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.

Revisiting Mr. Stoddard’s 1852 Subtraction

Posted by mark schwartz on September 29, 2016

Introduction

In this blog is a posting Mr. Stoddard Subtracts in 1852. If you haven’t read it, you don’t need to (but of course you can!). Mr. Stoddard presents an idea in subtraction which avoided the need for “borrowing”. For some reason, I was playing with a subtraction idea and after I had written out the entire algorithm, I realized that I basically had modified Mr. Stoddard’s; thus the title.

The Story

I’ll use a simple subtraction example to demonstrate the procedure, but I have examined much more sophisticated problems such as 20801 ̶ 278 and the procedure is still good.

Basically, treating ‘ab’ as a 2-digit number and ‘c’ as a single digit number, in the problem “ab ̶ c”, if c > b, the answer to ‘b ̶ c’ is 10  ̶  ( c ̶ b ) and then add 1 to the 10s place value in the subtrahend. For example, 12 ̶ 8 gives 10 ̶ (8 ̶ 2), or 4, then add 1 to the 10s place value in the subtrahend, giving 1 ̶ 1 or 0, which isn’t written.

Here’s why it works. In essence, it could be said that borrowing has happened but it’s hidden as well as not written!

In essence, 10  ̶  ( c ̶ b ) is borrowing, but it’s hidden. The ‘10’ in the 10 ̶ (8 ̶ 2) could be said to have been borrowed from the 10s column in the minuend. Given that, that ‘10’ can be said to have been subtracted from the10s column in the minuend. It’s known that if the same value is subtracted (or added) from both the minuend and subtrahend of a subtraction problem, the answer will be the same. Thus, adding a 1 to the next place value in the subtrahend adds a value which will be subtracted.

There it is. It’s a mild modification to Mr. Stoddard, but my ‘aha’ moment with 10 ̶  ( c ̶ b ) may well have been his idea incubating all this time. Try it with other problems – like 20801 ̶ 278 and after a while it becomes as automatic as doing the problem using borrowing.

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