# Example of a very simple math statement in old literature which is (verbatim) a pain to understand

As you know, before symbolic notations were introduced and adopted by the mathematical community, even simple statements were written in a very complicated manner because the writer had (nearly) only words to describe an equation or a result.

What is an example of a very simple math statement in old literature which is, verbatim, a pain to understand?

• "written in an old paper" so you want modern examples? – 6005 Aug 12 '16 at 3:03
• @6005 thanks. I removed that part. – Cauchy Aug 12 '16 at 3:07
• Try almost anything on calculus or mechanics written by Isaac Newton, if you want to see what these subjects look like in the language of Euclidean geometry. – DanielWainfleet Aug 12 '16 at 3:57
• I don't understand the necessary languages to do this, but if anybody else does then it would be interesting to compare the Chinese Remainder Theorem in its thirteenth century Chinese form with Gauss' presentation in Disquisitiones Arithmeticae, and then compare both to the (very) modern form: $$\mathbb{Z}/N \mathbb{Z} \cong \mathbb{Z}/n_1 \mathbb{Z} \times \mathbb{Z}/n_2 \mathbb{Z} \times\cdots\times \mathbb{Z}/n_k \mathbb{Z},$$ where $N = n_1n_2\cdots n_k$ and the $n_i$ are pairwise coprime (and we know the isomorphism). – Will R Aug 12 '16 at 19:09
• Pythagoras' theorem? – user207421 Aug 14 '16 at 2:25

## 15 Answers

Here's Proposition 2 from Book 5 of Euclid's Elements:

If a first magnitude and a third are equal multiples of a second and a fourth, and a fifth and a sixth are equal multiples of the second and fourth, then the first magnitude and fifth, being added together, and the third and the sixth, being added together, will also be equal multiples of the second and the fourth, respectively.

Or in modern notation: $a(x + y) = ax + ay$.

• If $$\exists b ~\bigg( a_1 = b a_2 \land a_3 = b a_4 \bigg)$$ and $$\exists c ~\bigg( a_5 = c a_2 \land a_6 = c a_4 \bigg)$$ then $$\exists d ~\bigg( (a_1 + a_5) = d a_2 \land (a_3 + a_6) = d a_4 \bigg)$$ Wow. – DanielV Aug 12 '16 at 3:31
• At the risk of sounding like a complete idiot, I'll bite the bullet and ask: why couldn't he just say, "A multiple of the sum of two numbers is equal to the sum of the same multiple of each individual number"? Or something like that? – Mehrdad Aug 12 '16 at 8:24
• @Mehrdad Euclid, following the traditions of the day (and particularly Eudoxus), was skeptical about "number". Mathematical proof was, for him, geometry. So he avoids "number" but uses "magnitude", which might be the length of a line segment (etc). He understands and is quite strict about the axiomatic method. But he doesn't know what the axioms of numbers are, and it probably hasn't occurred to him to define axioms for number theory. So everything must be cast in geometric terms. – James K Aug 12 '16 at 16:51
• @JamesK: I mean, replace it with "magnitude" if you want. Or whatever he liked. That doesn't really address my question. – Mehrdad Aug 12 '16 at 17:35
• Another translation, keeping the geometric notion of proportions, might be: If $a_1:a_2=a_3:a_4$ and $a_5:a_2=a_6:a_4$ then $(a_1+a_5):a_2=(a_3+a_6):a_4$ – Hagen von Eitzen Aug 13 '16 at 11:47

See Tartaglia's formula for the cubic equation, which is a poem with over 20 lines. (http://www.maa.org/press/periodicals/convergence/how-tartaglia-solved-the-cubic-equation-tartaglias-poem)

Admittedly, the cubic formula isn't "simple", but it is still simpler in the modern notation compared to ancient notation.

1. When the cube with the cose beside it $\langle x^3+px\rangle$
2. Equates itself to some other whole number, $\langle =q\rangle$
3. Find two other, of which it is the difference. $\langle u-v=q\rangle$
4. Hereafter you will consider this customarily
5. That their product always will be equal $\langle uv=\rangle$
6. To the third of the cube of the cose net. $\langle p^3/3\mbox{, instead of }(p/3)^3\rangle$
7. Its general remainder then
8. Of their cube sides, well subtracted, $\langle\sqrtu-\sqrtv\rangle$
9. Will be the value of your principal unknown. $\langle =x\rangle$
10. In the second of these acts,
11. When the cube remains solo, $\langle x^3=px+q\rangle$
12. You will observe these other arrangements:
13. Of the number $\langle q\rangle$ you will quickly make two such parts, $\langle q=u+v\rangle$
14. That the one times the other will produce straightforward $\langle uv=\rangle$
15. The third of the cube of the cose in a multitude, $\langle p^3/3\mbox{, instead of }(p/3)^3\rangle$
16. Of which then, per common precept,
17. You will take the cube sides joined together. $\langle\sqrtu+\sqrtv\rangle$
18. And this sum will be your concept. $\langle =x\rangle$
19. The third then of these our calculations $\langle x^3+q=px\rangle$
20. Solves itself with the second, if you look well after,
21. That by nature they are quasi conjoined.
22. I found these, & not with slow steps,
23. In thousand five hundred, four and thirty
24. With very firm and strong foundations
25. In the city girded around by the sea.

For comparison, the solution of $x^3+px=q$ (lines 1–9) given in modern notation in the above article is:

$x = \sqrt{ \sqrt{\Big({\frac{q}{2}}\Big)^2 + \Big({\frac{p}{3}}\Big)^3}+\frac{q}{2}} - \sqrt{ \sqrt{\Big({\frac{q}{2}}\Big)^2 + \Big({\frac{p}{3}}\Big)^3}-\frac{q}{2}}$

• Anyone who says modern notation isn't better should be handed this and asked to say what it's about. – DanielWainfleet Aug 12 '16 at 4:00
• @user254665: and then, for maximum fun, once they have correctly identified the answer, handed lyrics to a R.E.M. song and asked to prove the theorem stated therein. – Tobia Tesan Aug 12 '16 at 7:08
• It appears that Tatraglia's separation into different cases is due to an avoidance of or ignorance of negative numbers as actual objects of arithmetic. – DanielWainfleet Aug 12 '16 at 7:20
• So Tartaglia explains the solution of $x^3 + px=q$ in about 60 words. The modern accounts are maybe half as many words, plus algebraic notation. A surprisingly small savings. – zyx Aug 12 '16 at 15:52
• @zyx: I take it you count lines 1–9, the first case; by a quick count I make that 51 words in Italian, 66 in the English translation above. The modern solution given in another part of the same article I count as 43 symbols or in TeX source 163 non-blank characters. I agree that Tartaglia doesn’t come out badly! – PJTraill Aug 14 '16 at 21:19

This is the computation of the weight of a copper icosahedral shell (that is, the weight of 20 equilateral triangular copper plates, all of the same thickness), done in during the Kaassite period in Mesopotamia, in the second half of the second millennium BC. For comparison, in modern notation we would just write something like $$20\times (\text{area per plate}) \times (\text{assumed thickness of each plate}) \times (\text{assumed density of the material}),$$ and proceed to perform the computation.

The image comes from an article in the AMS Notices, A Remarkable Collection of Babylonian Mathematical Texts.

• I'm not sure what impresses me more, the fact that they were able to do this already at the time or the fact that someone actually figured out what the content on this tablet means... I guess it's a draw. – Philipp Aug 12 '16 at 8:21
• @Mariano, lovely post, but the tablet is not a computation of the volume of the solid icosahedron. It assumes that the outer shell of a large icosahedron is made of 20 equilateral triangle copper plates, and calculates the weight of copper as the product (20 plates) x (area per plate) x (assumed thickness of each plate) x (assumed density of the material). Calculating the volume of a solid icosahedron is much more complicated than that. – zyx Aug 12 '16 at 16:16
• It is the weight of a thin shell surrounding the icosahedron; surface area of the icosahedron multiplied by a nominal thickness of the plates making the surface. Not the weight or volume of the 3-dimensional solid polyhedron. In authors' words (p.1085), "the combined volume V of the 20 triangular copper plates, all 1 finger thick", where 1 finger is considered small relative to the dimensions of the icosahedron. – zyx Aug 12 '16 at 17:58
• @zyx is correct and I would suggest that the OP modify the answer accordingly. There is no evidence from that paper that the Babylonians were able to calculate the solid volume of an icosahedron. Rather, they only calculated the volume of its "1 finger thick" shell. – Matt Aug 12 '16 at 19:54
• @William: it seems reasonable to give the author the chance to do it first, in whatever way they prefer (or to refute it), but by now I agree the commenters should edit — but now I’ve edited it (proposed) anyway, assuming they know what they are talking about. – PJTraill Aug 14 '16 at 21:46

Classical Chinese Mathematics was usually presented completely using text, though diagrams and rod numerals were sometimes used. As an example, here is the first problem from 海島算經, "The Sea Island Mathematical Manual" As with all texts prior to the twentieth century, it is written in Literary Chinese, distinguished from vernacular Chinese by its conciseness (like Latin), pro-dropness and lack of comprehensibility to commoners. There is also no punctuation, so experience with Literary Chinese is needed to read the original text without struggling. Chinese text (with punctuation) is here.

Translated:

Now we see a sea island, erect two pillars, of uniform height three 丈 ($\ell$), forward and backward a thousand 步 between them ($d$), making the rear pillar and the front pillar collinear [with the island]. Walking backwards from the front pillar by one hundred and twenty three 步 ($a$), with eye on the ground looking at the peak of the island, the end of the pillar meets [the peak]. Walking backwards from the rear pillar by one hundred and twenty seven 步 (steps) ($b$), with eye on the ground looking at the peak of the island, the end of the pillar also meets [the peak]. We ask how tall ($h$) and how far ($x$) from the pillar is the island?

Answer saith: the island is four 里 fifty five 步 tall ($h$); one hundred and two 里 one hundred and fifty 步 from the pillar ($x$).

Method saith: take the pillar height multiplied by the pillar separation as the numerator, the difference [in backward walking distance] as the denominator, and divide these. To the result add the pillar height, to obtain the island height. To find the distance from the front pillar to the island: take the front pillar backward walk multiplied by the pillar separation as the numerator, the difference [in backward walking distance] as the denominator. Divide these, to obtain the distance to the island from the pillar.

In today's notation, the result for the island height is

$$h = \frac{\ell \cdot d}{b-a} + \ell$$

while the distance to the island from the front pillar is

$$x = \frac{a \cdot d}{b-a}.$$

Expressing the solution is already complicated enough; I wonder how it was derived. If you wish to work with the actual quantities in this problem, the relevant conversions are $里 = 300 步$, $3丈 = 5步$.

@WillR While the Chinese Remainder Theorem would be very interesting to compare, it is too big a task (blocks 4 to 25 here) for me to do now. Maybe some other time.

• As remarked on the question, Alfred J. van der Poorten’s entertaining Notes on Fermat’s Last Theorem claims to reproduce some form of the Chinese Remainder Theorem in Chinese. – PJTraill Aug 14 '16 at 21:34
• Very interesting! I might make a reference request for some good sources to learn about the history of non-Western mathematics. All I really know about Asian mathematics is that there are some things called "Sangaku," the Chinese did a fair amount of stuff, Brahmahgupta was the first to write down things about $0$ as a number and there's a famous proof-without-words of the Pythagorean theorem which is supposedly of Indian origin. By contrast, I know a fair amount about what the ancient Greeks did, and the maths of Descartes, Pascal, Fermat, Newton, the Bernoullis, Euler, Gauss, and so on. – Will R Aug 15 '16 at 1:06

Here is how Euclid describes a way (actually the only way) to get an even perfect number.
Theorem:
If $2^n-1$ is prime then $2^{n-1}(2^n-1)$ is perfect.

"If as many numbers as we please beginning from a unit are set out continuously in double proportion until the sum of all becomes prime, and if the sum multiplied into the last makes some number, then the product is perfect"

mind-boggling...

• – Frenzy Li Aug 14 '16 at 14:30
• What it actually says is: if $\sum_{i=0}^k2^i$ is prime, then $(\sum_{i=0}^k2^i)2^k$ is perfect. – Marc van Leeuwen Aug 15 '16 at 10:09

Dehn's solution in 1900-1901 of the Hilbert 3rd problem lacked the notation and theory of tensor products. This complicates things because the solution is completely tensor-theoretic: view the polyhedron as an element of (space of edges) $\otimes$ (group of dihedral angles).

About a year ago I happened to get a little known book that in one place deals tangentially with what you're asking. I don't remember how I even learned about this book, but I managed to find a copy to buy on the internet.

William Hope-Jones. A Memoir, privately published by Richard Clay (The Chaucer Press), 1968, x + 70 pages.

William Hope-Jones (1884-1965) taught mathematics for a long time at Eton College (beginning in 1906 or 1907; the book does not appear to give an ending year) and he was president of the Mathematical Association (England) in 1938. On the 1906 Tripos exam, Hope-Jones was in a 3-way tie for the last three Wrangler positions (#31-33).

The following is from pp. 48-49 of this book and is the first part of a well received talk titled Shams that Hope-Jones gave in 1948.

Probably not many of you have learnt the Propositions in Euclid's 2nd Book of the Elements of Geometry. In the modern world we make these far more concise and intelligible by the use of Algebra, which I define as "a way of making things more workable by giving them short names". But the earliest Algebra that has come down to us dates from Diophantus, who lived some seven centuries after Euclid; and for want of a way of making things more workable by giving them short names, Euclid enunciated portentously indigestible propositions such as "If a straight line be bisected and produced to any point, the square on the whole line thus produced and the square on the part of it produced are together double of the square of half the line bisected and of the square of the line made up of the half and the part produced."* [Note: Footnote * is $(2a+b)^2 + b^2 = 2\left\{a^2 + (a+b)^2 \right\}.$] Fifty years ago we were expected not only to learn this sort of stuff (which I never succeeded in doing) but also to do questions on it in the same style. And this is how the questions were set, done and dealt with. First, the master did a simple piece of Algebra secretly. Second, he dressed it up in the asphyxiating geometrical language which was thought at that time to be "consistent with the dignity of the subject". Third, he set it in this form to the boys. Fourth, the boys translated it into Algebra so as to find out what it meant. Fifth, they solved the thing very easily by Algebra. Sixth, they translated their solutions into geometrical language consistent with the dignity of the subject. Seventh, they showed up their finished product to the master. Eighth, the master took their solutions away and translated them back into Algebra, being the only way in which he could find out what they were about. Of these eight processes, you will see that four are mere translation, not concerned with the inventing, solving or correcting of the question, but only with satisfying some arbitrary convention of decency which the master probably no more believed in than we did $[\;\cdots \;]$ In fact all this abysmal inefficiency and the boredom which it induced was nothing but a gigantic human sacrifice to the god of Sham.

Note: The November 1891 Examination for Admission into The Royal Military Academy at Woolwich [SEE: Woolwich Mathematical Papers for Admission into The Royal Military Academy for the Years 1891-1900, edited by Eldred John Brooksmith (1856-1932), 1901], Section I. Euclid, problem #4, states the following (see top of p. 27 of the .pdf file):

Enunciate the two propositions represented by $(2a \pm b)^2 + b^2 = 2\left\{a^2 + (a \pm b)^2 \right\}.$ AB is bisected at C and produced to D; AE, which is at right angles to AB, is divided into two equal parts at F and into two unequal parts at G; CH, BI, DKL are drawn parallel to AE; and FK, GHL, EI parallel to AD. Show that the sum of the squares on GD and IL is double of the sum of the squares on FC and HK.

This is not completely related, but in the same spirit: before the standard notation of exponentation, Robert Recorde suggested the phrase zenzizenzizenzike to represent eighth powers. He wrote that it "doeth represent the square of squares squaredly" - the term essentially is just the word for square three times.

• This reminds me of the British names for note values (lengths) in music. An eight note is a quaver, a sixteenth note is a semiquaver… a two hundred fifty-sixth note is a demisemihemidemisemiquaver. – Akiva Weinberger May 12 '17 at 0:52

Much of basic linear algebra was discovered in the absence of adequate language and notation.

The multiplication property of determinants was published in 1812 by Cauchy and Binet. This was 40 years before the appearance of matrices, 66 years before the axiomatization of vector space, and about 100 years before the development of multilinear algebra.

There was no notation at the time to write $\det(AB)=\det(A)\det(B)$ since "$AB$" was a "multiplication" of things that were not codified yet. The volume interpretation of determinant was not available. Thus neither the proper statement nor the intuitive proof of that statement were possible.

Euler had a lot of work on what we would now call orthogonal matrices, quadratic forms and sum of squares identities. Much of this was abstruse-looking but becomes almost trivial with the ability to write words like "quaternions" or use matrix notation and some linear algebra.

Plato’s Parmenides, written 370 B.C.E.: First Hypothesis, Second Deduction (154b-d) has Parmenides questioning Aristotle on the form of time:

b) If one thing is in fact older than another, it would be impossible for it to become older still by an amount greater than its original difference in age; and again, what is younger cannot become still younger. For adding equals to unequals, in time or anything else whatever, always makes the difference equal in the amount by which the unequals originally differed.

Of course.

c) So what is would never be becoming older or younger than the unity which is, since the difference in age is ever equal: it is and has become older, and the other younger, but is not becoming so.

True.

So unity, since it is, is never becoming older or younger than the other things which are.

No.

Consider then whether it is becoming older and younger in this way.

In what way?

Insofar as unity appeared older than the others and the others older than unity.

Well?

When unity is older than the others, it has come to be, I take it, for a time greater than the others.

Yes.

d) Consider then again: if we add equal time to greater and less time, will the greater differ from the less by an equal (proportional) part or a smaller one?

By a smaller one.

So whatever difference in age there was to begin with between unity and the others will not continue into the future, but since unity takes time equal to the others it will always differ from them in age less than before. Agreed?

Yes.

In modern notation: as two people age together, the difference in their ages will be a constant, but the ratio of their ages will tend towards $1:1$. So letting $a$ and $b$ be their ages with $a>b$, and $e$ the equal time added to both, then

$$e:e<a:b \quad \text{and} \quad e:e<(a+e):(b+e)<a:b$$

Some old formula-based subjects like special functions, umbral calculus and $q$-identities have statements that are often hard to understand or prove in the original, but simple to explain using algebraic theories and notation that developed in the 20th century.

George Abram Miller (31 July 1863 – 10 February 1951) was an early group theorist whose many papers and texts were considered important by his contemporaries, but are now mostly considered only of historical importance. Much of his work consisted of classifying groups satisfying some condition, such as having a small number of prime divisors or small order or having a small permutation representation, or enumerating the possible finite groups which satisfy given conditions such as: the prime factors which divide the order, the orders of two generating permutations and their product; the types of subgroups; or the degree of a representation as a permutation group. Several papers investigate groups generated by two elements satisfying given conditions. For example he considered groups generated by two elements of order three whose product is of order four or three or six. He also considered permutation groups of small degree, groups having a small number of conjugacy classes, multiply transitive groups, and characteristic subgroups of finite groups. He found the list of all possible groups of order 1909 to 1919 inclusive. His Collected Works were published in five very large volumes. His style however, was very much aimed at using words, rather than symbols and formulas. No wonder that his collected works are so voluminous. He also use old group theory terminology (an element of a group is called an operator, a normal subgroups is called invariant, and more, etc.). See for instance this: Hilbert's own formulation of his three theorems on commutative algebra (Nullstellensatz, basis theorem, syzygy theorem).

• What were these formulations? – paf Aug 15 '16 at 23:22
• @paf: see 'Ueber die Theorie der algebraischen Formen' by Hilbert (they are called Theorem I,II,III respectively) – syzygy Aug 15 '16 at 23:56
• Thanks for the reference but I don't have this book and I don't speak German... – paf Aug 15 '16 at 23:57
• Hilbert is not from old litterature , he is alive in many heads – user354674 Aug 16 '16 at 0:34

The statement of the Pythagorean Theorem always seems to confuse high school students (at least the ones in my classes): "The square of the hypotenuse is equal to the sum of the squares of the other two sides."

• just insert after hypotenuse that it is the side opposed to the right angle and all will be clear ... – user354674 Aug 16 '16 at 0:33

Euclid - "a point is that which has no part"

• How is this translated to modern notation? – miracle173 Aug 15 '16 at 3:22
• From all the things that Euclid wrote, this seems the more clear of all. – ypercubeᵀᴹ Aug 15 '16 at 7:16