sqrt question

04062017, 05:46 PM
Post: #6




RE: sqrt question
(04062017 04:51 PM)KeithB Wrote: pier4r: You are mixing mathematical rules, and you are also ignoring order of operations. 1. \( \sqrt{a} \) is usually only defined for \( a \ge 0 \). The multiplicative property \( \sqrt{a}\cdot \sqrt{b} = \sqrt{a\cdot b} \) likewise only applies to nonnegative values of \( a \) and \(b \). Another way of putting it is that this property is a result of the assumption that the domain of \( \sqrt{x} \) is the set of nonnegative real numbers. 2. The order of operations dictate that exponents are evaluated first. So \( \sqrt{4} \cdot \sqrt{9} = 6 \)  however this assumes that we are not using the \( \sqrt{x} \) function as described in #1. In fact, we have have implicitly assumed that the \( \sqrt{\quad } \) operator is defined for negative values, and that the evaluation is \( i\cdot \sqrt{x} \) 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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