Entropy as a state property

Question

The usual "proof" entropy is a state property is like that:

"Consider a system which undergoes a reversible process from state 1 to state 2 along path A, and let cycle be completed along path B, which is also reversible. Since the cycle is reversible we can write:

$$\int_1^2 \delta Q / T + \int_2^1 \delta Q / T = 0 $$

Now let cycle be completed along path C, but paths B and C represent arbitrary reversible processes. So $\int_2^1 \delta Q / T $ is the same for all reversible paths between states 2 and 1."

My question is, isn't the equation above already assume entropy is a state property? Only if it is a state property it can go around a cycle without changes. How can it be valid to prove entropy is state property if it is already assumed from the beginning?

Answer 1

The usual argument is a proof of the existence of a function whose differential is $\delta Q/T$. The argument usually proceeds as follows:

Fact. (Physics) $\int_\gamma \frac{\delta Q}{T} = 0$ for any closed path $\gamma$ in thermodynamic state space.

Once you have this fact, you can prove that

Claim 1. $\delta Q/T$ is conservative, namely $\int_{\gamma_1} \delta Q/T = \int_{\gamma_2} \delta Q/T$ for any two path segments $\gamma_1$ and $\gamma_2$ with the same endpoints.

And then you appeal to the following mathematical fact (which is non-trivial to prove):

Claim 2. A $1$-form (this is just a fancy term for the kind of mathematical object $\delta Q/T$ is) is conservative if and only if it is exact (exact means it can be written as the differential of a scalar function).

Combining these claims shows that

Desired Result. There exists a scalar function $S$ such that $dS = \delta Q/T$.

As you can see, there is no need to assume anything about entropy in the beginning, you just need the fact (or some equivalent) about $\delta Q/T$ integrating to zero around closed curves.