On MathOverflow, question Decidability of equality of elementary expressions was asked, and their elementary expression construction only involve constant $i$, operation $\times$, and functions $\exp$ and $\ln$. If we ask a more basic question - giving two arbitrary arithmetic expressions over real numbers $\mathbb{R}$, could we check the equality of the two by an algorithm?
We can realized the equality of two arithmetic expressions over rational numbers $\mathbb{Q}$ is decidable: Given the two expression, we can evaluate both of them into two rationals, and then find a common denominator, to convert the question to the equality of two natural numbers.
But for real numbers, does such an algorithm exists?
Finally, I want point out that this equality problem lays at the center of how we define arithmetic expressions. The only way to define an arithmetic expressions is purely a syntactical way. The semantic validity of the built expressions must avoid the zero-divisor cases which can be reduced to the question here.
Formally, the construction process of an arithmetic expression $a$ over a field $\mathbb{F}$ is a structure given by the following production rules: \begin{equation}\label{eq:productionrule} \begin{aligned} a &\longleftarrow x\\ a &\longleftarrow ( a + a )\\ a &\longleftarrow ( a - a )\\ a &\longleftarrow ( a \times a )\\ a &\longleftarrow ( a \div a ) \end{aligned} \end{equation}
where $x \in \mathbb{F}$. This process build a tree. But the process is syntactical, we may not so easy to check the validity of the built expressions.
Then we can recursively define evaluation of $a$ which referred as $\nu(a)$:
- Constant leaf: for any $x \in \mathbb{F}$, $\nu(x) = x$.
- Compositional node by $+$: For any $(a + b)$, $\nu((a + b)) = \nu(a) + \nu(b)$.
- Compositional node by $-$: For any $(a - b)$, $\nu((a - b)) = \nu(a) - \nu(b)$.
- Compositional node by $\times$: For any $(a \times b)$, $\nu((a \times b)) = \nu(a) \nu(b)$.
- Compositional node by $\div$: For any $(a \div b)$, if $\nu(b) \neq 0$, then $\nu((a \div b)) = \nu(a) / \nu(b)$.
P.S.
Add three related questions here:
