02/20/2008, 11:01 PM
andydude Wrote:If we choose to hold \( a [N-1] a = a [N] 2 \) true for all N, then we get \( a [0] b = b + 2 \) for a == b. Both cannot be true, so we must make a choice.
My suggestion is to keep: a[s-1]a = a[s]2, like:
a^a = a[3]a = a[4]2 = a#2
a*a = a[2]a = a[3]2 = a^2
a+a = a[1]a = a[2]2 = a*2
a°a = a[0]a = a[1]2 = a+2
GFR

