I was asked by a colleague yesterday about real solutions to equations, i.e.,

sqrt(1-x)-sqrt(x-3)= 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(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.

Enjoy.