Machine Epsilon

Question

I'm trying to evaluate the machine epsilon of my computer (see below). I wrote this:

eps = 1.0;
p = 0;
While[(1.0 + eps) > 1.0, eps = eps/2.0; p += 1];
eps
p
eps*2

and got:

1.42109*10^-14
46
2.84217*10^-14

which is not what I expected (IEEE754). So I did this

EngineeringForm[1.0 + 2 * eps, 20]
EngineeringForm[1.0 + eps, 20]
EngineeringForm[1.0 + (eps/2.0), 20]
EngineeringForm[1.0 + (eps/4.0), 20]
EngineeringForm[1.0 + (eps/16.0), 20]
EngineeringForm[1.0 + (eps/32.0), 20]

and got that:

1.000000000000028
1.000000000000014
1.000000000000007
1.000000000000004
1.000000000000001
1.

which seems to contradict the termination condition of the 'while' loop. So I would like to understand what's going on under Mathematica's hood. I tried to look for some information on Mathematica's internal float representation without success. Any help to explain these results is welcome. I'm sorry if my question is a bit vague.

Eric

Context:

SystemInformationData[{"Kernel" ->
    {"Version" -> "9.0 for Mac OS X x86 (64-bit) (January 24, 2013)", 
    "ReleaseID" -> "9.0.1.0 (4055646, 4055073)", ...}]

EDIT

It was suggested the following program that works directly for some reason:

eps = 1.0;
n = 0;
While[(1.0 + eps) - 1.0 > 0.0,
  eps = eps/2;
  n++;
];
eps*2
n
Log[2, eps*2]

Answer 1

There is a tolerance Internal`$EqualTolerance that is applied when testing Real numbers. If the numbers agree up to the last Internal`$EqualTolerance digits, then they are treated as equal. Try this:

eps = 1.0;
p = 0;
Block[{Internal`$EqualTolerance = 0.},
 While[(1.0 + eps) > 1.0, eps = eps/2.0; p += 1];
 ]
eps
p
eps*2
(*
  1.11022*10^-16
  53
  2.22045*10^-16
*)

See How to make the computer consider two numbers equal up to a certain precision and also this answer by Alexey Popkov.