# CTK Insights

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17 Oct

### Elementary Problems that Beg for Generalization

In a well known puzzle, a father willed to his three sons $17$ camels, with the proviso that $\frac{1}{2}$ of the inheritance should go to the oldest among them, with $\frac{1}{3}$ being due to the middle one and $\frac{1}{9}$ to the youngest. Shortly after the father's death, a wise man riding on his camel through the village noticed the three brothers in a quandary. He added his camel to the inherited $17,$ thus getting a herd of $18$ animals. He gave $\frac{1}{2}$ of these (i. e. $9$ camels) to the oldest brother, $6$ $(= \frac{18}{3})$ to the middle one, and $2$ $(= \frac{18}{9})$ to the youngest. $1$ camel remained $(1 = 18 - 9 - 6 - 2),$ which he climbed up and rode away on. To the great satisfaction of all brothers, each of them received more than was willed by their father.

Among those who found interest in the puzzle, one may try solving it before reading the solution; another may read and enjoy the solution; somebody else may think up related problems. Paul Stockmeyer - a professor emeritus of computer science at the College of William and Mary - did just that (Math Horizons, Volume 21, Number 1, September 2013 , pp. 8-11), posing a generalized question and even listing several related research problems.

Are there situations that include two brothers? other problems with three? problems with four, etc. How many camels could be shared in such a manner? Below are several identities that answer some of the questions, if only partially. Rather obviously the problem involves the unit fractions:

Two sons, three camels: $\frac{1}{2} + \frac{1}{4} = 1 - \frac{1}{4}$

Two sons, five camels: $\frac{1}{2} + \frac{1}{3} = 1 - \frac{1}{6}$

Three sons, seven camels: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = 1 - \frac{1}{8}$

Three sons, eleven camels: $\frac{1}{2} + \frac{1}{4} + \frac{1}{6} = 1 - \frac{1}{12}$

Three sons, eleven camels: $\frac{1}{2} + \frac{1}{3} + \frac{1}{12} = 1 - \frac{1}{12}$

Three sons, nineteen camels: $\frac{1}{2} + \frac{1}{4} + \frac{1}{5} = 1 - \frac{1}{20}$

Four sons, 1805 camels: $\frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{43} = 1 - \frac{1}{1806}$

There are many more examples. But could an estate consist of 9 or 21 or of an even number of camels. Are there divisions where the oldest son is willed one third of the inheritance and not one half? What about one fourth?

This is amazing how many questions may be asked starting with such an elementary problem. Probably not all problem submit to such extensions, but many do. I've been recently referred to the site of an Online (more accurately, by correspondence) Russian Math Olympiad honoring Leonhard Euler. Here are some examples from this year's olympiad:

1. In the expression $\displaystyle\frac{1}{2}*\frac{2}{3}*\frac{3}{4}*\cdots *\frac{98}{99}*\frac{99}{100}$ replace each star with one of four arithmetic operations to make the result zero.

What comes to mind? First, an obvious generalization (that gives away a solution) is to end the product with $\displaystyle\frac{n^2-1}{n^2}.$ Then it's natural to ask if there are other cases. I am not sure of the answer yet. What I am sure about is that a solution - if exists - is never unique. Prove that.

2. In the island of knights (that always tell truth) and knaves (that always lie), $22$ fellows stand in a circle, all facing the center, and each declares that the ten to his left are all knaves. How many knaves are actually there?

Is the relation $1+10+1+10=22$ inherent for the solvability of the problem?

3. Draw five staright lines in a plane that intersect in exactly seven points.

What number of intersections is possible for $n$ lines?

09 Oct

### Undiluted Hocus-Pocus: The Autobiography of Martin Gardner

Martin Gardner refers to his latest, and - perhaps - last, book as a rambling (and also slovenly) autobiography, disheveled memoirs. It is anything but. It is the most sincere, unadulterated biography I ever read. He wrote it in 2009-2010 on an old typewriter while at an assisted living facility. "At ninety-five I still have enough wits to keep writing," he observed. At the time his library and files that once occupied the whole third story of a big house, have been reduced to a single file cabinet. So it must have been that he largely relied on his memory when bringing up names, episodes of people's lives, when describing events he witnessed or took a part in. The numbers of all those are huge. Martin Gardner lived an interesting life - he made it such.

For those who like me knew Martin Gardner from his books as a mathematics populariser, a pseudoscience debunker, a conjurer and a math magician, details of his life and personality exposed in the book help create a more complete picture of his fascinating person. It may surprise the reader that God and religion played an important role even in Gardner's adult life. So much so that he devotes the whole of a one page Preface to the Gods of the Old and New Testaments, mentions the question all throughout the book, explains his conversion in the "I lose my faith" chapter, and apportions the last chapter in its entirety to clarify his views. He ends his Preface with a paragraph that provides an inkling as to what they are:

The best-known remark of stand-up comedian Lenny Bruce was that people are leaving their churches and going back to God. What follows here is a rambling autobiography of one such person - me.

Gardner classified himself as a philosophical theist - a person who finds it comforting to believe in "a transcendent intelligence, impossible for us to understand ... somehow responsible for our universe and capable of providing, I do not know how, an afterlife."

Gardner wrote about H. G. Wells and G. K. Chesterton:

I have written elsewhere that if you can understand how I can admire both men, one a devout atheist, the other a devout Catholic, you can begin to understand my brand of theism.

But religiosity formed just one thread in Gardner's life and, consequently, the autobiography. In the book, Gardner emerges as an entertaining writer, well informed journalist, life's keen observer... very human, very open, very knowledgeable.

Martin Gardner had tremendous influence on several generations of young minds; his autobiography will help his fans appreciate how that came about. This is a book no one who ever heard his name would want to miss.

19 Sep

### Radical Simplification - Not That Simple!

In the previous post I proved an identity in radicals:

$\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}}=3$

which followed the method used in another post where another identity in radicals

$\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} = 1.$

has been derived. The latter post pointed to an earlier and perhaps a more exciting one where we established an unexpected identity

$\displaystyle\sqrt[3]{2 \pm \sqrt{5}} = \frac{1 \pm \sqrt{5}}{2}.$

This was verified by simply taking the cube of the two sides of the equation. In a private correspondence related to the previous post Bruce Reznick suggested a different way of showing that

$\sqrt[3]{18+\sqrt{325}}=\sqrt[3]{18 \pm 5 \sqrt{13}}=a+b\sqrt{13},$

with rational $a$ and $b.$ In Bruce Reznick's words:

So we look for $(a + b \sqrt{13})^3 = 18 + 5 \sqrt{13};$ that is, $18 = a^3 + 39 a b^2$ and $5 = 3a^2 b + 13 b^3.$ Hard to solve directly, but we can multiply the first by $5$ and subtract $18$ times the second to get

$0 = 5a^3 - 54 a^2 b + 195 a b^2 - 234 b^3,$

which your friendly computer algebra system (otherwise useless) will tell you equals $(a - 3b)(5 a^2 - 39 a b + 78 b^2).$ If $a = 3b,$ then $18 = 144 b^3$ and $5 = 40 b^3,$ so $b = 1/2$ and $a = 3/2$ and

$\displaystyle\sqrt[3]{18 + 5 \sqrt{13}} = \frac{3}{2} + \frac{1}{2}\sqrt{13}.$ Same with "$-$". This method works quite generally.

And would not it? With the prior knowledge that $\displaystyle\sqrt[3]{2 \pm \sqrt{5}} = \frac{1 \pm \sqrt{5}}{2}$ the latter result seems not that surprising, one can think. Bruce continues to show with another example that expectations can be easily shuttered in mathematics.

Again, our previous experience may suggest the existence of rational $a$ and $b$ such that

$\sqrt{2 + \sqrt{3}} = a + b \sqrt{3}.$

If we proceed as before, but now squaring both sides of the equation, and equating rational and irrational components, we get $2 = a^2 + 3b^2$ and $1 = 2 ab,$ so

$a^2 + 3b^2 - 2\cdot 2ab = (a - b)(a - 3b) = 0.$

If $a = b,$ then $2 = 4a^2,$ so $\displaystyle a = \frac{1}{\sqrt{2}} = b.$ This is not what we expected, but is at least true:

$\displaystyle 2 + \sqrt{3} = (\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}})^2 = \frac{1}{2} (1 + \sqrt{3})^2.$

What about the other factor? If $a = 3b,$ then $2 = 12b^2,$ so $\displaystyle b = \frac{1}{\sqrt{6}}$ and $\displaystyle a = \frac{3}{\sqrt{6}}.$ This gives

$\displaystyle 2 + \sqrt{3} = (\frac{3}{\sqrt{6}} + \frac{\sqrt{3}}{\sqrt{6}})^2.$

Strangely, this reduces to exactly the result due to the first (and different) factor:

$\displaystyle \frac{3}{\sqrt{6}} + \frac{\sqrt{3}}{\sqrt{6}} = \frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}}.$

In terms of quadratic fields, $\mathbb{Q}[\sqrt{m}]$, what we found may be expressed formally as $\sqrt{2 + \sqrt{3}}\not\in\mathbb{Q}[\sqrt{3}]$ but $\sqrt{2 + \sqrt{3}}\in\mathbb{Q}[\sqrt{2},\sqrt{3}]$ which, perhaps, helps enhance intuition of what goes on in Bruce's example but also adds to the mystery of $\sqrt[3]{18 + 5 \sqrt{13}}\in\mathbb{Q}[\sqrt{13}]$ or $\sqrt[3]{2 + \sqrt{5}}\in\mathbb{Q}[\sqrt{5}].$

15 Sep

Let's prove that

$\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}}=3.$

To this end, introduce $a=\sqrt[3]{18+\sqrt{325}}$ and $b=\sqrt[3]{18-\sqrt{325}}.$

Here's what we find: $a^{3}b^{3}=324-325=-1$ so that $ab=-1,$ and $a^3+b^3=36.$ If so,

$36=a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)((a+b)^2+3).$

This amounts to a cubic equation for $x=a+b:$ $x^3+3x-36=0.$ The sum of the roots of this equation is $-3,$ their product is $36,$ and the sum of their squares is $0.$ The latter implies that two of these are complex conjugates, say, $u\pm iv$ and one is a real number $w$ (that is supposed to be $3.)$

By direct verification, one can check that $x=3$ does solve that equation $(3^3+3\times 3-36=0.)$

This is how a similar identity has been proved elsewhere.

30 Aug

### Math Associations on a Field Trip

I am just back from a wonderful 10 day trip to the central Alaska. It was cloudy and rainy for the first two days; low clouds common to Alaska were covering mountain peaks and even tree tops. But then the skies have cleared, the visibility was perfect, and the imagery of mountain ridges and jagged portions of glaciers easily evoked the idea of fractals. On a flight from Anchorage to Coldfoot it was fascinating to observe that many clouds had perfectly smooth flat bottoms that at a distance projected into straight lines.

Well, once in the mood, math concepts kept creeping behind Alaska scenery and objects we came across. Here are some of evocative fragments (in alphabetical order).

06 Jul

### Weather Forecasting: A Story of Mathematical Triumph

Until very recently I knew precious little about applications of mathematics to weather forecasting. I had a notion that mathematics is bound to be involved, that there are difficulties inherent in processing vast amounts of data and sensitivity of solutions of multivariable differential equations due to their non-linearity, and that the current level of forecasting would not be possible without the tremendous technological progress achieved during the second half of the past century.

The new book Invisible in the Storm: The Role of Mathematics in Understanding Weather by Ian Roulstone and John Norbury (Princeton University Press, 2013) was an eye opener.

It's not only that the book shows how much modern weather forecasting depends on mathematics, the book outlines the many ways in which mathematics is applicable. Equations of motion, heat, fluids, vorticity, gravitational waves, laws of conservation and laws of change, numerical computations, theory of stability and chaos, and probability theory, are all being used in forecasting, often as competing utilities.

Indeed, mathematics employed in modern weather forecasting is formidable; the book lucidly describes its intuitive essentials as a thread in the story of the development of meteorology as a science in the past century. Formal equations, with two accidental exceptions (one pages 130 an 234), have been collected in clearly marked (and not too numerous) "tech boxes". The authors do an admirable job of translating if not the equations per se then at least their origins, significance, and utilization into plain language.

Not that it was always clear how - and even what - to predict, and which mathematics to use to this end.

Evolution of what is possible and what is not led eventually to a culture shift: Instead of trying to say exactly what the weather will be in the future, we instead try to predict what it will most probably be.

But naturally, mathematics was not evolving all by itself. The authors excel in presenting establishment of the science of meteorology as a human endeavor. The history of meteorology is rich in perseverance, sacrifice, enthusiasm, ingenuity, useful missteps, multinational collaboration ... and plain hard work. Authors' fluent recount makes the story all the more fascinating, even if math applications are only at the back of your mind. The book is a superior read.

19 Jun

### Climbing Pyramidal Slopes

On a pyramid having a square base and edge length $a$ path ascends from one of the bottom vertices to the apex at a slope of $25\%.$ Determine the height $h$ of the pyramid, the length $c$ of the first section of the path, the lengths $b$ and $a'$ as well as the total length $L$ of the path, all of them in terms of $a.$

(This problem, as I was informed by Peter Avxbybcbhybf (Petros Avxbybcbhybf), has been used at "the preparatory math camp for prospective IMO applicants to go to Korea for the IMO" (2000).)

It is not very steep and may be even tedious, but - at the end - the answer (summit) proves to be somewhat simpler than the climb that led there.

I slipped once and got an answer that included the golden ratio. Since the latter commonly pops up in unexpected situations, I was not at all surprised. However, I noticed in time that the golden ratio would lead the path downhill.

Tumbling into the golden ratio had its upside: if it were not for that, I would probably give up and return home without reaching the summit. As it happened, correcting the mistake took me all the way up there.

Solution

The easiest part of the solution is finding the height of the pyramid. This is the vertical leg of the right triangle with the base $\displaystyle\frac{a}{2}$ and hypotenuse $\displaystyle\frac{\sqrt{3}}{2}.$

What it comes out to is $\displaystyle h=\frac{\sqrt{2}}{2}a.$ For the rest of the solution I'll depend on the following diagram:

One thing known right away is that $\displaystyle u=\frac{c}{4}.$ From this and the Pythagorean theorem we can find the bases $v$ and $w$ in terms of $c$, and then express the latter as a function of $a.$

$\displaystyle v = \frac{\sqrt{15}}{4}c,$

$\displaystyle w = u\mbox{cot}(\angle EBD) =\frac{c}{4}\frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{12}c,\space \mbox{because}\space\angle EBD=60^{\circ},$

$\displaystyle a = v+w = \frac{3\sqrt{15}+\sqrt{3}}{12}c,$

$\displaystyle c = \frac{12a}{3\sqrt{15}+\sqrt{3}} = \frac{3\sqrt{15}-\sqrt{3}}{11}a,$

$\displaystyle w = \frac{\sqrt{3}}{12}\frac{3\sqrt{15}-\sqrt{3}}{11}a = \frac{3\sqrt{5}-1}{44}a,$

$\displaystyle b = 2w = \frac{3\sqrt{5}-1}{22}a,$

$\displaystyle a' = a - b = \frac{23-3\sqrt{5}}{22}a,$

$\displaystyle q = \frac{a'}{a}=\frac{23-3\sqrt{5}}{22},$

$\displaystyle \frac{1}{1-q} = \frac{3\sqrt{5}+1}{2},$

$\displaystyle L = \frac{c}{1-q} = \frac{(3\sqrt{15}-\sqrt{3})(3\sqrt{5}+1)}{22}a=2\sqrt{3}a.$

11 Jun

### Parallel Chords in Conics

CutTheKnotMath facebook page serves as a place to post math facts, questions or new problems that do not absolutely fit as comments to existing pages at the Interactive Mathematics Miscellany and Puzzles site. One such question has been posted by Cõ Gẫng Lên:

Points $A,B,C,D$ lie on ellipse such that $AB||CD.$ $AC$ and $BD$ meet in $H$. $AD$ and $BC$ meet in $G.$ Prove that $GH$ crosses $AB$ and $CD$ in their midpoints, $E$ and $F.$

Why that is so? A short answer is truly cryptic: Because this is true for circles.

For circles the statement holds because of the symmetry: every diameter (a line through the center) perpendicular to a chord passes through the midpoint of that chord (and vice versa.) Now, an ellipse is an affine image of a circle. Perhaps, more conventionally, ellipse is a projective image of circle. Both affine and projectiive transformations are known to preserve incidence and collinearity. The difference between the two stems from the nature of the objects the two geometries - projective and affine - deal with. The subject of the two-dimensional affine geometry are points and lines in a regular plane. Projective geometry deals with all these but, in addition, includes - for any direction, i.e., a family of parallel lines in a regular plane - a point at infinity and a line at infinity as a collection of such add-on points. In this sense, any two lines in projective geometry intersect - at a point at infinity if they are parallel in affine geometry. But if the point of intersection of two lines is finite and is projected to a finite point, the images of the two lines cross at a finite point.

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Angles and distances are not preserved under affine or projective transformations, but ratios of the lengths of segments on the same line are preserved by affine and, if points at infinity are mapped to points at infinity, by projective transformations. (In general, projective transformations preseve the cross-ratio of four collinear points.)

This is is why the statement also holds for parabola and hyperbola, the other two conic sections, all of which are projective images of circle. Under affine and projective transformations, the image of a diameter of a circle may cease to be perpendicular to the image of a chord at hand, but will continue passing through its midpoint and the center of the resulting conic. For two chords ($AB$ and $CD$ that remain parallel, all the incidences and collinearities also remain in place.

### References

1. D. A. Brannan, M. F. Esplen, J. J. Gray, Geometry, Cambridge University Press, 2002
2. H. Eves, A Survey of Geometry, Allyn and Bacon, Inc., 1972
3. F. Klein, Elementary Mathematics from an Advanced Standpoint: Geometry, Dover, 2004
4. D. Pedoe, Geometry: A Comprehensive Course, Dover, 1988
07 Jun

### Dynamic Software as Serendipity Enhancement

Several years ago I wrote a page with a Java illustration to a solution of problem 4 from the 1995 British Mathematical Olympiad. Earlier today I happened on that page and noticed that the solution refers to point $L$ that is not marked in the applet. Since the time I wrote that page, I moved to another computer and to using GeoGebra instead of writing Java applets, so much so that I do not even have my Java development environment on my present computer. To resolve the issue I decided to put together a GeoGebra applet with $L$ properly marked and replace the old Java applet. This took probably a couple of minutes. However, this is not what made my day.

At the end of the old page I observed that the configuration in the problem had additional features and pointed to two of them. Now, GeoGebra makes it easy to experiment - form and verify hypothesis, or just plain count on serendipitously stumbling on a dormant feature. It was a sin not to try. The configuration in the problem definitely had many more features waiting to be discovered. Simple though all it was, I am satisfied to have found some.

### Problem

$\Delta ABC$ has right angle at $C.$ The internal bisectors of angles $BAC$ and $ABC$ meet $BC$ and $AC$ at $P$ and $Q$ respectively. The points $M$ and $N$ are the feet of the perpendiculars from $P$ and $Q$ to $AB.$ Find angle $MCN.$

The angle was found to be $45^{\circ}.$ Checking the "Hint" box in the GeoGebra applet below will show the essential steps of the proof.

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Checking the "Extra" box will suggested a few more properties: $\angle MCN$ is not the only angle in the diagram that equals $45^{\circ}$ (e.g., $\angle ADN$ and $\angle CMD);$ $\angle NLM=90^{\circ};$ some intersections $(N,D,L,E,M)$ are concyclic; there are several similar triangles (e.g., $\Delta ALN$ and $\Delta NLD).$

There are probably other properties. Should you find any, please let me know.

05 Jun

### Naming Infinity - the book

My father was used to tell me a story of how in the late 1940s the famous Russian mathematician A. N. Kolmogorov slapped in the face another famous Russian mathematician N. N. Luzin. The occasion was a balloting at the Russian Academy of Sciences where Luzin blocked the candidacy of the third Russian mathematical luminary P. S. Alexandrov. The reason he explained was that Luzin had presented Alexandrov's long stays in Western Europe as an anti-Soviet activity. According to the book Naming Infinity by Loren Graham and Jean-Michel Kantor my father was wrong, and that on several accounts. In all likelihood, Graham and Kantor's version of the events is more trustworthy; I am satisfied to have learned at least a part of it, I am not sure I wanted to know all the details. Graham and Kantor write

Readers may wonder why we include such personal details in this discussion of an important chapter in the history of mathematics.

And subsequently the authors respond

We are not eager to denigrate anyone's accomplishments, and we have an immense respect for science, the most powerful means of knowing developed by human beings. But we think that our understanding of these episodes and, ultimately, our respect for the actors in them will be all the greater if we include the entire story and not just parts of it.

So, according to Graham and Kantor, Luzin - on being approached by Kolmogorov after the vote had been taken - rudely hinted at the alleged homosexual relationship between Kolmogorov and Alexandrov. I would not miss knowing the truth in that matter, but the other part of the story - not mentioned by my father, and which is probably neither gossip nor outright slander - presented the whole episode in entirely different light. In 1936, at the time of great purges, Kolmogorov and Alexandrov both denounced Luzin, their former teacher, to the Soviet authorities, which, if it were not for an intervention by Leonid Kapitsa, would have led to Luzin's imprisonment and likely execution. Understandably, Luzin carried a grudge against both Kolmogorov and Alexandrov.

Luzin was a religious person and vulnerable just for that reason. So were his teacher Dimitri Egorov and friend Pavel Florensky both of whom perished in the 1930s. It was a ghastly period in the Russian history, and, if nothing else, the book gives a broad picture of that time, with many terrible and beautiful details, lifting the curtain on human misery, ugliness, and bravery. Nikolai Chebotaryov resigned his professorship after learning that it was previously occupied by the fired Egorov; for this he was exiled to Kazan, where his doctor wife clandestinely cared for the dying Egorov who arrived there a few years later. Lev Shnirelman, who made a signficant contribution to solving the Goldbach conjecture, committed suicide at the age of 32, after an interrogation by the secret police, where he was apparently made to sign a fabricated confession that incriminated several of his friends.

It was a horrible time especially compared to the atmosphere of camaraderie that existed in the Moscow Mathematical School set up by Egorov and Luzin yet before the WWI. During the wartime and after the revolution, at the time of incredible deprivation, the school met on a regular basis at the Moscow University. The book tells us that at winter times students poured water in the corridors and, when it freezed, skated there.

The full title of the book is Naming Infinity: A True Story of Religious Mysticism and Mathematical Creativity; the above may give you an idea of why it really reads like a true story. This is a story in which mathematicians - mostly French and Russian - and not mathematics who play a central role. The (mathematical thread of the) story begins at the end of the nineteenth century with the turmoil that followed Cantor's introduction of hierarchy of infinities. The book describes what Hermann Weyl later called the foundational crisis in mathematics and its effects on the rational French school (Émile Borel, René Baire, and Henri Lebesgue) and ony the nascent Russian school (Dmitri Egorov, Nikolai Luzin and their associate Pavel Florensky - a student of mathematics and an ordained Russian Orthodox priest) with the characteristic tendency to interwine their work with societal, philosophical and ideological trends. The authors bring up many more names and illuminate many more lives, but these six are their main protagonists.

The Russian trio have been greatly influenced by the Name Worshiping sect in the Russian orthodoxy that evolved at about the same time. The adherents of the sect perceived the beginning of Genesis "And God said, Let there be light; and there was light" as affirming the existence of the word ("light" in this case) prior to the existence of the object or phenomenon that the word named. And so they worshiped the name of God as if it were God itself. Since God name is a linguistic idiosyncrasy, the sect's practice appeared to conflict with the established beliefs in the unity of orthodox God. The history of the sect, including the 1913 assault by tsarist marines on a Russian monastery on a Greek peninsula, was not a part of school curriculum, even in Russia. I was fascinated with the account.

This is an unusual book that eludes categorization. It's an outline of fundamental mathematical ideas cultivated by human beings, of mathematics as a human endeavor in the most candid sense of the word. It's a collection of biographical sketches - and not only of mathematicians - on a historic background, spread from the Dreyfus affair in France, and over the failed Russian revolution of 1905, the WWI, the October revolution, the Stalinists purges, the WWII, and post-Stalinist experimentations.

The book is a tangle of documented evidence and, likely, anecdotal testimony. It's warm, humane and makes an absorbing reading.