# Modular arithmetic: confusion one the $mod$ operator.

Suppose I have $a\equiv b \text{ (mod$c$)}$ and I just know $c$. I want to know, say $b$. Is this the same as $a$ mod $c$? If so, why?

I think I confuse the congruence with the equality symbol, because only recently I learned that sometimes $mod$ is used as an operator, and not as a mere part of the congruence symbol.

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One of the possible values for $b$ would be $a$ mod $c$, but there will be others too, because adding any multiple of $c$ to a solution yields another solution.

For example: $6 \equiv b$ (mod 3). We have (6 mod 3) = 0, and one of the possible values for $b$ is 0, but actually $b$ could be any multiple of 3.

Most of the time, in number theory, you don't really want to think about mod as an operator; you want to think of it as defining a relationship between numbers (i.e., a congruence). The congruence symbol is deliberately made to look like an equality symbol because many of the properties of equality also apply to congruences. For example, you can add, subtract, or multiply any integer to both sides of a congruence, and the congruence remains valid. (Just be careful with division.)

These properties are most cleanly stated when stated in terms of congruences, rather than using mod as an operator.

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Actually $a\equiv b\text{ ( mod$c$)}$ means $a-b=ck$, where $k\in\mathbb{Z}$. That is when a and b are divided by c in both cases we have same remainder.