Tuesday, 14 October 2014

geometry - Diagonal ladder length

Let $ ABCD $ be a square with side length $ a $. Let $ s $ be a staircase from $ A $ to $ C $ with total length $ l $ and number of steps $ n $. It consists of perpendicularly alternating lines of length $ \frac{a}{n} $, as pictured here.



We see that $ l $ can be expressed as follows:
$$ l = \frac{a}{n} \cdot n + \frac{a}{n} \cdot n = \frac{a}{n} \cdot n \cdot 2 = 2a $$

and as such stays constant at $ 2a $.



Now let us imagine that the amount of steps is infinite, e.g. $ n = \infty $. Per definitionem, the staircase should now be the diagonal of the square with length $ l=a\sqrt{2} $ according to Pythagoras. This is paradoxical! According to the equation pictured above, it should have the length $ 2a $, not $ a\sqrt{2} $.



My question is:



Does $ l $ equal to $ 2a $ or $ a\sqrt{2} $?

No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...