To fully understand this concept, you must first have a grasp on light's constant speed and frames of reference. All you need to know is that all frames of reference are considered to be valid in a system, unless they are accelerating, which they will not be in this case. Einstein's theory also proposed that the speed of light is constant in any frame of reference, and this must be taken as a given in order to explain this.
Imagine you're on a spaceship travelling at near-light speeds. If there were no windows on the spaceship, you would have no way of knowing if you were moving at all, therefore according to your frame of reference, you're standing still, and all the planets, stars, couches, etc are moving past you. To an observer on earth, or anywhere else, you are moving, and they are standing still. But when you shoot a beam of light ahead of your ship, both you and the observer see it moving at the speed of light, or c.
Now imagine your spaceship has some sort of laser that bounces off of one wall, and is recieved at the other by a timer which finds out how long it took the light to go from one end to another. If your spaceship is D wide, and the light travels at c, the amount of time it would take it to get from one side to the other again is 2D/c (See drawing): http://www.vvm.com/~wpeters/Images/h2g2images/r.jpg
But to an observer on Earth, you had moved forward from when the light was shot to when it was bounced. So by his view, the light travels in an angle throughout time, and by the Pythagorean theorem, you can calculate what that length will be. The length that the spaceship travels before the observer sees the light hit the mirror on the wall can be represented by L, and therefore, by the time he sees the light hit the receiver, the spaceship has travelled 2L, and the light, using the Pythagorean theorem, has travelled two times the square root of the quantity D squared plus L squared. (See next drawing): http://www.vvm.com/~wpeters/Images/h2g2images/o.jpg
This quantity divided by delta T, the time the observer measured, should yield c, so to solve for delta T and to find out how long it took the observer to see the light bounce, set this equation equal to c, square both sides, and solve for delta T, eventually getting the equation seen here : http://www.vvm.com/~wpeters/Images/h2g2images/e.gif
Putting in the very first variable we solved for, we can cancel out the 2D/c, and put the variable for the amount of time that you took to see it bounce back, in the numerator, yielding the following equation here: http://www.vvm.com/~wpeters/Images/h2g2images/e2.gif
Amazingly, according to this, the amount of time you spend on the spaceship divided by the square root of one minus the quantity your velocity squared over c squared, will tell you how much time has elapsed to a standing observer. Plugging in a few numbers, you can see that for every five years you spend in a spaceship travelling at 999/1000's the speed of light, 112 years pass to a stationary observer, meaning you have essentially travelled 107 years into the future! People who test this stuff have accelerated a short-lived particle to high speeds and observed it living longer, at least to them. Its high speed slowed down time for it (in comparison to the scientists) and allowed the particle to decay more slowly. As you may also have noticed, when you are travelling at the speed of light, this equation ends up with a zero in its denominator, yielding undefined, or infinity if you wish, meaning that most likely, the speed of light is unattainable. Eventually, time dialation may muddle things up quite a bit when our society begins higher speed travelling. Wars might be over before fleets even arrive there because of its effects, but now, as our cars travel around 90 km/hour, there isn't much to worry about.