Hold on we can do a bit better.

For base sqrt(2) = a using sheldon's y and z.

a^^0 is defined by

a^^0 ^ a^^0 = 2^4 = 4^2 = 16.

( 2 and 4 are a^^oo or the fixpoints )

So a^^0 = ssqrt(16).

From sheldon we then get

a^(y z) = 16.

subst a = sqrt(2) [ for example , this idea works for other bases too ]

=> y z = 8.

Now notice r = -(r+1) for r = -1/2.

So a^^(-1/2) ^ 2 = 8.

So a^^(-1/2) = sqrt( 8 ) = 2 sqrt(2).

Since we have a^^0 and a^^(-1/2) we actually have

a^^(K/2) for all integer K.

Maybe this idea can be extended to a unique analytic solution ...

¯\_(ツ)_/¯

Regards

tommy1729

The master

For base sqrt(2) = a using sheldon's y and z.

a^^0 is defined by

a^^0 ^ a^^0 = 2^4 = 4^2 = 16.

( 2 and 4 are a^^oo or the fixpoints )

So a^^0 = ssqrt(16).

From sheldon we then get

a^(y z) = 16.

subst a = sqrt(2) [ for example , this idea works for other bases too ]

=> y z = 8.

Now notice r = -(r+1) for r = -1/2.

So a^^(-1/2) ^ 2 = 8.

So a^^(-1/2) = sqrt( 8 ) = 2 sqrt(2).

Since we have a^^0 and a^^(-1/2) we actually have

a^^(K/2) for all integer K.

Maybe this idea can be extended to a unique analytic solution ...

¯\_(ツ)_/¯

Regards

tommy1729

The master