• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Poweroids andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/19/2009, 11:01 PM (This post was last modified: 12/19/2009, 11:03 PM by andydude.) (12/18/2009, 08:13 AM)andydude Wrote: $(\mathbb{R}^{+}-\{1\},\ \mathbb{R}^{+},\ \uparrow,\ 1)$ forms a poweroid. ... 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. « Next Oldest | Next Newest »

 Messages In This Thread Poweroids - by andydude - 12/18/2009, 08:13 AM RE: Poweroids - by andydude - 12/19/2009, 10:55 PM RE: Poweroids - by andydude - 12/19/2009, 11:01 PM

Users browsing this thread: 1 Guest(s)