Marveling At The Historical

Math Oldies But Goodies

  • About This Blog

    This blog is mostly about math procedures in textbooks dated from about 1825-1900. I’m writing about them because some of the procedures are exquisite and much more powerful, and simpler, than some of the procedures in current text books. Really!

    I update this blog as frequently as possible ... every 2-3 days. And, if you are a lover of old texts and unique procedures, you might want to talk to me about them, at markdotmath@gmail.com. I’m not an antiquarian; the books I have are dusty, musty, brown-paged scribbled-in texts written by authors with insights into how math works. Unfortunately, most of their procedures have vanished. They’ve been overcome by more traditional perspectives, but you have to realize that at that time, they were teaching the traditional methods.

Archive for the ‘duodecimal’ Category

An 1851 Use of Duodecimal

Posted by mark schwartz on June 23, 2016

Introduction.

In Hutton’s 1851 A Course of Mathematics he shows how the duodecimal system can be used to determine the area of a rectangular space where the dimensions are feet and inches. His algorithm is messy and I spent a lot of time realizing that his rules as stated left out some critical directions. So, in the story below is his algorithm and my comments about it. By the way, I would not recommend using this today. We have a simpler way to do it. However, I think it’s interesting to see how he did it using duodecimal math. Be thankful we weren’t in school in 1851.

The Story.

Using Hutton’s example and his notation, find the area of a rectangular garden “4 f, 7 inc. by 6 f. 4 inc.”

Our current algorithm for addressing this is to convert both dimensions to inches, multiply these two values and then divide the product by 144. The result is the area in inches and feet. For Hutton’s problem,

4 f, 7 inc. = 55 inches; 6 f. 4 inc. = 76 inches. Then 55•76 = 4180. Dividing this by 144 gives 29 f., 1/2 inc. Fairly simple.

Hutton’s algorithm is not as simple. Here are his words, except I’ve added paragraph numbers. After looking at his rules and my comments, I have included his example from the book.

  1. “Rule. ̶ Set down the two dimensions, to be multiplied together, one under the other, so that feet stand under feet, inches under inches, &c.
  2. Multiply each term in the multiplicand, beginning at the lowest, by the feet in the multiplier, and set the result of each straight under its corresponding term, observing to carry 1 for every 12, from the inches to the feet.
  3. In like manner, multiply all the multiplicand by the inches and part of the multiplier, and set the result of each term one place removed to the right hand of those in the multiplicand; omitting however what is below part of inches, only carrying to these the proper number of units from the lowest denomination.”

My comments: paragraph 1 is standard form for multiplying two values.

1st, Paragraph 2. It may have seemed unnecessary to him but he might have stated that the product of the feet values are the answer to which the carry is added.

2nd, Paragraph 2. He doesn’t state that if the product from paragraph 2 is divided by 12 to find the number of feet, the quotient is the carry and the whole number remainder is the answer to write down “under its corresponding term”, meaning in the inches column.

3rd, Paragraph 2. He doesn’t note that the product of the feet values is a whole number value to which the carry is added. It is the standard multiplication “carry” but I think it needs to be stated.

1st, Paragraph 3. “In like manner” means to start by multiplying the multiplier-inches value by the multiplicand-inches value. Again, although this is the standard multiplication algorithm, I believe it needed to be stated.

2nd, Paragraph 3. “…each term one place removed to the right hand of those in the multiplicand…” means just that but what he doesn’t say is that this result also needs to be converted to feet and inches. So, after placing the result to the right of the inches column, see my next comment.

3rd, Paragraph 3. He didn’t specify that the product of inches by inches is to be divided by 12 and again, the quotient is the carry to the next column and the whole number remainder stays in the column and is the answer. Place the quotient in the inches column and leave the remainder in that column as a fraction of the remainder over the divisor. This column to the right of the inches column will be a fraction.

4th, Paragraph 3. “…omitting however what is below part of inches…” implies to ignore fractions which is true except for my comment above. Don’t omit a fraction in the column to the right of the inches column; it’s part of the answer.

5th, Paragraph 3. Again, “In like manner” means that when you multiply the multiplier-inches by the multiplicand-feet, the result is placed in the inches column and then divide by 12 and the quotient is the carry to the feet column and the whole number remainder is the answer, which stays in the inches column. Then just add the values as you would in traditional multiplication. If the inches column is more than or equal to 12, convert this to feet, add it to the feet column leaving the rest in the inches column.

Wow … a lot of words in his rules and my comments, so let me show you how this works by using the example he presented in the book with his notation, referring to his rules and my comments as needed. It is

Multiply 4 f. 7 inc.

By 6     4

27     6

   1     6 1/3

Answer   29     0 1/3

First, 6×7 = 42; 42 ÷ 12 = 3r6 (carry the quotient 3 to the feet column; the remainder 6 is the answer that goes in the inch column.

Then, 6×4 + 3 = 27 in the feet column

Then, 4×7 = 28 in the column to the right of the inch column (which isn’t seen in his problem); 28 ÷ 12 = 2r4; the 2 is a carry which is entered in the inch column and the remainder 4 stays in that column as a fraction 4/12 = 1/3.

Then, 4×4 = 16 in the inch column; 16 ÷ 12 = 1r4; the 1 is a carry which is entered in the feet column and the remainder 4 stays in the inch column.

(Note: for the above two statements, you have a 2 1/3 and a 4 that both go in the inches column; he didn’t show this, rather he summed them and posted the sum of 6 1/3. )

Add the feet results and inches results and you get 28 f., 12 1/3 inc. The 12 inc. comes from totaling the 6, 2 and 4 in the inch column and the 1/3 from the column to the right of the inch column. Finally, convert 12 1/3 to 1 f. 0 inc. and the answer is 29     0 1/3.

I told you his algorithm was messy. Even with my comments, which I hoped help, it still is a complicated scheme, which again, I wouldn’t recommend using. However, if you followed it, you might have realized that if the numbers are small, you can do some of the work in your head, which reduces the messiness. I do, however, have to give Hutton applause for his creative use of the duodecimal system. If you want to play with one yourself, here it is …

Multiply 4 f. 7 inc.

By 9     6

 

Answer 43     6 1/2

 

 

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