For which stuff exactly would you like a proof?
Great! Just for fun I ordered the book.
There are not many books about the stuff you don’t learn in school.
I usually just find them, use them and then realise they in the middle ages they would use this kind of calculation.
Hi Kinma,
There was a post from TorstenBerg but it has gone now (?). Anyway he referenced Greymatters application of dividing by a digit that ends in 9 eg 29, 39 etc. Torstenberg refers to it as the “threshold” system. Basically the system involves rounding up to the next whole number and dropping the zero so 29 would become 3(0), that makes the calculation a lot easier. I’ll see if I can find the link.
Yes, you are right. I try to reconstruct what I wrote earlier.
Your method of using the next „round“ number in division instead, i.e. the next threshold, is not new. It is nicely described in Karl Menninger’s german book „Rechenkniffe“ and the German tern he uses is „Schwellenteilung“, meaning division using a threshold. The book was originally published in Germany in 1931 and had had many editions since then (I have one from 1972). There is a translation into english called <a href=https://www.amazon.com/dp/B0006BN0OG/ref=cm_sw_r_tw_dp_x_nfUNzbQS06J2H>Calculator’s cunning: The Art of Quick Reckoning.
Further, the leapfrog division method presented by Grey Matters is „just“ this threshold division in a special case, but still very helpful.
Hm, sometimes I feel we should have a standard vocabulary to refer to the various calculation methods. That would make it easier to recognize the basic principles especially in many of the derived techniques.
Torsten
Mea culpa! It was me.
I was replying to Torsten’s post and accidentally wiped his text. Auch! Sorry!
I think you refer to this one, which is the last of a series of 5 articles about this.
I see a lot of these kinds of ‘systems’, read a lot of different blogs, etc.
Even Vedic mathematics has a lot to say about it!
They all strike me as… too difficult.
In Vedic mathematics blogs usually only a way of how to do the calculation is given without the explanation why this is.
Let’s work on a general way of dividing wherein all these systems and shortcuts of division are contained and where we can deduct an working algorithm.
As an example let’s take something difficult, like the 73/139 division from:
https://owlcation.com/stem/Divide-Numbers-fast-and-easy-using-Vedic-Mathematics
73 does not go into 139, so we start with 730 and deal with the fact that the answer is about 0.5 later.
Now the way I do this - like I wrote here - I visualize 730 apples that need to be divided over 139 baskets. To make this easier I start with 140 baskets.
730 apples into 140 baskets results in 5 apples per basket with a remainder of 30.
However, we need only 139 baskets, so we take the 5 apples from the last basket and add them to the remainder.
In other words:
( \frac{730}{140} = 5), (remainder = 30 )
( \frac{730}{139} = 5), (remainder = 35 )
( \frac{730}{138} = 5), (remainder = 40 )
etc.
It also works the other way round.
Consider: ( \frac{730}{141})
We have 730 apples, and we divide them over 141 baskets. We start with putting 5 apples in each of the 140 baskets. This takes 700 apples, so the result is 30 apples and one empty basket. We fill the 141st basket with 5 apples - just like all the others - from the remaining 30, so now there are 25 apples left and all baskets are filled.
( \frac{730}{141} = 5), (remainder = 25 )
and
( \frac{730}{142} = 5), (remainder = 20 )
etc.
The other way of dividing can be visualized like this.
Again, we have 730 apples and 139 baskets.
We guess that we can fill maximum 5 apples per basket.
We fill the first 100 baskets with apples. This takes 500 apples, so we are left with 230 and 39 empty baskets.
Then we do another 30 baskets. This takes 150 apples. Remainder is 80.
There are now 9 baskets left to be done.
This takes 45 apples from the remaining 80, so the remainder is 35.
Again:
( \frac{730}{139} = 5), (remainder = 35 )
Now how to generate an algorithm for division by 139?
Recall:
( \frac{730}{140} = 5), (remainder = 30 )
and
( \frac{730}{139} = 5), (remainder = 35 )
In both these examples, the quotient is 5; there are 5 apples in each basket.
If we first divide by 140 we obtain a quotient, then the correction for dividing by 139 instead of 140 is to add the quotient to the remainder (see the apples go from the 140th basket and add them to the remaining apples).
In other words, if:
( \frac{a}{b} = q), (remainder = r )
then:
( \frac{a}{b-1} = q), (remainder = r + q )
more general:
( \frac{a}{b-c} = q), (remainder = r + cq )
and:
( \frac{a}{b+c} = q), (remainder = r - cq )
Thank you Kinma and well explained as usual and you are forgiven for deleting that other post.
I think I never answered this question.
You mean why the nine proof and eleven proof works?
This was proven by Carl Friedrich Gauss in 1801 or so in his book Disquisitiones Arithmeticae, laying the foundations for number theory.
As an introduction start by reading this:
In general,
if (ab) = d, then (ab) modulo e = d modulo e
also:
if (a+b) = d, then (a+b) modulo e = d modulo e
You can turn a division into a multiplication.
You can also turn a subtraction into an addition.
Because of this the modulo calculation can also be used on division and subtraction.
Long story short, you can use this for all 4 basic operations.
The reason we use 9 and/or 11 is that the modulo of a number can be quickly calculated and thus are suited for quickly determining if your answer is likely correct.