(Proof) Associativity of Summation of Real Number

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Definition of $\Bbb R$

For two sequences $\alpha,\beta : \Bbb N \rightarrow \Bbb Q$

Define $\alpha \sim \beta$ when $\forall e\;\;$ there $\exists N$ s.t. $\forall i \ge N$, $\lvert \alpha(i)-\beta(i)\rvert \lt e$

Since $\sim$ is equivalence relation of the set $\Bbb F = \{\alpha: \Bbb N \rightarrow \Bbb Q$},

call the each element of quotient set, $\;\;F/\sim\;\;$, real number.

Definition of Summation among Real Numbers

Define $[\alpha]+[\beta] = [\alpha+\beta]$ (*)

$\alpha + \beta$ is defined as $i \mapsto \alpha(i) + \beta(i)$

Question

I had already proved that above definition of summation is well-defined.

Now I want to prove whether this operation holds the associative property.

Proof

for $\alpha, \beta, \gamma : \Bbb N \rightarrow \Bbb Q$,

we need to check below two conditions:

1) $([\alpha] + [\beta]) + [\gamma] = [\alpha] + [\beta] + [\gamma]\;\;$ and

2) $[\alpha] + ([\beta] + [\gamma]) = [\alpha] + [\beta] + [\gamma]$

for 1), since $[\alpha]+[\beta] = [\alpha+\beta]$,

LHS of 1) = $[\alpha+\beta] + [\gamma] $ and

$[\alpha+\beta] + [\gamma] =[\alpha+\beta+\gamma]$ by defintion of (*)

for 2), since$\;\; [\beta]+[\gamma] = [\beta + \gamma]$,

LHS of 2) = $[\alpha] + [\beta + \gamma]$ and

$[\alpha] + [\beta + \gamma] = [\alpha + \beta + \gamma] $ by definition of (*)


Question

I am not familiar to logical thinking.

Is above proof is sufficient or logically clear under the notion of checking the associative property?

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No, you don't have to check your 1) and 2). Note that $[\alpha]+[\beta]+[\gamma]$ is undefined at the moment. Instead you have to check that $$\bigl([\alpha]+[\beta]\bigr)+[\gamma]=[\alpha]+\bigl([\beta]+[\gamma]\bigr)\ .\tag{1}$$ For the proof use that $[\alpha]+[\beta]$ is represented by $\alpha+\beta$, hence $$\bigl([\alpha]+[\beta]\bigr)+[\gamma]=[\alpha+\beta]+[\gamma]=[\alpha+\beta+\gamma]\ .$$ Similarly for the RHS of $(1)$.