Help to understand material implication This question comes from from my algebra paper:
$(p \rightarrow q)$ is logically equivalent to ... (then four options are given). 
The module states that the correct option is $(\sim p \lor q)$. That is: 

$$(p\rightarrow q) \iff (\sim p \lor q )$$

but I could not understand this problem or the solution. Could anybody help me?

As it was suggested in a comment above, drawing a truth-table, especially when there are only two or three variables (i.e. atomic sentences) can really help to illustrate exactly when two given expressions are equivalent.  
In this case, we see that $(\sim p \lor q)$ is true exactly when $(p\rightarrow q)$ is true, and it is false exactly when $(p\rightarrow q)$ is false. That is, $(p\rightarrow q) \equiv (\sim p \lor q)$. Alternatively, we can recognize the equivalence of the two expressions simply by comparing column of truth-values corresponding to each expression and see that the two columns are identical, and hence, the expressions are logically equivalent.  
Logical Equivalence : (p → q) $\equiv$ (¬p ∨ q) $\;\;$ Note that $"\equiv"$ is equivalent to $"\iff"$ 



Another way to use the truth-table above is to see that the implication $(p \rightarrow q)$ is false if and only if the truth-value of $p$ is true and value of $q$ is false. Symbolically, we can express that fact by asserting that for the implication to be true, then (it cannot be the case) that $(p \land \sim q)$; in other words, $\sim (p \land \sim q)$.  This conveys exactly the same information as the material implication $(p \rightarrow q)$. Note that $\sim (p \land \sim q) \equiv (\sim p \lor q)$, by De Morgan. 
As for understanding that when $(p \rightarrow q)$, then if $p$ is true, we must have that $q$ is true: perhaps the following analogy will help. 
In many respects, the proper inclusion (proper "is a subset of") relation corresponds to material implication, where $\subset$ corresponds to the $\rightarrow$ relation.  For example, suppose $A \subset B$.  Then if it is true that $x \in A$, then it must be true that $x \in B$, since $B$ contains $A$.  However, if $x \notin A$, that does not mean that $x \notin B$, since if $A \subset B$, then $B$ contains elements that $A$ does not contain.
