Okay, so we are going to use the base set of numbers [i], which contains all possible cases of ai, where a is any real number. Here are 4 possible groups on this set --> [i,*]... [i,+]... [i,/] (division)... [i,-], where *, +, /, and - are operations SIMILAR to and including the common arithmetic operations assigned to those symbols.

The four basic properties of a group:
-Closure under the given operation
-Associativity under the given operation
-There exists an identity object E in the set being operated on, such that for any object A in the set, A op E = A (where op is the given operation)
-There exists an inverse object I in the se being operated on, such that for any object A in the set, A op I = E

Now, from what I've read, a ring is an extension, so to speak, of a group. It includes a set and two operations + and * similar to those symbols meaning in common arithematic. Using S as our set, a ring is defined as [S, *,+] with the following properties:
-S is a group under addition
-S is commutative under addition
-S is associative under multiplication
-S is distributive under multiplication

Now, if [S,,+] is a ring, S is closed under addition because it is a group, but is it closed under multiplication? It seems like there would be some serious complications.

Intuitively, I feel like [S,/,-] could also be a ring. Well, actually, know that I go over the properties of a ring, I've noticed that [S,/,-] DOES satisfy all the properties of a ring if addition is replaced with subtraction and multiplication is replaced with division. Does anyone see any errors in this?

A short question: Take for example [i,
]. Is [i,*] the group, or is [i] (under the property of multiplication the group)? Essentially, does the group include both the set of elements and the operation(s) operating on it, or does it only include the set of elements?
One more thing, is my definition of a ring (as well as mine of a group I guess) accurate? Thanks.

  • $\begingroup$ But $\ (a/b)/c \ne a/(b/c),\ $ and $\, a/0\,$ is not defined, etc. $\endgroup$ – Bill Dubuque Aug 28 '14 at 5:37
  • $\begingroup$ Rings don't necessarily include the identity property for multiplication (and hence division)..but you may be right about the associativity. Let me read the other answers... @BillDubuque $\endgroup$ – Colin Michael Flaherty Aug 28 '14 at 6:16

Analysing the closedness of operations may be interesting from some points of view, but I believe that, in an introductory level, the best is to use Binary Functions: A binary function on a set $S$ is simply a function of the form $f:S\times S\to S$, that is, with domain $S\times S$ and codomain $S$.

We can use binary functions to define groups, for example.

Definition A group is a set $G$ equipped with a binary operation $*:G\times G\to G$ satisfying the following axioms:

1.(Associativity) $*(x,*(y,z))=*(*(x,y),z)$ for every $x,y,z\in G$.

2.(Identity) There exists a (unique) element $e\in G$ such that, for every $x\in G$, $*(x,e)=*(e,x)=x$.

3.(Inverses) For every $x\in G$, there exists some $y\in G$ such that $*(x,y)=*(y,x)=e$.

However, it is much more convenient to just denote $*(x,y)=xy$. Using this notation, the three axioms above may be rewritten in the following way:

1.(Associativity) $x(yz)=(xy)z$ for every $x,y,z\in G$.

2.(Identity) There exists a (unique) element $e\in G$ such that, for every $x\in G$, $xe=ex=x$.

3.(Inverses) For every $x\in G$, there exists some $y\in G$ such that $xy=yx=e$.

which are the axioms as usually stated.

Note that, with this definition, a group is is imediatelly closed by multiplication: If $x,y\in G$, then $xy=*(x,y)$ belongs to the codomain of $*$, which is $G$.

Since a group consists of a set equipped with an operation, then, to be precise, the group would have to be the pair $(G,*)$. However, we usually ommit the operation, refering just to the group $G$.

We could define a ring by a similar approach:

Definition A ring is a set $S$ equipped with two binary operations $+:S\times S\to S$ and $m:S\times S\to S$ (henceforth refered to as addition and multiplication) satisfying the following axioms:

  1. $(S,+)$ is a group.

  2. $+(x,y)=+(y,x)$ for every $x,y\in S$.

  3. $m(x,m(y,z))=m(m(x,y),z)$ for every $x,y,z\in S$.

  4. For every $x,y,z\in S$, $m(x,+(y,z))=+(m(x,y),m(x,z))$ and $m(+(x,y),z)=+(m(x,z),m(y,z))$.

If we write $+(x,y)=x+y$ and $m(x,y)=xy$, we again obtain the usual axioms for a ring. Also, a ring, by this definition, consists of the triple $(S,+,m)$ (or $(S,m,+)$, as you prefer), but again we usually ommit the operations and refer only to the ring $S$.

Finally, neither division nor subtraction are associative: $x-(y-z)=x-y+z$, whereas $(x-y)-z=x-y-z$ (the sign of $z$ is different), and similarly for division, so the sets with the operations you cited in the next-to-last paragraph are not rings.

I'm going to explain briefly the meanings of cartesian products and functions:

Given two objects $a$ and $b$ (which may be sets, elements of a sets, doughnuts...), we can construct (in a set-theoretic manner) what we call the ordered pair $(a,b)$. The ordered pairs have the following property: Given four objects $a,b,c,d$, we have $(a,b)=(c,d)$ IFF $a=c$ and $b=d$. (The exact definition of an ordered pair is not at all important. This is the property of ordered pair that matters. You can even study theories in which ordered pair are given as a certain axiom.)

Given two sets $A$ and $B$, we can form the set $A\times B$, which consists of all ordered pair of the form $(a,b)$ for $a\in A$ and $b\in B$.

Now, a function has another meaning: Given two sets $A$ and $B$, a function with domain $A$ and codomain $B$ can be tought as a "rule" which, for every element $a$ of $A$, gives us an unique element of $b$. Since it is unique, we may denote this element associated to $a$ as $f(a)$. Again, there is a proper definition for functions (which in fact uses cartesian products), but this definition is not at all important (in this discussion). We denote $f:A\to B$ to say that $f$ has domain $A$ and codomain $B$.

We may picture it in the following manner: let's think of a set as a box containing its elements. Given two sets $A$ and $B$, a function from $A$ can be tought as a machine that gets elements of $A$ and transforms them in elements of $B$. However, if the machine gets thesame element of $A$ several times, the result will always be the same element of $B$ (by uniqueness).

One thing to keep in mind is that domains and codomains are important. To be precise again, a function would have to be a triple like $(\text{rule},\text{domain},\text{codomain})=(f,A,B)$, which is written as $f:A\to B$.

In particular, a binary function/operation in a set $S$ is simply a function $f$ with domain $S\times S$ and codomain $S$. Notice that $S$ does not contain pair $(x,y)$ for $x,y\in S$, but $S\times S$ does.

This is my preferred approach to binary operations. If an operation was just "putting two elements in an ordered form" it could get a little strange. For example: if $e$ is the unit of a group, then $ee=e$. But the former is a string with two digits while the latter is a string with just one digit. So $xy$ is simply a way of writing "the result we obtain when we multiply $x$ with $y$" (or $x+y$, or $x\square y$, or whatever your operation symbol is).

A really nice book which discusses this elementary set theory is Halmos' Naive Set Theory (not naive at all).

  • $\begingroup$ Oh okay! So, for the binary operations are, by definition, closed, right? $\endgroup$ – Colin Michael Flaherty Aug 28 '14 at 6:17
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    $\begingroup$ Yes. The result of operations are in the codomain, which is the (underlying set of) the ring/group. $\endgroup$ – Luiz Cordeiro Aug 28 '14 at 6:18
  • $\begingroup$ You write "Binary Function" but link to "binary operation". Your definition of binary function if the definition of binary operator in wikipedia. $\endgroup$ – miracle173 Aug 28 '14 at 6:51
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    $\begingroup$ @Colin I added a little explanation about functions and cartesian products. Hope it helps a little. Also, binary functions or operations are the same thing, it's just a matter of preference (the same way that functions are also called operators, maps, etc...). $\endgroup$ – Luiz Cordeiro Aug 28 '14 at 17:16
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    $\begingroup$ @Luiz Cordeiro: That was so clear!!! It all just clicked for me (while I was in a Chemistry Class) as I read the sentence "In particular, a binary function/operation in a set S is simply a function f with domain S×S and codomain S". Thank you so much!! Mathgasm! Haha $\endgroup$ – Colin Michael Flaherty Aug 29 '14 at 19:09

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