3
$\begingroup$

Define a simple symbolic expression a, then let b be the square root of its square:

a = A + Sqrt[A + B];

a^2 // Expand

A + A^2 + B + 2 A Sqrt[A + B]


b = Sqrt[%]

Sqrt[A + A^2 + B + 2 A Sqrt[A + B]]

The difference, when simplified under the assumption that both A and B are positive reals should give zero:

a - b

A + Sqrt[A + B] - Sqrt[A + A^2 + B + 2 A Sqrt[A + B]]


FullSimplify[%, A > 0 && B > 0 && A ∈ Reals && B ∈ Reals]

A + Sqrt[A + B] - Sqrt[A + A^2 + B + 2 A Sqrt[A + B]]

Yet in the symbolic form it doesn't. What am I missing?

$\endgroup$
  • 1
    $\begingroup$ FullSimplify[a == b, A > 0 && B > 0 && A \[Element] Reals && B \[Element] Reals] returns True, so it is probably just a little too hard for Mathematica $\endgroup$ – mikado Sep 18 '17 at 20:22
0
$\begingroup$

Maybe you can "cheat" as follows:

FullSimplify[
    A+Sqrt[A+B]-Sqrt[A+A^2+B+2 A Sqrt[A+B]] /. B->c-A,
    0<A<c
]

0

Here is another variation of this idea:

FullSimplify[
    A+Sqrt[A+B]-Sqrt[A+A^2+B+2 A Sqrt[A+B]],
    A+B==c && A>0 && B>0 && c>0
]

0

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.