# Exponent e of 5 bits alias 16

I was reading my notes and came across the following

    The sign of the exponent e usually is not encoded by a
complement, but the so-called bias N (also referred to as
excess-N). This means that e = N stands for 0, all values
e > N for positive exponents and all values e < N for
negative exponents.


It was followed by this example:

Example: An exponent e of 5 bits, bias 16.

value of e = meaning

$00000_{2}$ = $-16_{10}$

....

$01111_{2}$ = $-1_{10}$

$10000_{2}$ = $0_{10}$

....

$11111_{2}$ = $15_{10}$

I understand that the above table is represented in 2's complement using 5 bits but how is it that $11111_{2}$ is now equivalent to $15_{10}$? When I convert $11111_{2}$ to decimal I get 31 at http://www.mathsisfun.com/binary-decimal-hexadecimal-converter.html.

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I think you're mixing up the number that is stored and the number it represents. $11111_2$ is indeed equivalent to $31_{10}$, and then you apply the excess-16 to get $31-16=15$. – Peter Phipps Apr 20 '12 at 12:13
Thank you so much it makes perfect sense now! – methuselah Apr 20 '12 at 13:04
@PeterPhipps: please post as an answer so it can be accepted. It's a good one and seems to satisfy methusaleh. Thanks – Ross Millikan Apr 20 '12 at 13:23

$11111_2$ is indeed equivalent to $31_{10}$, and then you apply the excess-16 to get $31−16=15$.