- #1

climbhi

We know the shortest distance b/n two points is a straight line. So, if we have a right triangle with legs of unit length, the shortest distance between the endpoints is the hypotenuse (which will have length [tex]\sqrt{2}[/tex]).

Okay, now let's walk the distance b/n the two points in several different ways:

1.) Walk 1 unit East, then 1 unit North.

**Total distance**= 1 + 1 =

**2**.

2.) Walk 1/2 unit East, then 1/2 unit North, then 1/2 unit East, then 1/2 unit North.

**Total Distance**= 1/2 + 1/2 + 1/2 + 1/2 =

**2**.

3.) Walk 1/3 unit East, then 1/3 unit North, 1/3 unit East, then 1/3 unit North, 1/3 unit East, then 1/3 unit North.

**Total distance**= 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 =

**2**.

I'm sure you see where I'm going by now. So, my question is, why doesn't the sum of this "stair step" method approach [tex]\sqrt{2}[/tex] as your step size approaches 0? Is there something fundamentally different about a diagonal line? I just can't see what the difference is b/n a "stair step" of infinitesimal step size and a diagonal line...

This has been bugging me for a while, so someone please set me straight (pun intended).