Question

Convert each of the integers in Exercise 6 from a binary expansion to a hexadecimal expansion.

a) (1111 0111)2

b) (1010 1010 1010)2

c) (111 0111 0111 0111)2

d) (1010 1010 1010 101)2

Short answer

$$\begin{gathered} a)(F7F{)}_{16} \\ b)(AAA{)}_{16} \\ c)(777{)}_{16} \\ d)(5516{)}_{16} \end{gathered}$$

Step-by-step solution

Step 1:

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

in hexadecimal notation

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

The hexadecimal expansion has base b = 2

$\begin{array}{r}(11110111{)}_2={1.2}^7+{1.2}^6+{1.2}^5+{1.2}^4+{0.2}^3+{1.2}^2+{1.2}^1+{1.2}^0\\ \\ =128+64+32+16+4+2+1\\ \\ =247\end{array}$

You obtain binary expansion of the decimal expansion of an integer by consecutively dividing the integer by 16 until you obtain 0

$\begin{array}{r}247=16.15+7\\ \\ 15=16.0+15\\ \\ =(F7F{)}_{16}\end{array}$

Step 2:

b)

The hexadecimal expansion has base b = 2

$\begin{array}{r}(101010101010{)}_2={1.2}^{11}+{0.2}^{10}+{1.2}^9+{0.2}^8+{1.2}^7+{0.2}^6+{1.2}^5+{0.2}^4+{1.2}^3+{0.2}^2+{1.2}^1+{0.2}^0\\ \\ =2048+512+128+32+8+2+1\\ \\ =2731\end{array}$

You obtain binary expansion of the decimal expansion of an integer by consecutively dividing the integer by 16 until you obtain 0

$\begin{array}{r}2731=16.170+11\\ \\ 170=16.10+10\\ \\ 16=16.0+10\\ \\ =(AAA{)}_{16}\end{array}$

Step 3: 

c)

The hexadecimal expansion has base $b=2$

role="math" localid="1668491270303" $\begin{array}{r}(111011101110111{)}_2=1\cdot 2^{14}+1\cdot 2^{13}+1\cdot 2^{12}+{0.2}^{11}+{1.2}^{10}+1\cdot 2^9+1\cdot 2^8+0\cdot 2^7+1\cdot 2^6+1\cdot 2^5+1\cdot 2^4+{0.2}^3+1\cdot 2^2+1\cdot 2^1+1\cdot 2^0\\ \\ =16384+8192+4096+1024+512+256+64+32+18+4+2+1\\ \\ =30582\end{array}$

You obtain binary expansion of the decimal expansion of an integer by consecutively dividing the integer by 16 until you obtain 0

$\begin{array}{r}30582=16.1911+6\\ \\ 1911=16.119+7\\ \\ 119=16.7+7\\ \\ 7=0.16+7\\ \\ =(777{)}_{16}\end{array}$

Step 4:

The hexadecimal expansion has base b = 2

$\begin{array}{r}(101010101010101{)}_2={1.2}^{14}+{0.2}^{13}+{1.2}^{12}+{0.2}^{11}+{1.2}^{10}+{0.2}^9+{1.2}^8+{0.2}^7+{1.2}^6+{0.2}^5+{1.2}^4+{0.2}^3+{1.2}^2+{0.2}^1+{1.2}^0\\ \\ =16384+4096+1024+256+64+16+4+2+1\\ \\ =21847\end{array}$

You obtain binary expansion of the decimal expansion of an integer by consecutively dividing the integer by 16 until you obtain 0

$\begin{array}{r}21847=1365\cdot 16+25\\ \\ 1365=85\cdot 16+5\\ \\ 85=5\cdot 16+5\\ \\ 5=0.16+5\\ \\ =(5516{)}_{16}\end{array}$