Clarification of the concept "less resistance means less heating" in a wire So my textbook says that the reason cables that are suppose to carry high currents, are thicker that those that are meant to carry lesser current, is that "less resistance (of the wire) means less heating..."? Is this even true? 
Isn't CURRENT the reason wires heat up? If we decrease resistance, more current flows, and that should produce more heating!
 A: A secondary effect that the other answers don't talk about: thicker wires have more surface area through which to dissipate the warmth generated by the electric current running through the wire.
If you double the diameter of a wire, there will also be twice as much surface area, so twice as much heat can dissipate, given the same wire temperature. If the wire is in thermal equilibrium with its surroundings, the heat dissipated must be the same as the power generated inside the wire, so doubling the diameter of the wire will also double the power generation that the wire can handle while staying at a particular temperature. Alternatively: the double-diameter wire will only be half as much hotter than the environment than the single-diameter wire, if both wires generate the same amount of power.
E.g. if the environment is 300K, and the single-diameter wire is 330K, then the double-diameter wire will only be 315K, at the same power generation.
This effect is inversely linear in the diameter of the wire, rather than the quadratic behaviour of the actual power being generated inside the wire, so it's a smaller effect in general. But if power generation is proportional to $1/r^2$ (see the other answers on why this is the case) and temperature difference with the environment at constant power generation is $1/r$, then total temperature difference with the environment should be $1/r^3$. (Here, $r$ is the radius of the wire.)
A: They’re describing the situation where the wires are carrying power to a load. It’s the load that (mostly) determines the current in the wires leading to it. 
A $1200$W oven on $120$V needs $10$A. 
Once the load has determined the current, the heat in the wires is given by their resistance via $I^2 R_{wire}$. 
A $0.02$ ohm wire to the oven will have $2$W of heat; a $0.01$ ohm will have less: $1$W.
That difference in wire resistance doesn’t change the current much because the current is really controlled by the ~$10$ ohm heater resistance. But it changes the wire heat a lot. 
A: You can regard a power supply in series with a load as a constant current source. At a constant current, the $I^2R$ losses in the cable reduce with resistance.
However, your concerns are valid in the event of a short circuit, and this is a thing in the design of resilient power distribution networks. Thicker cables mean higher short circuit currents, and it's not a question of if, but when, a short is going to happen somewhere. An effort is therefore made not to over-size any feeder or transformer with respect to its fuses, so that when the inevitable short occurs, the heat is spread out over a large part of the distribution network, rather than being concentrated in the lowest capacity parts.
A: The rate of heating of a wire is the power dissipated in the wire which is
$$P=I^{2}R$$
The resistance of wiring is much less than the resistance of the loads that the wiring supplies current to. Therefore the size of the wire has little effect on the current supplied to the loads (within reason). In other words, we can consider the current to be constant for different wire sizes within a limited range of sizes.
For a fixed current load ($I$ = constant), the greater the wire (conductor) resistance the greater the heating of the wire, the less the resistance the less the heating of the wire. So a larger conductor for a fixed load current produces less heat because its resistance is lower. 
Finally, overheating of undersized conductors is a concern because it can cause the failure of the insulation on the conductors due to melting or long term thermal degradation if temperatures exceed the insulation temperature rating. Since insulation may be relied upon to reduce the risk of fire and electric shock, overheating of the insulation increases the risk of fire and electric shock.
Hope this helps.
A: Well as far as i can i see, you cannot change the current passing through the wire. This is because they are "suppose to carry high current"(as per the question) . So the only possible way to reduce heat losses is to decrease resistance(heat loss is proprtional to resistance of wire).
That being said transmission cables for long distance generally operate at high voltages(not high current) to minimize heat loss
A: Not necessarily. This is because it is not always the current rather a coupling of both which causes the heating effect to occur. What this means is that for wires carrying high currents, lower resistance leads to less dissipated power, according to P=I^2*R, where I is the current in the wire, and R is the resistance. This means that wires which are thicker, which have lower resistance than thinner wires provided other factors remain constant, will not have as much heat produced, since lesser power is dissipated. Therefore, it need not always hold true that higher current leads to more heat being generated.
A: 
Isn't CURRENT the reason wires heat up? If we decrease resistance, more current flows, and that should produce more heating!

The key thing you're missing here is that current is (almost) constant and is (almost) not dictated by the wire.  The "ideal" wire has no resistance and the current through it is solely set by the load you have connected at the other end of the wire.
In the real world, that's not quite true.  Sometimes we really do need to consider the effect of the wire - for example, if you have 100km of transmission line carrying power from a power station to a nearby city, losses in the wire will certainly become significant.  Also if you put AC down a wire, the AC behaviour as frequency increases is certainly something you need to think about.
In practise though, the point of choosing wires that are thick enough is that the resistance in the cables is very small compared to whatever load is connected.  We can therefore usually ignore losses in the cables when we think about the circuit.
The only time where your statement is true is if the wire shorts out the power supply - for example, if you happen to connect a wire directly from live to neutral or earth in a mains cable.  In that case there is no other load, only the resistance of the wire.  A lot of current flows, and the wire heats up very very fast!  In your house, the mains supplies go through circuit breakers which trip if too much current is drawn; and depending on where you are in the world, your mains plug may also contain a fuse.  The purpose of these is to cut power rapidly in the event of an electrical short, so that the wires don't heat up and set your house on fire.
A: 
If we decrease resistance, more current flows

You're talking about the resistance of the transmission wires that carry electric current from the generating station (or other power supply) to the load. Normally those things are sized such that the power dissipated in the transmission line is much less than the power dissipated in the load.
Yes, decreasing the resistance of the transmission line will increase the total current IF the load is purely resistive*, but even if you could decrease it to zero, it only would increase the total current by a small amount because the resistance of the load dominates the equation.
$I_{total} = \frac{V_{supply}}{R_{line}+R_{load}}$, where $R_{line}\ll R_{load}$
At the same time, decreasing the transmission line resistance relative to the load resistance will increase the fraction of the total power that is delivered to the load, which generally is what we want.
$P_{load}={I_{total}}^2R_{load}$, and $P_{line}={I_{total}}^2R_{line}$

* Some loads, including motors, and electronic equipment with switching power supplies, may behave differently from a resistor (i.e., do not obey Ohm's Law.)
