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.

Definitions:

**c** = speed of light

**v** = speed of rocket as measured on Earth

**t _{Earth}** = time interval between clicks as measured on Earth

**t**= time interval as measured on board the rocket

_{rocket}
The light beam travels from one mirror to the other across a path of length **ct**_{rocket} *as measured on board the rocket*. While it does so, the rocket moves forward by the distance **vt**_{Earth} 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 **t**_{rocket} must be less than **t**_{Earth}. 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:

(ct_{rocket})^{2} + (vt_{Earth})^{2} = (ct_{Earth})^{2}.

Rearrange:

(ct_{rocket})^{2} = (ct_{Earth})^{2} - (vt_{Earth})^{2}.

Divide each term by c^{2} :

(t_{rocket})^{2} = t_{Earth}^{2} - (v^{2}/c^{2}) t_{Earth}^{2} = (1 - v^{2}/c^{2}) t_{Earth}^{2}.

Take the square root of both sides :

t_{rocket} = t_{Earth} √(1 - v^{2}/c^{2}) = t_{Earth} √[1 - (v/c)^{2}]

The square root factor is less than 1, so **t**_{rocket} (any time interval on the rocket) must be less than **t**_{Earth} (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 **t**_{rocket} would be only 14% of **t _{Earth}**. 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.