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My source is Ayres, *Modern Algebra", $1965$ ed , ch. $7$, § "additive inverses" , p.$68$.

A cut in $\mathbb Q$ is defined as a non empty proper subset $C$ of the rationals such that (1) if $c$ belongs to $C$, then every rational number $a$ less than $c$ also belongs to $C$ and (2) if $c$ belongs to $C$ there is some rational number $d$ greater than $c$ that also belongs to $C$.

The question deals with positive cuts ( a positive cut being a cut having at leat one strictly positive rational number as element).

The author first defines the sum of two positive cuts as :

$C_1 + C_2 = \mathbb Q^- \cup \{0\} \cup \{ c_1+c_2 | c_1\in C_1 , c_2 \in C_2 , c_1\gt 0 , c_2\gt 0 \} $ ,

in words" the sum of two poitive cuts is the set having as elements all the negative rational numbers , together with $0$ and all possible sums of two members of the positive parts of these cuts".

Then he states that the above definition is esquivalent to :

$C_1 + C_2 = \{ c_1 + c_2 | c_1\in C_1 , c_2 \in C_2 \}$.

Since the author does not provide a proof, my question is : how to justify this equivalence? I assume it does not simply go without saying

The proof amounts to proving that, for all $x$

$x\in \mathbb Q^- \cup \{0\} \cup \{ c_1+c_2 | c_1\in C_1 , c_2 \in C_2 , c_1\gt 0 , c_2\gt 0 \} \iff x\in \{ c_1 + c_2 | c_1\in C_1 , c_2 \in C_2 \}$.

I'm already stuck for the $\implies $ part, see (3) below :

(1) if $x\in \mathbb Q^-$ , then , $x = q_1 + q_2 , $ for some $q_1$ and $q_2$ such that both $q_1$ and $q_2 \in \mathbb Q^- $, implying that both $q_1$ and $q_2$ belong to $C_1$ and $C_2$, and that, in particular $q_1 \in C_1,$ and $q_2 \in C_2$ ;

(2) if $x\in \{ c_1 + c_2 | c_1\in C_1 , c_2 \in C_2 \}$ the consequence follows ipso facto ;

(3) but if $x=0$ , either $ x= 0+0$ and the consequance follows; or $x = q_1+q_2 $ with , say $q_1 \lt 0$ and $q_2 \gt 0$ and $ q_2 = - q_1$ ; however, one cannot - apparently - rule out the possibility that $q_1$ is so small that its opposite $q_2$ is too big to belong to $C_2$ , implying that $x$ is not the sum of two elements of , respectively, $C_1$ and $C_2$.

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    $\begingroup$ I'd start by showing that $C_1+C_2$ is, in fact, a cut, under both definitions. Then, to handle the third case, try using the fact that if $c\in C$ for some cut $C$ and $a\in\Bbb Q$ with $a<c,$ then $a\in C.$ $\endgroup$
    – Cameron Buie
    Commented May 4 at 0:16
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    $\begingroup$ Bear in mind all negative rationals and $0$ are all in both $C_1$ and $C_2$. if $x \in \mathbb Q^-$ then $x \in C_1$ and $0 \in C_2$ and so $x=x+0\in \{ c_1 + c_2 | c_1\in C_1 , c_2 \in C_2 \}$. And if $x = 0$ then $x=0 + 0$ and $0 \in C_1$ and $0 \in C_2$ so $0\in \{ c_1 + c_2 | c_1\in C_1 , c_2 \in C_2 \}$. I'd say the only two significant things to show are If $q= c_1 + c_2 > 0$ but we don't know the signs of $c_1,c_2$ that there exist $d_1,d_2>0$ so that $d_1+d_2=q$ . Do you need help with that. $\endgroup$
    – fleablood
    Commented May 4 at 0:28
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    $\begingroup$ Oops. I originally meant that we had to prove as a second thing that if $q > 0$ and $q < c_1+c_2$ for some $c_1,c_2$ then $q = d_1+d_2$ for some $d_1 + d_2$ but I decide that was both confusing to write and redundent. So I omitted it but forgot to edit my comment about "two things". $\endgroup$
    – fleablood
    Commented May 4 at 1:08
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    $\begingroup$ The only significant thing to prove is that $\{ c_1 + c_2 | c_1\in C_1 , c_2 \in C_2 \}\cap \{q\in \mathbb Q|q > 0\}$ is equal to $\{ c_1+c_2 | c_1\in C_1 , c_2 \in C_2 , c_1\gt 0 , c_2\gt 0 \}$. We can do that: If $0 < w = c_1 + c_2$ but $c_1 \le 0$ then $w \le c_2$ so $w\in C_2$. So $w = 0 + w$. As $0\in C_1$ there is teensy $r > 0$ that is in $C_1$ and $w = r + (w-r)$. And $w-r < w$ so $w-r \in C_2$. So $r\in C_1, w-r \in C_2$ and we are basically done. $\endgroup$
    – fleablood
    Commented May 4 at 1:24
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    $\begingroup$ The easiest way to think about cuts is that a cut is all the rationals up to but not including a particular point. So if a cut has a positive rational in it, then it must have all negative rationals, it must have $0$ and it must have an infinite number of arbitrarily small positive rationals. $\endgroup$
    – fleablood
    Commented May 4 at 1:55

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