Imagine a rocket moving at speed v away from the Earth. Aboard the rocket, two mirrors face each other so that a beam of light can bounce back and forth between them along a line at right angles to the motion of the rocket. This may now be thought of as a clock, where each bounce of the light is a "tick" of the clock.
c = speed of light
v = speed of rocket as measured on Earth
tEarth = time interval between clicks as measured on Earth
trocket = time interval as measured on board the rocket
The light beam travels from one mirror to the other across a path of length ctrocket as measured on board the rocket. While it does so, the rocket moves forward by the distance vtEarth as measured on Earth. During this same interval the light beam must therefore move across the longer path ct as seen from Earth.
Recall, speed (v) = distance (d) / time (t)
Therefore, d = vt
But the speed of light c is the same for all observers. Therefore, a glance at the triangle shows that trocket must be less than tEarth. In other words, corresponding time intervals are shorter on board the speeding rocket than on Earth.
Here's the quantitative solution, using only a few lines of simple algebra and the Pythagorean Theorem for a right triangle:
(ctrocket)2 + (vtEarth)2 = (ctEarth)2.
(ctrocket)2 = (ctEarth)2 - (vtEarth)2.
Divide each term by c2 :
(trocket)2 = tEarth2 - (v2/c2) tEarth2 = (1 - v2/c2) tEarth2.
Take the square root of both sides :
trocket = tEarth √(1 - v2/c2) = tEarth √[1 - (v/c)2]
The square root factor is less than 1, so trocket (any time interval on the rocket) must be less than tEarth (the corresponding interval on Earth).
The effect is enormous for speeds approaching that of light. For example, if you travel at 99% the speed of light, then 1 - (0.99)2 = 0.02, the square root of this is 0.14, so your trocket would be only 14% of tEarth. So if a space traveler left Earth at 99% the speed of light and returned 100 years later as measured on Earth, she would be only 14 years older by her biological clock.