Here's a simple question. It's no longer high priority that I know this, but it's something that can come in handy later on.
Simplify[a > Sqrt[b]/c + d, a < Sqrt[b]/c + d]
or
Simplify[Sequence @@ ({a > n, a < n} /. {n -> Sqrt[b]/c + d})]
or
n = Sqrt[b]/c + d;
Simplify[a > n, a < n]
all outputs:
Out: a > Sqrt[b]/c + d
while
Simplify[a > m, a < m]
outputs
Out: False
How come? This isn't an issue of not considering the negative root. The expressions are identical. As demonstrated by Simplify[a > m, a < m], replacing m
with more complex expressions aside from having the Sqrt
function.
In fact, if I use Surd
, no matter the nth root, even or odd, or square root, Simplify
will evaluate completely into False.
Why? Is this something that I can fix using Upvalues? -- Finally remember what the "overloading" feature was called.
Simplify[a > Sqrt[b]/c + d && a < Sqrt[b]/c + d]
, which returnsFalse
. $\endgroup$Simplify[a > Sqrt[b]/c + d, a <= Sqrt[b]/c + d]
. Even if you specify that all variables belong to theReals
, there doesn't seem to be a built-in transformation to deal with the trichotomy law in this case, perhaps becauseSqrt[]
is assumed to be complex-valued (unlikeSurd[]
). $\endgroup$