proving unity is unique First The textbook I am using (Contemporary Abstract Algebra) already gave a proof for this, but I am trying a different way. I am unsure how to go about one of the cases or if the proof is correct. I would appreciate your help.
Claim: If $R$ is a ring with unity, it is unique. If $\alpha \in R$ is a unity, then its multiplicative inverse is unique.
My proof 
Let $R$ be a ring with unity. Suppose $\beta_1, \beta_2 \in R$ both satisfy the properties of being a unity. Then $\forall \alpha \in R$ we have 
$1) \alpha\beta_1 = \beta_1\alpha = \alpha$
$2) \alpha\beta_2 = \beta_2\alpha = \alpha$
Hence from $(1)$ and $(2)$ we have 
$\alpha\beta_1 = \alpha\beta_2 \Rightarrow \alpha\beta_1 - \alpha\beta_2 = 0 \Rightarrow \alpha(\beta_1 - \beta_2) = 0$ 
This means that either $\alpha = 0$ or $\beta_1 - \beta_2 = 0$. 
If $\alpha = 0$ (here is my struggle case, anything multiplied by 0 is 0 so I dont know what to say, or how this case implies that $\beta_1 = \beta$)
If $\alpha \neq 0$ then $\beta_1 - \beta_2 = 0$ which means $\beta_1 = \beta_2$. Hence unity is unique.
The other part im okay with. Thank you 

The trick is to choose $\alpha=\beta_1$ or $\alpha=\beta_2$: 
$$\beta_1 = \beta_1 \beta_2  =\beta_2.$$
First equality because $\beta_2$ is a unit, second is because $\beta_1$ is a unit. 
