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Question 10 (Addition of numbers with base 2)

Question 10 : Find the sum of ​\( (111)_2 and (10)_2 \)​.

a ​\( (1001)_2 \)​               b) \( (1011)_2 \)​             c) \( (1101)_2 \)​                  d) \( (1111)_2 \)

Solution:

 

First of all we arrange these numbers in column and row style;

Here we start by adding entries in right column, then we will add entries in center column and lastly we will deal with left column entry which is 1 here.

How to add ?

We see that numbers are with base 2 !

To take sum of entries in each column , we will assume here that 2 is equal to 10 – i-e whenever we move two counts ahead we will assume it as 10.

To understand it  we see that;

1- In right side column there are entries 1 and 0 . Taking sum i-e 1+0 = 1  which we write below line.

\[ \begin{equation} \frac{ \begin{array}[b]{r} \left( 111 \right)_2\\ \ + \left( 10 \right)_2 \end{array} }{ \left( 1 \right) } \end{equation} \]

2- In the center column there are entries 1 and 1 . Taking sum 1+1 = 2, but as said above we assume that 2 is equal to 10 ! So, 1+1=10 . The we write 0 below line –  and carry 1 on last or left column

\[ \begin{equation} \frac{ \begin{array}[b]{r} \left( 111 \right)_2\\ \ + \left( 10 \right) \end{array} }{ \left( 01 \right) } \end{equation} \]

3- In the left coloumn there is only one entry i-e 1. We add this one with the carried 1 and again we get 1+1=2 but 2 is assumed to be equal to 10 so 1+1=10 which we write below line as below;

\[ \begin{equation} \frac{ \begin{array}[b]{r} \left( 111 \right)_2\\ \ + \left( 10 \right)_2 \end{array} }{ \left( 1001 \right)_2 } \end{equation} \]

Note:  Ignore ‘[b] r’ written up on left !

Option a is correct one here !

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