics completed
---------------------------------------------

Finally, the last (but not the least ☺) natural arithmetics axiom is the induction axiom:

(Axiom 12) INDUCTION:  For arbitrary subset  A  of  N,  if it satisfies two conditions:

  • unit  1  belongs to  A;
  • for every  a  in  A  also  a+1  belongs to  A;
then  A=N.

Induction in mathematics, especially in the number theory, is ubiquitous. Here is the first application of the induction:

Theorem 2      a+b =/= 1  for arbitrary  a b  belonging to  N.

Proof   This is true when. a=1,  i.e.  by Axiom 3

          1 + b  =/=  1    for arbitrary  b  in  N

Thus  a = 1  belongs to the subset  A  of  N  of  all natural numbers  a  such that

           a + b =/= 1      for arbitrary  b  in  N

hence the first induction's condition (Axiom 12 just above) is satisfied. Now, consider an arbitrary element  a  of the just defined set  A.  Then for arbitrary  b  in  N  we have

                       (a+1) + b  =  1 + (a+b)  =/= 1
i.e.
                       (a+1) + b =/= 1

We see that  a+1  belongs to  A. Now, the induction axiom tells us that  A = N,  i.e.  a+b=/=1  for arbitrary a b  in  N.     End of Proof

=======================================

Divisibility
========

Whenever  a b c are natural numbers such that  a*b = c,  numbers  a b  are said to be divisors of  c, and we can write   a | c  as well as  b | c.

While inequality  a < b  is an order in  N  related to  +,  divisibility  a | b  is an order in  N  related to multiplication  *:

Definition 1
                             a | b   <=:=>   there exists  x  in  N  such that  a*x = b

for arbitrary  a b  belonging to  N.  

Theorem 3     Divisibility  |  enjoys the following two properties:
  •   (a | b      &      b | c)    ==>    a | c

  •       a | a
for arbitrary  a b c  belonging to  N.

-----

For instance:
  • 1 | 6   &   2 | 6   &   3 | 6   &   6 | 6  --  thus  6  admits four different divisors, and not more.
  • 2 | a+a    &    a | a+a
  • a | a^2
for arbitrary  a  in  N.

-------

The last property (that a | a)  of  Theorem 3  tells us that the divisibility order  |  is not sharp. This is in a contrast to ordering  <  -- the ordering  <  is sharp meaning

                       a < a   for no  a  in  N

However, there is also a non-sharp variation of  <  namely ordering  \<  defined as follows:

                        a \< b  <=:=>  ( a = b   or   a < b )



                 UNDER CONSTRUCTION

























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