Division of Number Bases

Solution

In Binary Division, you first convert to base 10, do the division and then convert the answer to the given base.

(i) \( \scriptsize 1111.01_2 \div 1101_2 \)  

convert to base 10

i.e. \( \scriptsize 1111.01_{2} \\ \scriptsize = 1 \: \times \: 2^3 \:+\: 1 \:\times\: 2^2 \:+ \:1 \: \times \: 2^1 \: +\: 1 \: \times \: 2^0 \\ \scriptsize \: + \: 0 \: \times \: 2^{-1}\: + \: 1 \: \times\: 2^{-2}\)

:> \( \scriptsize 8 \: + \: 4 \: + \: 2 \: + \: 1 \: + \: 0 \: + \: 0.25 = 15.25 \)

\( \scriptsize 1101_{2} \\ \scriptsize = 1 \:\times \: 2^3 \: + \: 1 \: \times\: 2^2 \: + \: 1 \: \times \: 2^1 \: + \: 1 \: \times \: 2^0 \)

:>\( \scriptsize 8 \: + \: 4 \: + \: 2 \: + \: 1 = 13 \)

i.e \( \scriptsize 1111.01_2 \div 1101_2 \equiv 15.25 \div 13 \)

= \( \scriptsize 1.173 \)

Convert 1.173 to base 2

i.e. 0.173 = 0.001₂

Therefore, \( \scriptsize 1.173 = 1.002_2 \)

Solution

(ii) \( \scriptsize 242_5 \div 14_5 \)  

First convert the given numbers to base 10.

\( \scriptsize 2\: \times \: 5^2 \: +\: 4 \: \times \: 5^1 \: + \: 2 \: \times \: 5^0 = 72_{10}\)

\( \scriptsize 1 \:\times \:5^1 \:+ \:4 \:\times\: 5^0 = 9_{10} \)

\( \therefore \frac {72_{10}}{9_{10}} \scriptsize = 8_{10} \)

We then convert 8₁₀ back to base 5

Therefore, \( \scriptsize 8_{10} = 13_{5} \)

Solution

(iii) \( \scriptsize 83A6_{16} \: \div \: 8_{16} \)

Convert both numbers to base 10

\( \scriptsize 83A6_{16} \: \div \: 8_{16} \\ \scriptsize = 8 \: \times \: 16^3 \: + \: 3 \: \times \: 16^2 \: + \: 10 \: \times \: 16^1 \: + \: 6 \: \times \: 16^0 \\ \scriptsize = 32768 \: + \: 768 \: + \: 160 \: + \: 6 \\ \scriptsize = 33,702\)

\( \scriptsize 8_{16} \\ \scriptsize = 8 \: \times \: 16^0 \\ \scriptsize = 8 \: \times \: 1 \\ \scriptsize = 8 \)

i.e \( \scriptsize 83A6_{16} \: \div \: 8_{16} \equiv 33,702 \: \div \: 8 \\ \scriptsize = 4212.75 \)

Convert back to base 16

Whole number part

Decimal part

Whole number part = \( \scriptsize 1074 \)

Decimal part = \( \scriptsize C \)

\( \scriptsize \therefore \: Answer = 1074.C_{16} \)