- #1
Rijad Hadzic
- 321
- 20
I'm trying to prove that a^0 is = 1
So if I define a^1 to be = (a)(1)
and a^n to be = (1)(a)(a)...(a) with the product being taken n times
and a^m to be = (1)(a)(a)...(a) with the product being taken m times
a^n * a^m would then = (1)[(a)(a)...(a) with the product being taken n times * and a^m to be = (1)(a)(a)...(a) with the product being taken m times]
which clearly gives a^n * a^m = a^(n+m)
if m = 0, a^n * a^0 = a^(n+0) = a^(n), so a^0 = 1
for some reason this does make sense to me but I have a feeling the result is not satisfying enough.
So if I define a^1 to be = (a)(1)
and a^n to be = (1)(a)(a)...(a) with the product being taken n times
and a^m to be = (1)(a)(a)...(a) with the product being taken m times
a^n * a^m would then = (1)[(a)(a)...(a) with the product being taken n times * and a^m to be = (1)(a)(a)...(a) with the product being taken m times]
which clearly gives a^n * a^m = a^(n+m)
if m = 0, a^n * a^0 = a^(n+0) = a^(n), so a^0 = 1
for some reason this does make sense to me but I have a feeling the result is not satisfying enough.