Conversion from Hexadecimal to Binary

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:

We have that:
 * $0 \le a_j < 16$

and so:

and so:

Hence the result.