I think you would get sqrt(x^2) = x, if x belonged to the natural domain of sqrt, which is a Riemann surface, that may also be defined using the language of "sheaves". I don't know how to connect this to the article or Mathematica.
replyit's literally the prototypical example for `Assuming`
replyThat's not what it simplifies to using a real or complex number domains for x, it's abs(x). CAS need type inference assumptions and/or type qualifiers to be more powerful.
replyEdit: Fixed stuff.
For x = -i, square(x) = -1, sqrt(square(x)) = i. Meanwhile, abs(x) = 1. You're right that it simplifies to abs(x) for real x, but that no longer holds for arbitrary complex values.
replyfor arbitrary complex values sqrt() gives 2 answers with +- signs
replyso sqrt(square(-i)) = +-i, one of which is x
I've never seen a CAS that gives two answers for sqrt. Mathematica doesn't, sympy doesn't, and IIRC Maxima also doesn't.
replyThe sqrt function returns the principle square root, not both. That’s true for all numbers, positive, negative, and complex alike.
replyIt's abs(x) only over the reals, for complex numbers it's more complicated.
replyThat abs(x) (or |x| as we wrote it) used to catch out so many of us in HS trig and algebra.
replyRight, that's why you need further assumptions on x in order for that simplification to hold.
replyIt's not a simplification, it's wrong. Sqrt(square(x)) equals abs(x).
replyIt also equals x with appropriate assumptions (x > 0).
replyWell, then sin(x) = x if x is infinitely small
replyso there's an unconditionally correct answer (it's also equal to abs(x) for x>0), and then there is an answer that is only correct for half the domain, which requires an additional assumption.
replysqrt(square(i)) != abs(i)
replySo no, it’s not unconditionally correct either.
Not in general. As people have pointed out elsewhere, it's true if x is real. That isn't always a helpful assumption. (When x is real you can plug that assumption into Mathematica. Then Mathematica should agree with you.)
replyBut consider sqrt(i) = sqrt(exp(i\pi/2)). That's exp(i\pi/4). Your rule would give 1 as the answer. It's not helpful for a serious math system to give that answer to this problem.
When I square 1 I don't get i.