You don't need to forward bias the B-E junction. Whether you do depends on what you want the transistor to do. To keep it off, you want to not forward bias the B-E junction. As for the C-B junction, keeping that reverse biased is fundamental to how BJTs operate.
A BJT is basically a reverse biased junction that can be made selectively leaky. You apply a voltage accross C-E and with the base open nothing happens. The reverse biased junction doesn't allow any (except for small leakage we will ignore) current to flow. However, the special property of a BJT is that a little current thru the base messes up the insulating capability of the reverse biased junction. The gain, and hence the useful properties, of a BJT come from the fact that is only takes a little current to muck up the reverse biased junction such that it allows a lot more C-E current to flow. This ratio of C-E current to B-E current is the basic gain of the transistor. It can be as low as 5-10 in big mongo power transistors and 100s in high gain signal transistors.
The B-E junction also looks like a diode to the external circuit. It will have a forward voltage drop when conducting just like a regular diode. In silicon, this is 500-750 mV for most non-extreme applications.
If you want to use the transistor as a switch (either as full off or full on as you can make it) then you have to make sure there is no base current in the off case, and plenty enough to support the desired collector current in the on case. Driving the base to the emitter voltage is a good way to make sure the transistor is off. To turn it fully on, you need to provide at least 1/gain of the desired collector current.
In other cases, a BJT might be used in "linear" (it's often rather non-linear, but this is the term used to mean in-between mode or not-switch mode) mode, like a audio amplifier. In that case you want to always keep it somewhat on and have the input signal change its operating point. If done right, this can amplify the signal. Different configurations give you voltage gain, or current gain, or some combination. In these cases, biasing the transistor refers to keeping it somewhere in the middle of the operating range so that a little input signal can change the output both ways. Biasing is basically setting up the DC operating point.
In a nutshell, bipolar junction transistors work because of the physical geometry of the two junctions. The base layer is very thin, and the charge carriers that are flowing from the emitter to the base do not recombine right away — most of them pass right through the base altogether and enter the depletion region of the reverse-biased base-collector junction. Once this happens, the strong field in this region quickly sweeps them the rest of the way to the collector terminal, becoming the collector current.
Best Answer
Ground is just a point of your chosing to call the 0 reference for other points. Voltages can be negative, especially since the choice of what to call 0 is arbitrary.
Read the book more carfully, and you'll probably see this section was talking about the common base configuration. In that case the base is usually held at a fixed voltage, the input is the current drawn from the emitter, and the output is the collector. Common reasons for using this configuration are:
The current transfer ratio of a common base stage is 1 at first approximation. The emitter current (the input) is the collector current (the output) plus the base current. But the ratio of collector current to base current is the gain of the transistor. For high enough gains, the additional base current in the emitter current is small enough to ignore.
For example, let's say the gain of the transistor is 50. In this example, the output of the common base (the collector current) is 50 mA. That means the base current is 1 mA and the emitter current (the input) is 51 mA. The current transfer ratio in this example is 50 mA / 51 mA = .98. Like I said, basically 1.
In general:
current transfer ratio = gain / (gain + 1)