## 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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