tag:blogger.com,1999:blog-1572154923785712186.post6237555554164328323..comments2018-12-09T16:08:40.914-05:00Comments on Professor Quibb: Inverses of the Hyper OperatorsLouishttp://www.blogger.com/profile/15382160997783595665noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-1572154923785712186.post-22530354237225909212017-07-13T06:53:05.911-04:002017-07-13T06:53:05.911-04:00The last above comment is the first and only place...The last above comment is the first and only place on the internet, I have found, which makes the point that one inverse is enough for a complete formalism of hyperarithmetic. That's what I believe and think, at least. So I'd give you the credit for this.beautifulmathuncensoredhttp://www.beautifulmathuncensored.denoreply@blogger.comtag:blogger.com,1999:blog-1572154923785712186.post-28646608943260311452009-05-25T15:46:15.567-04:002009-05-25T15:46:15.567-04:00Thank you for the comment. But a root, rather tha...Thank you for the comment. But a root, rather than being an inverse of exponentiation, is merely a fractional exponent. For example, the square root of x is the same as x^(1/2) while the cube root of x is the same as x^(1/3). Although the square and the square root cancel each other out, they aren't generally inverse operations. Therefore, the log is the only inverse of exponentiation in its general form.<br /><br />For the hyper 4 operator, I have attempted to define the hyperlog, although my results are not universally accepted. The hyperroot, however, is simply the hyper 4 operator of its inverse. For example the "hyperroot" of x hyper 4 y is simply x hyper 4 1/y. For each subsequent operator, such as the hyper 5 and hyper 6 operators, there are separate inverses, and I do not believe that the degree of the operator is its own variable.Louishttps://www.blogger.com/profile/10746982398555711955noreply@blogger.comtag:blogger.com,1999:blog-1572154923785712186.post-81681566482294007792009-05-25T02:16:25.756-04:002009-05-25T02:16:25.756-04:00Just as there can be two inverses to exponentiatio...Just as there can be two inverses to exponentiation, log and root, there can be two inverses to any hyper operator, hyperlog and hyperroot.<br /><br />This is because functions have inverses, but operators, I mean binary operators with two variables (x^y takes x and y) can be made 'functional' and thus invertible either by holding x constant or y constant.<br /><br />Your hyper operators really are sort of trinary; ie 'x hyper 4 y' or 'x hyper y z', there are three arguments.<br /><br />Therefore there should be 3 ways to invert.<br /><br />Sorry to be replying to this so late, I am reading everything which I discover via google alerts very late because I am much slower at reading the alerts than they are at coming!Dranhttps://www.blogger.com/profile/08714941811085673416noreply@blogger.comtag:blogger.com,1999:blog-1572154923785712186.post-6740861267434666112009-02-18T21:13:00.000-05:002009-02-18T21:13:00.000-05:00Louis,I am not able to comment on this post, at le...Louis,<BR/><BR/>I am not able to comment on this post, at least right now, but I may have a query for you to pursue. Something about dating a possible mentioning of an eclipse in Homer's Odyssey. I will give you the article to ponder. Best, Dr. McGayAnonymousnoreply@blogger.com