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 equation $\alpha(\beta_1-\beta_2)=0$ does not imply $\alpha=0$ or $\beta_1-\beta_2=0$ in a general ring $R$. (For example, in $Z_6$ we have $2\cdot 3=0$. We say that $2$ and $3$ are zero divisors because they are nonzero elements which divide $0$. Rings with identity and no zero divisors are called integral domains, and in an integral domain the implication above works, but not every ring is an integral domain.)
Instead you should apply properties 1) and 2) with $\alpha=\beta_1$ and $\alpha=\beta_2$. If we use $\alpha=\beta_2$ in property 1), we find $\beta_2\beta_1=\beta_2$. If we use $\alpha=\beta_1$ in property 2), we find $\beta_2\beta_1=\beta_1$. Therefore $\beta_1=\beta_2$.
