# Addition modulo

Now here we are going to discuss a new type of addition, which is known as “addition modulo m” and written in the form $$a{ + _m}b$$, where $$a$$ and $$b$$ belong to an integer and $$m$$ is any fixed positive integer.

By definition we have

$$a{ + _m}b = r,\,\,for\,0 \leqslant r < m$$

Here $$r$$ is the least non-negative remainder when $$a + b$$, i.e., the ordinary addition of $$a$$ and $$b$$ is divided by $$m$$.

For example, $$5{ + _6}3 = 2$$, since $$5 + 3 = 8 = 1\left( 6 \right) + 2$$, i.e., it is the least non-negative reminder when $$5 + 3$$ is divisible by $$6$$.

Thus to find $$a{ + _m}b$$, we add $$a$$ and $$b$$ in the ordinary way and then from the sum, we remove integral multiples of $$m$$ in such a way that the remainder $$r$$ is either $$0$$ or a positive integer less than $$m$$.

When $$a$$ and $$b$$ are two integers such that $$a – b$$ is divisible by a fixed positive integer $$m$$, then we have $$a \equiv b\left( {\bmod m} \right)$$. This is read as $$a$$ is concurrent to $$b$$ $$(mod m)$$.

Thus,$$a \equiv b\left( {\bmod m} \right)$$ if and only if $$a – b$$ is divisible by $$m$$. For example $$13 \equiv 3\left( {\bmod 5} \right)$$ since $$13 – 3 = 10$$ is divisible by $$5$$, $$5 \equiv 5\left( {\bmod 5} \right)$$, $$16 \equiv 4\left( {\bmod 6} \right)$$, $$ – 20 \equiv 4\left( {\bmod 6} \right)$$.

anjaneya

February 27@ 7:40 pmIs it correct to say

In multiplication modulo the product of two element should be = OR < the Group order

In addition modulo the addition of elements should not exceed the Group order.

This way the closure property is maintained