Number Systems
Their are many different number systems. Base 10 is what we use. But we have base 2 (binary),
base 3, etc.
Examples:
Base 10 place value system (powers of 10):
... 104 103 102 101 100:
_____ _____ _____ _____ _____
10000 1000 100 10 1
The number 237 is:
_____ _____ ____2 ____3 ____7 or 2x100 + 3x10 + 7x1
10000 1000 100 10 1
Note: There are 10 symbols that we use (0 1 2 3 4 5 6 7 8 9)
Base 2 (binary) place value system (powers of 2):
... 24 23 22 21 20:
_____ _____ _____ _____ _____
16 8 4 2 1
The number 101102 is 22 in our base 10 system:
____1 ____0 ____1 ____1 ____0 or 1x16 + 0x8 + 1x4 + 1x2 + 0x1 = 22
16 8 4 2 1
The number 101112 is 23 in our base 10 system:
____1 ____0 ____1 ____1 ____1 or 1x16 + 0x8 + 1x4 + 1x2 + 1x1 = 23
16 8 4 2 1
The number 111112 is 31 in our base 10 system:
____1 ____1 ____1 ____1 ____1 or 1x16 + 1x8 + 1x4 + 1x2 + 1x1 = 31
16 8 4 2 1
Note: There are 2 symbols that we use (0 1)
The int memory location uses 32 bits (the left most bit is a sign bit).
The left bits are not shown above.
If the left most bit is a zero, the number is positive else negative.
Negative numbers are stored in twos complement form.
00000000000000000000000000000110 represents the number 6 as an int.
11111111111111111111111111111111 represents the number -1 as an int.
(to convert to a positive 1, flip all the bits and then add 1.)
00000000000000000000000000000001 represents the number 1 as an int.
Integer.toBinaryString(int value base 10) returns a String in base binary
Base 3 place value system (powers of 3):
... 34 33 32 31 30:
_____ _____ _____ _____ _____
81 27 9 3 1
The number 001203 is 15 in our base 10 system:
____0 ____0 ____1 ____2 ____0 or 0x81 + 0x27 + 1x9 + 2x3 + 0x1 = 15
81 27 9 3 1
The number 002203 is 24 in our base 10 system:
____0 ____0 ____2 ____2 ____0 or 0x81 + 0x27 + 2x9 + 2x3 + 0x1 = 24
81 27 9 3 1
The number 002213 is 25 in our base 10 system:
____0 ____0 ____2 ____2 ____1 or 0x81 + 0x27 + 2x9 + 2x3 + 1x1 = 25
81 27 9 3 1
Note: There are 3 symbols that we use (0 1 2)
Integer.parseInt(String num, int base) returns the converted num as an int
Integer.parseInt("101",2) would return 5
Base 8 (Octal) place value system (powers of 8):
... 84 83 82 81 80:
_____ _____ _____ _____ _____
4096 512 64 8 1
The number 000258 is 21 in our base 10 system:
____0 ____0 ____0 ____2 ____5 or 0x4096 + 0x512 + 0x64 + 2x8 + 5x1 = 21
4096 512 64 8 1
The number 000378 is 31 in our base 10 system:
____0 ____0 ____0 ____3 ____7 or 0x4096 + 0x512 + 0x64 + 3x8 + 7x1 = 31
4096 512 64 8 1
The number 002378 is 159 in our base 10 system:
____0 ____0 ____2 ____3 ____7 or 0x4096 + 0x512 + 2x64 + 3x8 + 7x1 = 159
4096 512 64 8 1
Note: There are 8 symbols that we use (0 1 2 3 4 5 6 7)
Integer.toOctalString(int value) returns a String in base 8
Integer.parseInt(String num, int base) returns the converted num as an int
Integer.parseInt("52",2) would return 42
Base 16 (Hex or Hexadecimal) place value system (powers of 16):
... 164 163 162 161 160:
_____ _____ _____ _____ _____
65536 4096 256 16 1
The number 0002516 is 37 in our base 10 system:
____0 ____0 ____0 ____2 ____5 or 0x65536 + 0x4096 + 0x256 + 2x16 + 5x1 = 37
65536 4096 256 16 1
The number 0002B16 is 43 in our base 10 system:
____0 ____0 ____0 ____2 ____B or 0x65536 + 0x4096 + 0x256 + 2x16 + 11x1 = 43
65536 4096 256 16 1
The number 0002F16 is 47 in our base 10 system:
____0 ____0 ____0 ____2 ____F or 0x65536 + 0x4096 + 0x256 + 2x16 + 15x1 = 47
65536 4096 256 16 1
The number 000AC16 is 172 in our base 10 system:
____0 ____0 ____0 ____A ____C or 0x65536 + 0x4096 + 0x256 + 10x16 + 12x1 = 172
65536 4096 256 16 1
The number 0001016 is 16 in our base 10 system:
____0 ____0 ____0 ____1 ____0 or 0x65536 + 0x4096 + 0x256 + 1x16 + 0x1 = 16
65536 4096 256 16 1
Note: There are 16 symbols that we use (0 1 2 3 4 5 6 7 8 9 A B C D E F)
A represents 10 objects, B represents 11 objects, etc.
Integer.toHexString(int value base 10) returns a String in base 16
Integer.parseInt(String num, int base) returns the converted num as an int
Integer.parseInt("1A",2) would return 26
There is a short cut to convert base 16 numbers into base 8 or base 2.
In order to convert into binary, rewrite each hex digit as a 4 bit binary number.
In order to convert into base 8, regroup the binary bits into groups of 3, and
then write the base 8 digit for each group of 3 bits (right to left)
For example, since A16 is 1010 in base 2 and C16 is 1100 in base 2,
we can write AC16 = 101011002
A C 2 5 4
1010 1100 --> 10 101 100
| Hexadecimal | Binary | Binary | Base 8 | Base 10 |
| AC16 | 1010 11002 | 10 101 1002 | 2548 | 172 |
| BF16 | 1011 11112 | 10 111 1112 | 2778 | 191 |
| FF16 | 1111 11112 | 11 111 1112 | 3778 | 255 |
Hex Binary Place Value Base 10
0 0000 0x8 + 0x4 + 0x2 + 0x1 0
1 0001 0x8 + 0x4 + 0x2 + 1x1 1
2 0010 0x8 + 0x4 + 1x2 + 0x1 2
3 0011 0x8 + 0x4 + 1x2 + 1x1 3
4 0100 0x8 + 1x4 + 0x2 + 0x1 4
5 0101 0x8 + 1x4 + 0x2 + 1x1 5
6 0110 0x8 + 1x4 + 1x2 + 0x1 6
7 0111 0x8 + 1x4 + 1x2 + 1x1 7
8 1000 1x8 + 0x4 + 0x2 + 0x1 8
9 1001 1x8 + 0x4 + 0x2 + 1x1 9
A 1010 1x8 + 0x4 + 1x2 + 0x1 10
B 1011 1x8 + 0x4 + 1x2 + 1x1 11
C 1100 1x8 + 1x4 + 0x2 + 0x1 12
D 1101 1x8 + 1x4 + 0x2 + 1x1 13
E 1110 1x8 + 1x4 + 1x2 + 0x1 14
F 1111 1x8 + 1x4 + 1x2 + 1x1 15
Run Binary Math Converter Applet (Java JRE Required)