Prove or disprove, that there is a partial order in N that is also an equivalence relation.

Question

I gave my answer based on the impossibility of a relation to be both symmetric and antisymmetric at the same time. I was wondering how to give a proper formulated answer to this.

Answer 1

$$ x\preceq y \,\,\text{ if and only if }\,\,x=y. $$ Then $\preceq$ is a partial order and an equivalence relation on $\mathbf{N}$.

Answer 2

Let's assume we have a partial order $?$ that is also an equivalence relation. Let's see what we can deduce about $?$.

Suppose there are $a$ and $b$ such that $a ? b$. Then since $?$ is symmetric, we have $b ? a$. Combining these, since $?$ is antisymmetric, we have $a = b$. Thus if we ever have that $a ? b$, we must have that $a = b$. But since $?$ is reflexive, if $a = b$ then $a ? b$. Thus $?$ is the same relation as $=$, and we can check that $=$ is both a partial order and an equivalence relation.