Prove that $\alpha(\beta+\gamma) =\alpha\beta+\alpha\gamma$ for all Dedekind cuts $\alpha, \beta, \gamma$.

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I'm in this math course and I did not understand dedekind cuts, someone please help!!!

Prove the distributive law in $\mathbb{R} $ as constructed using Dedekind Cuts. Recall that a cut $\alpha$, subset of $\mathbb{Q}$ that satisfies the following:

  1. $\alpha \neq \emptyset$ and $\alpha \neq \mathbb{Q}$
  2. If $p\in \alpha, q\in \mathbb{Q}$ and $q<p$ then $q\in \alpha$.
  3. If $p\in \alpha$, then $p<r$ for some $r\in \alpha$.

For simplicity, let $\alpha, \beta, \gamma$ be positive cuts, we want to show that $\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$

Where $\alpha+\beta = \{a+b|a\in \alpha,b\in \beta\}$ and $\alpha\beta = \{p\in \mathbb{Q}\mid p < ab, a \in \alpha, b \in \beta,\text{ and } a,b>0\}$. Remember that these are sets that we want to show are equal, so we need to show that

$\alpha(\beta + \gamma) \subseteq \alpha\beta + \alpha\gamma$ and $\alpha(\beta + \gamma) \supseteq \alpha\beta + \alpha\gamma$

Best Answer

If $r \not \in \alpha(\beta + \gamma)$. Then for any $a \in \alpha$ and any $c \in \beta + \gamma$ then $r > ac$.

Now let $e \in \alpha\beta + \alpha\gamma$. Then there are $f\in \alpha\beta$ and $h\in \alpha\gamma$ so that $e = f+h$. And there are $a_1, a_2 \in \alpha, b\in \beta, g\in \gamma$ so that $f= a_1b$ and $h = a_2g$ and so $e= a_1b + a_2g$. And there is an $f_2 = \max(a_1,a_2)b\in \alpha\beta$ and an $h_2 = \max(a_1,a_2)g\in \alpha \gamma$ and so thre is an $e_2 = f_2 + h_2 \in \alpha\beta + \alpha\gamma$ and $e_2\ge e$.

But $e_2 = \max(a_1,a_2)b + \max(a_1,a_2)g= \max(a_1,a_2)(b + g) \in \alpha(\beta + \gamma)$. So $r > e_2\ge e$ so $r \not \in \alpha\beta + \alpha\gamma$.

So $\alpha\beta + \alpha\gamma\subset \alpha(\beta + \gamma)$.

The opposite is similar.

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