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.,
sqrt(1x)sqrt(x3)= 0
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(1x) and y=sqrt(x3) 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 TI8X calculators give the thirdglance result (1) and produces a graph from inf to +inf (the typical cube root function).
Mathematica gives the secondglance result (complex value), and a graph from 0 to +inf accordingly.
Back Ontrack, slightly:
my HP15 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 nonhp) 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.
Enjoy.
