Definition 1.1:
i) Let Armstrong be me.
ii) Let {Armstrong} be the set containing me.
Proposition 1.2:
P({Armstrong})={{},{Armstrong}}
Remark: I like to think of myself as an element of my own power set... It's less lonely with the empty set for company.
Definition 1.3:
i) Let Eden be Eden.
ii) Let Chuck be Chuck.
iii) Let Brutal Snake be Brutal Snake.
iv) Let Boozy be Boozy.
v) Let Sparky be Sparky.
vi) Let Number Six be Number Six.
Lemma 1.4:
Number Six is not a number.
Proof: c.f. The Prisoner.
Proposition 1.5:
The integers do not exist.
Proof: Assume that there is an integer with the existence property, (i.e. the integer exists), let's call this integer x. Either x=6 or x is greater than 6 or x is less than 6.
In the first case we immediately have a contradiction, since x is not a number, (Lemma 1.4).
Consider the case that x is greater than 6, we show first that 7 is not an integer, for if it were, then 7 - 1 would also be an integer, yet we have already demonstrated that this is not the case.
Furthermore, we observe that if k is not an integer then neither is k+1, and we invoke the principal of mathematical induction to show that no integers greater than 6 can exist.
The case of x less than 6 is left as an exercise
Remark: If the integers do not exist, then my degree may prove to be pretty worthless. We must therefore conclude that our definitions are not self-consistent, and reluctantly abandon all rigour in future blogs.
Subscribe to:
Post Comments (Atom)
2 comments:
"we show first that 7 is not an integer, for if it were, then 7 - 1 would also be an integer"
Why? Are we not constructing the integers here? Or are we assuming the closure of the integers under addition already?
OK, to clarify...
By an integer I mean an element of a set where the following axioms hold:[1]
1. a+b=b+a
2. (a+b)+c=a+(b+c)
3. There exists an element, (called 0), s.t. 0+a=a+0=a
4. For each a, there exists a unique integer, (called -a), s.t. a+(-a)=(-a)+a=0
5. a.b=b.a
6. (a.b).c=a.(b.c)
7. There exists an element, (called 1), s.t. 1.a=a.1=a
8. a.(b+c)=a.b + a.c
9. (a+b).c=a.c + b.c
10. The set contains a non-empty subset N s.t. each element of the integers belongs to exactly one of the sets N, {0} or -N, where -N denotes the set {-x:x an element of N}, and for all a,b in N we have both a+b and a.b in N.
11. If U is a subset of N s.t. 1 is in U, and a+1 is in U whenever a is in U, then U=N, (i.e. the principal of mathematical induction holds).
In my 'proof', the assumption of the existence of a set with the above properties was meant to be implicit. I believe it to be a valid assumption, for if the set does not exist, then the integers do not exist and we are done.
[1] R.B.J.T. Allenby; Rings, Fields and Groups
Post a Comment