Is is possible to transform the conjunction of inequalities into one single equation?

Question

I am wondering if there is a way to transform the conjunction of inequalities into one single equation.

For example, let $x_i$ be variables and $a_i$ be fixed integers. The following condition $$(x_1 = a_1) \wedge (x_2 = a_2)\wedge \cdots \wedge (x_n = a_n) $$ can be transformed to

$$(x_1 - a_1)^2 + (x_2 -a_2)^2 + \cdots + (x_n -a_n)^2 =0. $$

Now consider the condition consisting of the conjunction of inequalities: $$(x_1> a_1) \wedge (x_2 > a_2)\wedge \cdots \wedge (x_n > a_n) .$$

Is is possible to transform it to a single equation? Like the above example. Such a single equation should not contain $\wedge$ or $\vee$.

Edit 1: For two-dimensions, $(x_1> a_1) \wedge (x_2 > a_2)$,we can think about the angle of between the vector $(x_1-a_1,x_2-a_2)$ and the $x_1$ axis. Can it be extended to higher dimension?

Edit 2: To make it more clear. The question is to find an analytic function $f(x_i,a_i)$. Given the input $x_1,\cdots,x_n,a_1,\cdots,a_n$, let $g = f(x_1,\cdots,x_n,a_1,\cdots,a_n)$. The condition $$(x_1> a_1) \wedge (x_2 > a_2)\wedge \cdots \wedge (x_n > a_n) $$ is equavalent to "$g$ satisfies certaion conditions (e.g., $g >0$)".

Answer 1

Yes. The condition $(x_1>a_1)\wedge\cdots\wedge (x_n>a_n)$ is equivalent to $$ \min_i \left(x_i-a_i\right) > 0. $$

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If the idea is to plug it into a calculator and do it approximately, you can choose a sufficiently large $m$ and just impose

$$ \left(\frac{a_1}{x_1}\right)^m + \cdots + \left(\frac{a_n}{x_n}\right)^m < 1 $$

the equivalence is approximate this time, but it can be made as precise as needed by increasing the number $m$. It only works when all the $a_i$ and $x_i$ are positive, or from another point of view, it approximates the condition $|x_i|>|a_i|\;\forall i$.

Answer 2

You can't do it without resorting to non-analytic functions somewhere. And I'm pretty sure everything you would consider "normal" is analytic.

In particular, there is no real analytic function $f:\mathbb R^2\to\mathbb R$ such that $f(x,y)>0 \iff x>0 \land y>0$.

Proof: By continuity, such a function would need to satisfy $f(x,0)=0$ for $x\ge 0$ and therefore by analyticity also for $x<0$. Similarly $f(0,y)=0$ for all $y$.

Now let $n$ be the smallest integer such that $\dfrac{\partial^nf}{\partial y^n}(0,x) \ne 0$ for some $x$. Because $\dfrac{\partial^nf}{\partial y^n}$ is analytic and not identically zero, it is nonzero almost everywhere -- in particular there are both postive and negative $x$ it is nonzero for.

But if $n$ is even, this means that there are positive $x$ where $f(x,y)$ have the same sign on both sides of the $x$-axis, so $\{(x,y)\mid f(x,y)>0\}$ is not the set we want.

On the other hand, if $n$ is odd, there are negative $x$ where $f(x,y)$ is positive on one side of the $x$-axis, again meaning that $\{(x,y)\mid f(x,y)>0\}$ is the wrong set.