# Definition of multiplication of real numbers from product of positive dedekind cuts and absolute value

--let $A, B \in \mathbb{R}$, with $0\leq A$ and $0 \leq B$, $A \star B :=\{q\in\Bbb Q\mid q<0\}\cup\{a\cdot b\mid a\in A\wedge b\in B\wedge a\ge 0\wedge b\ge 0\}$"

this definition is correct:

--let $A, B, C \in \mathbb{R}$, $C$ is product of $A$ and $B$, $C \triangleq A \cdot B$, if $C=\begin{cases} |A| \star |B|, & \mbox{if } (0 \leq A \wedge 0 \leq B )\mbox{ OR }(A \leq 0 \wedge B \leq 0 )\\ -(|A| \star |B|), & \mbox{if } (A \leq 0 \wedge 0 \leq B )\mbox{ OR } (0 \leq A \wedge B \leq 0 ) \end{cases}$

???

P.S.= OR, the question is edited!

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There's a major problem with your original definition: $A\star B$ will not be a real number as written! Instead, you need $$A\star B:=\{q\in\Bbb Q\mid q<0\}\cup\{a\cdot b\mid a\in A\wedge b\in B\wedge a\ge 0\wedge b\ge 0\}.$$ (Edit: This has been fixed.)

There's also a bit of an issue with your general definition, as it doesn't cover the case $0\cdot 0.$ The XORs rule out exactly this case. Switch them to $\vee$s and that issue will go away.

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true.. mmm!!!Thanks...but how to include the case $0 \cdot 0$??.. :) I corrected partially my question!! –  mle Aug 12 '13 at 16:18
if $A\star B:=\{q\in\Bbb Q\mid q<0\}\cup\{a\cdot b\mid a\in A\wedge b\in B\wedge a\ge 0\wedge b\ge 0\}$, then $A \star B=0$ iff $A=0 \vee B=0$.. is correct? –  mle Aug 12 '13 at 16:30
There's really no need for an XOR at all. If you change them to $\vee,$ that will do the trick. I'd also recommend that you alter the second condition to make it say that one of $A,B$ is negative and the other is positive. It isn't a necessary change, but it is a change that will make it simpler to prove that the operation is well-defined. Let me know if you aren't sure how to do this. –  Cameron Buie Aug 12 '13 at 16:46
It is correct that $A\star B=0$ if and only if $A=0$ or $B=0$. Can you prove it? –  Cameron Buie Aug 12 '13 at 16:47
Actually, the XORs rule out exactly this case. The following statement is false: $$(0\le 0\wedge0\le0)\text{ XOR }(0\le0\wedge0\le0)$$ This is because a statement P XOR Q will be true when and only when exactly one of P, Q is true. –  Cameron Buie Aug 12 '13 at 17:25