Is $f:\mathbb{Z}_{30}\longrightarrow\mathbb{Z}_{30}$ defined by $f([a])=[7a]$ well defined?

Question

To tell the truth, I'm not even sure what this means.

The professor gave an example saying that $\mathbb{Z}_m=\{[0],[1],[2],\dots,[m-1]\}$, and I sort of understand that.. but I have no idea what this means.

Well, I have maybe some idea, but I don't know how to determine if it's well defined or not. It looks like the function defines something to do with mod 7, but don't ask me to elaborate beyond that.

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Edit:

OK, so my current understanding of the problem is as such:

$\mathbb{Z}_{30}$ is a set of 30 equivalence classes such that 0~30, 1~31, 2~32, etc... It breaks up the integers into groups such that $x\equiv x+30\ (mod\ 30)$.

$\mathbb{Z}_{30}=\{[x]\mid a\in\mathbb{Z},\ x\sim a\iff x\equiv a\ (mod\ 30)\}$

$f([a])=[7a]$ is eluding me a bit though... Is seems like it's saying that when you give $f$ an equivalence class, it transforms it to the equivalence class $7$ times the one given.. so $f([2])=[14]$, $f([29])=[203]$... But what do I do with that? $203\ mod\ 30=23$, which is not the same equivalence class as $[29]$.

Answer 1

For a function to be well-defined, you need it to be defined on each element in the domain and to give a unique element in the range as its result. Your function will be on the equivalence classes $\pmod {30}$ and the result will be an equivalence class, perhaps not the same one. The threat to being well-defined would be if it took two representatives of the same class to representatives of different classes. To take your example, you would think $f([29])=[23]$. But $29, 59, 89,$ and $329$ are all in $[29]$. If you multiply them all by $7$ and take the result $\pmod {30}$ you will get $23$, so we haven't found a counterexample.

To prove it, let $[n]$ be one of our equivalence classes. An arbitrary element of this class is $n+30k$ for $k$ some integer. Then that element is sent to $7n+210k$, which is in the same class as $7n$, so it doesn't matter which representative you take, you get the same result.

Answer 2

First we need to understand what $\mathbb Z_{30}$ is. As your professor said $\mathbb Z_{30}=\{[0],[1],[2],\ldots,[29]\}$ is a set containing $30$ elements.
The formal definition of this set is as equivalence classes (the congruence classes) of an equivalence relation (the Congruence relation) on $\mathbb Z$. The elements of $\mathbb Z_{30}$ are actually sets.
The element $[i]\in\mathbb Z_{30}$ is the subset of $\mathbb Z$ containing all integers which leave remainder $i$ when they are divided by $30$. For example: $\{-120027,-57,3,33,63,370293\}\subseteq[3]$ since all elements of this set leave remainder $3$ when they are divided by $30$. The number $125\not\in[3]$ because the remainder of $125$ divided by $30$ is five and not three; therefore $125\in[5]$.
What is $[0]$? It is the subset of $\mathbb Z$ containing all integers with no remainder when they are divided by $30$ (i.e. they are divided evenly by $30$).
The elements of $\mathbb Z_{30}$ look like this: \begin{split} [0]&=\{\ldots,-60,-30,0,30,60,90,\ldots\},\\ [1]&=\{\ldots,-59,-29,1,31,61,91,\ldots\},\\ [2]&=\{\ldots,-58,-28,2,32,62,92,\ldots\},\\ \vdots \ & \hspace{19ex} \vdots\\ [29]&=\{\ldots,-31,-1,29,59,89,119,\ldots\}.\\ \end{split} Sometimes we use different names for the same elements of $\mathbb Z_{30}$. For example $[1]=[31]=[-29]$ and in general $[i]=[k]\iff k\in[i]$ for all $0\leq i\leq29$.

In general $\mathbb Z_n=\{[0]_n,[1]_n,\ldots,[n-1]_n\}$ (when is necessary we use the subscripts to avoid confusion) is the set containing $n-1$ elements, each element being a subset of $\mathbb Z$. The element $[i]_n\in\mathbb Z_n$ is $$[i]_n=\{k\in\mathbb Z: \text{the remainder of $k$ when is divided by $n$ is $i$}\}=\{rn+i:r\in\mathbb Z\}.$$ Another example is $\mathbb Z_2=\{[0],[1]\}$ with $[0]$ being the subset of $\mathbb Z$ containing all integers divided evenly by $2$, i.e. all even integers and $[1]$ is the rest of $\mathbb Z$, the odd integers. Therefore \begin{split} [0]_2&=\{\ldots,-46,\ldots,-24,\ldots,-4,-2,0,2,4,\ldots,36,\ldots\},\\ [1]_2&=\{\ldots,-45,\ldots,-23,\ldots,-3,-1,1,3,5,\ldots,37,\ldots\}. \end{split} Now a function $f:\mathbb Z_n \longrightarrow \mathbb Z_m$ between these sets is a relation which takes as input an element (a set) $[i]_n\in\mathbb Z_n$ and give as output an element of $\mathbb Z_m$.

What does it mean "$f:\mathbb Z_n\longrightarrow\mathbb Z_n$ defined as f([a])=[3a]"? It means $f$ maps the set represented by $[a]$ to the set represented by $[3a]$.

When we ask if such $f:\mathbb Z_n \longrightarrow \mathbb Z_m$ is well defined actually we ask: If we take different representatives $[a],[b]$ of the same set, then $f([a])=f([b])$? Since $[a]=[b]$ is the same thing then for $f$ to be well defined (to be function) we must have $f([a])=f([b])$.

Here are some examples of well defined and not well defined functions.

• $f:\mathbb Z_n \longrightarrow \mathbb Z_n$ with $f([a])=[a]$.
  Is well defined because if $[a]=[b]\in\mathbb Z_n$ then $f([a])=[a]=[b]=f([b])$.
• $f:\mathbb Z_2 \longrightarrow \mathbb Z_2$ defined as follows $f([a])=[x]$ where $x=r+1$ with $r$ the last digit of $a$.
  So $ \ f([34])=[5], \ f([-13])=[4], \ f([-1])=[2]$ etc. Is $f$ well defined? Lets see. The set $\mathbb Z_2$ contains $2$ elements $[0]$ and $[1]$. If $[a]=[b]$ then we have two possibilities. Either $[a]=[b]=[0]$, so $a,b$ are both even, or $[a]=[b]=[1]$ and $a,b$ are both odd integers. If $a,b$ are even then their last digit is one of $\{0,2,4,6,8\}$, therefore if $f([a])=[x]$ and $f([b])=[y]$ then $x,y\in\{1,3,5,7,9\}$ so $[x]=[y]=[1]$. Equivalently $f([a])=f([b])=[1]$. In a similar way if $a,b$ are odd $f([a])=f([b])=[0]$.
  It follows that $f$ is well defined.
• $f:\mathbb Z_3 \longrightarrow \mathbb Z_3$ defined as follows $f([a])=[x]$ where $x=r+1$ with $r$ the last digit of $a$.
  In this case $f$ is not well defined: $[1]=[13]$ and $f([1])=[2], \ f([13])=[4]\neq[2]$.
• $f:\mathbb Z_{47} \longrightarrow \mathbb Z_{47}$ defined as follows $f([a])=[3a]$.
  Let's say that $[a]=[b]\in \mathbb Z_{47}$. This means that $a$ and $b$ leave the same remainder when they are divided by $47$. Therefore $a=47k+i$ and $b=47m+i$ for some $k,m\in\mathbb Z$ and $0\leq i\leq46$. We want to check whether $f([a])=f([b])$; equivalently whether $[3a]=[3b]$. Since $a=47\cdot3k+3i$ and $b=47\cdot3m+3i$ it follows that the remainder of $a$ divided by $47$ is the same as the remainder of $b$ divided by $47$. Both are the remainder of $3i$ divided by $47$. Therefore $f([a])=[3a]=[3b]=f([b])$ and $f$ is well defined.
• $f:\mathbb Z_{30} \longrightarrow \mathbb Z_{30}$ defined as follows $f([a])=[7a]$...
  

Answer 3

If you consider $\mathbb Z_{30}$ as a ring, you want to prove that the ring structure is maintained. For example, is $f(2\cdot 3 + 8)\equiv f(2) \cdot f(3) + f(8) \pmod {30}?$ Let's try this one: $f(2 \cdot 3 + 8)=f(14)=98 \equiv 8 \pmod {30}.$ But $f(2) \cdot f(3) + f(8)=14\cdot 21+56=350 \equiv 20 \pmod {30}$. Failure. If you consider just the group structure of $\mathbb Z_{30}$ you will succeed.