ans = 
Does isAlways Make an Unwarranted Assumption that a Variable is real?
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Define a sym variable
syms v
isAlways can't prove that v is real. Makes sense.
isAlways(in(v,'real'))
Now make an assumption
assume(v > 0); % (1)
Is that assumption a sufficient condition to imply that v is real?
isAlways(in(v,'real')) %(2)
Apparently it does.
But we don't see that v is real in the assumptions
assumptions(v)
as we would if stated explicitly
assumeAlso(v,'real');
assumptions(v)
Now define v as a complex number, which clears all of the assumptions
v = sym(1+1i);
assumptions(v)
Here, v satisfies assumption (1) because symbolic gt only compares the real parts of both sides (though the doc page does not state that explicitly)
isAlways(v > 0)
But satisfying assumption (1) in this case does not imply the truth of condition (2) (thankfully)
isAlways(in(v,'real'))
Seems like the correct way for the software to interpret (1) would be Re(v) > 0, in accordance with the de facto definition of symbolic gt, which would provide no information for evaluating (2).
9 comentarios
Matt J
el 28 de En. de 2026
In my opinion, the thing that is wrong is,
>> isAlways( sym(1+1i) > 0)
ans =
logical
1
I never really understood this convention even for numeric variables, but certainly not for symbolic variables.
Paul
el 29 de En. de 2026
Matt J
el 29 de En. de 2026
That's my opinion, yes.
Paul
el 30 de En. de 2026
the cyclist
el 31 de En. de 2026
Editada: the cyclist
el 31 de En. de 2026
I guess I would have expected
1 + i > 0
to return NaN, since ordering is not defined on complex numbers.
Matt J
el 31 de En. de 2026
x >= x % returns false?
I can see the case for returning true iff the inequality is satisfied with equality.
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