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Let $r \in \mathbb{N}$. Observe the equivalence relation $\sim \subseteq \mathbb{Z} \times \mathbb{Z}$ defined as $x \sim y :\Leftrightarrow (\exists k \in \mathbb{Z})(y=x+kr)$. Show that an operation $'+'$ on $\mathbb{Z}/\sim$ given by $[x]_\sim + [y]_\sim := [x+y]_\sim$ is well-defined.

My attempt at a proof:

Let $x\sim x', y\sim y'$, then we know that $(\exists k_1 \in \mathbb{Z})(x=x'+k_1r)$ and $(\exists k_2 \in \mathbb{Z})(y=y'+k_2r)$. Adding $x$ and $y$, we see that $x+y=x'+y'+(k_1+k_2)r$. Because $\mathbb{Z}$ is closed under addition, $k_1+k_2 \in \mathbb{Z}$, and eventually $x+y\sim x'+y'$.

Unfortunately, I'm not even sure whether I'm on the right track here. As far as I can see, I've shown that adding two elements of (different) equivalence classes yields an element, whose equivalence class is in $\mathbb{Z}/\sim$. Is that sufficient to show that $'+'$ is well-defined?

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up vote 8 down vote accepted

Yes, you're on the right track. "Well-defined" just means that it should be an actual operation that takes 2 equivalence classes and sends them to a (uniquely determined) equivalence class. The way the operation is written makes use of specific representatives of the equivalence classes. $[x+y]$ makes sense as an element of $\mathbb{Z}/\sim$, but you need to know that it is determined by the classes of $x$ and $y$, rather than depending on the choices of representatives. To do so, you suppose that another arbitrary choice of representatives is made (your $x'$ and $y'$), and show that the sum $[x'+y']$ is the same. That is what your work shows.

The language in your second to last sentence isn't entirely accurate. What you've shown is that adding 2 equivalence classes determines an equivalence class. The key word is "an" (singular); you've shown that you couldn't get different equivalence classes by "adding" in different ways.

To contrast, suppose $r=3$ and we want to "define" an operation by $[x]^{[y]}=[x^y]$. If we try to combine $[2]$ and $[1]$ using this "operation", we get $[2]^{[1]}=[2^1]=[2]$. But $[1]=[4]$, so we also get $[2]^{[1]}=[2]^{[4]}=[2^4]=[16]=[1]$. Thus this "operation" is not well-defined. (Of course, there are other problems, like defining powers of $0$ or negative powers, but the point here is the non-uniqueness, even if it did make sense.) Better examples of non-well-defined "operations" arise when you study groups. There you study equivalence classes called cosets of subgroups and try to combine them analogously to how equivalence classes mod $r$ are added in your problem. You find that unless the subgroup satisfies a certain property called being "normal", the "operation" is not well-defined.

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thank you very much, your answer was very helpful. – Greg S Nov 21 '10 at 10:18

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