
Pythagoras Puzzle
Hi!
So, I thought of something the other day that my friend once told me when I was in high school. I still remember it, but can't remember what he taught me about it, and can't find anything on Google. I thought I'd post it here to see if anyone had any thoughts.
Basically, imagine you had a square with sides equal to $\displaystyle x$, and you want to travel diagonally from the top left corner of the square, $\displaystyle A$, to the bottom right, $\displaystyle B$.
Now, the shortest route would be to go diagonally, and you can calculate that from Pythagoras' theorem as $\displaystyle \sqrt{2}x$.
Alternatively, you could go across from $\displaystyle A$, to the top right corner, then down to $\displaystyle B$, and will obviously have travelled $\displaystyle 2x$.
Now, instead, imagine going halfway between $\displaystyle A$ and the top right corner, then down to the centre, then across to the centre of the right hand side vertical line, then down to $\displaystyle B$. You'll again have travelled $\displaystyle 2x$. If you then repeated this, but going only a quarter of the way, then down $\displaystyle \frac{1}{4}x$, then across $\displaystyle \frac{1}{4}x$, and so on, you'll again travel $\displaystyle 2x$ towards $\displaystyle B$. If you repeated this to infinity, your path would eventually tend towards the hypotenuse, but the distance you travel would still be $\displaystyle 2x$ according to the pattern I described above, when it should tend towards $\displaystyle \sqrt{2}x$.
I wondered if anyone had any ideas about that. My friend was really really into maths, so it might be a famous problem that people have thought about, but I can't seem to find any information about it.
Thanks.

It's a very nice example.
The horizontal distance x is divided into n equal parts.
The vertical distance x is also divided into n equal parts.
Hence the distance travelled, for any number of subdivisions n is
$\displaystyle \displaystyle\ n\left(\frac{x}{n}+\frac{x}{n}\right)$
You will always know the distance travelled,
however, if n is very large,
another observer may not spot the divisions for extremely large n
and conclude that the distance travelled is the straight diagonal.

That path is never same as the diagonal. No matter how fine it is made it will never be differentiable.

Sure it would! If the path you travel is equal to
$\displaystyle n\left (\frac{2x}{n} \right)$
and $\displaystyle n$ tends towards infinity, then it would become the diagonal. This is the nature of infinity.

Actually no, thinking about it. You can also zoom in infinitely and you'd get the same picture of steps. So you're right.
Interesting thing to think about though.