Let $m$ be any fixed positive integer. For each integer $j$, $0\le j \lt m$, let $\Bbb Z_j=\{x \in \Bbb Z\,|\, x-j=km \text{ for some } k \in \Bbb Z\}$.

The set notations for $\Bbb Z_0$, $\Bbb Z_1$, $\Bbb Z_2$ would be as below. Then how do we list elements of $\Bbb Z_0$, $\Bbb Z_1$, $\Bbb Z_2$ from the set notations below?

$\Bbb Z_0= \{x \in \Bbb Z\,|\, x=km \text{ for some } k \in \Bbb Z\}$

$\Bbb Z_1= \{x \in \Bbb Z\,|\, x=km+1 \text{ for some } k \in \Bbb Z\}$

$\Bbb Z_2= \{x \in \Bbb Z\,|\, x=km+2 \text{ for some } k \in \Bbb Z\}$


"Definition 6 Let F be an arbitrary family of sets. The union of the sets in F, denoted by $\bigcup\mathscr F$, is the set of all elements that are in A for some $A\in\mathscr F$.​

$\bigcup\limits_{A \in \mathscr F}A$={$x\in U$|$x \in A$ for some $A\in \mathscr F$}"

"Definition 7 Let F be an arbitrary family of sets. The intersection of sets in F, denoted by $\bigcap\limits_{A\in\mathscr F}A$ or $\bigcap\mathscr F$, is the set of all elements that are in A for all $A \in\mathscr F$.​ " $\bigcap\limits_{A\in\mathscr F}A$={$x\in U$| x∈A for all $A\in \mathscr F$}

Source: Set Theory by You-Feng Lin, Shwu-Yeng T. Lin.

  • $\begingroup$ I would prefer the naming convention $\mathbb{Z}_{j,m}$ since the definition is clearly dependent on $m$. Then e.g. $\mathbb{Z}_{1,m}=\{km+1 | k\in\mathbb{Z}\}$ which should look familiar ("modulo $m$"). $\endgroup$ – Dr_Be Jan 12 '16 at 8:52
  • 1
    $\begingroup$ In my view it is better to write $m\mathbb Z$, $1+m\mathbb Z$ etc. instead of $\mathbb Z_0$, $\mathbb Z_1$ etc. The notation $\mathbb Z_n$ is allready commonly used for $\{r+n\mathbb Z\mid r=0,1,\dots,n-1\}$. Where did you meet these notations? $\endgroup$ – drhab Jan 12 '16 at 8:53
  • $\begingroup$ @drhab It's just another curiosity sprang from math.stackexchange.com/questions/1607816/… $\endgroup$ – buzzee Jan 12 '16 at 9:00

Are you looking for something like this? $$\Bbb Z_0 = \{0, \pm m, \pm2m, \pm3m, \ldots\}\\ \Bbb Z_1 = \{1, \pm m +1, \pm2m + 1, \ldots\}\\ \Bbb Z_2 = \{2, \pm m +2, \pm2m + 2, \ldots\} $$

I'll do a concrete example, so you can see how it works. Say $m = 3$. Then $$ \Bbb Z_0 = \{x \mid x = 3k\text{ for some }k\in \Bbb Z\} $$ It is spelled out "$\Bbb Z_0$ is the set of all $x$ such that $x = 3k$ for some $k \in \Bbb Z$". This means that $15 \in \Bbb Z_0$, since $15 = 3\cdot 5$ (in that case, $k = 5$). However, $43 \notin \Bbb Z_0$, because there is no $k\in \Bbb Z$ so that $43 = 3k$. We get $$ \Bbb Z_0 = \{0, \pm 3, \pm 6, \pm 9, \ldots\} = \{0, 3, -3, 6, -6, 9, -9,\ldots\} $$ On the other hand, we have $$ \Bbb Z_1 = \{x \mid x = 3k+1\text{ for some }k \in \Bbb Z\} $$ which means that a number $x$ is an element of $\Bbb Z_1$ iff there is a $k\in \Bbb Z$ such that $x = 3k+1$. For instance, $43 \in \Bbb Z_1$, since there is a $k\in \Bbb Z$ that makes $43 = 3k + 1$ (it's $14$). However, $15 \notin \Bbb Z_1$, since there is no integer $k$ such that $15 = 3k + 1$. This gives $$ \Bbb Z_1 = \{1, \pm3 + 1, \pm6 + 1, \ldots\} = \{1, 4, -2, 7, -5, \ldots\} $$ Lastly, in exactly the same way, we get $$ \Bbb Z_2 = \{2, \pm3 + 2, \pm 6 + 2, \ldots\} = \{2, 5, -1, 8, -4,\ldots\} $$

  • $\begingroup$ since it's "for some k $\in Z$", not 'for every k $\in Z$' Shouldn't it be $$\Bbb Z_0 = \{k, 2k, 3k, \ldots\}\\ \Bbb Z_1 = \{k +1, 2k + 1, 3k+1\ldots\}\\ \Bbb Z_2 = \{k +2, 2k+ 2, 3k+2\ldots\} $$? $\endgroup$ – buzzee Jan 12 '16 at 9:08
  • 1
    $\begingroup$ @buzzee No, the $k$ are the numbers I've colored red here: $$ \Bbb Z_1 = \{\color{red}0m + 1, \color{red}{\pm 1} m +1, \color{red}{\pm2}m + 1, \ldots\} $$ $\endgroup$ – Arthur Jan 12 '16 at 9:12
  • $\begingroup$ Do you regard "for some k $\in Z$" as a meaning of 'k can be any integer'? $\endgroup$ – buzzee Jan 12 '16 at 9:32
  • 1
    $\begingroup$ The notation $$\Bbb Z_j = \{x \mid x = km + j\text{ for some }k \in \Bbb Z\}$$means, more or less by definition, "$x \in \Bbb Z_j$ iff there is some $k \in \Bbb Z$ such that $x = km + j$", so no, "... for some $k$" means "there exists a(t least one) $k$ such that...". $\endgroup$ – Arthur Jan 12 '16 at 9:42
  • 1
    $\begingroup$ @buzzee No. What you have written is "The set of all $x$ such that $x = 3k$ for all $k \in \Bbb Z$". There are no members in that set, since there is no $x$ that is simultaneously equal to all possible different $3k$. You might be thinking of something different, namely $\{3k \mid k \in \Bbb Z\}$. It is the same set, but the notation is used a bit more directly. It is "the set of all numbers of the form $3k$ as $k$ ranges over the integers". $\endgroup$ – Arthur Jan 12 '16 at 17:52

Start with another notation: $r+m\mathbb Z:=\{r+mk\mid k\in\mathbb Z\}$.

This instead of $\mathbb Z_r$.

Now define $\mathbb Z/m\mathbb Z:=\{r+m\mathbb Z\mid r=0,1,\dots,m-1\}$.

Then $\mathbb Z/m\mathbb Z$ is a list of the sets you mention.

Note that - defined like this - $\mathbb Z/m\mathbb Z$ is a partition of $\mathbb Z$.

Another commonly used notation for $\mathbb Z/m\mathbb Z$ is $\mathbb Z_m$.

However, that notation is unfortunately allready in use in your question.

Following your notation (I dislike it) we have $\mathbb Z/m\mathbb Z:=\{\mathbb Z_0,\dots,\mathbb Z_{m-1}\}$.

  • $\begingroup$ Unfortunately the definition of $\mathbb Z_j$ in the question is not the same as this commonly used notation. $\endgroup$ – hmakholm left over Monica Jan 12 '16 at 10:12
  • $\begingroup$ @HenningMakholm Yes, and indeed unfortunately. I have chosen for the other option ($\mathbb Z/m\mathbb Z$) now to diminish the chance on confusion. $\endgroup$ – drhab Jan 12 '16 at 10:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.