Question

Show that the hexadecimal expansion of a positive integer can be obtained from its binary expansion by grouping to-gather blocks of four binary digits, adding initial zeros if necessary, and translating each block of four binary digits into a single hexadecimal digit.

Short answer

Each block of 4 binary digits represents a unique hexadecimal digit

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 $a_k\dots a_2{a}_1{a}_0$such that

$n=a_k\cdot 2^k+a_{k-1}\cdot 2^{k-1}+\dots ..+a_7\cdot 2^7+a_6\cdot 2^6+a_5\cdot 2^5+a_4\cdot 2^4+a_3\cdot 2^3+a_2\cdot 2^2+a_1\cdot 2+a_0$

it is safe to assume that k + 1is a multiple of 4( if not, then we add zero terms in front of $a_k$until the number of digits In the binary representation increased by 1is a multiple of 4)

$\begin{array}{r}=\left(a_k\cdot 2^k+a_{k-1}\cdot 2^{k-1}+a_{k-2}\cdot 2^{k-2}+a_{k-3}\cdot 2^{k-3}\right)\dots \dots \\ \\ +\left(a_7\cdot 2^7+a_6\cdot 2^6+a_5\cdot 2^5+a_4\cdot 2^4\right)\\ \\ +\left(a_3\cdot 2^3+a_2\cdot 2^2+a_1\cdot 2+a_0\right)\end{array}$

Fact out power of 2 out of each block of 4 terms

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

we then note that each block $a_i\cdot 2^3+a_{i-1}\cdot 2^2+a_{i-2}\cdot 2^1+a_{i-3}\cdot 2^i$is a hexadecimal digit

$=h_{(k-3)/4}\cdot 2^{k-3}+\dots ..+h_2\cdot 2^8+h_1\cdot 2^4+h_0$

The corresponding hexadecimal expansion of n is then

$h_{(k-3)/4}\cdot 2^{(k-3)/4}\dots \dots h_2{h}_1{h}_0$

Step 2:

Each block 4 binary digits represents a unique hexadecimal digits