solve(x==x, x) returns [x == r1]

[pelegm]

Not sure if it's a bug or a problem with the documentation, but without any assumptions on x, solve(x==x, x) returns [x == r1]. The documentation does not state what r1 is, but gives the following example:

   If there is a parameter in the answer, that will show up as a new
   variable.  In the following example, "r1" is a real free variable
   (because of the "r"):

      sage: solve([x+y == 3, 2*x+2*y == 6],x,y)
      [[x == -r1 + 3, y == r1]]

However, without assumptions on x, there's no reason to believe that x is real.

This ticket also fixes a grammar issue introduced in #21554 (see comments).

Depends on #21554

Component: symbolics

Keywords: solve, days79

Author: Ashutosh Ahelleya

Branch/Commit: 224e064

Reviewer: Bryan Gin-ge Chen

Issue created by migration from https://trac.sagemath.org/ticket/21946

[sagetrac-aashu12]

Branch: u/aashu12/documentation

[sagetrac-aashu12]

Author: Ashutosh Ahelleya

[sagetrac-aashu12]

New commits:

c7acfd6	Fixes #21554
e8760d5	Fixed: #21946

[sagetrac-aashu12]

Commit: e8760d5

[dimpase]

Dependencies: #21554

[dimpase]

comment:6

what do you mean to say by

+    In case one of the solutions while solving an equation is a real number::

First of all, I would have written

+    In case one of the solutions of an equation is a real number::

Still, it's unclear what the following sequence of assumptions following this line has
to do with the one of solutions being real. Do you mean to say that in order to make sure that one (rather, every?) solution is real, you need to make the following assumptions? Something else?

[sagetrac-aashu12]

comment:7

Replying to @dimpase:

what do you mean to say by

+    In case one of the solutions while solving an equation is a real number::

First of all, I would have written

+    In case one of the solutions of an equation is a real number::

It was a part of ticket #21554 which has already been merged.

Still, it's unclear what the following sequence of assumptions following this line has
to do with the one of solutions being real. Do you mean to say that in order to make sure that one (rather, every?) solution is real, you need to make the following assumptions? Something else?

According to the documentation provided earlier, the solution of the equation described in the issue is r1, which is a real number (That is what the documentation says!). But the solution to this equation can be a complex number too! So, I just changed the documentation and redefined r1 to be any arbitrary constant. You can refer to this conversation: https://groups.google.com/forum/#!topic/sage-support/_XWjrYjk_3A