Sum of Rational Cuts is Rational Cut
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Theorem
Let $p \in\ Q$ and $q \in \Q$ be rational numbers.
Let $p^*$ and $q^*$ denote the rational cuts associated with $p$ and $q$.
Then:
- $p^* + q^* = \paren {p + q}^*$
Thus the operation of addition on the set of rational cuts is closed.
Proof
From Sum of Cuts is Cut, $p^* + q^*$ is a cut.
Let $r \in p^* + q^*$.
Then:
- $r = s + t$
where $s < p$ and $t < q$
Thus:
- $r < p + q$
and so:
- $r \in \paren {p + q}^*$
Hence:
- $p^* + q^* \subseteq \paren {p + q}^*$
$\Box$
Let $r \in \paren {p + q}^*$.
Then:
- $r < p + q$
Let:
- $h = p + q - r$
- $s = p - \dfrac h 2$
- $t = q - \dfrac h 2$
Then:
- $s \in p^*$
- $t \in q^*$
Hence:
- $r = s + t$
and so:
- $r \in p^* + q^*$
So:
- $\paren {p + q}^* \subseteq p^* + q^*$
$\Box$
Hence the result by definition of set equality.
$\blacksquare$
Sources
- 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.28$. Theorem: $\text {(a)}$