Subfield of a field F Question: Let $F$ be a field and let $K$ be a subset of $F$ with at least two elements. Prove that $K$ is a subfield of $F$, if for any $a,b \left ( b\neq 0 \right )$ in $K$, $a-b$ and $ab^{-1}$ belong to $K$.


This is an odd question but let's see.

It is necessary to investigate the case where $a=0$ and $a\neq 0$.

Let's proceed with the case where $a=0$ first.

Suppose $a=0, b\neq 0 \in K$. Let $K=\left \{ a,b \right \}$
Then, $a-b=0-b=-b$.
It doesn't seem as though I can proceed on with what I have here.
The additive inverse of $b$ is contained in $F$ by definition of a field but it is not necessary true that this additive inverse must be contained in $K$.

Any help is appreciated. Thanks in advance.
Let $x \in K.\,$ Then $x-x\in K$, hence $0 \in K$.

Let $x \in K.\,$ Then since $0 \in K$, we get $0 - x \in K$, hence $-x \in K$.

Let $x,y \in K.\,$ Then $-y \in K$, hence $x-(-y) \in K$, so $x+y \in K$.

Since $K$ has at least two elements, $K$ contains at least one nonzero element, $x$ say. Then $(x)(x^{-1}) \in K$, hence $1 \in K$.

Let $x \in K,\;x \ne 0$. Then since $1 \in K$, we get $(1)(x^{-1}) \in K$, hence $x^{-1} \in K$.

Let $x,y \in K$.

$\qquad$If $y \ne 0$, then $y^{-1} \in K$, so $(x)((y^{-1})^{-1}) \in K$, hence $xy \in K$.

$\qquad$If $y=0$, then $xy=0$, hence $xy \in K$.

Either way, $xy \in K$.

It follows that $K$ is a field.
It's ambiguous as to whether $a$ and $b$ must be distinct for those condition to hold.  

If $a = 0$ then we have $b$ and $0-b = b$ and hmmm, if $b = -b$ but $b \ne b^{-1}$ then we have $0, b$ and that's it.  Not a field.

But if distinction is not implied then:

You have $a, b, a-b$ so you have $(a-b) -a = -b \in K$.  We have $a - (-b) = a+b$ so addition is closed.  And $b*b^{-1} = 1$ and and $1* b^{-1} = b^{-1}$ and $a(b^{-1})^{-1} = ab$ so multiplication is closed.  So a field.

But distinction can not be implied.