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.

Yet Another Subtraction Algorithm!

Posted by mark schwartz on November 4, 2016

Introduction

I recently posted Revisiting Mr. Stoddard’s 1852 Subtraction. In that posting I modified Mr. Stoddard’s idea by introducing a procedure which allows for subtraction without borrowing. This posting modifies that modification.

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.

What I didn’t note clearly are two things. First, if in that example, b > c, then write down that value as the answer. Do not add 1 to the next place value in the subtrahend. However, if c > b, then the algorithm as noted is to be used. And here’s the modification – continue with this algorithm!

Here’s an example in slow-motion math. Using the problem 7234 ̶ 567 as a traditional ‘vertical’ problem, we hav

7234
–567

In the 1s column, 7 is greater than 4, so the answer is 10 ̶ ( 7 ̶ 4) which is 7. Add 1 to the 6 in the subtrahend 10s column. Then in the tens column, 7 is greater than 3, so the answer is 10 ̶ ( 7 ̶ 3), which is 6. Add 1 to the 5 in the subtrahend 100s column. Then in the 100s column, 6 is greater than 2, so the answer is 10 ̶ ( 6 ̶ 2), which is 6. Add 1 to the zero in the subtrahend 1000s column. Then in the 1000s column, 7 is greater than 1, so the answer is simply the difference of 6. The solution looks like this:

7234
– 567
6667

There are many subtraction algorithms posted in this blog and most of them focus on avoiding the need to borrow, so if you feel like trolling through the entire blog and compiling them, you might find one you like.

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One Response to “Yet Another Subtraction Algorithm!”

  1. sarablack said

    Reblogged this on Sarablack's Blog.

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