Indeterminate: the hidden power of 0 divided by 0 @Mathologer

Indeterminate: the hidden power of 0 divided by 0 @Mathologer

Uploaded December 2016 | Updated September 2026, 2 weeks ago

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You've all been indoctrinated into accepting that you cannot divide by zero. Find out about the beautiful mathematics that results when you do it anyway in calculus. Featuring some of the most notorious "forbidden" expressions like 0/0 and 1^∞ as well as Apple's Siri and Sir Isaac Newton.

In his book “Yearning for the impossible” one my favourite authors John Stillwell says “…mathematics is a story of close encounters with the impossible and all its great discoveries are close encounters with the impossible.” What we talk about in this video and quite a few other Mathologer videos are great examples of these sort of close encounters.

For those of you desperate to get hold of the t-shirt check out this link: shirt.woot.com/offers/how-natural-selection-works?ref=cnt_ctlg_dgn_1

Thank you very much to Danil Dmitriev the official Mathologer translator for Russian for his subtitles.

Enjoy :)

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$Ptolemy’s Theorem and the Almagest: we just found the best visual proof in 2000 years We are making history again by presenting a new visual proof of the 2000+ years old Ptolemys theorem and Ptolemys inequality. 00:00 Introduction 04:27 Geometry 101 08:19 Applications 14:46 Ptolemys inequality 18:34 LIES 25:35 Animated proofs 28:57 Thank you! 30:53 Degenerate Easter Egg There are some other proofs of Ptolemys theorem/inequality based on scaling and aligning suitable triangles. However, none of them is as slick, beautiful and powerful as Rainers new proof. In particular, check out the animated scaling proof on the wiki page for Ptolemys theorem (and this https://youtu.be/ZK08Z5A9xH4) and check out the scaling proof by Claudi Asina and Roger Nelson: Proof Without Words: Ptolemy’s Inequality in Mathematics Magazine 87, (2014), p. 291. https://www.jstor.org/stable/10.4169/math.mag.87.4.291 Rainer was inspired by a classic scaling based proof of Pythagoras theorem that I presented here https://youtu.be/p-0SOWbzUYI?si=GeGzZ0R_Dj1AsXqR&t=371 You can find a couple of full text versions of the Almagest here https://www.wilbourhall.org/index.html#ptolemy https://classicalliberalarts.com/resources/PTOLEMY_ALMAGEST_ENGLISH.pdf For more background info check out the very comprehensive wiki pages on: Ptolemy’s theorem https://en.wikipedia.org/wiki/Ptolemy%27s_theorem Ptolemy’s inequality https://en.wikipedia.org/wiki/Ptolemy%27s_inequality Claudius Ptolemy https://en.wikipedia.org/wiki/Ptolemy The Almagest https://sco.wikipedia.org/wiki/Almagest Trigonometric identities https://en.wikipedia.org/wiki/List_of_trigonometric_identities Cyclic quadrilateral https://en.wikipedia.org/wiki/Cyclic_quadrilateral The optic equation https://en.wikipedia.org/wiki/Optic_equation There are very interesting higher-dimensional versions of Ptolemys theorem just like there are higher-dimensional versions of Pythagoras theorem. I did not get around to talking them today. Google ... Highly recommended: T. Brendan, How Ptolemy constructed trigonometry tables, The Mathematics Teacher 58 (1965), pp. 141-149 https://www.jstor.org/stable/27967990 Tom M. Apostol, Ptolemys Inequality and the Chordal Metric, Mathematics Magazine 40 (1967), pp. 233-235 https://www.jstor.org/stable/2688275 https://demonstrations.wolfram.com/PtolemysTableOfChords/ an interactive exploration of Ptolemys table of chords Ptolemys theorem made a guest appearance in the the previous Mathologer video on the golden ratio: https://youtu.be/cCXRUHUgvLI Here is a nice trick to make Ptolemy counterparts of Pythagorean triples. Take any two sets of Pythagorean triples: 5² = 3² + 4², 13² = 12² + 5², and combine them like this: 65² = 13² × 5²= 13²(4² + 3²) = 52² + 39²= 5²(12² + 5²) = 60² + 25². Now combining the two right angled triangles 52-39-65 and 25-60-65 along the common diagonal in any of four different ways gives a convex quadrilateral with all sides integers. Note that you automatically get 5 integer lengths and then Ptolemys theorem guarantees that the remaining side is a fraction. Scaling up everything by the denominator of that fraction gives one of the special integer-everywhere quadrilaterals. See also Brahmagupta quadrilaterals. Here is a nice application of Ptolemys theorem to a International Maths Olympiad problem https://www.youtube.com/watch?v=NHjtHOE1lks In a cyclic quadrilateral the ratio of the diagonals equals the ratio of the sums of products of the sides that share the diagonals end points: https://www.geogebra.org/m/XQr5jJQg This extension of Ptolemys theorem is part of the thumbnail for this video. T-shirt: cowsine :) Music: Floating branch by Muted and I promise by Ian Post. Enjoy, burkard$ $What does this prove? Some of the most gorgeous visual shrink proofs ever invented Bit of a mystery Mathologer today with the title of the video not giving away much. Anyway it all starts with the quest for equilateral triangles in square grids and by the end of it we find ourselves once more in the realms of irrationality. This video contains some extra gorgeous visual proofs that hardly anybody seems to know about. 0:00 Intro 0:47 First puzzle 2:24 Second puzzle 3:50 Edward Lucas 4:41 Equilateral triangles 13:15 3d & 3rd puzzle 19:52 30 45 60 29:31 Credits Here are links to/references of some of the things I mention in the video: Joel Hamkins blog posts that inspired this video: http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/ http://jdh.hamkins.org/no-regular-polygons-in-the-hexagonal-lattice/ There is also a whole chapter about all this and much more related maths in his new book https://www.amazon.com/Proof-Mathematics-Joel-David-Hamkins/dp/0262539799 Here is another really good article which includes a nice characterisation of the triangles that can be found in square grids plus a very good survey of relevant results: Michael J. Beeson, Triangles with Vertices on Lattice Points, The American Mathematical Monthly 99 (1992), 243-252, https://www.jstor.org/stable/2325060?seq=1 Scherrers and Hadwingers articles: Scherrer, Willy, Die Einlagerung eines regulären Vielecks in ein Gitter, Elemente der Mathematik 1 (1946), 97-98. https://tinyurl.com/y45p64t7 https://gdz.sub.uni-goettingen.de/id/PPN378850199_0001?tify={%22pages%22:[101]} Hadwiger, Hugo Über die rationalen Hauptwinkel der Goniometrie, Elemente der Mathematik 1 (1946), 98-100. https://tinyurl.com/yx98kkqt https://gdz.sub.uni-goettingen.de/id/PPN378850199_0001?tify={%22pages%22:[102],%22view%22:%22info%22} Another, nice paper on rational (and algebraic) cosines https://arxiv.org/pdf/1006.2938.pdf Here is a solution to the first puzzle (one way to find the general formula): https://nrich.maths.org/657/solution The music in this video is by Chris Haugen, Fresh Fallen Snow (playing in the video) and Morning Mandolin (for the credits) A couple of remarks: 1. Probably the simplest way to deduce the sin and tan parts of the rational trig ratio theorem is to realise that they follow from the cos part via the trigonometric identities: sin(x)=cos(90-x) and tan^2(x) = (1-cos(2x))/(1+cos(2x)). Note that the second identity implies that if tan(x) is rational, then cos(2x) is rational (if tan(x)=c/d, then tan^2(x)=c^2/d^2=C/D and cos(2x)=(D-C)/(D+C)). 2. Bug report. a) Here I redefine cos(120◦) = 1. https://youtu.be/sDfzCIWpS7Q?t=1362 Remarkable :( b) This transition to the good stuff I clearly did not think through properly. https://youtu.be/sDfzCIWpS7Q?t=1018 Its possible to make this work for all regular n-gons. There is only one complication that occurs for ns that are of the form 2 * odd. For the corresponding regular n-gons, if you pick up the edges in the order that they appear around the n-gon and assemble them into a star, things close up into (n/2)-stars. For all other n, things work exactly as I showed in the video. Having said that you can also assemble the edges of one of the exceptions into stars. Have a look at this https://imgur.com/68A3fEe and youll get the idea. Anyway lots more nice side puzzles to be explored here if you are interested :) Enjoy! Burkard Two ways to support Mathologer Mathologer Patreon: https://www.patreon.com/mathologer Mathologer PayPal: paypal.me/mathologer (see the Patreon page for details) 14. Sep. 2021: Thank you very much Michael Didenko for your Russian subtitles.$

Indeterminate: the hidden power of 0 divided by 0