korters said:
I don't understand how a scalar can be negative.
Because it's defined that way.
Math is different from physics because you can choose what kind of rules you want to use as long as your reasoning is logical. Since math is so abstract, names we give things are pretty arbitrary. Real numbers are called real because they were originally considered to be the values that lengths could take in the real world.
A scalars is a scientist's term for a real number. The two are (in most contexts) completely synonymous. That's the reason they can take on positive, negative, and zero values. If you ask 'why', the answer is because that's how we've defined it.
It seems you are confused about the relation between scalars and vectors. Vectors are funny things. You can represent them in a few different ways:
Cartesian Coordinates: (x, y) is a point given two real numbers x and y. Move x units to the right then y units up.
Polar Coordinates: (r, theta) is a point in space given by a real number r, representing the distance from the origin ("r" for radius), and a real number theta, representing the angle by the triangle between the point, the origin and the point at (1, 0) on the x-axis.
Cartesian coordinates are very simple, and each point has exactly one representation. If x != x' or y != y', then (x, y) != (x', y'). Very simple.
But polar coordinates are a bit more confusing. First, r can be positive or negative. When it's negative, it's the same as doing a 180º change in direction. That means that (r, theta) = (-r, theta + 180º).
It actually gets worse. Angles are not unique either. If you spin 360º, you wind up looking the same direction. So (r, theta) = (r, theta+360º).
For this reason, it's sometimes convenient to put restrictions on r and/or theta. In this situation, r would be restricted to a non-negative real and theta would be restricted to a real between 0 (inclusively) and 2pi (exclusively). And you still have to be careful with the 0 vector! (0, theta) = (0, theta') for all theta and theta', so we might also want to put a third restriction that says if r = 0, theta must also be 0.
These restrictions are a pain sometimes, though, so depending on the problem, we might just let r and theta be whatever, and correct for it whenever we do a comparison. It's pretty easy to convert any vector (r, theta) to a "normalized" one which obeys the restrictions above.
But the take home lesson is, in math, you can define things however you want as long as you keep the definition. You'll run into this in other areas as well, such as the infamous 0^0 issue. Everyone agrees on what x^y means when x > 0, but what about for x=0 and y=0? Does 0^0 = 0? 1? Or is it undefined entirely?
The truth is that ^ means whatever the hell you want it to mean! If you want it to equal 1, polynomial equations come out nicely. But then the function f(x) = 0^x is not continuous. If you want 0^0 to mean 0, then g(x) = x^0 is not continuous. If you want to make it so ^ is not defined for 0^0, then both f and g are continuous at all points in their domain, but their domain is now R - {0}, making ^ a partial function. It's a case of pick your poison.
(I really hate it when "brainiacs"try to explain something using terms and mathematical ideas that are way beyond my education level in such a way that doesn't even coherently make any sense).
Don't make such negative assumptions about people when you ask them for help. It doesn't reflect well on you to say things like this. Sometimes the help you get isn't helpful, but you just ask for a more appropriate explanation, and hopefully someone on the other end will help guide you.