# Is there a way to confirm that a homomorphic division (multiplication with inverse) using ElGamal produced the correct result?

By running some tests I observed that if I perform a homomorphic division (multiplication with the multiplicative inverse) between two values using the ElGamal scheme, I get the correct result when the division produces a whole number.

For example, D(E(10) / E(2)) = 5 which is the expected result (here "/" denotes division applied homomorphically).

But if I try D(E(10) / E(3)) the result is a very large positive or negative number.

This might be expected since, I think, ElGamal supports homomorphic division only when the dividend is divisible by the divisor. My question is given only the decrypted result of such operation, is there a way to know whether the division "succeeded" in the sense that it is the result of a division where the dividend is divisible by the divisor?

I've included some discussion in general, but "El Gamal" here is a red herring. Your confusion is with the difference between "division" in $$\mathbb{R}$$, and "division" in $$\mathbb{F}_p$$. I discuss this more in detail at the end.
$$\mathsf{Enc}_{g, h}(m) = (g^y, mh^y)$$ Note that $$h = g^x$$, where $$x$$ is the secret key. This admits a (multiplicative) homomorphism, as you mention: $$\mathsf{Enc}_{g, h}(m_1)\ast \mathsf{Enc}_{g, h}(m_2) = (g^{y_1+y_2}, m_1m_2 h^{y_1+y_2})$$ I believe your confusion stems from confusion over what $$m_1m_2h^{y_1+y_2}$$ means. Like many cryptosystems, El Gamal is defined over what is known as a "finite field" (in particular $$\mathbb{Z}/p\mathbb{Z}\cong\mathbb{F}_p$$). This is a number system that allows you to add, subtract, multiply, and divide (just like with $$\mathbb{R}$$), but the operations are somehwat different than you are used to. Fix $$p = 5$$ (although any prime number works). Then addition is defined "mod 5", so: $$3 + 3\bmod 5 \equiv 6\bmod 5\equiv 1\bmod 5$$ So in this number system, $$3 + 3 = 1$$. Multiplication is similarly defined "mod 5". So: $$3 \times 3\bmod 5 \equiv 9\bmod 5 \equiv 4\bmod 5$$ So $$3\times 3 = 4$$ in this number system.
What does "division" mean in this number system? Well, one can show that for any (non-zero) $$x\in\mathbb{Z}/p\mathbb{Z}$$, there is a unique $$y$$ such that $$xy \equiv 1\bmod p$$. Over $$\mathbb{R}$$ this is true as well (instead with the equation $$xy = 1$$) --- if $$x = 2$$, then $$y = 1/2$$. But in $$\mathbb{Z}/5\mathbb{Z}$$, if we have that $$x = 2$$, then $$y = 3$$ means that $$xy\bmod 5\equiv 6\bmod 5\equiv 1\bmod 5$$. So, in $$\mathbb{Z}/5\mathbb{Z}$$, we have that $$2^{-1} = 1/2\equiv 3\bmod p$$.
This is now when we get to your point --- if we have $$x,y\in\mathbb{Z}/p\mathbb{Z}$$, then $$xy/y\equiv xyy^{-1}\equiv x\bmod p$$, so "when $$y$$ divides $$xy$$, division is like you would expect over $$\mathbb{R}$$". But the situation can be more complicated in general.