How To Really Eliminate The Staircase Paradox?

The staircase paradox has made me think a lot lately. Ever since I wrote the article “Can You Really Solve The Staircase Paradox?”, there have been quite a few hard technical discussions in the comments section.

When I originally wrote that article, I aimed to avoid a strict mathematical approach and keep things simple. By doing this, I thought it would make the topic accessible for anyone (which, in fairness, turned out to be the case). Having said this, as the comments and criticism grew more and more technical, I realized that the original article was ill-equipped to do justice to the mathematical finesse behind this paradox.

In this article, I aim to address the technical challenges behind this phenomenon. To start, I’ll briefly recap what this paradox is about. Then, I’ll report some of the common technical push-backs/criticisms that the original article received and give my take on them. Finally, I will employ a rigorous mathematical approach to demonstrate how to really eliminate this paradox.

In doing so, I hope to address all open questions and concerns. If this is not the case, consider this article a technical conversation starter. One way or another, science shall move forward.

This essay is supported by Generatebg

Recap — What is the Staircase Paradox?

The staircase paradox is a geometrical problem in which you are trying to reach the top of a staircase from a diagonally opposite starting point. The path you are to trace is up and along the stairs. It is important to note that we are interested in the distance covered along the stairs in both the horizontal direction as well as the vertical direction.

To aid convenient calculations later on, let us consider a staircase that is 4 Centimetres (cm) long and 4 cm high. Each equidistant step is (4/n) cm long and (4/n) cm high, where n is the total number of stairs. Now, we are interested in the total distance covered by the stairs (both horizontally and vertically). Henceforth, I’ll refer to this distance as the total length of the stairs.

Here, you can see that the leftmost staircase has 8 stairs. The centre staircase has 16 stairs, and the rightmost staircase has 32 stairs. Although we have doubled the number of stairs twice from 8, we see that the total length of the stairs remains constant at 8 cm, which is explained by the following generalized equation for any number of stairs (n):

Total length of stairs = [Sum of horizontal components] + [Sum of vertical components] = [n*(4/n)] + [n*(4/n)] = 4 + 4 = 8 cm.

This equation also shows that the total length of the stairs is seemingly independent of n.

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What Happens at The Limit of Infinity?

Something that you might have already taken note of is the fact that as n increases, the stairs start to resemble a line. It is important to note here that they only resemble a line as long as n is very large, yet finite. In fact, they are arguably a bad approximation of a line at this stage.

Although the total length of the stairs is seemingly independent of the number of stairs (n), something very peculiar happens when we enforce a limit of infinity. As n tends to infinity, the stairs seem to transform into a straight line, which is also a hypotenuse to the right-angled triangle that represents the staircase at this point.

When we calculate the length of the hypotenuse using the Pythagorean theorem, we get a length of 4√2 cm instead of 8 cm. How did we jump from 8 cm to 4√2 cm? This is essentially the problem statement of the paradox.

Total length of the stairs = √(4² + 4²) = √(16 + 16) = 4√2 cm.

Now that we have covered the basic problem statement, let’s jump directly into some of the criticisms that the previous article received from readers and math enthusiasts.

Comparing Apples and Oranges

alvaro ferreira suggested that this is a pseudo paradox because we are comparing apples with oranges. According to his argument, on the one hand, Curve A is the composition of n-segments of the length and the height (4/n and 3/n) of the triangle. On the other hand, Curve B is the union of n-segments of the hypotenuse (5/n) of the Pythagorean triangle. Please note that the numbers quoted here vary because different numbers were used in the original article.

“For sure (and naturally), the corresponding lengths ought to conduct to different values. QED”

— alvaro ferreira

Here is another argument along a similar vein:

“What paradox?

You have two measures of the length after n divisions. By one measure the length is n*7/n = 7, the other is n*5/n=5. These numbers don’t change as n goes to infinity. In other words

lim n->infinity (n*7/n)/(n*5/n) = 7/5, not 1.

Your paradox is akin to arguing that:

7×0=5×0 proves that 7=5.”

– Roeland Rengelink

My favourite critique came from brian vant-hull

“Your segments never approach a tangent to the line, unlike Archimedes approximation of the circumference of a circle.”

— brian vant-hull

This is my favourite critique because it knocks at the door of the core issue with the paradox. It is a great start that I would have liked to build further upon. brian vant-hull, however, concludes by saying that the fact that the stairs do not tangentially approach the hypotenuse is sufficient to say that the lengths will never converge. According to brian vant-hull, this can be considered a sufficient solution to the paradox.

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Summarising Critique and Laying Foundations for Proof

All of these critiques can be summarized in three points:

1. The so-called staircase paradox compares two different measurement systems (metrics) that are bound to produce different lengths. — C1

2. The stair-based measurement system (metric) does not reduce the error term, that is, the total length of the stairs does not approach the length of the hypotenuse. — C2

3. The approximation of the hypotenuse only works accurately if the perimeter function approximating it does so tangentially to the hypotenuse. This way, the error term can reduce to zero as n tends to infinity. With stair-like sawtooth perimeter functions, this is not possible. — C3

Henceforth, I will be referring to these points as C1, C2, and C3 respectively. Furthermore, I will be referring to the stair based measurement system as M1 (metric-1) and the hypotenuse based measurement system as M2 (metric-2).

Laying the Foundations for Proof

In order to challenge all of the critique points listed above, I have to somehow manage to find a way for M1 to converge smoothly to M2 as n tends to infinity.

If I accomplish this, the following results would occur:

1. M1’s convergence to M2 as n tends to infinity would disprove the claim that the M1 is bound to produce a different length as compared to M2. That is, c1 would be disproved.

2. M1’s convergence to M2 as n tends to infinity would mean that the error term in approximation diminishes to zero. That is, C2 would be satisfied.

3. M1’s convergence to M2 as n tends to infinity would mean that tangential geometry is not necessary and sufficient for accurate approximation. That is, C3 would be disproved.

4. Finally, M1’s convergence to M2 as n tends to infinity would also establish a way to eliminate the staircase paradox — our main goal here!

Now that I have laid the foundations for potential proof, let us see where I get with this.

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Flipping the Staircase Paradox on its Head

I’ll start by dissecting the geometrical anatomy of each stair in relation to the hypotenuse. To do this, let us first zoom into a section of the stairs and then rotate this section such that the hypotenuse appears horizontally to us.

Now that we are all zoomed in, note that each stair forms 2 triangles with 3 measurable quantities each: the base (from the hypotenuse) — b, the height perpendicular to the hypotenuse — h, and the one remaining side of the triangle of length — l.

The base of the triangle is given by the length of the hypotenuse divided by 2n. This is because we have split each stair into two equal triangles. There are 2n base segments in total.

b = (4√2)/2n cm

Using geometric symmetry, we can prove that the height of the smaller triangle is the same as its base.

This is because, this triangle forms one quarter of a projected square. When reflected, the base and height form the diagonals of the square.

h = b = (4√2)/2n cm.

However, there is a subtle but important detail we need to note here. The base b is solely a function of M2 (the hypotenuse based measurement system) and n (total number of stairs). The height h is a function of M1 (the stair based measurement system), M2, and n. The weird thing about the total length of the stairs (henceforth denoted as M1_n) is that it need not be 8 cm long; 8 cm is just a special case. It can be both bigger than 8 cm and smaller than 8 cm (for any given n) depending upon what we choose as the M1 perimeter function.

To take this effect into consideration, let us define ‘s’ as a variable (s is restricted to positive numbers only) to represent variable height (note that variable height here does not refer to ‘h’, but rather the fact that sum of all h terms for any given n is variable depending upon the choice of the perimeter function).

I take this step is to generalize the solution for any M1 stair-like sawtooth perimeter function we may choose. Consequently, we obtain the following height:

h = s/2n cm.

Now, using the Pythagorean theorem, we can calculate the length of the side l as follows:

l = √[((4√2)/2n)² + (s/2n)²] = (1/2n)* {√[(4√2)² + s²]} cm.

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Computing M1_n — The Total Length of Stairs

Now, we know that the shortest length that connects the bottom end of the staircase to the top end is the hypotenuse (4√2 cm). Therefore, by logical deduction, the path along the stairs has to be longer than the hypotenuse. That is, M1_n has to be greater than M2_n, where M2_n denotes the length of the hypotenuse for n stairs.

In order to calculate the total length of the stairs (M1_n), we need to multiply the length ‘l’ we just calculated by 2n since we have 2n such segments.

M1_n = (2n)* (1/2n)*{√[(4√2)² + s²]} = √[(4√2)² + s²] cm.

If 0 < s < infinity, then 4√2 < M1_n < infinity

It turns out that the total length of the stairs, M1_n, does not depend on n. This is not a surprising result. The ‘s’ term in there can be considered the error term. For an ‘s’ value of 4√2 cm, we get the total length of the stairs for any n value as M1_n = 8 cm (the special case).

The fact that M1_n is not dependent on n goes to show that whenever n increases, the height h drops by the perfect proportion to keep the total length M1_n constant. This also seems to align with the argument as to why M1 and M2 never converge.

Bounded Area and Boundless Perimeter

We can summarise our mathematical findings so far into 3 points:

1. M1_n = √[(4√2)² + s²], where 0 < s < infinity and 4√2 < M1_n < infinity

2. M1_n does not depend on n.

3. Height of each stair, h = s/2n

The first significant point to notice here is that M1_n is greater than 4√2 for any n as long as ‘s’ is positive.

The second significant point is that the area under the stair-like M1-perimeter function would approach the area of the triangle as n increases.

The third significant point to note here is that M1_n can be as large as we want by defining an ‘s’ as large as we want (both have an upper bound of infinity).

Consequently, the area of the stair-like sawtooth perimeter function is bounded by the minimized error term plus the area of the staircase triangle (that is completed by the hypotenuse), whereas its perimeter is boundless (depending upon our choice of ‘s’). This is a vital point to consider when trying to eliminate this paradox.

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How to Really Eliminate the Staircase Paradox?

We know that each stair’s height is given by h = s/2n. On the other hand, the total length of the stairs at n divisions (stairs) is given by M1_n = √[(4√2)² + s²].

Based on this, we know that M1_n would approach M2_n (4√2 cm) only when ‘s’ approaches 0. If s = 0, the height, h = s/2n = 0 as well. So, setting ‘s’ to zero is not really an option.

However, there is another subtle, yet important realization here. The choice of the variable ‘s’ matters. With h = s/2n, there is already an argument that with a very small choice of s, at the limit of infinity, h would tend towards zero, and thereby M1_n would smoothly converge to M2_n (4√2).

However, we can get even more rigorous here! Since l = (1/2n)* √[(4√2)² + s²], we have more freedom of choice when it comes to the variable ‘s’.

What if we consider non-linear sequences for s that decrease at a faster rate towards zero as n increases linearly towards infinity?

Final Stretch

Let this class of s-variable be represented by s_n (as we now choose ‘s’ as a function of n). Some examples of such sequences include 1/n!, 1/n², 1/2^n, etc. Let us consider that s_n = 1/n.

Then, the respective base, height, and length of the small triangle are calculated as follows:

Here are the respective dimensions represented graphically along with the triangle from each stair:

Based on this, M1_n is calculated as follows:

When we apply the limit, that is, as n tends to infinity, 1/n² (the error term!) tends to zero. This version of M1_n then smoothly approaches M2_n.

By proving this, I have established the following:

1. C1 is disproved — M1 is NOT necessarily bound to produce a different length as compared to M2.

2. C2 is satisfied — The error term diminishes to zero as n tends to infinity.

3. C3 is disproved — Tangential geometry is NOT necessary and sufficient for accurate approximation. Update post publishing: This statement is incorrect and is retracted, as proven by the argument from brian vant-hull (please refer to quote below).

4. The staircase paradox is eliminated!

Update Post Publishing — C3 is NOT Disproved!

“In showing an example of triangles for which the height decreases faster than 1/n, you’ve actually created triangles that approach the tangent as n increases! So you did not in fact demonstrate that tangential functions are not necessary.”

— brian vant-hull

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The Mathematics Behind Eliminating the Staircase Paradox

Here is a summary of all the mathematical steps that I took to eliminate the staircase paradox:

1. I first conducted the geometrical anatomy of each stair.

2. I established that the base is solely a function of M2 and n, whereas the height and length of each stair are functions of M1, M2, and n. I also established the corresponding mathematical relations.

3. I established that choice of the stair-like sawtooth perimeter function via the variable ‘s’ matters. Consequently, the height (sum of all perpendicular distances from the tip of each stair to the hypotenuse) plays a significant role.

4. I then demonstrated that ‘s’ can be chosen as a function of n such that it diminishes to zero at a faster rate than the rate at which n increases (linearly) towards infinity.

Key Takeaway

The primary takeaway from this exercise is that whenever it comes to building approximation systems, minimizing the error term smoothly and establishing non-linear error-minimization (acceleration) are of paramount importance.

Final Remarks

It is important to note that with the original version of the stairs we started with, M1_n would never approach M2_n.

What I have shown here is that there exists a class of stair-like sawtooth functions that enable M1_n to smoothly approach M2_n as n increases. So, in my opinion, we have just figured out a way of eliminating this paradox and not solving it outright.

I do not claim that I am correct with this analysis. I have tried my best and am willing to be wrong. Normally, work like this would be scientifically peer-reviewed. In this case, it is not. I am the only person who has worked on it, and I have tried my best to verify everything. There could be mistakes. So, one could consider this article as an appeal to open science.

Constructive criticism is welcome. However, I implore mathematical rigour from anyone who wishes to challenge the contents of this article.

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Credit: My approach to attacking this paradox was inspired by the work of Mr. Hassan Sedaghat.

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Further reading that might interest you: How To Solve The Dartboard Paradox? and Why Do You See Mirrors Flipping Words?