# FullSimplify used with Assuming giving unexpected results

Can someone shed some light on why

Assuming[{x ∈ Reals}, x (Sign[x]^2 - 1)] // FullSimplify


returns

x (-1 + Sign[x]^2)


I know one can not expect FullSimplify to perform miracles but this example does seem pretty obvious.

• It returns zero for me. Version 9.0. Nov 8, 2013 at 12:36
• FullSimplify[x (Sign[x]^2 - 1), Assumptions -> {x \[Element] Reals}] gives 0 on version 9.0.1 Nov 8, 2013 at 12:37
• ver@cormullion, I have a 64 bit 9.0.1.0 student version on a 64 bit Win 8 machine and get what I posted. Nov 8, 2013 at 12:50
• Sorry @cormuullion, I posted my comment too quickly. Yes I also get zero for your modified example using Assumptions, but the question still remains for my original example. Nov 8, 2013 at 12:54
• @Anon, you get zero? Are you using a student version? Nov 8, 2013 at 12:55

Assuming[{x ∈ Reals}, x (Sign[x]^2 - 1)]


BEFORE simplifying.

The correct way to inform your assumptions to FullSimplify[] is:

FullSimplify[x (Sign[x]^2 - 1), Assumptions -> {x ∈ Reals}]


Which returns zero.

Or

Assuming[{x ∈ Reals}, FullSimplify[x (Sign[x]^2 - 1)]]

• Yep, makes sense. Cheers... Nov 8, 2013 at 13:25

1. The first reason is that Sign yields 0, so even assuming x ∈ Reals this expression:
Sign[x]^2 - 1 cannot be evaluated to 0.

2. The next problem is that Assuming[{x ∈ Reals}, x (Sign[x]^2 - 1)] is evaluated first, then the assumption imposed doesn't affect the simplification procedure since FullSimplify being outside Assuming doesn't know anything about x, thus the final result is correct.

Ad 1. In general Sign is a complex function, for a complex number $z\neq0\;$ it is equal to z/Abs[z], e.g. see its graphs of the real and imaginary parts:

GraphicsRow[ Table[ Plot3D[ f @ Sign[x + I y], {x, -3, 3}, {y, -3, 3},
ColorFunction -> "DeepSeaColors",], {f, {Re, Im}}]] One can impose global assumptions for a Mathematica session, e.g.

\$Assumptions = z ∈ Reals;


then one can do as it was assumed in the question:

z (Sign[z]^2 - 1) // FullSimplify

0


On the other hand you can use assumption restricted to FullSimplify only:

FullSimplify[ x (Sign[x]^2 - 1), x ∈ Reals]

0


For more detailed discussion see e.g. this question How to specify assumptions before evaluation?.

• I don't have the points to up vote your answer so thanks. Nov 8, 2013 at 13:27