Originally posted by mtthw
Since at no point does it define what it means by "finite number", I'd say it's not a very satisfying proof.
I'd say that the
proof that the book gives is complete bollocks. Even the question is complete bollocks. Just like you say what is a finite number is never defined and we don't know what we have to assume that 7 is finite. If we were to go with the usual axioms of the set of real numbers I'd do something like this:
Let us assume that there is a set of numbers that we'll call the set of the real numbers from now on and denote by
R
Lets us say that there are two operations in this set of numbers and let us call them addition and multiplication.
Here are the axioms for the addition operation that for every two real numbers (a,b) makes a correspondende to another real number (c) a+b=c where + denotes the addition operation.
1 - a+b=b+a
2 - (a+b)+c=a+(b+c)
3 - for every real number there exists one and only one real number (n) which has the following property a+n=n+a=a
4 - for each real number exists one other real number that gives a+b=n.
For purposes of clarification let us call the real number that we denoted as n the number zero and write as 0. And let us call the number that summs up to 0 with another real number as the symmetric of the real number a and denote it as -a.
The axioms for the multiplication operation are:
1 - a.b=b.a where . denotes the multiplication operation.
2 - (a.b).c=a.(b.c)
3 - a.(b+c)=a.b+a.c
4 - there exists a real number e with the following property: a.e=e.a=a. Let us write 1 for e from now on and call this number one.
5 - For every real number there exists another real number with following property: a.b=b.a=1 and this number b will be called from now on as the inverse of the number a and an be denoted as a^(-1)
Now let us note that the numbers 0 and 1 can be different representations for the same number. But if that would be the case than the set of the real numbers would be composed of one number only and that wouldn't be very interesting. So let us say that indeed we have that 1 and o are different. (1!=0) Now we have the set of real numbers is composed of an infinite number of elements.
But what about the ordering relationships between 0 and 1? And by the way what are these ordering relationships I speak of? Let us define them by using the usual ordering signs.
I think that first we have to define positive and negative numbers. A number is said to be positive when a > 0 and negative otherwise. 0 is neither negative nor positive or if you wish he is negative and positive at the same time.
We have the following property for the any given two real numbers a < b, a > b , a = b and they are mutually exclusive.
1 - a < b implies a + c < b + c
2- a < b implies a.c < b.c if c is positive and a.c > b.c if c is negative.
So we are now able to proof that 1 > 0. Let us do this contradiction. 1 < 0 and then we can multiply each side of the equation by 1 so we have 1.1 >1.0 (remember that we assumed that 1 was a negative number in the beginning and so we are using property 2) which gives 1 > 0. But this is against our first assumption so it must be wrong. So now we have that indeed 1 > 0 and we have lots of real numbers to play around with one other thing that we can do is to define the inverse operations to addition and multiplication. This is easy to do and is just a case of
thinking backwards. Let us define 2=1+1. Now finite is taken to be an undefined term in this theory but we know that all real numbers are finite. An infinite number is a number that isn't a real one. This seems highly childlike but I really don't see any other way around. So 2 is finite. By the same token we can define seven as being 1+1+1+1+1+1+1=7 so seven is finite.
I think this way of approaching things is way better that the one the author of that bok used. It is more coherent and more mathematical so to speak. ut waht really pisses me of is his reasoning is this:so 3+4= 7 and since we agreed that 3 and 4 are finite 7 is too. But I never agreed that 3 and 4 are finite did I? And when was it proved that that a sum of two finite number yields a finite number as a result? But what I think is my most powerful argument is this: well 3=1+1+1 and 4=1+1+1+1 so 7 is indeed what he says we can use to prove 7 is finite. He's just masquerading that fact isn't he?
Had to edit to correct some text that was eaten out.