Poweroids

[andydude]

• \( (\mathbb{R}^{+}-\{1\},\ \mathbb{R}^{+},\ \uparrow,\ 1) \) forms a poweroid.
  

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Andrew Robbins

SCRATCH THAT! it is false.

\( \{1/2,\ 2\} \subset \mathbb{R}^{+}-\{1\} \), but \( \log_2(1/2) = -1 \) which is not a member of the positive reals. However, with the slight modification

• \( (\mathbb{R}^{+}-\{1\},\ \mathbb{R}-\{0\},\ \uparrow,\ 1) \) forms a poweroid.
  

I believe the statement is true.

The reason why 0 and 1 must be excluded is so that \( \sqrt[y]{x} \) is uniquely defined. If we include 1, 0 in the base domain, exponent domain respectively, then roots are not uniquely defined, at which point it stops being a poweroid.