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Confusion regarding the symbol '≡' (congruent to) in modular arithmetic
notation - Why do we use "congruent to" instead of equal to? - Mathematics Stack Exchange
modular arithmetic - Notation for modulo: congruence relation vs operator - Mathematics Stack Exchange
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There are several ways, choose the one you prefer.
\documentclass{article}
\usepackage{amsmath}
\renewcommand{\arraystretch}{1.5} % just to make the lines spread out
\begin{document}
\begin{tabular}{lll}
\verb|
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\verb|
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\verb|
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\verb|
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& (wrong)
\end{tabular}
\end{document}

Don't forget the braces: try a\equiv b \pmod 11 or a\equiv b \pmod pq and see why.
The last one is marked “wrong”, because the usage is improper: \bmod should be used for the “modulo” binary operation (the one that is often denoted by % in computing).
Use pmod n. It is known to work.
Hello everyone,
In modular arithmetic, if we know the remainder r when dividing a by m, we write it as:
a ≡ r mod m
As I understand it, r is the result of the operation a mod m.
However, in other formulas—like in RSA encryption—we often see something like:
y ≡ x^(e) mod m
This means that y is the result of the operation x^(e) mod n.
So to me, it would feel more intuitive to write:
x^(e) ≡ y mod n
since x^(e) mod n = y, and the expression being reduced appears on the left-hand side.
The way the modular expression is written can be a little confusing at first, but both forms describe the same relationship.
- I am a human being.
- You are a human being.
Therefore, I am you: right ? Well, as it turns out, the answer is no. It simply means that we belong to the same class. Likewise, , but
.
I prefer to use the congruence exclusively (I view the remainder operation as a source of headache even though I realize that it is a necessary evil in computing).
But I use both equal () and congruent to (
) signs together when processing a lengthy calculation in modular arithmetic. My freshman algebra students quickly catch on with my calculations like
$$
12^{3004}\equiv5^{3004}=5^{3000}\cdot 5^4=(5^6)^{500}\cdot 5^4\equiv 1^{500}\cdot5^4=25^2\equiv4^2=16\equiv2\pmod7.
$
=$ when there is an equality of integers between the steps, and
when I mean a congruence. The power of laws obeyed by congruences is apparent. Checking/following the progress of the calculation is easier this way. Of course, using
all the way is correct also. The
is there as a reminder that in this step we do something that only results in a congruence.
As the students become acquainted with the language of residue class rings, I gradually stop making the distinction between and
as well as, clarity of context permitting, the distinction between
and
.
Thinking about what that would look like when done by somebody who is only familiar with binary mod makes me shudder.