Conditionals in definition of Strictly Increasing Function I have a question concerning the definition of strictly increasing function, that I cannot really figure out. The definition reads:

Definition: A function $f : \mathbb{R} \to \mathbb{R}$ is strictly increasing if $$\forall a, b \ ( a > b \Longrightarrow f(a) > f(b) ).$$

My question is: Why do we have just one conditional and not a biconditional?
That is, why the definition is not the following?

A function $f : \mathbb{R} \to \mathbb{R}$ is strictly increasing if $$\forall a, b \ ( a > b \Longleftrightarrow f(a) > f(b) ).$$

That's how I see it: we don't want to have that $a > b$, and $f(a) \leq f(b)$ (fine with me); but I don't see why we accept to have in principle that $a \leq b$, and $f(a) > f(b)$.
Any feedback is most welcome.
Thank you for your time.
PS: Of course, this is just a particular case of a problem I have with definitions that involve conditionals.
 A: Take for example the function $f(x)=\cfrac {x^2-1}{x-1}$. On every point except $x=1$ it equals $x+1$ hence, it is strictly increasing on the definition.
But, if you take the iff definition you have given: $ 2>1$ means $f(x)$ has to admit:  $f(2)>f(1)$ in order to be strictly increasing. which is a meaningless statement. 
I'm not saying you can't expand the definition to an "if and only if" definition. You can, but then it will be much longer and in the end it will give the same result as the shorter one.
There is no rule in mathematics against saying Unnecessary things, it's just a waste of time and a headache sometimes. 
A: I think that the trouble is with the "asymmetry" of $<$. 
If we use $↔$, we have that the definition of increasing will be:

$f(x) ≤ f(y) ↔ x < y$

and that of strictly increasing will be: 

$f(x) < f(y) ↔ x < y$. 

We have to consider that $P ↔ Q$ is equivalent to $¬P ↔ ¬Q$, and thus the last one is equivalent to: 

$f(x) ≥ f(y) ↔ x ≥ y$. 

The case when $x=y$ is not interesting (because for any function : $f(x)=f(y)$, if $x=y$) and thus we have that the new definition of strictly increasing amounts to :

$f(x) ≥ f(y) ↔ x > y$

that coincides with (non-strict) increasing.
