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I'm really struggling to understand the literal arithmetic being applied to find a complete residue system of modulo $n$. Below is the definition my textbook provides along with an example.

Let $k$ and $n$ be natural numbers. A set $\{a_1,a_2,...,a_k\}$ is called a canonical complete residue system modulo $n$ if every integer is congruent modulo $n$ to exactly one element of the set

I'm struggling to understand how to interpret this definition. Two integers, $a$ and $b$, are "congruent modulo $n$" if they have the same remainder when divided by $n$. So the set $\{a_1,a_2,...,a_k\}$ would be all integers that share a quotient with $b$ divided by $n$?

After I understand the definition, this is a simple example provided by my textbook

Find three residue systems modulo $4$: the canonical complete residue system, one containing negative numbers, and one containing no two consecutive numbers

My first point of confusion is "modulo $4$". $a{\space}mod{\space}n$ is the remainder of Euclidean division of $a$ by $n$. So what is meant by simply "modulo $4$"? What literal arithmetic do I perform to find a complete residue system using "modulo $4$"?

Bobby B
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    a complete residue system modulo $4$ is ${0,1,2,3}$ or ${1,2,3,4}$, or ${0,\pm1,2}$; the first example holds because, for any integer $n$, $n\equiv0$ or $n\equiv1$ or $n\equiv2$ or $n\equiv3\pmod4$ – J. W. Tanner Feb 28 '21 at 03:07
  • What criteria is required for that first example to "hold"? So if we choose $n=6$ we have $6\equiv0$, $6\equiv1$, $6\equiv2$, $6\equiv3\pmod4$. Consider $6\equiv3\pmod4$. If my understanding of modulus is correct, this tells us that difference between 6 and 3 is divisible by 4. But if we're restricted to only considering integers, then we have $(6-3)/4$ as not possible. So I'm confused as to what property the set must satisfy. – Bobby B Feb 28 '21 at 03:29
  • $6\equiv2\bmod4$. $5\equiv1\bmod 4$. For any integer $n$, $n\equiv0$, or $n\equiv1$, or $n\equiv2$, or $n\equiv3\bmod4$ (by Euclidean division) – J. W. Tanner Feb 28 '21 at 03:30
  • Let me rephrase this question. What literal arithmetic operations did you perform to find that ${0,1,2,3}$ is a complete residue system of modulo 4? I don't understand how to do that because I don't know what "modulo 4" means nor do I understand what requirements a set has to satisfy to be a "complete residue system". – Bobby B Feb 28 '21 at 03:36
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    I took all the integers from $0$ to $4-1$, in accordance with the link I provided for Euclidean division – J. W. Tanner Feb 28 '21 at 03:38
  • There are $4$ possible remainders. You need an integer with remainder $0$. Okay, $39732$ has remainder $0$. You need an integer with remainder $1$. Okay, $29$ has remainder $1$. You need an integer with remainder $2$. Okay $2$ has remainder $2$. You need an integer with remainder $3$. Okay $7$ has remainder $3$. So my set is ${39732, 29,2, 7}$. This is a complere residue system if "every integer is congruent $\mod 4$ to exactly one of those. That is true because every integer has remainder $0,1,2$ or $3$ and thus will be congruent to exactly one element in the set. That is all. – fleablood Feb 28 '21 at 03:43
  • "So if we choose n=6 we have 6≡0, 6≡1, 6≡2, 6≡3(mod4). Consider 6≡3(mod4)" No. You don't have all four of those. You have exactly one of those. Since $6-3$ isn't divisible by $4$ you don't have $6\equiv 3$. And as $6-0, 6-1$ aren't divisible by $6$ either so you don't have $6\equiv 0$ or $6\equiv 1$. But you DO have $6-2=4$ is divisible by $4$. So you do have $6\equiv 2 \pmod 4$. You have $6\not\equiv 0; 6\not \equiv 1; 6\equiv 2; 6\not \equiv 3$. So $6 \equiv$ to exact one, and only one of ${0,1,2,3}$. (Namely $6\equiv 2$ ... an no other.) – fleablood Feb 28 '21 at 03:49
  • Now take another integer. Say $n= 31$. Then $31 \equiv $ one of $0,1,2,3$. Which one? Well $31-0=30$ is not divisible by $4$; $31-1=31$ is not divisible by $4$. $31-2=29$ is not divisibe by $4$ and $31-3=28$ is divisible by $4$. So $31\not \equiv 0; 31\not \equiv 1; 31\not \equiv 2$ but we do have $31 \equiv 3 \pmod 4$. It's simple really. $31$ has remainder $3$. So does $3$. So $31 \equiv 3$ and not any of the others. – fleablood Feb 28 '21 at 03:52
  • "... integers that share a quotient" Exact opposite. Any two integers would have different remainders. And every possible remainder will be represented... exactly once..... We can classify the integers by what remainder they have ${...,.... -8,-4,0,4,8,12,....}$ will be the class of integers with remainder $0$. ${....,-7,-3,1,5,9,...}$ will be the class integers with remainder $1$. ANd ${...,-6, -2,2,6...}$ with remainder $2$ and ${....,-5,-1,3,7,11...}$ with remainder $3$. So a complete residue class is a set of one from each. So take one from each ${12, -7,6,3}$ – fleablood Feb 28 '21 at 16:35

3 Answers3

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Prelim: The definition is confusing because it is not assuming $k = n$. You will be able to prove $k = n$ later but in mathematics we don't include anything in a definition that we can prove later.

The definition of a complete residue system is a collection of integers $\{a_j\}$ so that for any integer, that integer will be congruent (have the same remainder) with exactly one of those integers.

In Other words, and probably a much more straightforward definiton, For every possible remainder, there will be be exactly one integer with that remainder.

For instance if $n = 7$, the easiest and most obvious complete residue system would be simply $\{0,1,2,3,4,5,6\}$. Every integer will be have remainder $0,1,2,3,4,5,6$ and those are precisely the numbers in there.

Another complete system could be $\{63, 8, 15, -4, 32, 75, 146\}$. If an integer $b$ has remainder $0$ it is congruent to $63$. $63$ represents all the integers with remainder $0$. ANd if $b$ has remainder $8$ then $b$ is congruent to $8$. $8$ represents all the integers with remainder $1$.... And so on.

Every remainder is represented exactly once.

And that is what a complete residue system means. A residue is a representation of one class of remainders (all the integers with remainder $4$ for example are represented by $32\equiv 4 \pmod 7$, for example). And a complete residue system means every residue is represented.

And as there $n$ possible remainders there will be $n$ elements in the system so if the system is $\{a_1, ...., a_k\}$ then $k = n$. (If it were me, I wouldn't even bring up the idea this could be in doubt. It just confuses things the first time you see the definition.)

.....

Okay. To do a completer residue system $\pmod 4$ you need to find a $\{a_1, a_2, ..... , a_k\}$ were for every integer $-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,....$ is congruent to exact one of them.

So we need and $a_1\equiv -6\pmod 4$. Well $-6\equiv 2 \pmod 4$ so let's let $a_1 = 2$. And we need something $\equiv -5 \pmod 4$. We $-5\equiv 3 \pmod 4$ and $15 \equiv 3 \pmod 4$ and $15 \equiv -3 \pmod 4$ so lets use $a_2 = 15$.

And we need something $\equiv - 4 \pmod 4$. Well $-4 \equiv 0 \equiv 28\pmod 4$ so let's use $a_3 = 28$. And we need something $\equiv -3\pmod 4$ but $-3\equiv 1 \equiv 48321 \pmod 4$. SO lets let $a_4 = 48321$.

And we need something $\equiv -2\pmod 4$. But $-2 \equiv 2 = a_1$ so we already have it. In fact it looks like we have one of each.

So $\{2, 15,28, 48321\}$ seems to be complete.

If $b$ is an integer we have either $b = 4k$ or $4k + 1$ or $4k + 2$ or $4k + 3$.

If $b = 4k$ then $b \equiv 28\pmod 4$. ANd if $4k + 1$ then $b\equiv 48321$. And if $b = 4k + 2$ then $b \equiv 2\pmod 4$ and if $b= 4k + 3$ then $b\equiv 15\pmod 4$.

So ....

the definition is:

$\{2, 15,28, 48321\}$ is called a canonical complete residue system modulo n if every integer is congruent modulo $4$ to exactly one element of the set

Well, that is true so $\{2, 15,28, 48321\}$ is a complete residue system.

That's all.

fleablood
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  • Let me see if I understand some of the logic. So two integers are congruent if they have the same remainder. This remainder is found by the Euclidean division $a=bq + r$ where the remainder must exist within the bounds of $0\leq r \leq n-1$. For example, 63 has a remainder of $0$ because $63=3(21)+0$. The phrase "modulo $n$" tells us the information that our remainders must exist in the bounds $0\leq r \leq n-1$. So a residue system of modulo $n$ would be a collection of $n$-amount of integers that all have a remainder somewhere in the bounds of $0\leq r \leq n-1$. – Bobby B Feb 28 '21 at 23:53
  • Furthermore, a complete residue system is where there is exactly one integer in the set that represents exactly one possible remainder within the bounds. So we have an integer that correlates with remainder 0, another integer that correlates with remainder 1, so on until we have an integer that correlates with remainder $n-1$. Are these ideas correct? – Bobby B Feb 28 '21 at 23:57
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It may prove helpful to understand the general conceptual background that underlies this definition. Congruence is an equivalence relation (generalized equality) so it partitions the integers into equivalences classes, here $\,[a] = a + n\Bbb Z\,$ is the class of all integers $\equiv a\pmod{n}.\,$

Conceptually, modular arithmetic is actually working with these classes, e.g. the congruence sum rule $\,a'\equiv a,\,b'\equiv b\Rightarrow a'+b'\equiv a+b\,$ is equivalent to saying that $\,[a]+[b] = [a+b]\,$ is a well-defined addition on the classes. Ditto for multiplication, so these classes inherit the addition and multiplication from the integers. Further they inherit the ring laws from $\,\Bbb Z,\,$ i.e. associative, commutative, and distributive laws. Thus the classes enjoy the same algebraic (ring) structure as the integers $\Bbb Z,\,$ so we can perform arithmetic on them just as we do with integers.

But when calculating it is often more convenient to work with numbers than sets (classes), so we often choose convenient representatives ("normal forms") for each class. A complete system of residues is precisely that. The most common choice here is the least natural in $[a],$ which is simply its remainder ("residue") $= a\bmod n,\,$ so this complete system of rep's $\bmod n\,$ is $\,\{0,1,\ldots, n-1\}.\,$ Another common choice are reps of least absolute value, e.g. $\{0,\pm1,\pm2\}\bmod 5$.

It may help to consider the analogy with fractions. Here the congruence relation for fraction equivalence is $\,a/b \equiv c/d\iff ad = bc.\,$ and the common normal form reps are again the least terms reps, i.e. fractions with coprime numerator and denominator. The above sum rule is usually never proved at elementary level - which should make you ponder why you accepted its truth without proof. This will be rectified if you study fraction fields (or localizations) in abstract algebra (which also generalizes modular arithmetic in the study of general quotient (residue) rings, and congruences, and their relationship).

Remark $ $ The choice of the remainder as a normal form rep also works in polynomial rings over a field (e.g. this yields the complex numbers $\,\Bbb C \cong \Bbb R[x]/(x^2+1)\,$ as the complete system of reps of real polynomials mod $\,x^2+1),\,$ or in any domain with a (computable) Euclidean algorithm. But in general there may not be any (computable) natural way to choose such normal forms, so we may have no choice but to work with the classes themselves.

Bill Dubuque
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You can draw a picture for modulo 6. Get a piece of paper and place it with the longer dimension horizontal. With a dark pen and a ruler, draw seven horizontal lines completely across the page. In about the middle of the page write the integers from 0 to 5 ONE PER REGION, starting at the bottom.

Enter all the other integers (or at least a lot of them) as follows: In each region, repeatedly add 6 to the number on the left until you run out of paper.

And then return to the middle and repeatedly subtract 6 from every number on the right.

(A complete residue system will be 6 integers, no two being from the same region.)

David
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