Duality principle in arithmetic How do I show that if you have a valid rule in arithmetic that involves multiplication and addition, then you cannot interchange the signs of multiplication and addition and obtain a valid rule?
I read about duality and as from what I understood if I have an element, say $x$, then
$$ x = x + 0 = x + 0\cdot 1 = x \cdot 1 + x \cdot 0 $$
then I can interchange the signs of multiplication and addition and I get a valid rule. I'm so confused on how to show the above statement. Maybe Im not getting the question right myself. Please help.
 A: You need a counterexample.
Consider the distributive law:  $a \color{blue}{\cdot} (b \color{red}{+} c) = a \color{blue}{\cdot} b \color{red}{+} a \color{blue}{\cdot} c$.  
If the duality principle held, then interchanging the roles of multiplication and addition would produce a true statement.  If we do so for the distributive law, we obtain $$a \color{red}{+} b \color{blue}{\cdot} c = (a \color{red}{+} b) \color{blue}{\cdot} (a \color{red}{+} c)$$
However, this statement is false.  Let $a = 1$, $b = 2$, and $c = 3$.  Then 
\begin{align*}
a + bc & = 1 + 2 \cdot 3\\
       & = 7\\
       & \neq 12\\
       & = 3 \cdot 4\\
       & = (1 + 2)(1 + 3)\\
       & = (a + b)(a + c)
\end{align*}
so while multiplication distributes over addition, addition does not distribute over multiplication.  Therefore, the duality principle does not hold for addition and multiplication.  
The duality principle does hold for the intersection and union of sets. For instance, 
\begin{align*}
A \cap (B \cup C) & = (A \cap B) \cup (A \cap C)\\
A \cup (B \cap C) & = (A \cup B) \cap (A \cup C)
\end{align*}
