I've been doing some research into the basic definition of numbers, and I'd just like to make sure that I have everything right.
For the purposes of this post, natural number will be in bold, integers will be in italics, and rationals will be underlined.
We start be defining the natural number 0 to be the empty set. we can then inductively define the natural numbers N by saying that, for any natural number x, the subsequent natural number is the union of x and {x}. Let us call the natural number sucessor function S(x), such that S(x) is the union of x and {x}.
We say that a<b iff a is an element of b, and hence order the natural numbers.
We can then define natural number addition for all natural numbers x and y + such that x+0=x and S(x)+y=x+S(y). This gives us addition for all natural numbers, with additive identity of 0.
Finally, we can define natural number multiplication X such that, for all natural numbers a and b, aX0=0 and aX(S(b))=(aXb)+a. It emerges that the multiplicative identity is 1.
This gives each natural number as a set. To define the integers I, we define each natural number as an ordered pair of natural numbers, (a,b), such that (a,b)=(c,d) iff a+d=c+b.
We can define the integest succession function S(x) such that S((a,b))=(S(a),b)
We can define integer addition + to be such that (a,b)+(c,d)=(a+c,b+d). By doing this, it emerges that the additive identity, 0, is (0,0).
We can define integer subtraction - to be such that (a,b)-(c,d)=(a+d,b+c). It can then be seen that integer subtraction is the inverse of integer addition, with the same identity.
We can define integer multiplication X for all integers a,b to be such that aX0=0 and aX(S(b))=(aXb)+a, whereupon it can be seen that 1 is the multiplicative inverse.
We can now define a map from N to I such that for any natural number a, there exists integer b such that b=(a,0). This exists for all natural numbers, and each natural number has a corresponding integer, so we can define in integer natural numbers N such that a in N corresponds to (a,0) in N. Although N is not a subest of I, N is.
We define integer greater than > by saying that (a,b) > (c,d) iff a+d>b+c.
We can now define the rational numbers, Q. Once again, we define a rational number as an ordered pair, but it is an ordered pair of integers (a,b). We say that (a,b)=(c,d) iff aXd=cXb and bXd is not 0.
We define addition of the rational numbers + by (a,b)+(c,d)=((aXd)+(cXb),bXd). From this it emerges that (0,1) is an additive inverse. However, it alse emerges that there are many additive identity, of the form (0,a) for some integer a.
We define rational multiplication X by saying that (a,b)X(c,d)=(aXc,bXd). It can be observed that there are many multiplicative identities, each of which are equal to (1,1).
We can define integer subtraction - such that (a,b)-(c,d)=((aXd)-(cXb),bXd). Similarly, we define integer division / such that (a,b)/(c,d)=(a,b)X(d,c).
We say than a rational (a,b) is positive if aXb>0. We define rational greater than > by saying that a>b if a-b is positive.
We can define a map for all integers a, mapping them to a=(a,1). This is defined for all integers, so we can create subsets of Q, that integers and natural numbers I and N.
2-With the above, it is possible to divide by zero. We can't really do anything with numbers that have been divided by zero (can't use greater than, can't say it is equal to another number), but is still possible. Is this a grave mistake?
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