Question

Show that the binary expansion of a positive integer can be obtained from its hexadecimal expansion by translating each hexadecimal digit into a block of four binary digits.

Short answer

Each hexadecimal digit can be represented by a unique block of 4 binary digits.

Step-by-step solution

Step 1:

$n=a_k{b}^k+a_{k1}+b^{k1}+\dots .+a_{1}b+a_0$

in hexadecimal notation

$\begin{array}{r}A=10\\ B=11\\ C=12\\ D=13\\ E=14\\ F=15\end{array}$

Let n be an integer. the binary representation of n is then $h_k\dots h_2{h}_1{h}_0$such that

$n=h_k\cdot {16}^k+\dots .+h_1\cdot 16+h_0$

Each hexadecimal digit can be representation by a unique block of 4 binary digits

$\begin{array}{r}n=\left(a_{4k+3}{2}^3+a_{4k+2}{2}^2+a_{4k+1}{2}^1+a_{4k}{2}^0\right)\cdot {16}^k+\dots \\ \\ \left(a_7{2}^3+a_6{2}^3+a_5{2}^1+a_4{2}^0\right)\cdot 16\\ \\ +\left(a_3{2}^3+a_2{2}^2+a_1{2}^1+a_0{2}^0\right)\end{array}$

Rewrite 16 as a power of 12

$\begin{array}{r}n=\left(a_{4k+3}{2}^3+a_{4k+2}{2}^2+a_{4k+1}{2}^1+a_{4k}{2}^0\right)\cdot 2^{4k}+\dots \\ \\ +\left(a_7{2}^3+a_6{2}^2{a}_5{2}^1+a_4{2}^0\right)\cdot 2^4\\ \\ +\left(a_3{2}^3+a_2{2}^2{a}_1{2}^1+a_0{2}^0\right)\cdot 2^4\end{array}$

Use distributive property

$\begin{array}{r}n=a_{4k+3}{2}^{4k+3}+a_{4k+2}{2}^{4k+2}+a_{4k+1}{2}^{4k+1}+a_{4k}{2}^{4k}+\dots \\ \\ +a_7{2}^7+a_6{2}^6{a}_5{2}^5+a_4{2}^4\\ \\ +a_3{2}^3+a_2{2}^2{a}_1{2}^1+a_0{2}^0\end{array}$

The corresponding binary expansion of n is then $a_{4k+3}{a}_{4k+2}\dots \dots a_2{a}_1{a}_0$