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Let $(a,b),(x,y) \in\Bbb R\times\Bbb R$ and define $(a,b) \equiv (x,y)$ iff $a+b = x+y$.

a. Show that $\equiv$ is an equivalence relation.

Define the operation $\oplus$ on the equivalence classes as follows:

$$[(a,b)]\oplus[(x,y)]= \left[\left(a+\frac{x+y}2,b+\frac{x+y}2\right)\right]$$

b. Show that $\oplus$ is well-defined.

c. Show that $\oplus$ is a commutative and associative operation.

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  • $\begingroup$ Part (b) could be a bit difficult, since students often have a bit of trouble understanding what it means for an operation to be well-defined, but Part (a) is extremely straightforward, and part (c) is routine; you should at be able at least to make a solid start on these. Have you made any progress at all? $\endgroup$ – Brian M. Scott Oct 18 '13 at 14:35
  • $\begingroup$ @BrianM.Scott This does help me, I'm just stuck on part B $\endgroup$ – Danielle Oct 18 '13 at 14:51
  • $\begingroup$ Okay; I’ve written up an answer to try to get you started on (b). $\endgroup$ – Brian M. Scott Oct 18 '13 at 15:04
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Each equivalence class has infinitely many members. The definition of $\oplus$ refers to specific members of those classes:

$$[(a,b)]\oplus[(x,y)]= \left[\left(a+\frac{x+y}2,b+\frac{x+y}2\right)\right]\;.\tag{1}$$

It’s conceivable that if we started with different representatives of the classes $[(a,b)]$ and $[(x,y)]$, formula $(1)$ would produce a different result. For instance, $(3,7)\equiv(5,5)$ and $(-2,5)\equiv(0,3)$ so $[(3,7)]=[(5,5)]$ and $[(-2,5)]=[(0,3)]$, and therefore we certainly expect that

$$[(3,7)]\oplus[(-2,5)]=[(5,5)]\oplus[(0,3)]\;;$$

does definition $(1)$ actually guarantee this? If it does in all similar cases, we say that $\oplus$ is well-defined: if $C_1$ and $C_2$ are two $\equiv$-equivalence classes, the value of $C_1\oplus C_2$ does not depend on which representatives of $C_1$ and $C_2$ we substitute into $(1)$.

This means that in order to show that $\oplus$ is well-defined, you must show that if $[(a,b)]=[(a',b')]$ (which just means that $(a,b)\equiv(a',b')$) and $[(x,y)]=[(x',y')]$, then $$[(a,b)]\oplus[(x,y)]=[(a',b')]\oplus[(x',y')]\;,\tag{2}$$ where both sides of $(2)$ are calculated using the definition $(1)$.

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