Addition & Subtraction in Modular Arithmetic

The diagram above shows addition in modulo 6 in which an element on the extreme left is added to an element on the top row, in that order. You will observe that the first column on the extreme left and the top row contain the same set of integers.

Example 5.2.1:

\( \scriptsize 3 \: \bigoplus \: 2 \)

Solution

Look for the number 3 on row 4 and the number 2 on column 3, they intersect at number 5.

i.e.  \( \scriptsize 3 \: \bigoplus \: 2 = 5\:(mod\:6) \)

In the same vein, subtraction can be considered as the inverse operation of addition. So, in mod 6, we can use the addition table to work out subtraction:

Example 5.2.2:

\( \scriptsize 2 \: \circleddash \: 5 \)

Solution

\( \scriptsize 2 \: \circleddash \: 5 \) is a number y such that 5 + y = 2, i.e. 5 plus a number gives 2.

From the residue columns and rows, 5 and 3 intersect at 2, therefore y = 3.

It is important to know that in subtraction the first number is picked from the residues column and the 2^nd is picked from the residues row.

More Examples: Addition in mod 6

(i) \( \scriptsize 5 \: \bigoplus \: 4 = 3 \: (mod\: 6) \)

(ii) \( \scriptsize 4 \: \bigoplus \: 1 = 5 \: (mod \:6) \)

(iii) \( \scriptsize 3 \: \bigoplus \: 5 = 2 \: (mod \:6) \)

(iv) \( \scriptsize 1 \: \bigoplus \: 5 = 0 \: (mod \:6) \)

Example 5.2.3:

Evaluate \( \scriptsize 4 \: \bigoplus \: 7 \: (mod\:4) \)

Solution

Step 1: First step add 4 and 7 = 4 + 7 = 11

Step 2: Convert the answer to modulus 4 (Remember numbers in modulus 4 can only be 0, 1, 2, or 3)

\( \frac{11}{4} \)

= 2 remainder 3

= 3 (mod 4)

Example 5.2.4: