How is this proof correct (takes $a$ to be a specific number)?

Question

This is my professor's proof of uniqueness of additive identity:

assume there exists $0', 0 \in F$ where $a + 0 = 0 + a = a$ and $a + 0' = 0' + a = a$.

w.t.s that $0 = 0'$

taking $a = 0'$ and $a = 0$ we get $0 + 0' = 0'$ and $0 + 0' = 0$, therefore $0 = 0'$.

Why is she allowed to simply take $a = 0$ and $a = 0'$? Don't we want to show it for every $a$? This is what I don't get about this specific proof.

Answer 1

Hi I can maybe provide a "filled out" version of the proof with more details.

Assume there exists $0',0\in F$. By definition of identity, we know that $\forall a$, we have $a+0=0+a=a$ and $a+0'=0'+a=a$. Since these 2 equations are true for any $a$, they are true when $a=0$ and when $a=0'$ specifically. Substituting the 2 specific values of $a$ into the equations above, we will have 4 true equations:

1. $0+0=a=0$

2. $0+0'=0'+0=a=0$ (when $a=0$),

3. $0'+0=0+0'=a=0'$

4. $0'+0'=a=0'$ (when $a=0'$).

Out of the 4 true equations, 2 and 3 are the ones we need. We see that $0+0'=0$ and $0+0'=0'$. Combining the 2 equations, we get $0=0'$.