Prove the following properties of $\mathbb{Z}_n$ [duplicate]

The associative property is defined in the picture above. The screenshot is from Discrete Math and its Applications 8th edition, by Kenneth Rosen. Also, we have this definition for modular multiplication and addition.

Both of these images are on page 257 of the textbook.

My question is why is the associativity property valid? Every other property listed is very intuitive, but I don't know how I'd go about proving the associative property.

Hint:

The author proves

The $\mathbf{mod} \, m$ function has the property that

$\tag 1 [a \, \mathbf{mod} \, m] \, \mathbf{mod} \, m = a \, \mathbf{mod} \, m$

so, using corollary 2, this is also true,

$\tag 2 (a + b)\, \mathbf{mod} \, m = [a + b \, \mathbf{mod} \, m] \, \mathbf{mod} \, m$

Since there is an accepted completed answer, "here is how it is done strictly":

$\quad (a +_m b)+_m c =[(a +_m b) + c] \, \mathbf{mod} \, m = \big[(a + b) \, \mathbf{mod} \, m + c\big] \, \mathbf{mod} \, m =$
$\quad \quad \big[(a + b) + c\big] \, \mathbf{mod} \, m = \big[a + (b+c)\big] \, \mathbf{mod} \, m = \big[a + (b+c) \, \mathbf{mod} \, m \big] \, \mathbf{mod} \, m = $
$\quad \quad \big[a + (b+_mc)\big] \, \mathbf{mod} \, m = a +_m (b+_m c)$

The key is to show that $$a+_m(b+_mc)=[a+(b+c)]\mbox{ mod $m$}$$ (and similarly for $(a+_mb)+_mc$). Using the definition of modular arithmetic you've quoted this goes through the fact that $$a+(b+_mc)\equiv a+(b+c)\quad\mbox{ mod $m$}.$$ This is a little cringe-inducing since it mixes integers and integers-mod-$m$; this is in keeping with the definition you've written, but doesn't make sense if we define $\mathbb{Z}_m$ via equivalence classes instead (this is the standard approach, which is more complicated at the outset but ultimately much nicer), and then a slightly different reasoning is needed.

Note that this is all a bit slippery until we've shown that "$+_m$" is in fact well-defined - if you haven't yet, you should do that first.

At that point, associativity of $+_m$ follows immediately from associativity for $+$: we have $$a+(b+c)=(a+b)+c\implies [a+(b+c)]\mbox{ mod $m$}=[(a+b)+c]\mbox{ mod $m$}.$$ This is just a slightly messy example of the more general fact that $$x=y\implies x\mbox{ mod $m$}=y\mbox{ mod $m$},$$ that is, that "mod" is well-defined.