Tautological implication : what does $\alpha \to \beta$ mean? If $\alpha \to \beta$ is a tautology, it is clear that any assignment that satisfies $\alpha$ means that $\beta$ is always satisfied. But what about assignments that don't satisfy $\alpha$? Does this mean that $\beta$ will necessarily not be satisfied either?
 A: $\alpha \rightarrow \beta$ is (defined to be) logically equivalent to $\neg \alpha \vee \beta$, which is true exactly whenever either $\alpha$ is false or $\beta$ is true (or both).  When $\alpha$ is false the truth value of $\beta$ is completely irrelevant to the truth or falsity of the implication $\alpha \rightarrow \beta$: $\beta$ could be true, but it also could be false.
For example:


*

*"$2\text{ is odd} \rightarrow 2\text{ is even}$" is a true implication; and

*"$2\text{ is odd} \rightarrow 7\text{ is even}$" is also true implication.


In such cases we may say that the implication is vacuously true.
A: You are "mixing" two different (but related) concepts : that of tautology and that of tautological implication.
We say that $\alpha$ is a tautology : $\vDash \alpha$, iff :

$\alpha$ is true for all truth assigments.

We say that $\beta$ is a tautological consequence of $\alpha$, or that $\alpha$ tautologically implies $\beta$, : $\alpha \vDash \beta$, iff :

all truth assignments satisfying $\alpha$ also satisfy $\beta$.

The (obvious) relation between the two notions is :


$\alpha \vDash \beta$ iff $\vDash \alpha \rightarrow \beta$.


Thus, if $α→β$ is a tautology, we have that any truth assignment that satisfies $α$ must also satisfy $β$. But for an assignment that doesn't satisfy $α$, nothing follows regarding $β$. 
According to the truth-functional properties of $\rightarrow$, if $v$ is an assignment such that $v(\alpha)=0$, then $v(\alpha \rightarrow \beta)=1$, irrespective of the value which $v$ assign to $\beta$.
