Conversion from Hexadecimal to Binary

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Theorem

Let $n$ be a (positive) integer expressed in hexadecimal notation as:

$n = \sqbrk {a_r a_{r - 1} \dotso a_1 a_0}_H$


Then $n$ can be expressed in binary notation as:

$n = \sqbrk {b_{r 3} b_{r 2} b_{r 1} b_{r 0} b_{\paren {r - 1} 3} b_{\paren {r - 1} 2} b_{\paren {r - 1} 1} b_{\paren {r - 1} 0} \dotso b_{1 3} b_{1 2} b_{1 1} b_{1 0} b_{0 3} b_{0 2} b_{0 1} b_{0 0} }_2$

where $\sqbrk {b_{j 3} b_{j 2} b_{j 1} b_{j 0} }_2$ is the expression of the hexadecimal digit $a_j$ in binary notation.


That is, you take the binary expression of each hexadecimal digit, padding them out with zeroes to make them $4$ bits long, and simply concatenate them.


Proof

We have:

\(\ds n\) \(=\) \(\ds \sqbrk {a_r a_{r - 1} \dotso a_1 a_0}_H\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 0}^r a_j 16^j\) Definition of Hexadecimal Notation


We have that:

$0 \le a_j < 16$

and so:

\(\ds a_j\) \(=\) \(\ds \sqbrk {b_{j 3} b_{j 2} b_{j 1} b_{j 0} }_2\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^3 b_{j k} 2^k\) Definition of Binary Notation


and so:

\(\ds n\) \(=\) \(\ds \sqbrk {a_r a_{r - 1} \dotso a_1 a_0}_H\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 0}^r a_j 16^j\) Definition of Hexadecimal Notation
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 0}^r \paren {\sum_{k \mathop = 0}^3 b_{j k} 2^k} 16^j\) Definition of Binary Notation
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 0}^r \paren {\sum_{k \mathop = 0}^3 b_{j k} 2^k} 2^{4 j}\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 0}^r \paren {\sqbrk {b_{j 3} b_{j 2} b_{j 1} b_{j 0} }_2} 2^{4 j}\) Definition of Binary Notation
\(\ds \) \(=\) \(\ds \sqbrk {b_{r 3} b_{r 2} b_{r 1} b_{r 0} }_2 2^{4 r} + \sqbrk {b_{\paren {r - 1} 3} b_{\paren {r - 1} 2} b_{\paren {r - 1} 1} b_{\paren {r - 1} 0} }_2 2^{4 {r - 1} } + \cdots + \sqbrk {b_{1 3} b_{1 2} b_{1 1} b_{1 0} }_2 2^4 + \sqbrk {b_{0 3} b_{0 2} b_{0 1} b_{0 0} }_2\)
\(\ds \) \(=\) \(\ds \sqbrk {b_{r 3} b_{r 2} b_{r 1} b_{r 0} b_{\paren {r - 1} 3} b_{\paren {r - 1} 2} b_{\paren {r - 1} 1} b_{\paren {r - 1} 0} \dotso b_{1 3} b_{1 2} b_{1 1} b_{1 0} b_{0 3} b_{0 2} b_{0 1} b_{0 0} }_2\)

Hence the result.

$\blacksquare$


Examples

Example: $6 \mathrm C 5$

The integer expressed in hexadecimal as $6 \mathrm C 5_H$ can be expressed in binary as:

$6 \mathrm C 5_H = 110 \, 1100 \, 0101_2$


Sources

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