2011-03-15, 07:22 PM
Actually, what I meant was that if 0^0=0, then x^k = y must look exactly like y^(1/x) = k. After all, 0^0=0 should still be 0^0=0 when rearranged around.
In this case, k = 1/x. Notice that 1/x can never finitely be 0. In fact, 1/x diverges around the point 0. The only way to make k or x equal to zero is to make the other (+/-) infinity, which violates the initial condition that k=x=0.
I have yet to see a complete proof that 0^0 can ever equal 0. There's always a discontinuity there. People who claim that 0^0=0 only take one-sided limits, which is just isn't as strong as the numerous two-sided limit proofs that exist for 0^0=1. Actually, I would be surprised to see an everywhere continuous (except at 0) function that makes p^q=0 when approached from both sides of the point, (0,0).
In this case, k = 1/x. Notice that 1/x can never finitely be 0. In fact, 1/x diverges around the point 0. The only way to make k or x equal to zero is to make the other (+/-) infinity, which violates the initial condition that k=x=0.
I have yet to see a complete proof that 0^0 can ever equal 0. There's always a discontinuity there. People who claim that 0^0=0 only take one-sided limits, which is just isn't as strong as the numerous two-sided limit proofs that exist for 0^0=1. Actually, I would be surprised to see an everywhere continuous (except at 0) function that makes p^q=0 when approached from both sides of the point, (0,0).