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Art-of-Arith-art - 4/3/00


"The Noble Art of Arithmetic" by Jan van Seist.


NOTE: See also the files: alphabets-msg, measures-msg, measures-art, Latin-msg, Latin-msg, Latin-online-art, literacy-msg, med-letters-msg.





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                               Thank you,

                                    Mark S. Harris

                                    AKA:  Stefan li Rous

                                         stefan at florilegium.org                                        



This article was originally written for Storm Tidings, the newsletter of the

   Shire of Adamastor (Cape Town) in the Kingdom of Drakenwald.


The Noble Art of Arithmetic

By Jan van Seist


In days of old, when knights were bold, but by and large illiterates,

The merchant class made lots of brass, by treating them as idiots.



        A key feature of our historical period was the struggle for numerical supremacy in Europe.  Mediaeval Christendom enjoyed its XXXX, whereas Al-Islam believed that maths should be as easy as 1,2,3.   From the Xth century onwards, European academics fought furiously to persuade the West (or rather the North) of the advantages of arabesque algebra. Unfortunately for the learned lecturers, the North (or perhaps West is better after all) wasn’t interested.  The Roman system worked perfectly well for day to day life and all the exciting new possibilities of Arabism, like long division, required scrap paper for calculations.  Given that paper was worth substantially more than its weight in gold, the reticence of the merchants to waste it on their III times tables is perhaps understandable.


        In the XVth century with the advent of the printing press and cheap paper, the tide turned.  Texts preaching and teaching the new maths started rolling off the presses and a new breed of maths teacher strode forth to conquer Europe.   By the mid XVIth century, only a few small pockets of resistance remained (most notably the IX supporters of Amsterdam).   In order to help those among you who have not yet made the jump to modern mathematics, I have translated II problems for you from I of the more modern text books to help you get started.  Good luck.


*      *      *

Two problems from Christianus van Varenbraken’s  "Die Edel Conste Arithmetica” of 1530" translated by Jan van Seist


1)     {folio 150r (original), p.85* (Kool 1988)}


A lord had hired a man-servant, whom he was to give for a whole year 10 guilders and a tabard.  And that servant lived with the lord for 7 months and at the end of that time they had such a disagreement, that the lord dismissed him saying "Get out of my house and take the tabard with you, then I am left owing you nothing."


Now the question is:  What was the tabard worth? Do as follows: Mark how much that 7 months is less than a year.  That is 5 months.  And had that man-servant stayed there so long with his masters he would have had the tabard and 10 guilders.  Therefore, say as follows: “5 months would have won 10 guilders.  What have 7 months won?”  Then use the rule of three.


Eenen heere hadde eenen cnape ghehuert, dien hij alle jare gheven soude 10 guldenen ende eenen tabbaert.  Ende den selven knecht woonde 7 maenden met den heere ende tenden van dien, soo warden sij kivende, als dat hem die heere orlof gaf, segghende: “Gaet uut mijnen huuse ende draecht den tabbaert met u, soo ne blive ic u niet sculdich”.

Nu es die vraghe, wat den tabbaert weert was.  Doet aldus:  Merct hoe veel dat 7 maenden min es dan een jaer.  Dat es 5 maenden.  Ende hadde die knecht daer soo langhe bij sijnen meester ghebleven, soo soude hij den tabbaert ende x guldenen ghehadt hebben.  Daer om segt aldus: “5 maenden souden ghewonnen hebben 10 guldenen.  Wat hebben ghewonnen 7 maenden?”  Doet na den regel van drien.


2)     {folio 177v (original), p.149 (Kool 1988)}


A question for fun.

A drunkard drinks a barrel of beer alone in 14 days.  And if his wife drinks with him, they drink the barrel dry in 10 days.  Now the question is:  How much time would his wife take to drink it dry by herself?


Solution:  Subtract the shortest drinking period from the longest.  That is 10 from 14 and there remains 4.  And that is your divisor. Now say as follows:  4 gives 10.  What gives 14?  Do it according to the rule of three and there are 35 days and it is done etc.


Een questie uut ghenouchten

Eenen dronckaert drinct een tonnebiers alleene uut binnen 14 daghen.  Ende als sijn wijf met hem drinct, so drincken sij die tonne uut binnen 10 daghen.  Nu es de vraghe, binnen hoe veel tijdt dat sijn wijf die alleene uut drincken soude.


Solutio:  Subtraheert dat minste drincken van dat meeste, dats 10 van 14, ende daer blijft 4.  Ende dat es u divisor. Nu segt aldus:  4 gheven 10.  Wat gheven 14?  Maket naerden reghel van drien ende compt 35 daghen, ende es ghemaect etcetera.


*(M. Kool  1988 Christianus van Varenbrakens “Die Edel Conste Arithmetica” Mediaeval and Renaissance Texts and Studies. No. 21  SCRIPTA, Brussels. Pp. 181)


Copyright 1999 by Dr. Ian van Tets

                  Zoology department,

                  University of Cape Town,

                  Rondebosch, 7701

                  Republic of South Africa


Permission granted for republication in SCA-related publications, provided

the author is credited and receives a copy.


If this article is reprinted in a publication, I would appreciate a notice in

the publication that you found this article in the Florilegium. I would also

appreciate an email to myself, so that I can track which articles are being

reprinted. Thanks. -Stefan.


<the end>

Formatting copyright © Mark S. Harris (THLord Stefan li Rous).
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Comments to the Editor: stefan at florilegium.org