|Slightly OT: simple math dilemma|
Message #1 Posted by Chuck on 12 Apr 2011, 10:46 p.m.
I was asked by a colleague yesterday about real solutions to equations, i.e.,
Question: is 2 a solution? and at what level of math should it be accepted as a solution.
Most math texts will say it's not a solution: if y=sqrt(1-x) and y=sqrt(x-3) are graphed, they don't intersect in the real plane, yet the real number 2 is a solution (can't argue with that). So what constitutes a real solution. I claim you can't slide into the complex world and then back to real world; or since 2 isn't in the real domain of the functions, it can't be considered as a solution.
One thing led to another, and it got me thinking about the value of (-1)^(2/6).
At first glance I "want" it to be +1, since we have "even" powers, (sort of).
On second glance it could be 1/2 + sqrt(3)/2 i, (the principal sixth root of a negative is complex; and then squared).
But, on third glance, basic algebra says to reduce all rational exponents to lowest terms before evaluating, which gives us -1. Hmmmm.
The TI-8X calculators give the third-glance result (-1) and produces a graph from -inf to +inf (the typical cube root function).
Mathematica gives the second-glance result (complex value), and a graph from 0 to +inf accordingly.
Back On-track, slightly:
my HP-15 gives -1 (as expected). Haven't checked the others.
my 28S gives the complex answer (as does my Casio) and graphs for only x>0.
I've yet to find one that gives +1 (bummer,even powers and all).
What do other calculators (hp or non-hp) give,and what "should" it be?
I now can only presume TI's are for the less mathematically mature student, and HP's are for the more mathematically advanced. (Ouch!)
But I wait your opinions.