Write down the table of binary place holders depending on how many digits are in the number.
Multiply each digit's value by it's corresponding placeholder's value; OR multiply each digit by 2^n (which is equal to its placeholder value), where n = number of digits to the right of the digit that is going to be multiplied in the entire binary number.
Add each binary digit's denary value together.
Write down table of binary placeholders depending on how many digits are un the number, usually 8 or 16 bits are used.
Put a 0 in the place values larger than the denary number.
Put a 1 in the largest place value smaller than or equal to the denary number (which can only "go in" that place value once).
Subtract the place value from the denary number.
Repeat steps 2-4 for the subtraction result, do this continually until the subtraction result is 0, which is basically where the binary and denary numbers are equivalent.
Sense of scale: 1 character of text requires around ________ of storage. A full page of text requires around _________ of storage. 1 small digital colour photograph requires around ___________ of storage. Music CD capacity requires around _____________ of storage. DVD capacity requires around ______________ of storage. Hard disk capacity requires around ______________ of storage.
HTML/CSS Colour Codes
MAC Addresses
Assembly Codes and Memory Dumps
These contain 6 digits following after a # sign:
The first 2 digits show how much red (00 to FF has 255 possibilities) there is
The 3rd and 4th digits show how much green (00 to FF has 255 possibilities) there is
The last 2 digits show how much blue (00 to FF has 255 possibilities) there is
Example: #FF059A - this shows the R value for red is FF (255), the G value for green is 05 (5), and the B value for blue is 9A (154).
Hexadecimals are beneficial (over binary) because:
They are easier and faster to work with, taking up less screen space
Mistakes are less likely and easier to trace / debug
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 2 (denary) = 10 (binary) → unit digit of 0 but must carry the 1 over to the next placeholder value.
1 + 1 + 1 = 3 (denary) = 11 (binary) → unit digit of 1 but must carry the 1 over to the next placeholder value.
Generally:
Odd number of ones added together → unit digit = 1 and carry necessary amount of 1s over to the next placeholder(s) (depending on how many 2s or powers of 2s are created from the addition)
Even number of ones added together → unit digit = 0 and carry necessary amount of 1s over to the next placeholder(s) (depending on how many 2s or powers of 2s are created from the addition)