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How does one prove that addition on $\mathbb{R}$ is associative?

We can define addition to be a binary function $+ : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$, by that definition $\mathbb{R}$ is closed under addtition as for any $(a, b) \in \mathbb{R}^2$ we have $+(a, b) = a + b = c$ for some $c \in \mathbb{R}$.

But if we take $a, b, c \in \mathbb{R}$, how can we prove rigorously that $(a + b) + c = a + (b + c)$?

We know also that $(\mathbb{R}, +)$ is a group, but most proofs of this fact assume that addition in $\mathbb{R}$ is associative, the proofs do not actually prove that addition is associative.

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    At this level, it much depends on how you define $\mathbb{R}$ and $+\,$ to begin with. – 2017-02-08
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    On that note, checking out the Peano axioms is certainly worth your time :) – 2017-02-08
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    You can't. All you've said is addition is binary and closed. There are many binary closed operations that aren't associative. Let $a+b=a^b $ for instance. $2^{(3^4)}=2^{81}=2.7*10^{24} $ whereas $(2^3)^4=8^4=4096$. – 2017-02-08
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    By *definition* group addition is associative. – 2017-02-08
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    It is unnecessary to prove things that are definitions. You may have seen a verification that a certain set satisfies all the axioms and is therefore a model of the real numbers, which I think is the direction Ittay is going. But that is a proof the model satisfies the real number axioms, not a "proof that the real numbers satisfies its own axioms". Very different. – 2017-02-08

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Only once you give a construction of the reals can you prove these basic properties of them. Depending on the particularities that can be easy or more complicated. There are many constructions of the reals (e.g., this survey), with proofs of associativity of addition, which is typically easy. A very detailed construction, with all proofs, is given in this paper.