Last Sync: 2022-10-05 13:30:05

This commit is contained in:
tactonbishop 2022-10-05 13:30:05 +01:00
parent 1e39794eb5
commit aee5d83902

View file

@ -86,16 +86,19 @@ Look at the following equivalences
It is obvious that a pattern is maintained between the hexadecimal and binary numbers and that this pattern is obscured by the decimal conversion. In the first example the binary half-byte `1111` is matched by the hexadecimal `F00F`. It is obvious that a pattern is maintained between the hexadecimal and binary numbers and that this pattern is obscured by the decimal conversion. In the first example the binary half-byte `1111` is matched by the hexadecimal `F00F`.
Mathematically comparing hex `F` and binary `1111` Mathematically comparing hex `F` and binary `1111`:
$$ $$
\textsf{1111} = (2^{3} + 2^{2} + 2^{1} + 2^{0}) \\ \textsf{1111} = ((1 \cdot 2^{3}) + (1 \cdot 2^{2}) + (1 \cdot 2^{1}) + (1 \cdot 2^{0})) \\
= 8 + 4 + 2 + 1 = 8 + 4 + 2 + 1 \\
= 15
$$ $$
$$ $$
\textsf{F00F} = (15 \cdot 16^{4}) + (15 \cdot 16^{0}) \\ \textsf{F} = 15 \cdot 16^{0} \\
= 8 + 4 + 2 + 1 = 15
$$ $$
// TODO: Relation to binary and bytes ![](/img/hexadecimal-to-bytes.svg)
> Every four bits (or half byte) in binary corresponds to one symbol in hexadecimal. Therefore **a byte can be easily represented with two hexadecimal symbols, a 16-bit number can be represented with four hex symbols, a 32-bit number can represented with eight hex symbols and so on.**