Use of the Equivalence symbol in logic statements Is it correc to use the equivalence symbol to express that two statements are logically equivalent to one another? For example,
$$\forall (x_j,y_j) \in \mathbb N^{2}\,, \exists\,k \in \mathbb N\,:  y_j/x_j=1/k\equiv {\frac {x_{{j}}}{y_{{j}}}}- \biggl\lfloor {\frac {x_{{j}}}{y_{{j}}
}} \biggr\rfloor =0
$$
I guess the double-ended implicative arrow would be suitable, for that matter I would very much appreciate someone explaining the difference between a statements being equivalent to another, and two statements implying one another.
 A: First, let's add parentheses where they should:
$$\forall (x_j,y_j) \in \mathbb N^{2}\, \big( \exists\,k \in \mathbb N\,:  y_j/x_j=1/k\equiv {\frac {x_{{j}}}{y_{{j}}}}- \biggl\lfloor {\frac {x_{{j}}}{y_{{j}}
}} \biggr\rfloor =0 \big)
$$
Now, what you want to use here is the symbol for the material or logical biconditional, which is the symbol used for the truth-functional operator (or connective) that combines two logic formulas into one logic formula. You definitely do not want to use the symbol used for logical equivalence, since that is a meta-logical symbol, i.e. a symbol used to make a claim about logic statements. 
Now, confusingly, some texts use the $\equiv$ to express the material biconditional, while others use the $\equiv$ to express logical equivalence. It sounds like your text uses the $\leftrightarrow$ for the material biconditional, and that is uses $\equiv$ for logical equivalence. If so, the you should definitely use the $\leftrightarrow$, and not the $\equiv$
Indeed, if you want to use logical equivalence, you need to have two logic statements, but you have only one statement here. Sure, there are two component parts of that statement: 
$$\exists\,k \in \mathbb N\,:  y_j/x_j=1/k$$
and 
$${\frac {x_{{j}}}{y_{{j}}}}- \biggl\lfloor {\frac {x_{{j}}}{y_{{j}}}} \biggr\rfloor =0 $$
but those are formulas, and not statements. Statements have truth-values, but these formulas have free variables, and thus we cannot assign any truth-value to them.
Now, there actually is such a thing as logical equivalence of logical formulas, which is defined as:
$\phi$ and $\psi$ are logically equivalent if and only if for any interpretation $I$, and any variable-assignment $v$ that sets the free variables in $\phi$ and $\psi$ to the domain of $I$, we have that $I[v] \vDash \phi$ iff $I[v] \vDash \psi$.
But, we don't have that here, since we are free to interpret any of the functions you are using here in any way we want (we could, e.g. interpret $-$ as addition, and $\lfloor \rfloor$ as square, just to pick something silly)
So, at best these statement are mathematically/arithmetically equivalent, but they are certainly not logically equivalent, and so you have:
$$\exists\,k \in \mathbb N\,:  y_j/x_j=1/k \not \equiv \forall (x_j,y_j) \in \mathbb N^{2}\,{\frac {x_{{j}}}{y_{{j}}}}- \biggl\lfloor {\frac {x_{{j}}}{y_{{j}}}} \biggr\rfloor =0 $$
