I am having trouble understanding why a proof in my text book proceeds as it does, eventhough the steps involved are quite simple. If possible,I would appreciate some clarifications regarding the following theorem.
Theorem.(Quotient Algebra)
Suppose that $Q=(A,*,+,\dots)$ is an algebraic structure. Let ~ be a congruence relation on $Q$. Then there exist well-defined binary operations $\bigotimes, \bigoplus,\dots$on $A/$~ such that
- $[a]\bigotimes[b] = [a*b]$
- $[a]\bigoplus[b] = [a+b]$
for all $a,b \in A$. Therefore $R=(A/\text{~},\bigotimes,\bigoplus,\dots)$ is also an algebraic structure,called the quotient algebra.
What I thought was :
Let :
- $Q=(A,*,+,\dots)$ be an algebraic structure
- ~ be a congruence relation on $Q$.
Then :
(i)By definition,all of the binary operations $\bigotimes,\bigoplus,\dots$ preserve the relation ~,such that for all $a,b,c,d \in A$,$a \text{~}c \text{ and } b \text{~}d \Rightarrow (a*b) \text{~}(c*d)$.
(ii)To prove that the definition $[a]\bigotimes[b] = [a*b]$ produces a well-defined binary operation,the following two conditions must be satisfied:
[a]prove:
For each $([a],[b])\in \text{( A/~}\times \text{A/~) }$,there is some $[c] \in \text{A/~}$ s.t. $(([a].[b]),[c]) \in \bigotimes$
- Let $([a],[b])\in \text{( A/~}\times \text{A/~) }$.Since $*$ is a well-defined binary operation on A,then $a*b \in A$.Let $c\in A$ be such that $c=a*b$. Therefore $[c] \in \text{A/~}$ and hence $(([a],[b]),[c]) \in \bigotimes$.
[b] prove
If $(([a],[b]),[c]) \in \bigotimes$ and $(([a],[b]),[d]) \in \bigotimes$, then $[c]=[d]$.
Assume$(([a],[b]),[c]) \in \bigotimes$ and $(([a],[b]),[d]) \in \bigotimes$, then $[c]=[a] \bigotimes [b] = [a*b]$ and similarly $[d]=[a] \bigotimes [b] = [a*b]$.
Since $*$ is a well-defined binary operation on A,then $*(a,b)$ must be unique, hence $[c]=[d]$.
Therefore $\bigotimes$ is a well-defined binary operation ?
What the book says:
We must show that the definition $[a]\bigotimes[b] = [a*b]$ produces a well-defined binary operation.
Assume $[a]=[c]$ and $[b]=[d]$.We will show that $[a*b]=[c*d]$...Although i understand how the proof proceeds,i do not know where this assumption comes from and why the conclusion implies a well-defined binary operation. Would my attempt above be a viable alternative?
Help!
Thank you for your time.
To prove that the operation is well-defined you have to prove that it doesn't depend on which representatives of the equivalence classes you choose. That is, if $a~c$ and $b~d$ then for each operation $\circ$, $[a]\circ[b]$ and $[c]\circ[d]$ give the same value. In your $(i)$ you've asserted that this is true. It may well be obvious in this case, depending on your definition of "congruence relation." You can make the same definition with any equivalence relation however, and it isn't always obvious, or even true, that binary operation will be preserved.
So, the approach taken in the book is the general one.
EDIT
Added in response to a comment by the OP.
To show that something is well-defined, we generally have to show two things. First, that the object we claim to be defining actually exists; we are not "legislating it into existence." Second, that if there are choices made in the definition, we have to show that the value doesn't depend on those choices. For example, if we define the area of a triangle as half the product of the base and the height, we have to show that the definition doesn't depend on which side we choose to be the base.
In this problem, we are told that $[a]\bigotimes[b]=[a*b]$ so the existence problem doesn't really arise. Maybe this is more a matter of style. I can't say it's wrong to make any true statement in the course of a proof, but when the matter is this clear, I find it confusing, as if there might be something deeper that I'm not seeing. (There isn't.)
As this is, finally, a matter of style, I suggest you ask your instructor for his opinion, which is what counts, at least for the moment.