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 ‘subtraction’ Category

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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Posted in basic math operations, Historical Math, math instruction, mathematics, remedial/developmental math, subtraction | Tagged: , , , | 1 Comment »

Two Alternatives to “Borrowing” When Doing Subtraction

Posted by mark schwartz on July 4, 2016

Introduction

To do traditional subtraction one has to know the operation of “borrowing”. Most students can do it, but if there are double zeros (or multiple zeroes in either or both numbers), students find this troublesome. Further, most students don’t know the basis of borrowing. There is no sense of place value and also there is no awareness of what is being borrowed and what bundling and unbundling means. They simply follow (as best they can) the steps they were taught.

That’s one of the principle reasons why students are bothered by subtraction. And, sometimes, even after discussing place value and bundling and unbundling, there is still no significant change in students being able to do subtraction. Borrowing is bothersome.

But there are two alternative methods which don’t involve borrowing. Both of them involve an interesting operation and, in my view, this simplifies subtraction. The first alternative has been presented previously in the blog article Subtract by Adding but I wanted to present both of these ideas in the same article because, in essence, they are the same!

The Story

The first alternative: the 9s-complement

This is based on computer math. In the problem 203004 ̶ 044726, the subtrahend (the number being subtracted) will first be replaced with its 9s-complement. The 9s-complement is found by subtracting each digit in the subtrahend from nine. So 044726 becomes 955273. The leading zero in the subtrahend is included to assure that for that place value, it’s really 9 – 0 or 9.

The next step is to add one to 203004, making it 203005. This ‘one’ is the leading one that shows up in the answer. It is described as ‘dropping’ the leading 1’, but the reality is that it is added to the unit column in the minuend as the first step of the procedure. If this seems strange, I’ll demonstrate later why this is done.

 Now ADD these two numbers:  203005
                           + 955273 
                            1158278

In this case, drop the leading ‘1’, and you have the answer (“leading” means sticking out beyond the place values of the numbers in the problem). The mystery about dropping the one and the mystery of adding one to the minuend can be explained by demonstrating why this operation works. A problem with fewer digits will make it easier to follow the explanation. Use 312 – 67.

First add ‘zero’ to this problem in the following way: 1000 – 1000 + 312 – 67. Adding 1000 and subtracting 1000 doesn’t change the value of the problem. Replace 1000 with 999 + 1 and use the commutative property: 312 + 1 + 999 – 67 – 1000.

Continuing with the associative property and doing all the indicated operations:

(312 + 1) + (999 – 067) – 1000 = 313 + 932 – 1000 = 1245 – 1000 = 245.

This demonstrates (1) why 1 is added to 312, (2) why the 9s-complement of 067 is taken, and (3) why the leading ‘1’ is dropped.

The second alternative: the 10s-complement

The second alternative is very similar and uses the same complements method, but in this method the complement of 10 (not 9) is taken. This is based on operations on the abacus, not the computer complements.

So, back to the original problem, subtracting each digit in 44726 from 10 gives 66384. Then ADDING this 10s-complement gives:

           203004
          + 66384
           269388

In this procedure, there is no leading zero to include in the subtrahend and there is no 1 added to the minuend. In essence, the reason why it’s 66384 and not 966384 is because 10 ̶ 0 leaves a zero in that place value position.

We this isn’t the answer of 158278 which we got above. However, subtracting 1 1 1 1 1 0 from 269388 gives 158278 – the same answer as above. Again, this looks like magic but a demonstrate will again show why this operation works.

Applying this method to the simpler problem of 312 – 67, gives

        312
       + 43
        355

And subtracting 110 gives the answer of 245.

Or, as a student suggested, as each place value addition is done, subtract one from the answer, except in the unit column.

Why does this work? This is similar to demonstrating how the 9-complements works.

312 – 67

312 – 67 +110 – 110      (basically, add a zero in the form of +110 – 110.)

312 +110 – 67 – 110     (commutative property)

312 + (110 – 67) – 110   (associative property)

312 + 43 – 110

355 – 110

24 5

 

In different classes, we had interesting discussions comparing the two methods. The consensus was that the 10s-complement was easier, although it had two steps. It was easier because you didn’t have to remember to add one to the minuend or remember the leading zero in the subtrahend. But then I reminded them of what one student said about the 10s-complement, which supported their consensus.

Remember what the student suggested? He said “as each place value addition is done, subtract one from the answer, except in the unit column.” He realized that in the 9s-complement where each subtrahend value is subtracted from 9, this is exactly the same as in the 10s-complement when after subtracting each subtrahend value from 10, just subtract 1 more! Algebraically it’s (10  ̶  n)  ̶  1, giving 9 ̶ n. This student really understood both procedures!

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