I'm more familiar with the notation $a \equiv b \pmod c$, but I think this is equivalent to $a \bmod c = b \bmod c $, which makes it clear that we should put a $=$ instead of $\equiv$.
What's the reason for the change of sign? If it's to emphasize that modular equivalence is a congruence relation, why don't we use the $\equiv$ sign in both notations?
- 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, $3\neq7$, but $3\equiv7\bmod4$.
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 ($\equiv$) signs together when processing a lengthy calculation in modular arithmetic. My freshman algebra students quickly catch on with my calculations like $$ 12^{3004}\equiv5^{3004}=5^{3000}\cdot 5^4=(5^6)^{500}\cdot 5^4\equiv 1^{500}\cdot5^4=25^2\equiv4^2=16\equiv2\pmod7. $$ IOW, I use $=$ when there is an equality of integers between the steps, and $\equiv$ 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 $\equiv$ all the way is correct also. The $\equiv$ 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 $\equiv$ as well as, clarity of context permitting, the distinction between $n$ and $\overline{n}$.
Thinking about what that would look like when done by somebody who is only familiar with binary mod makes me shudder.
% for binary mod! It is such a clean and neat operator and directly applicable to real numbers without even changing its definition $x%y = x - \lfloor \frac{x}{y} \rfloor y$. You can then write things like $(a^k)%b = ((a%b)^k)%b$ for any positive reals $a,b$ and integer $k$. Nice? =D
– user21820
Aug 15 '15 at 12:53
% is a good idea. The power of congruence is that you really ever need a single mod per calculation. Actually you can just state that all the arithmetic is done modulo $n$ and be done with writing mod at all! Binary mod will be strictly needed, if there are several different moduli in the same calculation. But that is so unnatural. Mind you, I had to do just that for one of my contributions (a pilot signal offset scheme) to a 4G standardization body. It was later decided not to use such schemes at all (to my relief :-)
– Jyrki Lahtonen
Aug 15 '15 at 18:49
% for binary mod is that this symbol already has a different but strongly confuseable meaning in many programming languages. In Java, C#, C, and C++ (-9)%7 evaluates to $-2$, whereas in order to be sanely useful for modular arithmetic we must have $(-9)\bmod 7 = 5$.
– hmakholm left over Monica
Aug 15 '15 at 18:56
% is different from my definition but there is no choice if we want one consistent with mathematical use so I ignored the programming version. =D
– user21820
Aug 16 '15 at 03:38
MOD — I wonder whether that can be traced to the BASIC programming language … (Cont’d)
– Scott - Слава Україні
Aug 16 '15 at 03:57
REM (short for REMARK) as the token to begin a comment — and, as a result, millions of people believe (incorrectly) that they know what “modulo” means. (These people will often argue that a statement like $4 \equiv 7 \pmod 3$ is false or even nonsense.) I’m glad I’m not alone in my feelings.
– Scott - Слава Україні
Aug 16 '15 at 03:58
%. It is an extremely useful operator when used well; I've also seen people get themselves into all kinds of difficulty by misusing it.
– David K
Aug 16 '15 at 13:53
% if you ever use | in mathematical expressions.
– user21820
Aug 16 '15 at 14:02
%; I'm just saying it comes with baggage. In fact $a \bmod b$ has much the same baggage. The problem is not that one symbol has two meanings: for example, vertical bar can mean many different things in math, none of which will be confused with bitwise or. The problem is when you're using a symbol almost but not quite the way it is defined elsewhere. Then you should take extra care to ensure that people do not read the symbol the wrong way.
– David K
Aug 17 '15 at 20:17
It depends on how you define your notation. The standard definition of congruence is $a \equiv b \pmod c$ if and only if $c \mid (b-a)$. The standalone expression $b \pmod c$ is undefined, hence not equal to anything.
On the other hand, some people like to write "$b \bmod c$" to stand for the least nonnegative number $d$ that satisfies $d \equiv b \pmod c$. If you adopt this notation then it is true that $a \equiv b \pmod c$ if and only if $(a \bmod c) = (b \bmod c)$.
(mod 5) does look LISP-ish...
– Kevin
Aug 14 '15 at 20:29
The difference between the two is as follows:
Note that (in its usual usage) $a \mod c$ is a function that gives a particular answer, and has nothing to do with equivalence relations per se. For example, we might note that $$ \renewcommand{\mod}{\operatorname{mod}} 7 \mod 2 = 1 $$ That is, $7$ has a remainder of $1$ upon division by $2$. What we cannot say is that $$ 7 \mod 2 = 3 $$ because $1$ and $3$ are distinct values. We can say, however, that $$ 7 \equiv 3 \pmod 2 $$ Since both $7$ and $3$ have the same remainder upon division by $2$.
Why is it important? Well, we often make "substitutions" of equivalent values, and writing "mod" every time you make a substitution becomes unwieldy. For example, we can compute: $$ 5*7*9 - 2*3 + 5 + 3 \equiv 1*1*1 - 0*1 + 1 + 1 \equiv 3 \equiv 1 \pmod 2 $$ If we were to write this with mod as an operator, we'd have $$ [5*7*9 - 2*3 + 5 + 3] \mod 2 =\\ [(5 \mod 2)(7 \mod 2)(9 \mod 2) - (2 \mod 2)(3 \mod 2) + (5 \mod 2) + 3 \mod 2] \mod 2 = \\ [(1)(1)(1) - (0)(1) + (1) + 1 ]\mod 2 =\\ 3 \mod 2 = 1 $$
\mod in $\TeX$, but instead write \textrm{mod}. The increases whitespace on the left-hand side produced by \mod makes things look weird if used as a binary operation, and blurries the distribution between the binary operator $\textrm{mod}$, and the clause $\mod n$ which indicates that an equality holds in $\mathbb{Z}_n$, not $\mathbb{Z}$.
– fgp
Aug 16 '15 at 11:45
Equal to means that they are the (exact) same thing. Congruent to means that they have some important property in common, in this case the same remainder when divided by the modulus. The relation, however, must be reflexive, symmetric, and transitive.
I (like most people, I believe) move freely between $=$ and $\equiv$ in the following sense.
Saying $a \equiv b \pmod{n}$ is the same as saying $a = b$ as elements of $ \mathbb Z / n\mathbb Z$. At least this is what I say to my students to keep in mind in case I jump around in my notation.
"Why" questions about notation can range into non-mathematical territory: Notation is not always perfectly logical or consistent.
This isn't an explanatory answer, but an extended comment about writing $$ a \equiv b \pmod{c} \tag{1} $$ versus $$ a \bmod{c} = b \bmod{c}. \tag{2} $$
In (1), mod $c$ is grammatically an adverb, modifying the binary relation "is equivalent to". Consistency dictates writing something like $$ a \equiv_{\bmod{c}} b\qquad\text{or}\qquad a \equiv_{c} b. $$ I'm not a historian, but believe the widespread use of the notation (1) at the school level arose in the New Math era of the 1950s. (It's easy to imagine how this "afterthought" notation (mod $c$) would work psychologically for students raised on arithmetic drills involving integers. I haven't a shred of evidence, however, and will keep my speculations private.)
In (2), by contrast, mod $c$ functions grammatically as an adjective, modifying the symbols $a$ and $b$. This is a more sophisticated viewpoint: Instead of relating integers $a$ and $b$ (which are familiar to young students), (2) equates residue classes mod $c$. That's already one reason to avoid (2) in an introductory setting.
To give a completely different reason, writing "$a \bmod{c}$" to denote a residue class is clumsy. The generic equivalence class notation $[a]_{c}$ would be preferable, or even $[a]$ if the modulus $c$ were fixed during the discussion/computation. One does in fact see $[a] = [b]$ in introductory algebra at the university level.
Honestly, even square brackets are cumbersome. (They can also cause perplexing LaTeX errors when Cayley tables are typeset as arrays. Exercise: Why?) Between friends and in private, you might as well denote the residue class simply as "$a$" if the modulus is fixed, so that mod $7$ you have, e.g., $$ 5^{2015} \equiv -(2^{2015}) \equiv -(2^{3 \cdot 671 + 2}) \equiv -4 \equiv 3. $$
Ideally: Mathematical notation conveys concepts concisely, precisely, and consistently across fields and disciplines.
In practice: Changing historical perspectives and notational habits, field-specific conventions and emphases, and computer typesetting capabilities influence the ways we express mathematics in writing. Even sub-specialities of pure mathematics may use mutually-incompatible notation, to say nothing of discussions between pure mathematicians and engineers (say).
The answer to the question at hand may be as banal as "(1) is more compact than (2)" or "that's just the way everyone writes it (in 2015)". The take-away message is: "Mathematical notation is not always logical or consistent, and making notation consistent is much harder than it might first appear."
Answer to Exercise:
If a square-bracketed expression follows a newline command
\\with only white space in between, LaTeX reads the expression as an optional argument, expecting a length.
We could have $-1 \bmod 2=-1$ and $1\bmod 2=1$, so $1\neq -1$, but nevertheless we have $1\equiv -1 \bmod 2$. So the equality $a\bmod c=b\bmod c$ is not clear, because it depends of the choice of a residue system.
The $a$ in the expression $$a \equiv b \pmod c$$ and the thing you intuitively write as $a \mod c$ in the expression $$a \bmod c = b \bmod c$$ are different kinds of things.
The symbol $a$ denotes an integer—a variable of type $\mathbb{Z}$, as a programmer might say, or an element of the ring $\mathbb{Z}$, as a mathematician might say.
The thing you denote by $a \bmod c$ is an integer modulo $c$—a variable of type $\mathbb{Z}/c\mathbb{Z}$, or an element of the ring $\mathbb{Z}/c\mathbb{Z}$. You can think of an integer modulo $c$ as an "hour" on a "$c$-hour clock," as described at the link above.
There's a natural way to turn an integer $n$ into an integer modulo $c$: you put the clock hand at noon, and then tick it forward $n$ hours. The hour you end up at is the one you call $n \mod c$. A more common notation for it is $[n]_c$, which can be abbreviated to just $[n]$ if it's clear that you're working on a $c$-hour clock.
Saying that the integers $a$ and $b$ are congruent modulo $c$, written $$a \equiv b \pmod c,$$ is the same as saying that the hours $[a]_c$ and $[b]_c$ are equal, written $$[a]_c = [b]_c.$$ In fact, this is how many people define congruence. You should be able to convince yourself, if you think about it long enough, that the hours $[a]_c$ and $[b]_c$ are equal if and only if the integers $a$ and $b$ have the same remainder when divided by $c$, which is another common definition of congruence modulo $c$.
Typically $=$ is used to denote that two objects are equivalent in every single context. It is usually a primitive in logic, defined by textual replacement.
Other symbols, for example $\equiv$, are used for context sensitive equivalence. $9$ is equal to $4$ in the context of $\pmod 5$. Two binary trees may be equivalent in terms of their shape, but not where they are located in memory. A husband and wife may be equivalent for the sake of tax purposes, but not for their respective expectations of each other.
In more advanced math, you will find number systems where it is not possible (or possible, but not wise) to define $a\mod c$ in an obvious way, while you will be able to write it as $a\equiv b\pmod{c}$.
There is an additional problem with $a\mod c = b\mod c$, which is that it allows for $a\mod c = b\mod d$. But this is hardly ever useful.
There are ways to write it as equality, but it is pedagogically useful to become comfortable with congruence as an instance equivalence relation.
I don't see any problem in writing $a\bmod n = b\bmod n$ as long as one is familiar what it means.
Even "worse" is when dealing with decimal system, we write, for example, $1=0.\dot9$ even though they are not the same series - they merely converge to the same real number.
Now, to the point. We start with ring $\mathbb Z$ (i.e. a structure consisting of integers with familiar addition and multiplication) and fix some natural $n$. Then we define equivalence relation $\equiv$ such that $$ a\equiv b \iff n|b-a$$ Having the fact that one can form quotient set of equivalence classes with respect to $\equiv$, using the properties of $\mathbb Z$ (namely that multiples of $n$ form an ideal) it follows that the quotient set $\mathbb Z/\equiv$ is actually a ring itself and then we use more familiar notation $\mathbb Z/n\mathbb Z$. Very importantly, we get a surjective function $\pi:\mathbb Z\to\mathbb Z/n\mathbb Z$ that sends any $a$ to it's equivalence class $[a]$ satisfying $$\pi(a+b) = \pi(a) + \pi(b),\ \pi(ab) = \pi(a)\pi(b)\quad \text{(i.e. $\pi$ is a ring homomorphism)}$$ So, when we write $a \equiv b\ \pmod n$ we mean that $a$ and $b$ belong to the same equivalence class (which just means that they give the same residue when divided by $n$ in this context) and consequently $\pi(a) = \pi(b)$. We might as well write $\operatorname{mod} n$ instead of $\pi$ and apply convention $a\bmod n = (\operatorname{mod}\ n)(a)$.
Why don't we do it? Because it is tedious to write down calculations that way.
One usual way of writing down calculations is to write $\overline 1,\ \overline 2,\ldots$ for congruence classes when having fixed $n$. For example $\overline 1 = \overline 3$ in $\mathbb Z/2\mathbb Z$, instead of $1\equiv 3\pmod 2$. I wouldn't have problem even if one would write $1=3$ as long as it is clear that these merely represent same elements of the ring $\mathbb Z/2\mathbb Z$, and not equality of integers, just as $1 = 0.\dot 9$ as real numbers, but not as series.
Long story short, how you write depends on the context and most important thing is that it is not ambiguous to the reader.
Relating to @Omnomnomnom's answer, I think there is a subtle difference in the mathematical and computer science perspectives to this.
From a computer science perspective, a mod b looks a lot like a % b, which is a distinct operation with a single answer (try typing 7 % 3 into your Python interpreter, and you will always get the same answer).
I generally think of congruency in the following manner:
Congruency: small mod medium ≡ big if and only if small ≡ big mod medium
Equality: big mod medium = small (or in Python, big % medium == small), but the opposite direction does not hold (small mod medium ≠ big).
My experience with this topic is limited to single dimensional modular integer arithmetic used in cryptography, so I cannot speak to any subtleties that may arise in other contexts.
\bmod= binary mod, when you mean the operation that calculates the remainder of a division of one integer by another. – Jyrki Lahtonen Aug 14 '15 at 13:44