- #1
climbhi
Alright, I have a question that I know is very dumb, but despite my best efforts I've been unable to come up with an answer that satisfies me. Here it is:
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).
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).