I'm not sure what this is exactly asking Without using words of negation, write the meaning of : "f is not an increasing function"
I did:
$$"f\ is\ not\ an\ increasing\ function" \ \equiv\ "f\ is\ a\ decreasing\ function"$$
Is this what it's asking or am I completely missing it?
 A: No this is not true, for example any function from $\mathbb{R}$ to itself which increases on one interval and decreases on another is neither increasing nor decreasing.
It should be noted that "nonincreasing" is actually a perfectly valid adjective for describing a function, but that is probably not what it being sought after.
One way to write this, would to be to use the definition of increasing functions: a function $f$ is said to be increasing if $x>y\implies f(x)\geq f(y)$, or in other words $\forall x,y$ such that $x>y$, it follows that $f(x)\geq f(y)$.
The negation of this statement is therefore, $\exists x,y$ such that $x>y$, and $f(x)<f(y)$.
A: Well I'm guessing words of negation would be ones like not, did not, does not, was not, etc. So yes you have not used any words of negation in your answer. But it certainly is not correct. 
We are asked to negate the statement "$f$ is not an increasing function". Think about what should be satisfied for the statement to be wrong. Yes it would be wrong if $f$ is decreasing. But $f$ does not have to be decreasing. 
So your key word in the given sentence is not. See what you can do with it. 
A: What if you were asked to write the meaning of, "$f$ is an increasing function"?
I think the words "for all" might appear when the meaning of that term is written out in full.
Then you could use the fact that $\lnot\forall x.P$ is equivalent to $\exists\, x.\lnot P$
(but of course in your exercise you have to write $\lnot P$ without using the word "not").
A: The sine function on the whole line is certainly neither an increasing function nor a decreasing function, since it oscillates.  So the fact that a function is not increasing certainly does not mean that it is decreasing.
A: A common definition of increasing function, f, follows. 

For a, b in the domain, f(a)>= f(b) whenever  a >= b 

Relevant note: Strictly increasing drops the equality. 
So any function which breaks the definition of increasing is not increasing.
A: Alex Wong's answer from 40 minutes ago is essentially right.  The english construction is a tad bit awkward though.  Something like, the negation of f is increasing is:

there exists x,y element of R such that x <=y and f(x) > f(y)

If I was the professor and this was a 10 point question, I'd give 8 to 10 points for your answer.
Also, since you've got all the essential I'll ask about a picky note.  In some courses assuming the domain is all real numbers is incorrect.  Did the definition of increasing in your text or from your professor apply only to real numbers or to other types of ordered sets also?
If I was the professor, I would consider David K's incorrect because I think the intent of the question is to apply the concept of negation in practice, not merely with symbols.
Feel free to edit these into comments.
