4 ms·

A LOT of these have massive errors. I strongly recommend comparing all that to this paper: https://pdfs.semanticscholar.org/5ca1/fa0ffca55e9003053de2f5ef36a258

A LOT of these have massive errors. I strongly recommend comparing all that to this paper: https://pdfs.semanticscholar.org/5ca1/fa0ffca55e9003053de2f5ef36a25819109b.pdf https://pdfs.semanticscholar.org/5ca1/fa0ffca55e9003053de2f5... - Harald Hammarström - Rarities in Numeral Systems (2009) And this page: https://mpi-lingweb.shh.mpg.de/numeral/ https://mpi-lingweb.shh.mpg.de/numeral/ "Numeral Systems of the World's Languages" By the way: If anyone can find a digital copy of this paper from 1840 by Augustin-Louis Cauchy (yes, THAT Cauchy!): https://scholar.google.com/scholar?cites=6877129426190119313&as_sdt=2005&sciodt=0,5&hl=en https://scholar.google.com/scholar?cites=6877129426190119313... I'd very much appreciate it. Donald E. Knuth cites it in TAOCP while discussing Balanced Ternary: "Cauchy pointed out that negative digits make it unneccesary for a person to memorize the multiplication table past 5x5." [Comptes Rendus Acad. Sci. 11 (Paris, 1840), 789-798] But he doesn't elaborate on it further, and neither does anyone else from what I can tell, and I have yet to figure out the trick. (I currently try to teach myself to EFFICIENTLY do all my mental arithmetic in balanced ternary - not an easy task!)

joshuamorton 9y ago
>But he doesn't elaborate on it further, and neither does anyone else from what I can tell, and I have yet to figure out the trick 5 * 9 = 5 * (10 - 1) = 50 - 5 = 45 Specifically, you need to know everything in the 5x5 area, and how to multiply by 10. More generally, any digit 6-9 can be rewritten as (10 - (10 - digit)) and then you can apply distribution rules to clean things up. Here's a three by two digit example: 247 * 68 = (10 - 3) * (10 - 2) + // 7 * 8 10 * (10 - 3) * (10 - 4) + // 7 * 60 10 * 4 * (10 - 2) + // 40 * 8 10 * 10 * 4 * (10 - 4) + // 40 * 60 10 * 10 * 2 * (10 - 2) + // 200 * 8 10 * 10 * 10 * 2 * (10 - 4) // 200 * 60 evaluating we get 100 - 50 + 6 + 10 * (100 - 70 + 12) + 10 * (40 - 8) + 100 * (40 - 16) + 100 * (20 - 4) + 1000 * (20 - 8) And then to finish up: 00056 + 00420 + 00320 + 02400 + 01600 + 12000 = 16796, which is the correct answer. Edit: The fun thing to think about is conceptually why this works. If you consider numbers as just bits, all you're doing is converting the highest bit from base to sign, without actually changing it, doing the multiplications on the lower bits, and then tracking the sign to know whether or not to add or subtract later. Or to put it another way, you're shifting your set of numbers from being +0 - +10 to being -5 - +5. This makes the lookup table smaller, but requires you to do some additional intermediate lookups (up to four, one in each "quadrant" of the 10x10 table) to be as powerful. Fun to conceptualize that way.
- Numberwang 9y ago
  Is this the best way to do quick arithmetic, or are there other ways?
  - dahart 9y ago
    There are lots of ways, depending on what you want to calculate. Here's a fantastic book on the subject: https://www.amazon.com/Dead-Reckoning-Calculating-Without-Instruments/dp/0884150879 https://www.amazon.com/Dead-Reckoning-Calculating-Without-In...
- pierrebai 9y ago
  I think this is one instance where using a larger look-up table (i.e. multiplication table up to 10x10) yields clear benefit. The technqiue using only 5x5 increases the number of intermediary results to keep in one's head.
  - jacobolus 9y ago
    Most everything is easier with negative digits. Easier addition, easier rounding, easier multiplication, much easier subtraction. The one part that needs slightly more distinct techniques vs. what you are used to is division, but it's still probably easier overall. The grandparent poster's example only looks more complicated because we don't have standard glyphs for the negative digits.
xavieralexandre 9y ago
Comptes rendus hebdomadaires des séances de l'Académie des Sciences Tome XI, p.789 (16 novembre 1840) Found in Oeuvres complètes d'Augustin Cauchy 1ère série Tome V III. Notes et articles extraits des comptes rendus hebdomadaires des séances de l'Académie des Sciences. (Suite.) 105. Calculs numériques. sur les moyens d'éviter les erreurs dans les calculs numériques. https://www.e-rara.ch/zut/content/pageview/5702727 https://www.e-rara.ch/zut/content/pageview/5702727
- no_identd 9y ago
  Wow. I figured it'd sit online in some archive that I failed to find. Thanks a lot! PDF of Tome V here: https://www.e-rara.ch/download/pdf/5702285?name=Tome%2520V%46 https://www.e-rara.ch/download/pdf/5702285?name=Tome%2520V%4...
jacobolus 9y ago
If you normalize your numbers to be written with digits from -5 to +5 instead of 0 to 9 in each place, then the multiplication table is more symmetrical so you can memorize a table 1/4 as big and then just remember the signs of each quadrant.
tom_mellior 9y ago
> A LOT of these have massive errors. I strongly recommend comparing all that to this paper As you have already found those errors and done that comparison, could you share some of what you've found? EDIT: For whatever it's worth, I checked French on https://mpi-lingweb.shh.mpg.de/numeral/ https://mpi-lingweb.shh.mpg.de/numeral/, and it lists the same forms for 70 and 90 for French and Belgian French. Which is incorrect.

joshuamorton 9y ago

>But he doesn't elaborate on it further, and neither does anyone else from what I can tell, and I have yet to figure out the trick 5 * 9 = 5 * (10 - 1) = 50 - 5 = 45 Specifically, you need to know everything in the 5x5 area, and how to multiply by 10. More generally, any digit 6-9 can be rewritten as (10 - (10 - digit)) and then you can apply distribution rules to clean things up. Here's a three by two digit example: 247 * 68 = (10 - 3) * (10 - 2) + // 7 * 8 10 * (10 - 3) * (10 - 4) + // 7 * 60 10 * 4 * (10 - 2) + // 40 * 8 10 * 10 * 4 * (10 - 4) + // 40 * 60 10 * 10 * 2 * (10 - 2) + // 200 * 8 10 * 10 * 10 * 2 * (10 - 4) // 200 * 60 evaluating we get 100 - 50 + 6 + 10 * (100 - 70 + 12) + 10 * (40 - 8) + 100 * (40 - 16) + 100 * (20 - 4) + 1000 * (20 - 8) And then to finish up: 00056 + 00420 + 00320 + 02400 + 01600 + 12000 = 16796, which is the correct answer. Edit: The fun thing to think about is conceptually why this works. If you consider numbers as just bits, all you're doing is converting the highest bit from base to sign, without actually changing it, doing the multiplications on the lower bits, and then tracking the sign to know whether or not to add or subtract later. Or to put it another way, you're shifting your set of numbers from being +0 - +10 to being -5 - +5. This makes the lookup table smaller, but requires you to do some additional intermediate lookups (up to four, one in each "quadrant" of the 10x10 table) to be as powerful. Fun to conceptualize that way.

Numberwang 9y ago
Is this the best way to do quick arithmetic, or are there other ways?
- dahart 9y ago
  There are lots of ways, depending on what you want to calculate. Here's a fantastic book on the subject: https://www.amazon.com/Dead-Reckoning-Calculating-Without-Instruments/dp/0884150879 https://www.amazon.com/Dead-Reckoning-Calculating-Without-In...
pierrebai 9y ago
I think this is one instance where using a larger look-up table (i.e. multiplication table up to 10x10) yields clear benefit. The technqiue using only 5x5 increases the number of intermediary results to keep in one's head.
- jacobolus 9y ago
  Most everything is easier with negative digits. Easier addition, easier rounding, easier multiplication, much easier subtraction. The one part that needs slightly more distinct techniques vs. what you are used to is division, but it's still probably easier overall. The grandparent poster's example only looks more complicated because we don't have standard glyphs for the negative digits.

xavieralexandre 9y ago

Comptes rendus hebdomadaires des séances de l'Académie des Sciences Tome XI, p.789 (16 novembre 1840) Found in Oeuvres complètes d'Augustin Cauchy 1ère série Tome V III. Notes et articles extraits des comptes rendus hebdomadaires des séances de l'Académie des Sciences. (Suite.) 105. Calculs numériques. sur les moyens d'éviter les erreurs dans les calculs numériques. https://www.e-rara.ch/zut/content/pageview/5702727 https://www.e-rara.ch/zut/content/pageview/5702727

no_identd 9y ago
Wow. I figured it'd sit online in some archive that I failed to find. Thanks a lot! PDF of Tome V here: https://www.e-rara.ch/download/pdf/5702285?name=Tome%2520V%46 https://www.e-rara.ch/download/pdf/5702285?name=Tome%2520V%4...

jacobolus 9y ago

If you normalize your numbers to be written with digits from -5 to +5 instead of 0 to 9 in each place, then the multiplication table is more symmetrical so you can memorize a table 1/4 as big and then just remember the signs of each quadrant.

tom_mellior 9y ago

> A LOT of these have massive errors. I strongly recommend comparing all that to this paper As you have already found those errors and done that comparison, could you share some of what you've found? EDIT: For whatever it's worth, I checked French on https://mpi-lingweb.shh.mpg.de/numeral/ https://mpi-lingweb.shh.mpg.de/numeral/, and it lists the same forms for 70 and 90 for French and Belgian French. Which is incorrect.