When to use the $\equiv$ symbol (such as in $5^{6}$ $\equiv$ 1 mod 7) vs =

Question

Can anybody explain why we would use the $\equiv$ symbol in the statement $5^{6}$ $\equiv$ 1 mod 7 ? I understand the $\equiv$ symbol means equivalence, but it seems like it would be more appropriate to use the = symbol.

Answer 1

for example, with $=$ you have to write: $$5^6 \mod (7) = 1$$

With $\equiv$ you can write: $$5^6\equiv 1\mod(7)$$

It has some abuse of notation sometimes, but is used naturally for understanding theorems.

Answer 2

It is clear that $5^6$ is not equal to $1$. But from a certain point of view it is equivalent to $1$, and in particular that point of view is when looking at equivalence classes modulo $7$.

Answer 3

$5^6\equiv 1 (\mod 7)$ means $5^6$ is of the form $1+7n$ where $n$ is an integer ( in this case, it is positive integer). As you can see that both $7n+1$ and $5^6$ leave remiander 1 when divided by 7. This is why we say $5^6$ and $1$ are "equivalent" under the modulo operation 7. and thats why we use equivalent symbol instead of equality symbol.

If you are trying to write $5^6=1 (\mod 7)$ you are actually saying that n=0.

Answer 4

Typically, we use the $\equiv$ symbol to denote that one element is related to another under some specific relation, when we are not specifically working with the equivalences classes of the relation as algebraic objects.

For example, "congruence mod 7" is a relation on integers, So let's label it $R$, which would be defined as:

$$ x \; R\; y \text{ if and only if } x = 7k + y \text{ for some k} \in \mathbb{Z} $$

and we can say things like, $3 \; R \; 3$, and $10 \; R\; 3$.

Now, we usually don't label such a relation as $R$, instead, we typically use $\equiv$ to denote equivalence relations. In the case of modular arithmetic, we use $\text{(mod 7)}$ in a parenthetical to remind the reader of the particular equivalence relation we are referring to.

In contrast, we use $=$ only for algebraic objects that are equal.

For example, if we are working in the space $\mathbb{Z}_7$, the members of this set are exactly the equivalence classes (sets) of integers modulo 7.

The elements of this set are $\overline{0} = \{\dots, 0, 7, 14, \dots\}$, $\overline{1} = \{\dots, 1, 8, 15, \dots\}$, and so on.

In this sense, it would be entirely correct, given two members of $\mathbb{Z}_7$, say $\overline{x}, \overline{y}$, that $\overline{x} = \overline{y}$ if they are the same set (equivalence class).

In short

• $x \equiv y \; \text{(mod n)}$ means that $x$ is related to $y$ under the relation "congruence modulo $n$"

• $\overline{x} = \overline{y}$ means that $\overline{x}$ and $\overline{y}$ are the same set or equivalence class, when these equivalence classes are being treated as objects.

Answer 5

Perhaps you are confusing the relation vs. operator form of $\,\rm{mod =}$ modulo.

$\quad a\, \equiv\, b\pmod n\ $ means $\ n\mid a-b,\,$ i.e. $\,n\,$ divides $\,a-b$

$\quad a\, =\, b\ {\rm mod}\ n\ $ means the above $ $ and $\ 0\le a < n,\,$ i.e. $\,a\,$ is the remainder of $\,b\div n$

$\rm\,mod\,$ is a ternary relation in the first form, but a binary operation in the second. See here for much further discussion on this topic.

Answer 6

If you know some abstract algebra, this another way to look at it.

$5^6 \equiv 1 \mod 7$ denotes equality between the equivalence classes represented by $5^6$ and $1$. This is usual denoted as '$[5^6]=[1]$ in $\mathbb{Z}/7\mathbb{Z}$'.

For consistency, one could argue to use this whenever you quotient by a equivalence relation. Number theory is filled with arithmetic and system of equations of these equivalence classes, see the Chinese Remainder Theorem, so these brackets would 'drown' the notation.

Another reason why they use $\equiv$ in number theory is history. Abstract Algebra is new compared to modular arithmetic.