Talk:Long division

message: The proof in the article does not seem clear to me.
It does not prove the fact that \beta_{i} < b. Also, why does it not use the Euclidean division?
Here is a proof:
d_{i} can be divided by  m, see Euclidean division. So there exist q and r such that d_{i}=mq+r and 0 \leq r < m.
If one shows that q < b, then we have found \beta_i and r_i as  \beta_i=q and r_i=r.
We proceed by induction. At i=0, d_i is formed by the first l+1 digits of n, so it has l+1 digits, which is the total number of digits of m. Then, q cannot be equal or larger than b, since, qm would be a number with l+2 or more digits, and it would be d_i < qm. This contradicts r \geq 0.
So, we must have  q < b.
Now, assume d_i=m\beta_i+r_i with 0 \leq r_i < m. Then, by definition, d_{i+i}=br_{i}+\alpha_{i+1+l}. That is, d_{i+1} is r_i shifted to the left with the \alpha_{i+1+l} digit added to the new place to the right.  
Since, r_i < m, the number of digits of r_{i} is less than or equal to those of m, that is l+1. The hardest case to check is when its digits are exactly l+1. Let's write down the digits of r_i, d_{i+1} and m
:r_i=aa \ldots as \ldots
:m  =aa \ldots at \ldots
:d_{i+1}  =aa \ldots as \ldots \alpha_{i+1+l}
where we assume that some initial digits of r_i and m may be the same, but surely at some point, some digit must differ and, in particular, s < t since  r_i < m.
Notice that d_{i+1} has l+2 digits.
If one multiplies m by b, m shifts to the left and the t digit goes at the same position of the s digit of  d_{i+1}, since now both of them have l+2 digits and since d_{i+1} is, in its initial part, equal to r_i. Thus, it follows that bm > d_{i+1}. So, chosing a number b or greater, one cannot divide d_{i+1} by m. Since we can (by the Euclidean division), the quotient must be q < b.
The other case where the number of digits of r_i is less than those of m is handled similarly to the i=0 case:
multiplying by b would shift m and yield a too large number.
Alopumo (talk) 17:25, 20 September 2019 (UTC)
thread_title: Proof of existence and uniqueness of <math>\beta_{i}</math> not clear

message: Equation d_{i}=br_{i-1}+\alpha_{i+l-1} should be d_{i}=br_{i-1}+\alpha_{i+l} instead.
In fact, if you see the expression of r_{-1}, the last \alpha digit is \alpha_{l-1}, so the first to be added in d_0 is \alpha_{i+l}=\alpha_l, which is, rightly the successive one.
Alopumo (talk) 17:25, 20 September 2019 (UTC)
thread_title: Wrong subscript in equation

message: Under the section on conventions in Eurasia, the Belgian standard is said to be the French one:
     127|4    
    −124|31,75
       30
      −28
        20
       −20
         0
However, I do not recognise this form, yet I'm from Belgium. The notation I was taught, both in elementary school as well as in high school and then in university, is what the article assigns to Cyprus (and strangely, also France):
6     	3     	5     	9	17
− 5   	1     	      	 	374
1     	2     	5     	 	 
− 1   	1     	9     	 	 
 	 	6     	9	 
      	−     	6     	8	 
      	      	 	1	 
My guess is that either of the following is the case: either Belgium was misplaced, or this has something to do with the Flemish/French division running through the central latitude of the country. I'm based in Flanders, and I do believe everyone here is taught the latter of the two notations. I'm not editing the article because I don't know which of my guesses is correct, but it might be worthwhile for others to chime in. MewTheEditor (talk) 19:56, 7 November 2020 (UTC)
thread_title:  Two halves of Belgium 

message: the given example mentions 15 inc left over which is 12+3 while the column top about inch mentions 9
further in the first cq miles column the multiplier is given 1760, but in the yardscolumn the multiplier VALUE is ommitted which is e for 22x3=66, just as in the feet column.
there are MANY mistakes in this example!!!  — Preceding unsigned comment added by 85.149.83.125 (talk) 14:46, 10 April 2023 (UTC)
thread_title:  Mixed mode long division