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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} $?

rogerl
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MedwedianPresident
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  • The diagonal length should be $a\sqrt2$ not $\sqrt{2a}$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Dec 03 '17 at 16:24
  • The paradox lies in your choice of the Cantor module; a ring or a group. You have used a group splitting field for the length, with respect to the ring of its area. You'll find the induction equal to their quotient and would thus require a modular order of 2 for each a | l (mod n) i.e a vacuous residue of order $ 2^{n} $. This method will be satisfying for defined values, however complete analytical continuation takes you to p-adics. – McTaffy Dec 03 '17 at 16:27
  • The steps distance will always be $\sum_n 2a/n=2a$ and the diagonal will always be $\sum_n \sqrt{2}a/n=\sqrt{2}a$ for any $n$. There is not any paradox at all. – Brethlosze Dec 03 '17 at 16:27

1 Answers1

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This is not paradoxal. The total staircase distance is, as you have shown, of length $2a$. However, this does not mean that the diagonal distance is the same.

No matter how small a right-angled triangle you take (in this case as $n \to \infty$), the hypotenuse (diagonal) will always be shorter than the sum of the other two sides (staircase).

Hence for an isosceles right-angled triangle of length $\frac an$, the hypotenuse is $$\sqrt{\left(\dfrac an\right)^2+\left(\dfrac an\right)^2}=\dfrac an \sqrt2.$$ Multiply this by $n$ and you get $a\sqrt2$, which is not equal to $2a$.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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