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I need to understand what guarantess closure in $\mathbb{Z}_n$ under the operations

$[a]_n + [b]_n := [a + b]_n$

$[a]_n \cdot [b]_n := [a \cdot b]_n$

Different manuals state this:

It is clear that the rule $[a]_ + [b]_n := [a + b]_n$ yields an element of $\mathbb{Z}_n$, but the uniqueness of this result needs to be verified.

These properties are an imnmediate consequence of the definition of + and * in $\mathbb{Z}_n$.

I need to understand why it is so obvious (not for me) that closure is implied in the definition.

Previously this theorem was stated

If $a ≡ b \mod n, c ≡ d \mod n$ Then $a + c ≡ b + d \mod n$ and $ac ≡ bd \mod n$

I think that $a ≡ b \mod n, c ≡ d \mod n$ implies $a + c ≡ b + d \mod n$ and this implies that $[a+c]$ is in $\mathbb{Z}_n$.

But the manual seems to suggest that closure is implied in the definition of the operations I given above. I'm not interested to prove that those are well-defined for the moment.

Nicolas FRANCOIS
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Giuliano Malatesta
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  • The definition says that to add two classes, pick an integer rep in each class, say $a$ and $b$ then return the class containing their sum $,a+b.,$ This yields a well-defined operation $!\iff!$ the result does not depend on the arbitrary rep choices, which is exactly what the quoted theorems say (the Congruence Sum & Product Rules). – Bill Dubuque Dec 17 '20 at 12:10
  • There exists a class containing their sum because a congruence is an equivalence relation on $\Bbb Z.,$ An equivalence relation is equivalent to a partition into classes. In particular every integer $,a,$ is a member of exactly one class, viz. $,a\in [a]_n = a + n,\Bbb Z\ \ $ – Bill Dubuque Dec 17 '20 at 12:11
  • $\Bbb Z_n,$ is by definition the set of classes $[c]_n$ for all integers $,c,,$ so $,[a+b]_n\in \Bbb Z_n,$ (i.e. closure) is immediate. – Bill Dubuque Dec 17 '20 at 12:25

2 Answers2

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The key idea is that each equivalence class such as $\,[a]_n\,$ can be specified by any of its elements. That is, $\,[a+kn]_n = [a]_n\,$ for all integers $\,k.\,$ The definition $\,[a]_n + [b]_n = [a+b]_n\,$ for addition of equivalence classes automatically implies closure, but the problem is to prove that it is well defined. That is, do you get the same result no matter which element you pick for both equivalence classes? Yes, because $\, (a+jn) + (b+kn) = a+b+(j+k)n.\,$ A similar proof holds for multiplication of equivalence classes.

Somos
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Since $a$ and $b$ are just integers, $a+b$ and $ab$ are also integers, and therefore $[a+b]_n$ and $[ab]_n$ are in $\mathbb{Z}_n$ by definition (if your definition of $\mathbb{Z}_n$ is the set of equivalence classes of integers modulo n).

Richard Jensen
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  • I could make the syllogism: Since a and b are just integers, a/b and ab are also integers. But that's not true in general. – Giuliano Malatesta Dec 17 '20 at 11:25
  • I don't understand your point. We only define addition and multiplication specifically because integers don't have inverses in general. And all we need for an equivalence class of a number to be in $\mathbb{Z}_b$ is that said number is an integer.

    For example. Why is $[4+5]_7$ is $\mathbb{Z}_7$? Well, $4+5$ is an integer, and therefore $[4+5]_7$ is in $\mathbb{Z}$, where we have $[4+5]_7=[9]_7={...,2,9,16,... }$.

     – Richard Jensen Dec 17 '20 at 11:30
  • @Giuliano If your point is that the definition of addition and multiplication of classes depends on the fact that $,\Bbb Z,$ is closed under addition and multiplication then this is certainly true. But obvious things like this are usually not explicitly mentioned once one gets past the basics (algebra would be very tedious if every time we added ring elements we had to explicitly mention that the result lies in the ring, of if we had to explicitly mention associative, commutative and distributive laws when performing ring arithmetic. – Bill Dubuque Dec 17 '20 at 12:34