I'm reading through this book: Algebras of Linear Transformations. Here is an image of a short passage: 
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I'm struggling to show that the multiplication of equivalence classes ([$a_1$][$a_2$] = [$a_1 a_2$]) is well-defined. Note: I changed $x$ to be $a_1$ and $y$ to be $a_2$ from the book, because using the definition of $[x]$ gets confusing for me.
I know that I need to show that [$a_1$][$a_2$] = [$a_1^\prime$][$a_2^\prime$]. I'd really appreciate some direction regarding this.
I think you're not even trying to prove the right thing.
Given the definitions in the image, to prove that $[x][y]$ is well defined can be done by showing that, if $[a] = [a']$ and $[b] = [b']$ then $[ab] = [a'b']$ (the product doesn't depend on the choice of the elements in each equivalence class).
So suppose $[a] = [a']$ and $[b] = [b']$, that is, $a - a' \in \mathfrak{I}$ and $b - b' \in \mathfrak{I}$.
First, we see that $ab - a'b = (a - a')b \in \mathfrak{I}$, because $a - a' \in \mathfrak{I}$ and $\mathfrak{I}$ is an ideal.
So $[ab] = [a'b]$.
Similarly, $a'b - a'b' \in \mathfrak{I}$, whence $[a'b] = [a'b']$.
By transitivity, $[ab] = [a'b']$.