I am trying to make a loop in mathematica around:
  
  If  [Sign [fc] == Sign[fa], {a = c}, {b = c}]
      fa = f[a]
        fb = f[b]
          c = a + 0.5*(b - a)
              fc = f[c]

when I do:
While [n < 10,
  If  [Sign [fc] == Sign[fa], {a = c}, {b = c}]
      fa = f[a]
        fb = f[b]
          c = a + 0.5*(b - a)
              fc = f[c]; n++]

I am getting errors all around:
Set::write: Tag Plus in 0.0222862 +2 is Protected.

Set::write: Tag Times in -0.0112788\2.5 is Protected.

Set::write: Tag Times in -0.0112788\0.110031 is Protected.

General::stop: Further output of Set::write will be suppressed during
this calculation.


what am I doing wrong?

my full code is:
Clear[a]
Clear[b]
Clear[c]
Clear[b]
Clear[f]
f[x_]=(x*E^(-x))-0.16064
a=2
b=3
n=1
fa=f[a]
fb=f[b]
c=a+0.5*(b-a)
fc=f[c]
While [n < 10,
  If  [Sign [fc] == Sign[fa], {a = c}, {b = c}]
      fa = f[a]
        fb = f[b]
          c = a + 0.5*(b - a)
              fc = f[c]; n++]

You need to put semicolons at the end of each command.  If you need to
see the output, you can put in additional lines like
Write["stdout",fa]
to look at the values.

I dont think this is the problem.
using : 
 While[n <  10^2 ,
     If[Sign[f[a]] \[Equal] Sign[f[c]], 
      a = c,   b = c;
       f[b] = f[c];]
          c = a + 0.5*(b - a);
           n++;]
works without any problems

removing f[b] = f[c] makes it complain.

It's hard to guess what you are really trying to do, what's sure that
you are attempting some very ugly procedural programming in
Mathematica!
(I am looking at it)

well here's your working solution anyway, 4 bucks please :)

f[x_]:=(x*E^(-x))-0.16064;

a=2;
b=3;
n=1;
c=a+0.5*(b-a);

fa=f[a];
fb=f[b];
fc=f[c];

While [n<10,
  If [Sign[fc]==Sign[fa],a=c,b=c];
  fa=f[a];
  fb=f[b];
  c=a+0.5(b-a);
  fc=f[c];
  n++]

if you explain me what you really wanna do though I can make the code
twice as short at least

the 4$ is yours :) though I think you need to post it as an answer.
what I am trying to do is to find the root of the function between the
points 2,3 using the bisection method.
I have managed to run it without any problems with the following code:
f[x_] :=(x*E^(-x))-0.16064
a=2;
b=3
c=a+0.5*(b-a);
n=0;
While [n<10^2,
  If[Sign[f[a]] == Sign[f[c]],
  a = c, b = c;
    f[b] = f[c];]
     c= a+ 0.5 * (b-a);
      n++;]
Print[a]
Print[n]

now. when I take out the line : f[b] = f[c]; 
it complains again that its protected.
Why this line is so important?
more than that when I put the line back I still get the error and the
only way for me to fix it is to copy the whole if statement from the
previous version (the one I didnt remove f[b]=f[c] from ) and
miraclously it works.

HELP ME :)

f[b] = f[c]; 

doesn't make sense to me, it could mean your asking to assign 3 to 2 (2=3).
If you want to solve equation=0 then just use Solve[] it's 100 times
more efficient than some crappy procedural programming.

Find the complete Mathematica listing to your prob on:
http://math.fullerton.edu/mathews/n2003/Web/BisectionMod/BisectionMod.html

:)