why is 1+2 exactly 3 by the dedekind cut way of definding real number? Suppose by the definition of Dedekind cut, every element must be close downwards meaning it must add up every single element on $Q$ before the element. So when we define additions under real in this case, that is :
$$x+y = \{r\in Q :\exists t \in x,\exists s\in y,(r < t+s)\} $$
then how do we tell that 1+2 is exactly 3 by intuition?
thank you!
 A: Let $\mathbf{1}$, $\mathbf{2}$, and $\mathbf{3}$ denote the real numbers one, two, and three respectively, which have a different set-theoretic definition to the rational numbers $1$, $2$, and $3$. We wish to prove that $\mathbf{1+2=3}$, which amounts to showing that
$$
\{p\in\mathbb Q:p<1\}\,\mathbf{+}\,\{q\in\mathbb Q:q<2\}=\{r\in\mathbb Q:r<3\}
$$
Using the definition of $\mathbf{+}$ given in your question, we thus have to establish the following equality between two sets:
$$
\{p+q:p<1,q<2,p,q\in\mathbb Q\}=\{r\in\mathbb Q:r<3\} \tag{*}\label{*}
$$
Let $A$ and $B$ denote the LHS and RHS of $\eqref{*}$, respectively. It is trivial to prove that $A\subseteq B$: if $p+q\in A$ with $p<1$ and $q<2$, it follows that $p+q<3$, using the usual rules of arithmetic on the rational numbers (and in particular that $1+2=3$ in $\mathbb Q$). Hence $p+q\in B$. To prove that $B\subseteq A$, let $r\in B$, and put $k=3-r$. Setting $p=1-k/2$, $q=2-k/2$, we see that $p<1$ and $q<2$, so $p+q=r\in A.$ Thus $A=B$, completing the proof.
A near-identical argument can be used to prove more generally that if $a+b=c$ in $\mathbb Q$, then $\mathbf{a+b=c}$ in $\mathbb R$.
A: Every x < 3 is less than the sum of two numbers less than 1 and 2, for example 3-eps < (1-eps/3) + (2 - eps/3). Any x >= 3 is no such sum.
