Is $x = y \mod 7$ for a set of integers an equivalence relation?

Question

Equivalence relation is the relation which is reflexive, symmetric and transitive. I have read somewhere that modulo operator defines an equivalence relation. But for this relationship I cant find $(7,7)$. If $y=7$ then $x=0$ (because $7$ is completely divisible by itself). Then how can it be reflexive? and how can it be an equivalence relation?

Answer 1

What is meant by "modulo operator is an equivalence relation" is the following:

We define that $x$ is congruent to $y$ modulo $n$, denoted $x \equiv y \pmod n$, if $n$ is a divisor of $x - y$.

This definition states in a mathematically precise way that $x \equiv y \pmod n$ if $x$ and $y$ have the same remainder modulo $n$.

Can you now prove that this $\equiv \pmod n$ is an equivalence relation? It is a good exercise to familiarise yourself with the concept.

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Edit: It just occurred to me that you may be subconsciously bracketing the expression $x \equiv y \pmod 7$ in an unintended way. What is meant is:

$$(x \equiv y) \pmod 7$$

as opposed to:

$$x = (y \mathrel\% 7)$$

where $\%$ is the remainder operation. The former will be an equivalence relation. The latter won't, for $(7,7) \notin R$. I hope that clears the air for you.

The first notation $x \equiv y \pmod 7$ can alternatively be read as:

$$(x \mathrel\% 7) = (y \mathrel\% 7)$$

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Edit 2: A few worked examples to get familiar with the $\equiv$ notation.

• $7 \equiv 14 \pmod 7$? By definition, this holds if $7 \mid (7 -14)$. Since $7-14 = -7$, we conclude $7 \equiv 14 \pmod 7$.
• $23 \equiv 8 \pmod 6$? This holds if $6 \mid (23-8)$. Since $6 \nmid 15$, we conclude $23 \not\equiv 8 \pmod 6$ (that is: "it is not the case that $23 \equiv 8 \pmod 6$").
• $4 \equiv 4 \pmod{11}$? This holds if $11 \mid (4-4)$. Since $11 \mid 0$, it follows that $4 \equiv 4 \pmod{11}$.

Answer 2

It would be an equivalence relation if x - y is always divisible by 7.
In this case taking x = 7 we'd get x - y = 7n for some integer n and hence $x = 7 \equiv$ 0 mod 7 and $y = x - 7n = 7 - 7n = 7(1-n) \equiv$ 0 mod 7.
Hence x $\equiv$ y.
In general taking x = a mod 7, since x - y is divisible by 7, x - y = 7n for some integer n.
Hence y = x - 7n $\equiv$ a mod 7 - 7n $\equiv$ a mod 7.

If it's not always the case that x - y is divisible by 7, it's not an equivalence relation.
In this case there exists an x and a y such that there's no integer n for which x - y = 7n.
Thus x = a mod 7 and there's no integer n such tha y = x - 7n $\equiv$ a mod 7 - 7n $\equiv$ a mod 7.
Hence x isn't eqivalent to y mod 7.