sqrt question

04062017, 06:18 PM
(This post was last modified: 04062017 06:21 PM by Han.)
Post: #8




RE: sqrt question
(04062017 06:03 PM)KeithB Wrote: So to summarize, by assuming sqrt(4) is a valid operation we are automatically going to interpret it as 2i. No, the point was that in choosing to evaluate \( \sqrt{4} \cdot \sqrt{9} \) to be equal to 6 (by following the order of operations), we have implicitly made the assumption that \( \sqrt{ x } = i \sqrt{x} \) for \( x < 0 \). The conventional use of \( \sqrt{x} \) is that it is the positive square root of \( x \). There is no single convention, however, for \( \sqrt{x} \) if \( x < 0 \). EDIT: One could also argue that we may have actually implicitly assumed that \( \sqrt{x} = i \sqrt{x} \) for \( x < 0 \). We still get 6, but all the more reason that there is no single convention for \( \sqrt{x} \) for negative values of \( x \). Graph 3D  QPI  SolveSys 

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Messages In This Thread 
sqrt question  KeithB  04062017, 03:25 PM
RE: sqrt question  pier4r  04062017, 04:01 PM
RE: sqrt question  Namir  04062017, 04:02 PM
RE: sqrt question  KeithB  04062017, 04:51 PM
RE: sqrt question  Han  04062017, 05:46 PM
RE: sqrt question  pier4r  04062017, 05:15 PM
RE: sqrt question  KeithB  04062017, 06:03 PM
RE: sqrt question  Han  04062017 06:18 PM
RE: sqrt question  Claudio L.  04072017, 01:23 PM
RE: sqrt question  Han  04072017, 04:48 PM
RE: sqrt question  Claudio L.  04072017, 09:15 PM
RE: sqrt question  Han  04072017, 10:54 PM
RE: sqrt question  Claudio L.  04092017, 03:59 AM
RE: sqrt question  David Hayden  04242017, 09:36 PM
RE: sqrt question  Claudio L.  04262017, 03:08 AM
RE: sqrt question  Han  04282017, 06:09 PM
RE: sqrt question  nsg  04072017, 11:34 PM
RE: sqrt question  Vtile  04092017, 10:41 AM
RE: sqrt question  nsg  04092017, 05:26 PM
RE: sqrt question  Vtile  04092017, 11:07 PM
RE: sqrt question  nsg  04102017, 01:44 AM
RE: sqrt question  Vtile  04252017, 11:38 PM

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