0

I am looking for a formal proof for the following problem:

You travel from point A to point B on a right triangle only along its legs. For a 3 4 5 right triangle you would travel a distance of seven. Now imagine going in horizontal and vertical lines to point B while remaining within the right triangle (basically steps going from point A to B). You would again travel a distance of seven: Figure. My question is once you start decreasing the size of your vertical and horizontal lines until they are infinitely small, when do you travel the length of your hypotenuse (for this example five) instead of the length of the legs (for this example seven)? Does this also mean at some point you travel a distance of in between the length of both legs and hypotenuse? Figure 2

Snakeryan
  • 1
  • For a special case of this, suppose you divide the base into $n$ intervals and the height into $m$ intervals. Then the distance travelled, regardless of what order you take upwards/rightwards steps, is still $n(3/n)+m(4/m)=7$. – Semiclassical Feb 05 '21 at 19:57
  • I have asked a similar question, as have others, and the answer has always been that there is no final point where $ax+bx=c$, at least within mathematics as we know it. – poetasis Feb 05 '21 at 22:14
  • Hi, thank you so much for the help! Could I formally prove that it can never equal five with using the epsilon-delta definition of a limit? – Snakeryan Feb 06 '21 at 00:40

1 Answers1

1

This example illustrates that limits$^1$ need not commute, specifically here:

The limit of the lengths is not the length of the limit.

$^1$ Note that length of a curve is itself a limit

Hagen von Eitzen
  • 374,180