# The Rhind Mathematical Papyrus

The Rhind Papyrus is an ancient Egyptian mathematical papyrus which dates from about 1650 BC. This is around the time of ancient Egypt's middle kingdom (a bit later, it seems). The papyrus contains some (very) elementary problems of arithmetic and geometry - in particular, quantities are reckoned (somewhat arbitrarily!) in terms of /Egyptian fractions/, which are sums of unit fractions (fractions whose numerator is one):

en.wikipedia.org/wiki/Egyptian_fraction

In the mid-19th century one Alexander Rhind, a Scottish antiquarian, purchased the papyrus in a shady deal; this is where the texts' namesake comes from. Mr. Rhind died a few years later, and so the papyrus came to be in the possession of the British Museum, where it lives today.

I went to the library and I have an English edition (Chace) of the papyrus close at hand, so I'm going to go over it and post about it in this thread. It will be a very, very simple mathematical recreation, which yet has some historical interest, and then I can say I've been through a meme text. If you find this thread boring, you're welcome to hide it, but as we'll see, it has a perfect right to be on this board.

Comparison with my library book shows that this OP picture is problems 45-60, or thereabouts.

Very cool, I'll be following this thread.

Me too, sounds pretty cool. Pretty, pretty, pretty cool.

Unfortunately the Egyptians' attitude towards math was purely utilitarian. They did not contemplate theoretical stuff, but only sought methods to perform calculations directly connected to real life problems like measuring areas of polygon shapes etc.

The Rhind Papyrus was originally a single roll that was about 18 feet long by 13 inches high, and writing is found on both sides (although a large part of the backside is blank). A few fragments of this are kept in New York, now. This is the text I will drill into shortly, but I want to mention a few others first. You can skip ahead to the wiki if you like:

en.wikipedia.org/wiki/Rhind_Mathematical_Papyrus

Now, when Mr. Rhind got the papyrus, he also got this small, tough, rolled-up piece of leather that didn't get un-rolled until the 1920s (chemical treatment to do that), just in time to be photographed for this Chace edition that I've checked out. This is a small table of more Egyptian fractions, and it is called the Mathematical Leather Roll:

en.wikipedia.org/wiki/Egyptian_Mathematical_Leather_Roll

Moscow and Berlin each have a so-called (Egyptian) papyrus as well, and wiki lists a few other items along these lines. I wanted to see if any quadratics would be mentioned in the Rhind papyrus (I know ancients were aware of them), but at first glance it doesn't sound like it. Quadratics, however, were treated by Babylonians, Greeks, and very slightly in the above Berlin papyrus, with the Pythagorean babby example in the link:

en.wikipedia.org/wiki/Berlin_Papyrus_6619

So. Egyptian fractions, not very advanced stuff. Historical interest. Having set the stage, let us recapitulate the CONTENTS of the Rhind papyrus.

The AUTHOR of the text is one Egyptian scribe by the name of Ahmose (or Ahmes), as he identifies himself (along with a pharoah or two) in the TITLE PAGE.

This is followed by a table that does the following: take 2, divide it by all the odd numbers from 3-101, and then convert those fractions into some (arbitrary!) Egyptian fractions, showing work in each case. This is known as the "2/N TABLE", and I don't feel like going into detail on it, so here's a simple link of this part of the text instead, with what was done:

en.wikipedia.org/wiki/Rhind_Mathematical_Papyrus_2/n_table

This is followed again by a much shorter "1-9/10 TABLE" which gives particular Egyptian fractions for each of 1/10 ... 9/10. I haven't checked yet but I would imagine that both of these tables set the stage for the

PROBLEMS. According to Chace, I've come up with a total of 91 problems; these will be the focus of the remainder of my treatment. As for Chace's treatment, they are numbered as 1-87 inclusive, along with four additional items noted as 7B, 59B, 61B, and 82B. Note that this may vary slighly from wiki's treatment of the overall text (my library book is literally 90 years old. Conventions may have changed since then.)

And that's the Rhind Papyrus, in a nutshell.

Pic related is a nice look at the Chace treatment of the Egyptian, which isn't really what I"m interested in. He has relatively modern transliterations of the problems, which is where I'll spend the rest of my time. I reserve the right to summarize what's being done in even more modern terms.

The first 21 problems, problems 1-20, 7B are simple divisions and multiplications, which will probably have been reckoned in terms of the above "1-9/10 TABLE" once I finish going through them. So let me first simply give the modern version of this much smaller table, which comes right before the problems begin. Arbitrarily chosen Egyptian fractions!

$\displaystyle \frac{1}{10} = \frac{1}{10}, \\ \displaystyle \frac{2}{10} = \frac{1}{5}, \\ \displaystyle \frac{3}{10} = \frac{1}{5} + \frac{1}{10}, \\ \displaystyle \frac{4}{10} = \frac{1}{3} + \frac{1}{15}, \\ \displaystyle \frac{5}{10} = \frac{1}{2}, \\ \displaystyle \frac{6}{10} = \frac{1}{2} + \frac{1}{10}, \\ \displaystyle \frac{7}{10} = \frac{2}{3} + \frac{1}{30}, \\ \displaystyle \frac{8}{10} = \frac{2}{3} + \frac{1}{10} + \frac{1}{30}, \\ \displaystyle \frac{9}{10} = \frac{2}{3} + \frac{1}{5} + \frac{1}{30}. \\$

I should also mention that for whatever reason, the Egyptians/Ahmose allowed an exception to this "get everything into Egyptian fractions" condition: /he allows of the fraction $\frac{2}{3}$ ./ And it is used above.

For this very simple fragment, I did allow myself to check the original, and I rapidly made sense of the glyphs. Kinda neat.

Now the problems themselves. 1-6 are simple divisions, and so are simple repetitions of the previous table's information, with work shown to prove out the concepts.

Problems 1-6 are, respectively, "Divide 1,2,6,7,8 and 9 loaves among 10 men. Each man receives x [quotient]." In every case, the appropriate Egyptian fraction from the table just posted is given as an answer, being that quotient.

Some multiplications follow, but I want to check these first. This is where I start working through the thing.

7-20, 7B are prosaic multiplications, and 7B is a simple re-statement of 7 (as well as of 10, just with slightly different work shown), so now I understand what those "B" problems are about, they're just copies/errata. For whatever reason, Ahmose keeps using the same two multiplicands over and over again here, so I'll just tidy it up by starting with what he uses and making my own shorthand. Ahmose's original Egyptian-fractional rendering is assigned by me as two (modern) constants, S as in Seven-fourths, and E as in Eight-fourths, thus:

$\displaystyle \frac{1}{1} + \frac{1}{2} + \frac{1}{4} = \frac{7}{4} = S \;\;\; ; \;\;\; \frac{1}{1} + \frac{2}{3} + \frac{1}{3} = 2 = E .$

The problems, then, amount to the below. I think this will start getting less dreary soon, but in any event I'm committed now. No, there's no typos below (shouldn't be), he did write (effectively) the same problem thrice over. Notice that multiplier, multiplicand and product are all given as Egyptian fractions.

$\displaystyle P.7: \bigg( \frac{1}{4} + \frac{1}{28} \bigg)S = \frac{1}{2} \\ \displaystyle P.7B: \bigg( \frac{1}{4} + \frac{1}{28} \bigg)S = \frac{1}{2} \\ \displaystyle P.8: \frac{1}{4} E = \frac{1}{2} \\ \displaystyle P.9: \bigg( \frac{1}{2} + \frac{1}{14} \bigg)S = 1 \\ \displaystyle P.10: \bigg( \frac{1}{4} + \frac{1}{28} \bigg)S = \frac{1}{2} \\ \displaystyle P.11: \frac{1}{7} S = \frac{1}{4} \\ \displaystyle P.12: \frac{1}{14} S = \frac{1}{8} \\ \displaystyle P.13: \bigg( \frac{1}{16} + \frac{1}{112} \bigg)S = \frac{1}{8} \\ \displaystyle P.14: \frac{1}{28} S = \frac{1}{16} \\ \displaystyle P.15: \bigg( \frac{1}{32} + \frac{1}{224} \bigg)S = \frac{1}{16} \\ \displaystyle P.16: \frac{1}{2} E = 1 \\ \displaystyle P.17: \frac{1}{3} E = \frac{2}{3} \\ \displaystyle P.18: \frac{1}{6} E = \frac{1}{3} \\ \displaystyle P.19: \frac{1}{12} E = \frac{1}{6} \\ \displaystyle P.20: \frac{1}{24} E = \frac{1}{12} \\$

The egyptians were black too werent they?

No. Does their visual art make them look black?

Yes.

They're just tanned (those in your pic who even have darker skin, which is only like half of them all). Their facial features don't look negroid at all.

Ancient egyptians were black as in skin color but not negroid.

There's no such thing as negroid. if they were black Africans they were black Africans.

Do you think only africans can have black skin color retard ?

More proof that defining people by their skin colour is dumb as shit.

>anthropologists had been so wrong for centuries, thankfully tumblerites finally arrived and set anthropology straight

>19th century social theorists were never wrong
Funny how if Veeky Forums agrees with a theory it's the right one, if Veeky Forums doesn't agree with it (race, climate change, string theory) it's a big conspiracy by the joos to hide the real theory.

Not him but traditional visual art of royalty is a shitty 1:1 representative of the composition of a given nation's population.

Using this logic one would surmise there were virtually no native americans and very few blacks in america using traditional monument/statues and paintings as evidence.

Obviously this would be wrong since they did exist in decent numbers but had little representation in the ruling government.

Same shit could have happen in egypt which would make sense since Nubia did trade with them and eventually got it's shot in ruling over the nation for a bit.

Come on, the whole "race is skin-deep" nonsense is just political agenda that was pushed way futher than political agendas should ever be allowed to be pushed if science was to remain integrity. Anyone who's not a halfwit and not commited to that political agenda knows it.

Things get slightly more interesting in passing from the first 20 problems to the next 20; we pass now from arithmetic into /algebra/, and although "a quantity" is mentioned more than once in a given problem, all we are really doing throughout is solving for x in what all(?) ultimately simplify to linear equations. Ahmose seems to understand "solve for x" and his own manipulation methods for actually doing so, but he doesn't seem at this point to have quite our same notion of a linear equation.

That said, I am simply summarizing the mathematics. What I'm doing is to re-state the problems in modern form, then wade through each problems' "work" to pick out what looks like the solution, and check whether I agree. So far I haven't noticed a straight-up goof (Ahmose is supposed to have made these as well); I'm through problem 30 and I haven't caught a bad "final answer" yet, though perhaps there are goofs in his work-steps, which I'm not reading.

The /language/ of these problems' statement is quite similar to our abstraction in that "a quantity" is usually spoken of; only a few are concrete story-problems (divide these loaves, use those hekat...) A "hekat", by the way, is a volume unit of about 5 liters that they used for grain.

en.wikipedia.org/wiki/Hekat_(unit)

I shall present modern summaries of problems 21-30 in the sequel, followed by 31-40. Then the Rhind Papyrus passes into geometry, which should be a bit more interesting.

Actually, even the ordering of the material also makes some modern sense: rote tables -> arithmetic -> algebra -geometry -> presumably some slightly more complex problems at the end.

Here is a modern summary of problems 21-30. One item of interest here is that the lcm of the denominators used in the long problem 23 is precisely 360, a number we can thank Babylon for IIRC.

$\displaystyle P.21: \bigg( \frac{2}{3} + \frac{1}{15} \bigg) + x = 1 \;\;\; \rightarrow \;\;\; x = \frac{1}{5} + \frac{1}{15} \\ \displaystyle P.22: \bigg( \frac{2}{3} + \frac{1}{30} \bigg) + x = 1 \;\;\; \rightarrow \;\;\; x = \frac{1}{5} + \frac{1}{10} \\ \displaystyle P.23: \bigg( \frac{1}{4} + \frac{1}{8} + \frac{1}{10} + \frac{1}{30} + \frac{1}{45} \bigg) + x = \frac{2}{3} \;\;\; \rightarrow \;\;\; x = \frac{1}{9} + \frac{1}{40} \\ \displaystyle P.24: x + \frac{1}{7} x = 19 \;\;\; \rightarrow \;\;\; x = 16 + \frac{1}{2} + \frac{1}{8} \\ \displaystyle P.25: x + \frac{1}{2} x = 16 \;\;\; \rightarrow \;\;\; x = 10 + \frac{2}{3} \\ \displaystyle P.26: x + \frac{1}{4} x = 15 \;\;\; \rightarrow \;\;\; x = 12 \\ \displaystyle P.27: x + \frac{1}{5} x = 21 \;\;\; \rightarrow \;\;\; x = 17 + \frac{1}{2} \\ \displaystyle P.28: \bigg( x + \frac{2}{3} x \bigg) - \frac{1}{3} \bigg( x + \frac{2}{3} x \bigg) = 10 \;\;\; \rightarrow \;\;\; x = 9 \\ \displaystyle P.29: \frac{1}{3} \Bigg( \bigg( x + \frac{2}{3} x \bigg) + \frac{1}{3} \bigg( x + \frac{2}{3} x \bigg) \Bigg) = 10 \;\;\; \rightarrow \;\;\; x = 13 + \frac{1}{2} \\ \displaystyle P.30: x \bigg( \frac{2}{3} + \frac{1}{10} \bigg) = 10 \;\;\; \rightarrow \;\;\; x = 13 + \frac{1}{23} \\$

just as black as they are today

This is a really neat thread. That Wikipedia article (among other things ) encouraged me to start learning ancient Egyptian a while ago. I never really stuck with it though.

This is actually false. The Nubians would have been highly intermingled with Egypt at this time. There have been a few negroid groups that once lived in Egypt and have since been dispersed back into continental Africa.

Happy to hear that some anons appreciate the thread. I thought the thread might have a we wuz kingz spate (it did) but hopefully we're past that.

Grinding into this, my pace is slowing down, which is a nice sign. The remaining problems up through 40 are solutions of linear equations as expected, but a bit more "culture" is involved: a /ro/ is 1/320 of a hekat, as given above, and certain solutions are alternately expressed in terms of "Horus-eye" fractions. These are simply reciprocals of low powers of 2, which were culturally associated with parts of pic related. We are veering dangerously close to pseudoscience /x/ territory (in a modern context), but I make the mention since this is a historical document after all.

Problem 35 in particular involves finding the volume of an unknown unit measure, versus the established hekat measure, and has a somewhat confused original wording. I shall gloss over these to present 31-35, and eventually (once I've checked them), 36-40. Then the geometry, later.

This is the picture meant to go with this post, whoops. no harm done.

>Unfortunately the Egyptians' attitude towards math was purely utilitarian.
Yes, engineers are not scientists; they are not engineers who love to think that it is worthy dwelling in your mental proliferation.

Problems 31-35 can be paraphrased as follows. 31 has a particularly onerous solution, but Ahmose seems to like using factors of 28 and certain small primes for his examples. Or at least, his predecessors did - the scroll is described as repeating methods previously handed down.

$\displaystyle P.31: x + \frac{2}{3} x + \frac{1}{2} x + \frac{1}{7} x = 33 \;\;\; \rightarrow \;\;\; x = 14 + \frac{1}{4} + \frac{1}{56} + \frac{1}{97} + \frac{1}{194} + \frac{1}{388} + \frac{1}{679} + \frac{1}{776} \\ \displaystyle P.32: x + \frac{1}{3} x + \frac{1}{4} x = 2 \;\;\; \rightarrow \;\;\; x = 1 + \frac{1}{6} + \frac{1}{12} + \frac{1}{114} + \frac{1}{228} \\ \displaystyle P.33: x + \frac{2}{3} x + \frac{1}{2} x + \frac{1}{7} x = 37 \;\;\; \rightarrow \;\;\; x = 16 + \frac{1}{56} + \frac{1}{679} + \frac{1}{776} \\ \displaystyle P.34: x + \frac{1}{2} x + \frac{1}{4} x = 10 \;\;\; \rightarrow \;\;\; x = 5 + \frac{1}{2} + \frac{1}{7} + \frac{1}{14} \\ \displaystyle P.35: \bigg( 3 + \frac{1}{3} \bigg) x = 1 \;\;\; \rightarrow \;\;\; x = \frac{1}{5} + \frac{1}{10} \\$

btw I encourage checking of thes babby-problems.

Let me next present 36-39. I want to hold 40 separate, as it appears to ACUTALLY BE SOMEWHAT INTERESTING!!!

$\displaystyle P.36: \bigg( 3 + \frac{1}{3} + \frac{1}{5} \bigg) x = 1 \;\;\; \rightarrow \;\;\; x = \frac{1}{4} + \frac{1}{53} + \frac{1}{106} + \frac{1}{212} \\ \displaystyle P.37: \bigg( 3 + \frac{1}{3} + \frac{1}{3} \frac{1}{3} + \frac{1}{9} \bigg) x = 1 \;\;\; \rightarrow \;\;\; x = \frac{1}{4} + \frac{1}{32} \\ \displaystyle P.38: \bigg( 3 + \frac{1}{7} \bigg) x = 1 \;\;\; \rightarrow \;\;\; x = \frac{1}{6} + \frac{1}{11} + \frac{1}{22} + \frac{1}{66} \\ \displaystyle P.39: Given \;\; 6x = 4y = 50, find \;\; y-x. \;\;\ \rightarrow \;\;\; y-x = 4 + \frac{1}{6} \\$

Problem 40 concludes the "simple algebra" section, and has the added bonus of having enough stuff going on to be significantly more interesting than what has preceded it (in our accepted babby-terms).

After #40, the next several problems are the "geometry" section, and it is clear to me from skimming that this middle part (including this problem 40) is the "sexy" part of the text. Following the geometry section are some more prosaic "miscellaneous" bits.

Problem 40 has two phrasings because Chace provides both the original (much shorter) Egyptian transliteration, and a (relatively) modern English paraphrase of his own. But Chaces' paraphrase is complex/modern enough (sufficiently out of character with what has gone before) that I wanted to check the original, and there is a discrepancy in language, as I expected. Basically Ahmose was using a method and making assumptions that he doesn't "show work" on.

Problem 40 goes like this. Loaves are divided among men, and some bulleted conditions and questions are put on this.

1) 100 loaves are unevenly divided among 5 men according to the below,
2) the 5 mens' shares are in arithmetic progression, (according to Chace and Ahmoses' after-the-fact statement, this is "assumed", /not actually stated by Ahmose/, and fixed per 4), but we should consider a more general case for the fun of it)
3) 1/7 of the sum of the three largest shares is equal to the sum of the smallest two shares,
4) /The difference of the shares is "arbitrarily" fixed by Ahmose as 5 1/2, before he works the problem, almost after done stating it.

Simplifying this philology, the essential statement of the problem is as follows, where x_1 is the (presumably) largest share, x_5 the smallest:

$\displaystyle x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 100 \\ \displaystyle \frac{1}{7} \big( x_{1} + x_{2} + x_{3} \big) = x_{4} + x_{5} \\$

The arithmetic progression and Ahmoses' "5 1/2" are overlaid on this, which I will poke around with next.

But there was a hidden problem!

Problem 41:
n>2

a^n+b^n=c^n -> a=..., b=... and c=... ? It says proof is left to the reader but I can't find any simple proof.

There's more to that symbol

There's nothing dumb about using skin color as a statistical feature.

It has good predictive properties by itself and when combined with other features you have an even better prediction.

Leaving it alone for tonight, 40 has a few different angles to it and I'm drunk.

If you like this thread, be a lamb and bump it overnight.

Good stuff OP!

Don't be a Fer-memer.

en.wikipedia.org/wiki/Egyptian_fraction#Early_history

This wiki page says some interesting things about their notation, including:
>The Egyptians had special symbols for 1/2, 2/3, and 3/4 that were used to reduce the size of numbers greater than 1/2 when such numbers were converted to an Egyptian fraction series. The remaining number after subtracting one of these special fractions was written using as a sum of distinct unit fractions according to the usual Egyptian fraction notation.

So I guess if I had some fraction like 5/4 then I would write it as a sum of 3/4 + (stuff smaller than 1/2 in proper egyptian fractions).

The wiki also mentions some other special notation for numbers of the form 1/2^k that were used when dealing with real world stuff. Makes sense, sounds like how we use tenths, hundredths, thousandths, etc..

It would've helped your argument if you had actually posted a statue that didn't look like a black woman.

>thinks the argument is "race is skin-deep".
How embarrassing for you. Let me help you out user. Either learn population genetics or gb2/pol/.

I lol'd.

You're wrong. If you disagree with the fact that "race is skin-deep" and look for genetic variances you should gb2/pol/. Because we're all identical and phenotypes aren't real.

>This is what /pol/ actually thinks Veeky Forums actually thinks.

Racist science isn't science you bigot. Science must be politically correct without discrimination. And yes, this is what Veeky Forums thinks.

>moooooom come look at my awesome false flag!

Keep going retard. This is the best laugh I've had all day. You /pol/esmokers sure are dumb.

Are you retarded or something ? /pol/ is for political incorrectness, Veeky Forums is for science. Science can't be politically incorrect.
> false flag
> reaction image replies
You sure sound like /pol/ yourself.

The International Academy of Science, The National Academy of Science and the Human Genome Project have all officially rejected "Scientific Racism" as being pseudoscience.
They have provided evidence and proofs.

Deal with it. Your parents and community lied to you, as did Stormfront. Back to pol please.

>/pol/ is the only board with reaction images and image macros.

Replied to the wrong user but yes you are correct.

Political correctness is what we follow to avoid racial discrimination, which applies to science as well. Discussing inexistent pseudoscience like genetic variances and population genetics is what we call scientific racism.
see :

strange (becoz inconsistent) that they would write
$\displaystyle \frac{7}{10}= \frac{2}{3}+ \frac{1}{30}$
... rather than
$\displaystyle \frac{7}{10}= \frac{1}{2}+ \frac{1}{5}$
... ainnit?!

Population genetics has nothing to do with race. Race doesn't exist this isn't because of political correctness, but because genetics actually turned out to be much more complex.

"Racial science" and "scientific racism" are what tinfoil hat retards who can't into science (i.e. /pol/) use to justify their racism. Like you pointed out though, those things are now considered pseudoscience.

the dumb, it hertz

Population genetics is just a codeword for "racial science" or some other bullshit like that. Discussing genetic variances among ethnic groups is what causes discrimination and highlights differences between groups of people. Nobody cares if you call it something else than "race".
Why do you even care about it so much that you have to discuss it ?

OP can you Indian and Chinese math next?

I'm pretty sure OP went to sleep but I think the reason lies with this.
What surprises me more is that they didn't write 3/4 in 8/10 and 9/10. I wonder if that symbol just wasn't used as often.

That wiki page says some really interesting things under the equally distributing objects section. I can actually see why these Egyptian numbers are practical and not just a quirky limitation of their writing system.

en.wikipedia.org/wiki/Egyptian_fraction#Equally_distributing_objects

This.

I actually know two autistic /pol/ retards at my university. They constantly spout memes irl, talk about /pol/ and Veeky Forums, and rant racist shit with each other and at anyone who will listen.

In population genetics you can group populations in any way. For instance you can look at two different villages belonging to the same tribe as different populations and study their genetic distance.

Traditional notions of race on the other hand instead rely on physical features (nowadays considered shitty due to findings related to convergent evolution) and then try to make inferences about the most retarded shit.

They are different things, user. I don't think you could even abuse population genetics to study traditional notions of race if you wanted to.

>nooo, IAS, NAS, and HGP are all wrong
>La La La La Laaa, I can't heeeear you!
fgt pls

#40, cont.

Before I go further, the OBJECT of the problem is to find the difference of shares (which are in arithmetic progression, so you want a single number).

This is also the first situation where I feel compelled to comment on Ahmose's work. Upon review, I mis"spoke" earlier, and he does both "assume" /and actually state/ an expected arithmetic progression throughout. To solve, he takes a nicer version of his original problem that instead adds up to 60, and then simply scales it. He knows that this (scaling) works, but it passes without comment. Let's put on our modern-cap for a bit.

Let x_5, the smallest term, be 1. then just keep adding 5.5 (11/2) for each successive term, thus taking care of the arithmetic progression. x_1, the largest term, is 23 just now. The sum of the lot is 60, and the 1/7 relationship checks out.

Now to scale the lot up to 100, multiply by 5/3 (or, 1 + 2/3). Everything checks out again, and then you can convert into relatively easy egyptian fractions to wind up with the same thing that Ahmose did. But he doesn't state the solution as-such!

The answer is simply (11/2)(5/3) = 55/6, or, as the Egyptian would say, 9 + 1/6. An ugly-looking set of conditions turns out to be consistent, and may be expressed thus:

$\displaystyle P.40: x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 100 \\ \displaystyle \bigwedge x_{1} + x_{2} + x_{3} = 7( x_{4} + x_{5} ) \\ \displaystyle \bigwedge x_{1} > x_{2} > x_{3} > x_{4} > x_{5} \\ \displaystyle \bigwedge x_{1} - x_{2} = x_{2} - x_{3} = x_{3} - x_{4} = x_{4} - x_{5} = \Delta \\ \displaystyle \rightarrow \Delta = \frac{55}{6} = 9 + \frac{1}{6}$

>correlation proves things
Found the undergrad. Wait at least until after stats to post.

Out of curiosity, are the arguments in the text written out in prose or do they use some sort of syntax or what?

And yet, this single tersely-worked problem is unsatisfying to a modern for two reasons: it is the first problem in the set to suggest a generalization of any interest (which we present in pic related), and the author seems to have pulled his solution out of thin air (or perhaps chose his problem to coincide nicely wiht an already-worked solution). In other words, one ought to be able to derive a solution without having Ahmoses' specific-case method spoon-fed.

We can begin a modern solution by taking three of the above four conditions as a simpler starting point, and working towards the fourth. This is worked out to my satisfaction in pic related. The short version is that in Ahmoses' problem as stated his solution is not only correct, but more interestingly, unique.

THERE. Now I'm done with #40, and what feels like the first half of the Rhind Papyrus. 41-87 remain, being simple geometry and some miscellany (more algebra I guess).

The original Egyptian (just to skim it) and its modern English transliteration are not flowery. It's actually pretty to-the-point: set up problem, display calculations, declare answer. The hieroglyphs make the ideas being expressed look more complex than they really are. So far it's the usually the simple statement of the problem in sentence form (perhaps with some number-forms, just as we would do), followed by the calculations, just as a schoolboy would turn in his homework, but with more-adult motivational notes a la "we did this because this, okay now add that to this", etc.

For example, what follows is the literal modern-English transliteration of the problem #40 that I just spent some time with, and I agree that it captures the spirit of the original:

"PROBLEM 40: Divide 100 loaves among 5 men in such a way that the shares received shall be in arithmetical progression and that 1/7 of the sum of the largest three shares shall be equal to the sum of the smallest two. What is the difference of the shares?"

Even that treatment is too wordy. The Egyptian was more like "Loaves 100 for man 5, 1/7 of the 3 above to man 2 those below. What is the difference of share?"

So we're (originally) dealing in relatively tight language, there are some units of measure, and we even speak of "a (unknown) quantitiy" rather than using some concrete-noun every time. But we haven't quite (yet) advanced to the point of writing someting like " 3x = 2, solve for x".

Apart from the language itself, the most persistent mathematical syntax are the aforementioned units of measure, and the arbitrary conventions for how to express Egyptian fractions. The glyphs themselves and syntax end up expressing the same thing as "1/7", "1/8" or whatever unit fraction, and with about as much effort as it took me to write those.

I salute you for this quality thread, OP.

GEOMETRY: 41-60

Grain silo time. Problems 41-43 are calculations of volumes of (right circular) cylinders, which in itself is straightforward in modern terms.

These stick out for three reasons: first inexact and incorrect answers (approximating pi), first serious mathematical mistakes besides (#43), and tacking on more units after each "main" problem, so a little dimensional analysis. Thus we review both units and volume methods used. Yes, we're using literal cubits as units, now. :D

A CUBIT is a unit of LENGTH, from which a unit of VOLUME, a CUBIC CUBIT, may be derived in the usual way.

2/3 of ONE CUBIC CUBIT = 1 KHAR (another VOLUME unit)

A HEKAT (mentioned earlier, ctrl+f for the wiki, it's useful here) is a unit of VOLUME.

1 KHAR = 20 HEKATS, 2 HEKATS = 1 DOUBLE HEKAT, and 4 HEKATS = 1 QUADRUPLE HEKAT.

Note that wiki's treatment of khars doesn't apply here. It's one of those goofy things that changed as time passed.

The pertinent volumes are given below, and the expanded forms are meant to show the original methods. V41,42 is a nice one desu, it's within 1%. Since the problems work in terms of diamaters, we supply the exact version for comparison. the "Elsewhere" volume is something that Ahmose seemed to copying (from another text) for problem 43 to be cute and shortcut a certain unit conversion, but he fucked up by doing the 1/9 step, resulting in an 8/9^2 error term.

$\displaystyle V = \pi r^{2} h = \frac{ \pi }{4} d^{2} h \\ \displaystyle V_{41,42} = \bigg( d - \frac{1}{9} d \bigg)^{2} h = \frac{64}{81} d^{2} h\\ \displaystyle V_{elsewhere} = \frac{2}{3} h \bigg( d + \frac{1}{3} d \bigg)^{2} = \frac{32}{27} d^{2} h = \frac{3}{2} V_{41,42} \\ \displaystyle V_{43FUBAR} = \frac{2}{3} h \bigg( \bigg( d - \frac{1}{9} d \bigg) + \frac{1}{3} \bigg( d - \frac{1}{9} d \bigg) \bigg)^{2} = \frac{2048}{2187} d^{2} h = \bigg( \frac{8}{9} \bigg)^{2} V_{elsewhere} \\$

Modern treatment of problems 41-43. I only express the Egyptian fraction results given for the "units" of these, which are cubic cubits (the others generally seem to check out and I'm uninterested in them):

See if you can spot any goofs between this and the above setup post!

P. 41: "Given a cylinder with d = 9 cubits and h = 10 cubits, find V and V_{41,42}, and express the latter in terms of cubic cubits, khar and quadruple hekats."

$\displaystyle V = \frac{405}{2} \pi \;\; cubit^{3} \;\;\; ; \;\;\; V_{41,42} = 640 \;\; cubit^{3} = 960 \;\; khar = 4800 \;\; quad.hekat$

P. 42: "Given a cylinder with d = 10 cubits and h = 10 cubits, find V and V_{41,42}, and express the latter in terms of cubic cubits, khar and quadruple hekats. Further express the latter cubic cubit result in Egyptian fractions."

$\displaystyle V = 250 \pi \;\; cubit^{3} \;\;\; ; \;\;\; V_{41,42} = \frac{64000}{81} \;\; cubit^{3} = \frac{32000}{27} \;\; khar = \frac{160000}{27} \;\; quad.hekat \\ \displaystyle = 790 + \frac{1}{18} + \frac{1}{27} + \frac {1}{54} + \frac{1}{81} \;\; cubit^{3}$

Due to its extant mistakes and various forms of unit conversion, problem 43 has multiple angles which have been set up above and which I have paraphrased like so:

P.43: "Ahmose has made a mistake, and so let's check his existing work and fix it. Given a cylinder with d = 9 cubits

and h = 6 cubits, do the following:

a) Find V, $V = \frac{243}{2} \pi \;\; cubit^{3}$

Find V_{43FUBAR} and express it (thereby "verifying" what Ahmose did) in

b) cubic cubits, $= \frac{8138}{27} \;\; cubit^{3}$
c) khar, $= \frac{4096}{9} \;\; khar$
d) and quadruple hekats... $= \frac{20480}{9} \;\; quad.hekat$

...Then, fix Ahmoses' mistake by finding V_{elsewhere} and expressing it in

e) cubic cubits, $= 384 \;\; cubit^{3}$
f) khar, $= 576 \;\; khar$
g) and quadruple hekats. $2880 \;\; quad.hekat$

44 and 45 are mercifully simple, after this.

P.44: A cube measures 10 x 10 x 10 cubits. Find its volume in terms of quadruple hekats.

Ans: 7500, with straightforward conversions.

P.45: A cube's volume is 7500 quadruple hekats. Find its edge length.

Ans: 10 cubits. This is an exact reversal of 44.

P.46: A rectangular prism's volume is 2500 quadruple hekats. describe its three dimensions in terms of cubits.

Ans: This is of course amenable to an infinity of solutions, but " 10 x 10 x 10/3 cubits" is preferred, and given. I actually need to think about this one a bit more, so this post is slightly premature.

47-49

Problem 47 amounts to a small multiplication table, done in certain units (including the small /ro/, check the hekat wiki), all of which check out. A phrase suggests cylindrical vs. rectangular volume (whether in a rectangular or circular granary...), but this seems to be a way of indicating generalization on the "volume" concept of previous problems, and not a relevant restriction from a modern POV.

P.47: "Give the products when 100 is multiplied by 1/10, 1/20, 1/30 ... 1/100". Ans: (and, they are correctly given).

P.48: "Compare the area of a circle of diameter 9 and of its circumscribing square, which also has side length 9."

The comparison is given flatly as 64/81, as opposed to pi/4. This is the most explicit statement yet of (this) Egyptian method for finding the area of a circle; this slight overestimation is consistent with the method of problems 41-42.

Problem 49 is a rectangle computation, and introduces another unit of measurement, a khet, which is either a unit of area or length, and bears a happy "metric" relationship to either one. A khet is 100 square(?) cubits, or perhaps 100 cubits, and while we're at it, here's a helpful page on Egyptian units:

en.wikipedia.org/wiki/Ancient_Egyptian_units_of_measurement

The problem's phrasing is slightly goofy by our usage, and between this and wiki, . P.49: "A rectangle ("10 khets by 1 khet") is like so. Express this area in terms of square cubits". Ans: 1000 square cubits. The setup of course suggests a khet as a unit of length, but this problem is dumb-simple enough that I'm not worried about conversions or minutae, except to point out that "khet" may be interpreted as length or area depending on what's going on (wiki seems to want to say it's a unit of area).

Problem 50 strengthens Rhinds' internal logic that a khet is a unit of length. We have a round field of diameter 9 khet (length!) and we are to find its area. It is 64 setat (square khet). ~37 to go and I mean to finish it.

overnight bump. same encouragement.

Bump for OP

I'm loving these posts, OP

I wanted to hand-wave about 49, but I was unsatisfied with my treatment of it both because I didn't understand the units being used at-a-glance, and also because my answer of "1,000 square cubits" is clearly wrong.

Ignore wiki for the purposes of 49, and refer to pic related. For right now...

A KHET is a unit of LENGTH, equal to 100 CUBITS.

ONE SQUARE KHET, therefore, is equal to 10,000 SQUARE CUBITS, both obviously units of AREA.

There is another name for ONE SQUARE KHET, itself manifestly a unit of AREA. ONE SETAT is equal TO ONE SQUARE KHET, or 10,000 SQUARE CUBITS, as just stated.

Here is the confusion about 49, which Chace clears up, and which I misrepresented earlier. What's being counted are strips of area measuring 1 cubit by 100 cubits. For clarity, Chace dubs these "CUBIT-STRIPS", to differentiate them from the above. So let's try 49 again.

P.49 (take 2): A strip measures 10 khet by 1 khet. Express its area in terms of CUBIT-STRIPS. Ans: 1,000 CUBIT-STRIPS.

Pic related clears this up, it's basically alloting acreage or plots on a given block, or parcel, as we'd say today My confusion also seems indicitive of Egyptian confusion on concepts of length vs. area and their dimensional representations.

Now that we've got our units straight, 50 offers no confusion. It is just a repetition of the "64/81" rule for squares and circles that we've seen multiple times already, with a "unit conversion".

P.50: "A circle has a diameter of 9 khet. Express its area in setat." Ans: 64 setat, a nice 1/1 "unit conversion" as 1 setat = 1 khet^2.

Now, for problems 51-53, you can compare the original, especially the diagrams, by taking a GOOD CLOSE LOOK at the OP picture, which is decent quality albeit with some annoying watermarks. The areas in the middle of the picture contain 51 (triangle), 52 (trapezoid) and 53 (triangle with tiers on it), going down.

There seems to be a little confusion in 51, over whether the author had a grasp on the notion of a right triangle versus an isosceles, or scalene triangle. That drawing looks pretty close to a right-triangle, but not quite, and you can judge for yourself. :^)

I believe just from skimming that Ahmose uses a straightforward A = (1/2 b"h") argument in 51, taking one of the sides as the "height", (and the diagram would seem to bear this out) but I'll have to check.

Short version of 51:

P.51: A triangle has a base of 4 and an altitude of 10. Find its area. Ans: 20.

Long version of 51: pic related.

The motivation for this autism is that the figure almost, but not quite! suggests a right triangle. We then wonder just how well Ahmose understood finding areas of triangles, in general, and it prompts a review of certain things.

If 51 suggests some problems, 52 is a literal "Thanks Common Core!" trapezoid problem. Short version first, making certain assumptions and giving Ahmose the benefit of the doubt:

P.52: A trapezoid has two bases 6 (khet) and 4 (khet) , and an /altitude/ of 20 (khet). Find its area. Ans: 100 (setat, or square khets, via the usual method).

The issue is much the same as 51, pic related. I'm not going to beat this one to death just now like I did the other.

53 is of a piece with 51 and 52, has its own challenges, and I'm still thinking about it. In the interest of productivity, let me paraphrase the much simpler 54 and 55 (dumb-simple division problems masquerading as geometry), and get them out of the way.

P. 54: There are 10 different fields, and you have to carve equal areas out of each one, all 10 of which sum to 7 setat. What is the area of one (each) of them? Ans. 0.7 setat (expressed as 1/2 + 1/5).

P. 55: There are 5 different fields, and you have to carve equal areas out of each one, all 5 of which sum to 3 setat. What is the area of one (each) of them? Ans: 0.6 setat (expressed as 1/2 + 1/10).

53 now. It has a diagram, but is plagued by the same issues as 51 and 52.

"This simple (charitable!) paraphrase: "Given an isosceles triangle of altitude 14 and base 9/2, divided into sectors per the figure, find the two bases and the three areas of the sectors."

Ans: pic related.

My big personal takeaway from this recreation is that the Egyptians were just fine about manipulating fractions, but really sucked hard at geometry.

Problems 56-60 deal with, of all things... pyramids! And they have some fancy diagrams to yield up. However given my pace with the above and a real urge to finish this thread in one go for better or worse, I've made a decision in the interest of productivity for now to skip 55-60 for the time being, and skip ahead to 61-87, the miscellany.

61 is a very simple table of multiplications of arbitary (Egyptian) fractions, all of which is correct and checks out. As one representative example, " 2/3 x 1/3 = 1/6 + 1/18 ".

61B comes up, and is a rule for taking 2/3 of the reciprocal of an odd number, using 1/5 as a brief example (add 1/10 and 1/30, is what it amounts to), "and so for any other odd number". It's right in this case and demonstrates a correct method, and although odd, I can let it pass for now. He wants to express it in egyptian fractions is the issue, but the sum works.

62 is some linear combination involving a money-unit and a weight-unit of three precious things, in a heap. I'm probably not turned in yet but I'm quite serious about finishing this so any further overnight bumps are appreciated (thx guy !) I hope to dispense with the miscellany and be into the pyramids to wrap it up between tonight and tomorrow, but personal life intrudes, to your surprise.

overnight bump.

Page 4? Bumpitty bump bump then. This is the only Veeky Forums worthy thread I've seen in weeks.

Now I've got my head all the way around 61B (Ahmose barely writes anything, we have to discern context!)

P.61B: Give a general procedure for converting the product of 2/3 and the reciprocal of any odd number 2n+1 into an egyptian fraction of two terms, e.g.

$\displaystyle \frac{2}{3} \cdot \frac{1}{2n+1} = \frac{1}{p} + \frac{1}{q}$

With natural p and q.

In Ahmose's words, the procedure is: multiply 2 by your odd number (odd denominator) (which is p), then take the reciprocal of that. Also multiply 6 by your same odd number (q), and take the reciprocal of /that/. The sum of those two is your Egyptian fraction in this case. (And we can clearly see this in modern terms, with a few examples of Ahmoses' discovery, viz.

$\displaystyle \frac{2}{3} \cdot \frac{1}{2n+1} = \frac{2}{3(2n+1)} = \frac{1}{2(2n+1)} + \frac{1}{6(2n+1)} \\ \displaystyle \frac{2}{9} = \frac{1}{6} + \frac{1}{18} \\ \displaystyle \frac{2}{15} = \frac{1}{10} + \frac{1}{30} \\ \displaystyle \frac{2}{21} = \frac{1}{14} + \frac{1}{42} \\ \displaystyle \frac{2}{27} = \frac{1}{18} + \frac{1}{54} \\$

P.62:

A bag of three precious metals, gold, silver and lead, has been purchased for 84 sh'tay, which is a monetary unit. All three substances weigh the same, and a deben is a unit of weight (or mass if you like, but either way it's immaterial to this simple problem :^).

1 deben of gold costs 12 sh'tay, 1 deben of silver costs 6 sh'tay, and 1 deben of lead costs 3 sh'tay. Find the common weight W of any of the three metals.

Ans: divide the bag's purchase price by the sum of the unit price-per-weight for the three metals to find 4 deben, e.g.

$\displaystyle \frac{84 \;\; sh'tay}{ 21 \frac{sh'tay}{deben} } = W = 4 \;\; deben$

Ahmose is getting cleverer here

Yes, I thought so too once I took another look and understood what he did. He uses 5 in one example, and basically says " 2x5 with 6x5, and so on".

63 is another linear equation in disguise.

P.63: 700 loaves are to be divided unevenly among 4 men, in 4 unequal, weighted shares. The shares stand to one another in this relationship (I then give the solution in this same chunk of TeX:

$\displaystyle \frac{2}{3} : \frac{1}{2} : \frac{1}{3} : \frac{1}{4} \\ \displaystyle Soln \;\; (shares): \frac{800}{3} , 200, \frac{400}{3} , 100$

Since the fractions themselves do not sum to 1 but 7/4 (so we don't start out by just carving up 700), and since some x would act as a scalar on the fractions, preserving the relationships, Ahmose recognizes the need to solve for x, which is 400 (convenient next to the 700 and 7/4 terms). The solutions are then given as above.

I had hoped to have some content ready by now, but I've been busy with RL stuff today which went well - and which now frees me to get srs about this. I sat with the old Chace in my lap in an auto repair shop earlier today.

64 is a redux of 40 with an even number of x's, but I've stupidly been hung up on some details. 65 is another linear combination which I thought to skip ahead and check, but I haven't got it worked out just yet.

I appreciate all the positive comments on this treatise of dumb-simple elementary mathematical history, but I really must be producing more and completing this project. Coming back shortly.

Despite the "we wuz kingz" suggestion of the link in the pic, the diagram may prove helpful for the pyramid-stuff that I skipped.

P.64: 10 numbers in arithmetic progression sum to ten. The difference between consecutive terms is 1/8. Find this solution.

Ans: {7/16,9/16,11/16,...,25/16}.

The problem is a redux of 40, this time with an odd number of terms. The mathematics is mostly the same, and a generalization is (again) pic related.

P. 65. 100 cash-money-bread are divided unequally among 10 employees. 7 of the slaves (er, employees) each gets a single share, while the other 3 happier employees receive double shares. Compute each share.

This becomes an elementary linear algebra problem entailing 2 equations with 2 unknowns, which doesn't even need gaussian elimination, simple substitution suffices. Ahmoses' method is not to do the substition that we would gravitate towards, and instead to consider "13 (primitive) shares", a natural step, and then he gets it right. The peons get 100/13, and the bigwigs get 200/13.

P.65: Recall that the hekat is a unit of volume. There is also a little-bitty tiny volume of unit called a "ro", and their relationship is 1 hekat = 320 ro. Furthermore, the Egyptian is aware that a year is (commonly) 365 days so just assume 1 year = 365 days for this one.

The pharoah deigns to give out 10 hekat of fat over the course of a year to so-and-so, in an equal amount every single day. Express the fat given out in (arbitrary!) terms of hekat and ro as you see fit, just make sure your (Ahmoses') version is right.

(Ahmoses') Ans: Take into account the unit conversion of hekat and ro above, and also take into account the 1 yr = 365 day above. Do your unit conversions and so-and-so gets 2/73 hekats per day. Or (what checks out)...

$\displaystyle \frac{2}{73} \;\; hekat = \frac{1}{64} \;\; hekat + \bigg( 3 + \frac{2}{3} + \frac{1}{10} + \frac{1}{2190} \bigg) \;\; ro$

THE PREVIOUS PROBLEM WAS CLEARLY PROBLEM 66. A GOOF ON MY PART.

P.67. Shepherd-dude has to give the lord-conqueror-dude (2/3)(1/3) of his flock, which is 70, to the lord-dude as tribute. So figure out the size of shepherd-dude's original flock before paying the tribute.

Ans: solve the linear equation and the original flock was 315 mang.

P.68: 4 overseers have gangs of men, and the men in their gangs work at identical linear rates so go ahead and assume their labor is fungible, etc.

Anyway, working on the same interval of time these 4 overseers and their gangs have collectively produced 100 units (quad.hekat-of-such-and-such, doesn't matter).

One overseer has a gang of 12,
the next has a gang of 8,
the next has a gang of 6,
the next has a gang of 4.

How much did each gang produce during the interval (that the overseer can take credit for)?

Ans. Straightforward linear decomposition of the orignal 100, since "everyone's the same". 100/30 as a man-labor unit...

and so the gang of 12 produced 40 units,
the gang of 8 produced 80/3 units,
the gang of 6 produced 20 units,
and the gang of 4 produced 40/3 units,

which of course sum to 100 units.

The next few problems entail a new unit measure called a "pefsu", used in the context of food prep. These appear to entail mostly volumes and amounts, but I've been stuck on things before so let me not be premature.

Chace says that "pefsu" is supposed to denote "units of food-stuff that you get from such-and-such material", but we'll see.

if an 8ball of crack leaves houston on a bentley travelling hood miles per hour howdunna fast delivery to chicago and how much street value in that 8ball in chicago man dindu nuffin

Wait so they described angles in terms of length? Trigonometric ratios of lengths more accurately

Go to Anchorage instead for premium rates? I stayed in a motel room in Chicago recently with a bullet hole in the window.

P.69: 3 + 1/2 hekat of meal make 80 loaves. Give two things: meal per loaf, and the "pefsu".

Ans: the "pefsu" is given as the 80/3.5. This suggests a reciprocal of what I just said.

The meal per loaf is the flip of that, and he converts it into units easy enough that I bother to repeat them: 1/32 hekat + 4 ro.

I haven't drilled into that just yet (I skipped the pyramids problems for last), but Chace suggested something along those lines, thinking of angles or whatever in terms of units, being "palms", and maybe not making the abstract leap.

The starting point of math is to describe everything you can in terms of number, of quantify. So since length is in its own way coterminous with number, this should come as no surpise to you.

What else is 2π, but a length? What are 30, 45, 60, 90, but lengths in relation to each other and such-and-such?

P.70: 7 + 1/2 + 1/4 + 1/8 hekat of meal make 100 loaves of bread, give meal/loaf and the "pefsu", again.

Ans: The pefsu is given as loaves over hekat, or 800/63, circa 12.7-something. Meal/loaf: 63/800, given in the units, checks out (ro...)

Still loving it. Bump before bed.

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