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## DESCRIPTION

```       Most people use the decimal numbering system. This system
uses ten symbols to represent numbers. When those ten sym­
bols are used up, they start all over again and increment
the position just before this. The digit 0 is only shown
if it is the only symbol in the sequence, or if it is not
the first one.

If this sounds as crypto to you, this is what I've said in
numbers:

0
1
2
3
4
5
6
7
8
9
10
11
12
13

and so on.

Each time the digit nine should be incremented, it is
reset to 0 and the position before is incremented. Then
number 9 can be seen as "00009" and when we should incre­
ment 9, we reset it to zero and increment the digit just
before the 9 so the number becomes "00010". For zero's we
write a space if it is not the only digit (so: number 0)
and if it is the first digit: "00010" -> " 0010" -> "
010" -> "   10". It is not "   1 ".

This was pretty basic, you already knew this. Why did I
tell it ?  Well, computers do not represent numbers with
10 different digits. They know of only two different sym­
bols, being 0 and 1. Apply the same rules to this set of
digits and you get the binary numbering system:

0
1
10
11
100
101
110
111
1000

the decimal system. If we are talking about another num­
bering system, we'll have to make that clear. There are a
few wide-spread methods to do so. One common form is to
write 1010(2) which means that you wrote down a number in
the binary form. It is the number ten.  If you would write
1010 it means the number one thousand and ten.

In books, another form is most used. It uses subscript
(little chars, more or less in between two rows). You can
leave out the parentheses in that case and write down the
number in normal characters followed with a little two
just behind it.

The numbering system used is also called the base. We talk
of the number 1100 base 2, the number 12 base 10.

For the binary system, is is common to write leading
zero's. The numbers are written down in series of four,
eight or sixteen depending on the context.

We can use the binary form when talking to computers
(...programming...)  but the numbers will have large rep­
resentations. The number 65535 would be written down as
1111111111111111(2) which is 16 times the digit 1.  This
is difficult and prone to errors. Therefore we normally
would use another base, called hexadecimal. It uses 16
different symbols. First the symbols from the decimal sys­
tem are used, thereafter we continue with the alphabetic
characters. We get 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C,
D, E and F. This system is chosen because the hexadecimal
form can be converted into the binary system very easy
(and back).

There is yet another system in use, called the octal sys­
tem. This was more common in the old days but not anymore.
You will find it in use on some places so get used to it.
The same story applies, but now with only eight different
symbols.

Binary      (2)
Octal       (8)
Decimal     (10)

00110   6    6    6
00111   7    7    7
01000  10    8    8
01001  11    9    9
01010  12   10    A
01011  13   11    B
01100  14   12    C
01101  15   13    D
01110  16   14    E
01111  17   15    F
10000  20   16   10
10001  21   17   11
10010  22   18   12
10011  23   19   13
10100  24   20   14
10101  25   21   15

Most computers used nowadays are using bytes of eight
bits. This means that they store eight bits at a time. You
can see why the octal system is not the most preferred for
that: You'd need three digits to represent the eight bits
and this means that you'd have to use one complete digit
to represent only two bits (2+3+3=8). This is a waste. For
hexadecimal digits, you need only two digits which are
used completely:

(2)      (8)  (10) (16)
11111111 377  255   FF

You can see why binary and hexadecimal can be converted
quickly: For each hexadecimal digit there are exactly four
binary digits.  Take a binary number. Each time take four
digits from the right and make a hexadecimal digit from it
(see the table above). Stop when there are no more digits.
Other way around: Take a hexadecimal number. For each
digit, write down its binary equivalent.

Computers (or rather the parsers running on them) would
have a hard time converting a number like 1234(16). There­
fore hexadecimal numbers get a prefix. This prefix depends
on the language you're writing in. Some of the prefixes
are "0x" for C, "\$" for Pascal, "#" for HTML.  It is com­
mon to assume that if a number starts with a zero, it is
octal.  It does not matter what is used as long as you
know what it is.  I will use "0x" for hexadecimal, "%" for
binary and "0" for octal.  The following numbers are all
the same, just the way they are written is different:  021
0x11  17  %00010001

To do arithmetics and conversions you need to understand
one more thing.  It is something you already know but per­
haps you do not "see" it yet:
1 * 10^3
2 * 10^2
3 * 10^1
4 * 10^0

where ^ means "to the power of".

We are using the base 10, and the positions 0,1,2 and 3.
The right-most position should NOT be multiplied with 10.
The second from the right should be multiplied one time
with 10. The third from the right is multiplied with 10
two times. This continues for whatever positions are used.

It is the same in all other representations:

0x1234 will be

1 * 16^3
2 * 16^2
3 * 16^1
4 * 16^0

01234 would be

1 * 8^3
2 * 8^2
3 * 8^1
4 * 8^0

This example can not be done for binary as that system can
only use two symbols. Another example:

%1010 would be

1 * 2^3
0 * 2^2
1 * 2^1
0 * 2^0

It would have been more easy to convert it to its hexadec­
imal form and just translate %1010 into 0xA. After a while
you get used to it. You will not need to do any calcula­
tions anymore but just know that 0xA means 10.

To convert a decimal number into a hexadecimal one you
could use the next method. It will take some time to be
able to do the estimates but it will be more and more easy
when you use the system more frequent. Another way is pre­
sented to you thereafter.

First you will need to know how many positions will be
used in the other system. To do so, you need to know the
If a number is smaller than 65536 it will thus fit in four
positions.  If the number is bigger than 4095, you will
need to use position 4.  How many times can you take 4096
from the number without going below zero is the first
digit you write down. This will always be a number from 1
to 15 (0x1 to 0xF). Do the same for the other positions.

Number is 41029. It is smaller than 16^4 but bigger than
16^3-1. This means that we have to use four positions.  We
can subtract 16^3 from 41029 ten times without going below
zero.  The leftmost digit will be "A" so we have 0xA????.
The number is reduced to 41029 - 10*4096 = 41029-40960 =
69.  69 is smaller than 16^3 but not bigger than 16^2-1.
The second digit is therefore "0" and we know 0xA0??.  69
is smaller than 16^2 and bigger than 16^1-1. We can sub­
tract 16^1 (which is just plain 16) four times and write
down "4" to get 0xA04?.  Take 64 from 69 (69 - 4*16) and
the last digit is 5 --> 0xA045.

The other method builds the number from the right. Take
again 41029.  Divide by 16 and do not use fractions (only
whole numbers).

41029 / 16 is 2564 with a remainder of 5. Write down 5.
2564 / 16 is 160 with a remainder of 4. Write the 4 before the 5.
160 / 16 is 10 with no remainder. Prepend 45 with 0.
10 / 16 is below one. End here and prepend 0xA. End up with 0xA045.

Which method to use is up to you. Use whatever works for
you. Personally I use them both without being able to tell
what method I use in each case, it just depends on the
number, I think. Fact is, some numbers will occur fre­
quently while programming, if the number is close then I
will use the first method (like 32770, translate into
32768 + 2 and just know that it is 0x8000 + 0x2 = 0x8002).

For binary the same approach can be used. The base is 2
and not 16, and the number of positions will grow rapidly.
Using the second method has the advantage that you can see
very simple if you should write down a zero or a one: if
you divide by two the remainder will be zero if it was an
even number and one if it was an odd number:

41029 / 2 = 20514 remainder 1
20514 / 2 = 10257 remainder 0
10257 / 2 =  5128 remainder 1
5128 / 2 =  2564 remainder 0
2564 / 2 =  1282 remainder 0
1282 / 2 =   641 remainder 0
641 / 2 =   320 remainder 1
320 / 2 =   160 remainder 0
160 / 2 =    80 remainder 0

%101000000100 0101
%10100000 0100 0101
%1010 0000 0100 0101

Group %1010000001000101 by three and convert into octal:

%1010000001000101
%1010000001000 101
%1010000001 000 101
%1010000 001 000 101
%1010 000 001 000 101
%1 010 000 001 000 101
%001 010 000 001 000 101
1   2   0   1   0   5 --> 0120105

So: %1010000001000101 = 0120105 = 0xA045 = 41029
Or: 1010000001000101(2) = 120105(8) = A045(16) = 41029(10)
Or: 1010000001000101(2) = 120105(8) = A045(16) = 41029

At first while adding numbers, you'll convert them to
their decimal form and then back into their original form
after doing the addition.  If you use the other numbering
system often, you will see that you'll be able to do
arithmetics in the base that is used.  In any representa­
tion it is the same, add the numbers on the right, write
down the rightmost digit from the result, remember the
other digits and use them in the next round. Continue with
the second digits from the right and so on:

%1010 + %0111 --> 10 + 7 --> 17 --> %00010001

will become

%1010
%0111 +
||||
|||+-- add 0 + 1, result is 1, nothing to remember
||+--- add 1 + 1, result is %10, write down 0 and remember 1
|+---- add 0 + 1 + 1(remembered), result = 0, remember 1
+----- add 1 + 0 + 1(remembered), result = 0, remember 1
nothing to add, 1 remembered, result = 1
--------
%10001 is the result, I like to write it as %00010001

For low values, try to do the calculations yourself, check
them with a calculator. The more you do the calculations
yourself, the more you find that you didn't make mistakes.
In the end, you'll do calculi in other bases as easy as
you do in decimal.

I hope you enjoyed the examples and their descriptions. If
you do, help other people by pointing them to this docu­
ment when they are asking basic questions. They will not
only get their answer but at the same time learn a whole
lot more.

Alex van den Bogaerdt <alex@ergens.op.het.net>

1.0.45                      2002-02-26             BIN_DEC_HEX(1)
```

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