# Showing that a binary operation is well-defined

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• Mr Davis 97
In summary: Since ##G \cong \mathbb{R}/\mathbb{Z}## I agree with his suspicion, that those classes are probably meant. I just haven't seen it, before he mentioned it. So the better exercise is: Show that ##[x]\,\cdot \,[y] := [\,x+y - \lfloor x+y \rfloor \, ]## is well-defined.For your operation to be a binary operation, it has to be a map G\times G\to G. I think the point of the question is to verify that if x,y are in (0,1], then x\cdot y is
Mr Davis 97
Let ##G = \{x \in \mathbb{R} ~: 0 \le x < 1 \}##. Let ##x, y \in G##.I have to determine whether ##x \cdot y = x + y - \lfloor x + y \rfloor## is well-defined. However, it seems obvious. Since addition, subtraction, and the floor function are all well-defined functions in and of themselves, isn't it clear that the supposed operation in question is well-defined?

Well, yes, since the definition of the operation doesn't allow ambiguities. Formally, we will have to show that ##x+y-\lfloor x+y\rfloor \neq u+v-\lfloor u+v \rfloor## requires ##x \notin \{\,u,v\,\}## or ##y \notin \{\,u,v\,\}##. So assumed ##x=u## show that ##y \notin \{\,u,v\,\} ## and also ##x=v## implies ##y \notin \{\,u,v\,\}##. Written in this explicit way, it is far less obvious. At least it's a good exercise, in order to get used to the question. Obvious or not, it's always important to check.

You are right. Your argument is sufficient to show the dot operator is well-defined.

My guess is that they meant to ask a different question. The hint is that they use G for the set, which suggests they may be thinking of groups. I think they meant to specify that ##G=\mathbb R / \mathbb Z## where those two elements are considered as Abelian groups under addition. For G thus defined it follows from known theorems about quotient groups that the + operation is well defined, but it remains to be proved that the floor operation is, although proving that is pretty easy. It can either be proved explicitly for the quotient group, or by proving that the quotient group is isomorphic to the group [0,1) in the OP via the obvious isomorphism, and that the dot operation is well-defined on the group [0,1) as you showed above.

It is usually in the context of quotient entities that the question of well-definedness arises.

fresh_42 said:
Well, yes, since the definition of the operation doesn't allow ambiguities. Formally, we will have to show that ##x+y-\lfloor x+y\rfloor \neq u+v-\lfloor u+v \rfloor## requires ##x \notin \{\,u,v\,\}## or ##y \notin \{\,u,v\,\}##. So assumed ##x=u## show that ##y \notin \{\,u,v\,\} ## and also ##x=v## implies ##y \notin \{\,u,v\,\}##. Written in this explicit way, it is far less obvious. At least it's a good exercise, in order to get used to the question. Obvious or not, it's always important to check.
Is one way to write the property of a binary operation being well-defined the following? For ##x,~y,~x',~y' \in G##, if ##x = x'## and ##y = y'## then ##x \cdot y = x' \cdot y'##

This is how I have seen it. If this is the case, isn't it trivial to prove that the binary operation in question is well-defined, since we simply write
##x \cdot y = x + y - \lfloor x + y \rfloor## and then substitute ##x## with ##x'## and ##y## with ##y'## to get the result?

Mr Davis 97 said:
Is one way to write the property of a binary operation being well-defined the following? For ##x,~y,~x',~y' \in G##, if ##x = x'## and ##y = y'## then ##x \cdot y = x' \cdot y'##

This is how I have seen it. If this is the case, isn't it trivial to prove that the binary operation in question is well-defined, since we simply write
##x \cdot y = x + y - \lfloor x + y \rfloor## and then substitute ##x## with ##x'## and ##y## with ##y'## to get the result?
Sure. I just thought it's a good habit always to check.

@andrewkirk 's answer is a better example: assume ##x,y \in \mathbb{R}/\mathbb{Z}## with the same definition of multiplication. Then it really has to be shown. In this case, ##G## is just one possible set of representatives of ##\mathbb{R}/\mathbb{Z}##. He is also right, that usually this question is important if the elements are sets represented by one of its elements, e.g. equivalence classes such as cosets. Saying ##x=x'## and ##y=y'## in your example already finishes the proof, as there are no two ways for an element to equal another. But if the elements are ##[x]=x+\mathbb{Z}## then there is no unique representative; ##[x]=[x']## means only ##x-x' \in \mathbb{Z}##.

Since ##G \cong \mathbb{R}/\mathbb{Z}## I agree with his suspicion, that those classes are probably meant. I just haven't seen it, before he mentioned it. So the better exercise is: Show that ##[x]\,\cdot \,[y] := [\,x+y - \lfloor x+y \rfloor \, ]## is well-defined.

Mr Davis 97
For your operation to be a binary operation, it has to be a map $G\times G\to G$. I think the point of the question is to verify that if $x,y$ are in $(0,1]$, then $x\cdot y$ is also in $[0,1)$, and that we should not assume something else (about equivalence classes) is meant.

member 587159

## 1. What does it mean for a binary operation to be well-defined?

A binary operation is well-defined if it produces the same output when the same input is used, regardless of how the input is represented. This means that the output is consistent and does not depend on the specific representation of the input.

## 2. How do you show that a binary operation is well-defined?

To show that a binary operation is well-defined, you must demonstrate that it produces the same output for all possible inputs. This can be done by evaluating the operation on different representations of the same input and showing that the output is the same.

## 3. What are some common mistakes when trying to prove that a binary operation is well-defined?

One common mistake is assuming that a specific representation of the input is the only valid one. This can lead to incorrect conclusions about the well-definedness of the operation. Another mistake is not considering all possible inputs and only evaluating the operation on a few examples.

## 4. Can a binary operation be well-defined for some inputs but not others?

Yes, it is possible for a binary operation to be well-defined for some inputs but not others. For example, division by zero is not well-defined, but division by any non-zero number is well-defined. It is important to consider all possible inputs when determining the well-definedness of a binary operation.

## 5. Why is it important for a binary operation to be well-defined?

A binary operation must be well-defined in order for it to be considered a legitimate mathematical operation. If an operation is not well-defined, it can lead to inconsistencies and contradictions in mathematical reasoning. It is also important for ensuring the validity and accuracy of mathematical proofs and calculations.

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