
Re: Equivalent Definition of Exponentiation on N
Posted:
Nov 19, 2013 4:29 PM


Correction
Here is an informal development of exponentiation on N (including 0) using x^2 =x*x as a starting point.
1. For all x in N, we have x^2 = x*x.
2. For all x, y we have x^(y+1) = x^y * x.
The case for exponents greater than 2 follows directly from this these 2 rules. What about the exponents 0 and 1?
Two cases to consider: x=/=0, x=0
Case 1: x=/=0.
Applying (2), we work the pattern backwards to obtain:
x^2 = x^1 * x = x * x.
Then x^1 = x by right cancelability.
Working the pattern backwards once more, we have:
x^1 = x^0 * x = x = 1*x
Then x^0 = 1 by right cancelablity.
Case 2: x=0
We cannot "work the pattern backwards," because we cannot use right cancelability as above.
From (2), we do have:
0^1 = 0^0 * 0 = 0
where 0^0 can have any value.
Again, we have x^1 = x, but there does not appear to be any restriction on 0^0. Any value will work.
Therefore, we can define ^ as a binary function on N such that:
1. x^2 = x*x for all x in N 2. x^(y+1) = x^y * x for all x in N
This definition leaves 0^0 undefined in the sense that 0^0 is assumed to be a natural number, but no specific value is assigned to it.
From this definition, we can derive:
1. x^1 = x for all x in N 2. x^0 = 1 for all x in N, x=/= 0 3. all the usual Laws of Exponents (see my previous thread).
Dan Download my DC Proof 2.0 software at http://www.dcproof.com Visit my new math blog at http://www.dcproof.wordpress.com

