a² + b² = c²

But who actually discovered this formula, and how did they figure it out?

The following video presents the earliest proofs from China, India, and Greece (Euclid), as well as a few tangents to modern proofs including President Garfield’s proof and Albert Einstein’s proof.

Who Really Discovered The Pythagorean Theorem?

I spent a long time on the visuals and animations for the video, so I have not prepared a text version with static graphics as I normally do. I hope you can watch the video when you have time. Here are some time stamps that you might find useful:

0:00 Intro
1:50 China
5:00 India
7:35 Garfield
8:45 Pythagoras
9:24 Greece
15:18 Einstein

References

Community post
https://www.youtube.com/post/UgkxGSwX5dtLg6zWkm-rNohQdcgDduhp5Gi2

Babylon, Egypt
https://en.wikipedia.org/wiki/File:Plimpton_322.jpg
https://en.wikipedia.org/wiki/File:Papyrus_Berlin_6619.jpg
https://commons.wikimedia.org/wiki/File:All_Gizah_Pyramids.jpg
https://commons.wikimedia.org/wiki/File:Corda-a-12-nodi.svg
Ricardo Liberato, CC BY-SA 2.0 , via Wikimedia Commons

China (Gougu)
https://en.wikipedia.org/wiki/Xuan_tu
https://en.wikipedia.org/wiki/Zhoubi_Suanjing
https://en.wikipedia.org/wiki/File:Duke_of_Zhou.png
https://www.rediff.com/news/special/did-india-discover-pythogoras-theorem-a-top-mathematician-answers/20150109.htm

India (Baudhayan)
https://en.wikipedia.org/wiki/Baudhayana#/media/File:Bas-relief_of_Baudhayana_(cropped).jpg
https://archive.org/details/baudhyanasrautas01bauduoft/page/n5/mode/2up
https://en.wikipedia.org/wiki/Yajna
Arayilpdas at Malayalam Wikipedia CC BY-SA 3.0 via Wikimedia Commons
https://w.wiki/S$M5
https://x.com/ncert/status/2006602652638130291
https://ncert.nic.in/textbook/pdf/hegp202.pdf
https://bhaskarkamble.substack.com/p/the-sulbasutras-and-the-pythagoras

Greece (Pythagoras)
https://commons.wikimedia.org/wiki/File:Pythagoras_knapp.jpg
https://archive.org/details/cu31924008704219/page/n165/mode/2up

Greece (Euclid)
https://mathcs.holycross.edu/~little/Mont201516/EuclidPythagoras1.html
https://commons.wikimedia.org/wiki/File:Tysoe_Windmill_-_geograph.org.uk_-_7335000.jpg
Tysoe Windmill by Derek Harper, CC BY-SA 2.0 , via Wikimedia Common
https://www.cut-the-knot.org/pythagoras/euclid.shtml
https://webspace.ship.edu/mrcohe/inside-out/vu1/d_joyce/elements/bookVI/propVI31.html
http://aleph0.clarku.edu/~djoyce/elements/bookVI/propVI31.html
https://en.wikipedia.org/wiki/File:Visual_proof_of_the_Pythagorean_theorem_by_area-preserving_shearing.gif
https://www.desmos.com/calculator/zlf3dweki4
https://www.youtube.com/watch?v=eWNBw5GpqeY&t=611s

Einstein
https://archive.org/details/fractalschaospow0000unse/page/2/mode/2up
https://www.cut-the-knot.org/pythagoras/PythagorasBySimilarity.shtml
https://terrytao.wordpress.com/2007/09/14/pythagoras-theorem/

Impossible proof
https://www.tandfonline.com/doi/epdf/10.1080/00029890.2024.2370240?needAccess=true
https://maa.org/news/groundbreaking-pythagorean-research-featured-in-the-american-mathematical-monthly/
https://www.youtube.com/watch?v=juFdo2bijic

If you pick the answer to this question at random, what is the chance you will be correct?

(a) 25%
(b) 50%
(c) 0%
(d) 25%

As usual, watch the video for a solution.

Multiple choice question baffles the internet

Or keep reading.

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Answer To Multiple choice question baffles the internet

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

If you pick the answer to this question at random, what is the chance you will be correct?

(a) 25%
(b) 50%
(c) 0%
(d) 25%

Since you are selecting an option at random, each of the 4 choices is equally likely or a 25% chance to each. But this is more of a logic problem than a probability puzzle.

Suppose (a) 25% is the correct answer, then (d) 25% would also be correct. But then 2 of the 4 options would be correct, or a 50% chance of selecting the correct answer.

This means (b) 50% is the correct answer. However that would then mean only 1 of 4 choices is correct, or a 25% chance, which we already said could not be a valid answer.

So (a) 25%, (b) 50% and (d) 25% cannot be correct.

This leaves only (c) 0%, but that would be 1 of the 4 choices is correct for a 25% chance, again a contradiction.

The question has no correct answer because it is an example of a self-referential paradox.

This exact puzzle has made the rounds on the internet for at least 16 years.

But such paradoxes have fascinated philosophers for thousands of years.

Socrates once said, “I know that I know nothing.” But the very knowledge of one’s ignorance is knowledge itself, a contradiction to knowing nothing.

The ancient Indian Bhartrihari asked if objects can exist that cannot be described by language. If such a person existed–He Who Cannot Be Named–then we have applied a label to that person so the person can be labeled.

Bertrand Russell described a famous paradox of a barber. This barber shaves all of those, and those only, who do not shave themselves. But who shaves the barber? If he does, then he is shaving a person who shaves himself which is a contradiction. If not, then he is missing shaving a person who does not shave himself.

So these kinds of paradoxes have delighted humanity for millennia, and one can only imagine they will continue to be fun for centuries to come.

References

Math Files
https://x.com/Math_files/status/2084529394081730572

3 years ago
https://www.reddit.com/r/ProgrammerHumor/comments/15m9cia/ifeellikethisfitshere/
15 years ago
https://www.reddit.com/r/reddit.com/comments/jurc9/a_multiple_choice_question/

16 years ago
https://www.reddit.com/r/pics/comments/c64gp/multiple_choice_if_you_choose_an_answer_to_this/

Bhartrihari
https://en.wikipedia.org/wiki/Bhart%E1%B9%9Bhari#/media/File:Hindi_Manuscript_884_Wellcome_L0024558.jpg
CC BY 4.0 , via Wikimedia Commons

Socrates
https://en.wikipedia.org/wiki/I_know_that_I_know_nothing
Photograph by Greg O’Beirne. Cropped by User:Tomisti, CC BY-SA 3.0 , via Wikimedia Commons

Barber paradox
https://en.wikipedia.org/wiki/Barber_paradox

Self referential
https://en.wikipedia.org/wiki/Self-reference

I am thinking of 6 numbers. Each number is the product of the other 5 numbers. What could my 6 numbers be? Each number is a real number.

This puzzle is adapted from Reddit AskMath, which asks for the number of ordered sets of such numbers. The poster found the correct answer which was different from the listed choices!

As usual, watch the video for a solution.

6 Special Numbers

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Answer To 6 Special Numbers

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Let the numbers be x₁, …, x₆.

Define:

X = x₁…x₆

Write each number as the product of the others.

x₁ = x₂…x₆
x₂ = x₁…x₆
…
x₆ = x₁…x₅

Multiplying the equations together gives:

X = X⁵
X⁵ – X = 0
X(X⁴ – 1) = 0

Since each x_i is a real number, X must be a real number and we have 3 possibilities:

X = 0
X = 1
X = -1

But we can eliminate the last one. Start with the product equation:

x₁ = x₂…x₆

Multiply both sides by x₁ to get:

(x₁)² = X

By similar logic this is true for any of the variables so:

(x_i)² = X

Since x_i is a real number, we cannot have X = -1 and we can only consider X = 0 or 1. This means:

(x_i)² = 0 or 1
x_i = 0 or 1 or -1

Solving all cases

X = 0

This means some x_i = 0, and then any other term is equal to the product that includes 0, so every other term is 0 too. Thus we only have 1 possibility that all terms are identically 0.

(0, 0, 0, 0, 0, 0)

X = 1

No term can be 0, so each term can either be 1 or -1. We must have an even number of -1 terms to get a product of 1. The number of -1 terms can either be 0, 2, 4, or 6. Thus we have:

(1, 1, 1, 1, 1, 1)
(-1, -1, 1, 1, 1, 1)
(-1, -1, -1, -1, 1, 1)
(-1, -1, -1, -1, -1, -1)

So there are 5 possibilities without regard to the order of the numbers.

(0, 0, 0, 0, 0, 0)
(1, 1, 1, 1, 1, 1)
(-1, -1, 1, 1, 1, 1)
(-1, -1, -1, -1, 1, 1)
(-1, -1, -1, -1, -1, -1)

To count ordered sets, we can see there is only 1 combination for the sets of all 0s, 1s, or -1s. In the case of 2 terms of -1, there are 6C2 = 15 ways to place the -1 terms. Similarly in the case of 4 terms of -1 there are 6C4 = 15 ways to place the -1 terms. This gives a total of:

(0, 0, 0, 0, 0, 0) – 1 combination
(1, 1, 1, 1, 1, 1) – 1 combination
(-1, -1, 1, 1, 1, 1) – 15 combinations
(-1, -1, -1, -1, 1, 1) – 15 combinations
(-1, -1, -1, -1, -1, -1) – 1 combination

1 + 1 + 15 + 15 + 1 = 33 combinations

Special thanks this month to:

Lee Redden
Kyle
Daniel Lewis

Thanks to all supporters on Patreon and YouTube!

Reference

Reddit AskMath