Why is friction force negative in ice skater problem? 
A 68.5 kg skater moving initially at 2.40m/s on rough horizontal ice comes to rest uniformly in 3.52s due to friction from the ice. What force does friction exert on the skater? 

I am not really asking about the answer here, because I can calculate that, but more of an explanation. 
We find the acceleration with $\frac{V-V_0}{t}=a$, so the acceleration is: $-0.68$. 
$F=ma$, so force is $-46.7$ Newton, or N. 
First of all, why is it negative? Does the guy skating put an equal force on the ice? So the skater extends a force of 46.7 N on the ice, and the ice extends a force of -46.7 back at the skater? Why does this sound like the Normal force? 
I am getting really confused and I simply don't understand it, but I got the math right by just plugging in the numbers. 
 A: When the skater is on the ice, friction stops him/her in 3.52 seconds as you said. The molecules in the skates rub against the molecules in the ice, and the ice molecules absorb some of the skate molecules's energy, slowing the skater down. The reason the force is negative is because the friction is acting in the opposite direction of the skater's motion. The skater does apply the same force back towards the ice, this causes the ice to warm up. 
A: It's very common to get mixed up about signs. The only recommendation I can give is to establish a clear sign convention and stick carefully to it. To show what I mean let's consider your skater:

I'm going to use the convention that positive is to the right and negative is to the left. remember that quantities like velocity and acceleration are vectors, so they have a direction as well as a magnitude. According to my convention a vector pointing to the right is positive while one pointing to the left is negative.
The skater's velocity points to the right. We know the skater is slowing down, so the skater's acceleration points to the left. That means the acceleration must be negative. We know the force on the skater is related to the acceleration of the skater by:
$$ \vec{F} = m \vec{a} $$
and since mass is positive that must mean that $\vec{F}$ is negative, just as you concluded.
A: It's all a matter of how you've chosen your coordinate system. Think of how the velocities were calculated. They essentially looked at two points in time and the two corresponding points in space. If your positive direction is to the right, and the guy has advanced to the right, then its a positive number for velocity.
Force carries the same logic here. The only way you can have a decrease in velocity in this coordinate system is if a negative force is acting on you. If you were to draw a vector arrow that force would point towards negative infinity.
If you know about the work-energy theorem then a negative force acting across a positive displacement is doing negative work, which corresponds to a decrease in kinetic energy.
The guy does exert an equal and opposite force on the ice. This might sound to you like the normal force perhaps because it is a contact force. But keep in mind 'normal' is another word for orthogonal or 90º. So the normal force has to be 90º from the surface.
A: What everyone said, but also: the way the question is phrased, the model it uses, is a 1-dimensional model. Imagine a point travelling along the real number line/the x-axis on a cartesian plane. The model in the problem assumes there is no vertical movement, or 'side to side', but only 'pure' 1-D movement along a straight horizontal line. There is only that, and the mass of the skater, in the 'world' of the problem.
In the motion of a cannonball, you can separate the x-movement (horizontal) - relatively constant, slowed by air resistance etc - from the y-movement (vertical) - accelerating downwards due to gravity. In this skater problem, all y-components are assumed to be 0, or ignored, not even considered. (likewise the 'side to side' z-components)
It seems your confusion comes from not realizing that this is purely a question about the x-components. Friction will be either a force to the left (negative) or to the right (positive). Downward forces etc don't come into the model. That would be another problem (Maybe he breaks the ice below a certain velocity..) 
I'm no physics expert, but I hope that was helpful.
A: Other people have answered your questions about the signs; I want to address your comment, "why does this sound like the normal force?"
By Newton's third law, pairs of forces always come in equal and opposite pairs.


*

*If the ground exerts an upward normal force on you, you exert a downward normal force on the ground.

*If sliding on the ground exerts a leftward friction force on you, then you exert a rightward friction force on the ground. (Don't believe this? Rub your hands together and note the force on each hand.)

*If Earth's gravity pulls down on you, your gravity pulls up on the earth.


So on the face of it, there's no reason the friction force here is being any more 'like the normal force' than any other force. Except there actually is something that makes the normal force special!
In order to keep you from falling through the ground, the normal force must exactly counteract your weight. That is, it must be equal and opposite to the gravitational force. But this is totally different from the Newton's third law statements above -- the normal and gravitational forces are different forces, not a 3rd law pair.
In summary: you said the equal-and-oppositeness of the friction force here reminds you of the normal force. However, all forces come in equal-and-opposite pairs. The reason you specifically thought about the normal force was because it happens to be equal and opposite to the gravitational force, but this is for a different reason.
A: The force/ acceleration of the skater comes out as negative because: the force or acceleration is in the opposite direction of motion.
You can also think of it like this, frictional force always tries to oppose relative motion. In this case the skater is moving forward w.r.t the ice and the ice is moving backwards w.r.t the skater; so the ice exerts a backwards frictional force on skater and the skater applies an equal frictional force on the ice towards front. 
Considering the mass of ice to be very large its acc. is negligible.
Just a little clarification, the ice does not extend -46.7 N back on the skater; if you are  prefixing the force with (-), then you need not say backwards or if you say backwards you need not put the (-) sign before the force. It will be best if you draw the co-ordinate axes right from the begining, that would minimise your confusions.
