Addition on R (via Dedekind Cuts)

See Dedekind Cuts, Constructing R for a refresh of cuts. Let $A, B \in R$ :

Here $A + B$ should be a new set to the right of $B$ over some $A$ 'amount'.

This approach actually works:

A + B = {a + b | a \in A, b \in B}

Closure of Set Addition Operation in

R

This operation is closed, namely $A + B$ is always a cut. For example, for property 2, if we have $a + b \in A + B$ and $q \in Q$ where $q < a + b$ then $q - b < a$ so then $q - b \in A$ since $A$ is a cut. But:

q = (q - b) + b \in A + B

So then $q \in A + B$ as required.
The other properties would be proved similarly.

Commutivity and associativity come out from the commutativity and associativity of the $Q$ addition operation.

Commutativity and Associativity in

R

$R$ has commutativity and associativity properties when defined via Dedekind Cuts.

Proof
(Commutativity) Let $A, B \in R$ . We'll show $A + B = B + A$ since:

A + B = {a + b | a \in A, b \in B} = {b + a | b \in B, a \in A} = B + A

from the commutativity of addition on $Q$ .
(Associativity) Is shown in pretty much the same way.
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There's also the additive identity $O$ defined as:

O = {q \in Q | q < 0}

Let's show that any $A + O = A$ :

O

is additive identity

Let $A \in R$ . Then $A + O = A$ .

Proof
We'll show both $\subseteq$ and $\supseteq$ :

( $\subseteq$ ): Let $a^{'} \in A + O$ so then $\exists a \in A, q \in O$ such that $a^{'} = a + q$ . We want to show that $a^{'} \in A$ . Notice that since $q < 0$ then:

q < 0 \Rightarrow a + q < a \in A \Rightarrow a + q \in A

Thus since $a^{'} = a + q$ then $a^{'} \in A$ .

( $\supseteq$ ): Let $a^{'} \in A$ , so then $\exists a \in A$ such that $a > a^{'}$ .

A + O = {a + q | a \in A, q \in O}

Now notice that $a^{'} < a \Rightarrow a^{'} - a < 0$ so clearly $a^{'} - a \in O$ . Now notice:

a^{'} = \underset{\in A}{\underset{⏟}{a}} + \underset{\in O}{\underset{⏟}{(a^{'} - a)}} \in A + O

as required.

#todo/school : Finish this proof! 📅 2024-09-26 ✅ 2024-09-26
#todo/school : Check $(\supseteq)$ w/ prof ⏳ 2024-10-03 ✅ 2024-10-07

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What is the additive inverse $- A$ ? Let's think on the number line:

Thus we define it as follows:

- A

Define $- A = {q \in Q | \exists t \notin A s.t. t < - q}$ .

- A

gives rise to Ordered Field properties

Given the above $- A$ definition, then:

$- A$ is a cut (well defined)
$A + (- A) = O$

Pasted image 20240925223714.png

Proof

Let's prove $- A$ is a cut:
a. $A$ is a cut so $A \neq \emptyset$ and $A \neq Q$ . Thus $\exists t \notin A$ so then choose any $r < - t$ since then $- r > t$ so $r \in - A$ (for example, choose $r = - t - 1$ ). Thus $- A \neq \emptyset$ . Further we know $\exists a \in A$ so then $\forall t \notin A$ then $t > a$ . Thus since $- a \leq a$ then $- a \notin - A$ so $- A \neq Q$ .
b. Let $a \in - A$ and $q \in Q$ such that $q < a$ . WTS $q \in - A$ . Notice $\exists t \notin A$ where $t < - a$ . But notice that $- q > - a > t$ so then $q \in - A$ .
c. Let $a \in - A$ . Then $\exists t \notin A$ where $t < - a$ . If we construct $t^{'} = \frac{t - a}{2}$ then $t < t^{'} < - a$ so then since $t < t^{'}$ where $t \notin A$ we can choose $- t^{'} \in - A$ as a result.

#todo/school : Finish this part of the proof 📅 2024-10-08 ✅ 2024-10-08

Let's prove $A + (- A) = O$ :
( $\subseteq$ ): Let $a \in A$ and $a^{'} \in - A$ . Then $\exists t \notin A$ where $t < - a^{'}$ . Since $a \in A$ and $t \notin A$ then we must have it that $t > a$ (since otherwise it'd force $t \in A$ as a contradiction). Then notice that:

\begin{aligned} a^{'} < t & t > a \\ \Rightarrow - t < - a^{'} & t > a \\ \Rightarrow a - a^{'} < a - t & 0 > a - t \\ \Rightarrow a - a^{'} < a - t < 0 \\ \Rightarrow a - a^{'} < 0 \\ \equiv a - a^{'} \in O \end{aligned}

Thus $A + (- A) \subseteq O$ .

( $\supseteq$ ): Let $o \in O$ so $o < 0$ . Since $A$ is a cut then $\exists a \in A$ . Further $\exists q \in A$ where $q > a$ . Then $q - a > 0 > o$ so $q - a > o$ . Then $0 > a + (o - q)$ . This suggests showing that $o - q \in - A$ , since once we have that we are done.

Notice that $- a \notin - A$ , because for contradiction if $- a \in - A$ then $\exists t \notin A$ where $t < a$ . But by $A$ being a cut we have to have $t \in A$ which is a contradiction. As a result, since $- A$ is a cut and because $o - q < - a$ then $o - q \in - A$ , which finishes the proof. ☐