Talk:Mixed radix

Here are some subtopics that might help the exposition along once the proper references are dug up …

Why each radix must be an integer

For example Dollars/Quarters/Nickels/Pennies makes a perfectly good MRN system, but Dollars/Quarters/Dimes/Nickels/Pennies doesn't because the radix Dimes -> Quarters is 2.5, and 0₁₀0₄2_2.50₂4₅ (2-dimes,4-pennies) doesn't know whether to roll over to 0₁₀0₄2_2.51₂0₅ (2-dimes,1-nickel) or 0₁₀1₄0_2.50₂0₅ (1-quarter).

There is a perfectly legitimate 'base by' notation, which does support fractional integers. Binary-coded decimal is an example. In the case of us money, one would consider the quarter (1s 3d) to be 2 dimes (6d) + 1 nickel (3d), You could have a dime and a nickel, or two dimes, but no other combination. (I put money equivalents in so to follow what is going on.) The quarter is then 5 nickels, counted by intermediate dimes.

In the usual form, one might count in decimal to say, 71, and then roll over to 1,00. This would be 'base 71 by 10s'. Computers are quite at ease handling BCD, and one can express any base in a decimal, or other form, encoding, even when this does not divide. Bases like 2.61803398875 etc work by having a broken carry, and you can have complex bases by having a double column negative carry.

--Wendy.krieger (talk) 07:33, 17 November 2011 (UTC)

1 is a perfectly good radix

Radix 1 digits can act as place holders, but don't otherwise affect the MRN system.

Proper and improper MRNs

"Improper" MRNs have digits equal to or greater than their radices. They still specify integers, but these may not be limited to the product of the radices minus 1.

I don't know. A base of 3, with digits 3, 1, -1, works quite nicely. Wendy.krieger (talk) 08:32, 30 October 2011 (UTC)

Carry

Procedure to generate the proper MRN from an improper one if the integer represented is within range. Uses truncating division and the modulo operator. Overflow results if the integer is not in range.

Addition and scalar multiplication

Addition defined in the obvious way on two MRNs with the same base produces in general an improper MRN, which is rendered into a proper one using carry (with the possibility of overflow). Scalar multiplication on an MRN is just multiplying all the digits by a (non-negative) integer. Again carry is applied and overflow is a possibility.

long division, long multiplication are both implemented by scalar multiplication (or division), and addition. The method of practice allows one to multiply lengths in chains, yards and feet to get a result in acres, roods, and perches, with no reduction of unit. You can do things like find 3 3/8% of £21 10 6 with no reduction

   100%  costs  £ 21 10 6       0@    =   0
    10%  costs  £  2  3   .5    0     =   0
     1%  costs        4  3.65   3     =   1s   10.95d
   1/8%  costs           0.65   3     =         1.95d
                                          2s    0.90d  =   2s 3/4d  or 2s 1d

One sees in this example, a multiplication of mixed radicals, with little confusion. This is what i call 'invoice arithmetic' but i seen a reference in an old 'shilling arithmetic' as the method of practice. Wendy.krieger (talk) 08:42, 30 October 2011 (UTC)

Conversion between MRN systems

Obvious algorithms for doing this using the above tools are well known (so says I, research to be done :-) Doing these conversions efficiently seems to be a serious issue in Computer Science.

--J. W. McLeod 19:04, 27 Mar 2005 (UTC)

Factorial radix system

the table shows 0! as 0 whereas it should be 1. There is probably a good reason for this, but it's not obvious. Would s/o care to elaborate ? — Preceding unsigned comment added by 88.172.186.106 (talk) 23:36, 28 September 2011 (UTC)

The product of zero numbers leaves the original undisturbed, and is so 1. Even something like a × x⁰ = a, for all values of x, including x=0. Elsewise, one can argue that any number is equal to any other, by virtue that

  a = a.00,    by virtue that the LHS has zero values = 0
    = 0                   by placing 0^0 = 0

It's pretty much simple to demonstrate that whatever you set 0^0, like all x^0, equals 1, as is the product of an empty list.

So it's

  0! =    product()     = 1
  1! =    product(1)    = 1
  2! =    product(1,2)  = 2
  3! =    product(1,2,3)= 6
       &c.

Wendy.krieger (talk) 08:30, 30 October 2011 (UTC)