Question

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

Short answer

Each octal digit can be uniquely represented as a block of 3 binary digits

Step-by-step solution

Step 1:

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

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

$n=h_k\cdot 8^k+\dots .+h_1\cdot 8+h_0\mid$

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

$\begin{array}{r}n=\left(a_{4k+3}{2}^3+a_{4k+2}{2}^2+a_{4k+1}{2}^1+a_{4k}{2}^0\right)\cdot 8^k+\dots \\ \\ \left(a_7{2}^3+a_6{2}^3+a_5{2}^1+a_4{2}^0\right)\cdot 8\\ \\ +\left(a_3{2}^3+a_2{2}^2+a_1{2}^1+a_0{2}^0\right)\end{array}$

Rewrite 8 as a power of 12

$\begin{array}{r}n=\left(a_{3k+2}{2}^2+a_{3k+1}{2}^1+a_{3k}{2}^0\right)\cdot 2^{3k}\\ \\ +\left(a_5{2}^2+a_4{2}^1+a_3{2}^0\right)\cdot 2^3\\ \\ +\left(a_2{2}^2{a}_1{2}^1+a_0{2}^0\right)\end{array}$

Use distributive property

$\begin{array}{r}n=a_{3k+2}{2}^{3k+2}+a_{3k+1}{2}^{3k+1}+\dots \\ \\ +a_5{2}^5+a_4{2}^4+a_3{2}^3+a_2{2}^2+a_1{2}^1+a_0{2}^0\end{array}$

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

Step 2:

Each octal digit can be uniquely represented as a block of 3 binary digit.

Binary	Decimal	Octal
000	0	0
001	1	1
010	2	2
011	3	3
100	4	4
101	5	5
110	6	6
111	7	7