I am currently taking linear algebra and I am not sure what this notation means. I am asked to simplify a fractions within fractions all of the numbers are surrounded by square brackets like so [ ].
Ex: [4]/[3] among other fractions like this.
Edit: the previous question is about multiplicative inverses of Z11 if that helps
Best Answer
In the context of abstract algebra, group theory, equivalence relations, and modular arithmetic, square brackets around an element have the following meaning:
Suppose that $X$ is a set and suppose that $\sim$ is an equivalence relation on the set and $x$ an element of $X$, we define and notate the equivalence class of $x$ as:
$$[x]=\{y\in X~:~ x\sim y\}$$
Further, we notate the set of equivalence classes of $X$ as:
$$X/\sim ~~= \{[x]~:~x\in X\}$$
The set $\Bbb Z_{11}$ could have been defined as $\Bbb Z/\sim$ where $\sim$ is the equivalence relation over $\Bbb Z$ given by $x\sim y\iff 11\mid x-y$
We could alternatively have defined $\Bbb Z_{11}$ as the quotient group $\Bbb Z/11\Bbb Z$, or equivalently quotient ring, etc...
Regardless how you define it, the end result is that $\Bbb Z_{11}$ contains the following eleven elements: $\{[0],[1],[2],\dots,[10]\}$, each of which are themselves equivalence classes containing many elements themselves, e.g. $[2]=\{\dots,-9,2,13,24,35,\dots\}$. Some people prefer to ignore this aspect and treat the elements of $\Bbb Z_{11}$ as simply $\{0,1,2,\dots,10\}$ and choose to remember that "addition" and "multiplication" over this set don't function like normal.
For the specific problem of finding $[4]/[3]$ in the context of $\Bbb Z_{11}$, what it is really asking is to find $[4]\times [3]^{-1}$ because that is how "division" is defined: multiplication of the numerator by the multiplicative inverse of the denominator.
To do so, first recognize that $[4]\times [3]=[4\times 3]=[12]=[1]$ and so since $[4]\times[3]=[1]$ that implies that the multiplicative inverse of $[3]$ is $[3]^{-1}=[4]$.
So, reworded yet again, the problem is asking us what is $[4]\times[4]$.
The above manipulations follow the rule that $[a]+[b]=[a+b]$ and that $[a]\times[b]=[a\times b]$ but we prefer to write things with their usual representatives in the range $\{0,1,2,\dots,10\}$ when possible, remembering that $[a]=[a-11]$ in the context of $\Bbb Z_{11}$. It is worth checking the first time that you encounter modular arithmetic like this and quotient rings that such addition and multiplication really is well-defined and allowed and does not cause any errors or discrepancies.
We define multiplicative inverses the same way in the context of modular arithmetic as we do in other rings or fields. Given an element $x$ from a ring with unity $\textbf{1}$, it's multiplicative inverse, denoted by $x^{-1}$, is the unique element in the ring such that:
$$x\times x^{-1} = x^{-1}\times x = \textbf{1}$$
For example in the real numbers we have $2\times 0.5 = 1$ so we say that $2^{-1}=0.5$
It is the same concept when working in $\Bbb Z_{11}$. When we talk about $[7]^{-1}$ we are talking about that element that when multiplied by $[7]$ results in $[1]$. We could go the long way of consulting a multiplication table:
$\begin{array}{rl} [7]\times[0]&=[0]\\ [7]\times [1] &=[7]\\ [7]\times [2]&=[3]\\ [7]\times [3]&=[10]\\ [7]\times[4]&=[6]\\ [7]\times[5]&=[2]\\ [7]\times[6]&=[9]\\ [7]\times[7]&=[5]\\ [7]\times[8]&=[1]\\ [7]\times[9]&=[8]\\ [7]\times[10]&=[4]\end{array}$
Note: Do NOT bother trying to memorize multiplication tables. The above multiplication table is only useful for arithmetic in $\Bbb Z_{11}$. The tables will look differently when working in different $\Bbb Z_n$.
The above multiplication table is formed just how I described earlier. For example $[7]\times[4]=[7\times 4] = [28]=[28-11]=[17]=[17-11]=[6]$. First perform the arithmetic inside of the brackets like you normally would, and then continually subtract $11$ from the inside until you are left with a number between $0$ and $11$.
Note again that we can have something like $[17]=[6]$ and it be perfectly true. After all in this context we have $[17]=\{\dots,-14,-5,6,\color{red}{17},28,39,\dots\}=\{\dots,-14,-5,\color{red}{6},17,28,39,\dots\}=[6]$. They really are just two names for the same equivalence class.
You will find by consulting the above table that $[7]\times [8]=[1]$, so just like how $2\times 0.5=1$ in real numbers and we called $2^{-1}=0.5$, here since $[7]\times [8]=[1]$ we call $[7]^{-1}=[8]$.
Trying to find $[7]^{-1}$ from a table like we just did is long and tedious. A quick mind might have been able to spot it from a distance by thinking "Oh, what multiples of seven do I know of which are one more than a multiple of eleven? Oh, $56$ is one more than $55$! Okay, so then $[8]$ must be the multiplicative inverse of $[7]$!" which is what I did in the comments below.
If you cannot do such quick mental gymnastics and are not willing to go through the effort of creating and looking through a table for a multiplicative inverse, then you may use the Extended Euclidean Algorithm. We know that $\gcd(7,11)=1$ and the Euclidean Division Algorithm will verify that. The Extended Euclidean Division Algorithm will go a step further and show you the coefficients used for Bezout's Identity.
Bezout's Identity and the Euclidean Division Algorithm, together with the fact that $\gcd(7,11)=1$ together imply that there must be some $x,y$ that we can calculate such that:
$$7x+11y=1$$
The value of $x$ in the above is precisely going to be the multiplicative inverse of $7$ when working in $\Bbb Z_{11}$.
Indeed, if you were to go through the effort you will find that it is:
$$7\times 8 + 11\times (-5)=1$$