Binary to Hexadecimal

Date: 12/08/98 at 11:15:11
From: Ian hamilton
Subject: number base conversions

I am a mature student on a computing HND. We have just gone through 
the followung question in class, but I am still clueless. Convert 
110000001101 from base two to hexadecimal. 

Please help.

Date: 12/08/98 at 12:48:27
From: Doctor Rick
Subject: Re: number base conversions

Hi, Ian. A big part of the reason that we use hexadecimal is that it is 
relatively easy to convert between binary and hexadecimal. Hexadecimal 
numbers are closely related to binary, but they are shorter and easier 
to read than binary.

Group the binary digits into groups of 4 starting from the right:

  1100 0000 1101

Each group will correspond to one hexadecimal digit. The leftmost digit 
in a group is the 8's place; next come the 4's, 2', and 1's places. In 
each group, take the numbers for the places that have 1's in them, and 
add them together:

  8421 8421 8421
  1100 0000 1101
  \__/ \__/ \__/
  8+4    0  8+4+1
   12    0   13

I did the additions in base 10, so now we need to convert base 10 to 
base 16. Remember that in hexadecimal:

  A = 10; B = 11; C = 12; D = 13; E = 14; F = 15

So our hexadecimal number, reading from left to right, is: C0D.

Why does this work? Each place in a hexadecimal number represents a 
power of 16. Thus, the hexadecimal number C0D is 12*16^2 + 0*16 + 13. 
Each place in a binary number represents a power of 2. Since 16 = 2^4, 
you can write, for instance:

   11    3    8    3     2
  2   = 2  * 2  = 2  * 16

Therefore, 4 consecutive binary digits can be grouped and a power of 
16 extracted from each:

     11     10     9     8        3      2      1      0      2
  a*2  + b*2  + c*2  +d*2   = (a*2  + b*2  + c*2  + d*2 ) * 16

Since a 4-digit binary number is at most 15 (decimal), each group 
converts into one hex digit.

For another source on this topic, please see:

   Binary Conversion
   http://mathforum.org/dr.math/problems/stirling1.7.98.html   

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/