# Complementary Numbers

~ on ~ 0 commentsBefore discussing the theory of self-complementing codes, we must have an idea about *complementary numbers*. Consider a number system with radix *r*. The basic elements of this family are numbers from 0* *to* r ‒ *1*.*

*A number in a system with radix 2 is said to be the complementary of another number in the same system, if the sum of the two numbers amounts to the highest number (i.e., r ‒*1

*) in that system.*The first number is then said to be the (

*r ‒*1)’s complement of the second number. If we add a ‘+1’ to the first number, we then get the

*r*’s complement. The following steps will help in determining the complement of a given number.

**1.**Find the radix of the given number. Let it be

*r*.

**2.**Find the basic elements of this number system. These will be 0, 1, …,

*r*

*‒*1.

**3.**Find the highest basic number, which is

*r*

*‒*1.

**4.**To find the complement of a given number, subtract it from

*r*

*‒*1. Thus, if

*k*is a given number in the system with radix

*r*, then its complement will be (

*r*

*‒*1)

*‒k*=

*r*

*‒*1

*‒ k*.

We now illustrate the use of the steps given above with some examples.

**Example 19:**Obtain the complement of decimal number 1.

**Solution:**In this example, we have to get the complement of

*k*= 1.

Using 1, for decimal number system

*r*= 10.
Using 2, the basic elements of this scheme are digits 0 to 9.

Using 3, the highest number is

*r*‒ 1 = 10 ‒ 1 = 9.
Using 4, complement of 1 = 9

*‒*1 =**8**.**One’s (1’s) Complement**

In the binary number system, 1’s complement and 2’s complement play important roles in the subtraction of numbers. In this section, we explain the idea of 1’s complement using examples.

**Example 20:**Obtain the complement of binary numbers

**0**and

**1**.

**Solution:**In this example, we have to get the complement of

*k*=

**0**and

*k*=

**1**.

Using 1, for decimal number system

*r*= 2.
Using 2, the basic elements of this scheme are digits

**0**and**1**.
Using 3, the highest number is

*r*‒ 1 = 2 ‒ 1 =**1**.
Using 4, complement of

**0 = 1***‒**k***=****1 ‒***0 = 1*
Using 4 again, complement of

**1 = 1***‒*1 = 0
We thus find that

*the**complement of***1**is**0**and the complement of**0***is***1**. The advantage of this scheme is that any binary number has its complement easily obtained by simply replacing the**1**s in it with**0**s and**0**s in it with**1**s.**Example 21:**Obtain the complementary number of

**10110.**

**Solution:**Replacing the

**0**s and

**1**s in

**10110**with

**1**s and

**0**s, respectively, we obtain the complementary number of

**10110**=

**01001**.

**r**

*’s*and 2’s Complements
The

*r*’s complement of a given number = (*r*‒*1)’s complement + 1. Thus, 2’s complement of a binary number = 1’s complement of that number + 1.***Example 22:**Obtain the 2’s complement of

**10110.**

**Solution:**Replacing the

**0**s and

**1**s in

**10110**with

**1**s and

**0**s, respectively, we obtain the 1’s complement of

**10110**as

**01001**. Adding bit

**1**to

**01001**yields

**01010**. Thus

2’s complement of

**10110**=**01010**
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