Chapter 4 - Number Representation and Calculation - 4.3 Computation in Positional Systems - Exercise Set 4.3 - Page 235: 39

The operation’s value is, \[{{1000110}_{\text{two}}}\].

Consider the expression, \[{{10110}_{\text{two}}}+{{10100}_{\text{two}}}+{{11100}_{\text{two}}}\] This can be written as, \[\begin{align} & \ \ {{10110}_{\text{two}}} \\ & \ \ {{10100}_{\text{two}}} \\ & +{{\underline{11100}}_{\text{two}}} \\ \end{align}\] Begin by adding the numbers of the one’s column as, \[{{0}_{\text{two}}}+{{0}_{\text{two}}}+{{0}_{\text{two}}}={{0}_{\text{two}}}\] \[\begin{align} & \ \ {{10110}_{\text{two}}} \\ & \ \ {{10100}_{\text{two}}} \\ & +{{\underline{11100}}_{\text{two}}} \\ & \text{ 0} \\ \end{align}\] Add the numbers of two’s column as, \[{{1}_{\text{two}}}+{{0}_{\text{two}}}+{{0}_{\text{two}}}={{1}_{\text{two}}}\] \[\begin{align} & \ \ {{10110}_{\text{two}}} \\ & \ \ {{10100}_{\text{two}}} \\ & +{{\underline{11100}}_{\text{two}}} \\ & \text{ 10} \\ \end{align}\] Add the numbers of three’s column as, \[{{1}_{\text{two}}}+{{1}_{\text{two}}}+{{1}_{\text{two}}}=3\] 3 is not a digit symbol in base two. It can be express as one group of 2 and one 1 left over as, \[\begin{align} & {{1}_{\text{two}}}+{{1}_{\text{two}}}+{{1}_{\text{two}}}={{3}_{\text{ten}}} \\ & =\left( 1\times 2 \right)+\left( 1\times 1 \right) \\ & ={{11}_{\text{two}}} \end{align}\] Write 1 in the three’s column and carry forward 1 to four’s column. \[\begin{align} & \text{ 1} \\ & \text{ }{{10110}_{\text{two}}} \\ & \text{ }{{10100}_{\text{two}}} \\ & +{{\underline{11100}}_{\text{two}}} \\ & \text{ 110} \\ \end{align}\] Add the numbers of four’s column as, \[{{1}_{\text{two}}}+{{0}_{\text{two}}}+{{0}_{\text{two}}}+{{1}_{\text{two}}}=2\] 2 is not a digit symbol in base two. It can be express as one group of 2 and zero ones left over as, \[\begin{align} & {{1}_{\text{two}}}+{{0}_{\text{two}}}+{{0}_{\text{two}}}+{{1}_{\text{two}}}={{2}_{\text{ten}}} \\ & =\left( 2\times 1 \right)+\left( 0\times 1 \right) \\ & ={{10}_{\text{two}}} \end{align}\] So, write \[0\] in four’s column and carry forward\[1\] to the five’s column. \[\begin{align} & \text{ 11} \\ & \text{ }{{10110}_{\text{two}}} \\ & \text{ }{{10100}_{\text{two}}} \\ & +{{\underline{11100}}_{\text{two}}} \\ & \text{ 0110} \\ \end{align}\] Add the numbers of five’s column as, \[1+1+1+1=4\] 4 is not a digit symbol in base two. It can be express as one group of 4 and two zeros left over as, \[\begin{align} & {{1}_{\text{two}}}+{{1}_{\text{two}}}+{{1}_{\text{two}}}+{{1}_{\text{two}}}={{4}_{\text{ten}}} \\ & =\left( 4\times 1 \right)+\left( 2\times 0 \right)+\left( 1\times 0 \right) \\ & ={{100}_{\text{two}}} \end{align}\] Therefore, the desired sum can be written as, \[\begin{align} & \ \ \ \ {{10110}_{\text{two}}} \\ & \ \ \ \ {{10100}_{\text{two}}} \\ & +{{\underline{\ \ 11100}}_{\text{two}}} \\ & {{1000110}_{\text{two}}} \\ \end{align}\] Hence, the operation’s value is, \[{{1000110}_{\text{two}}}\].