## Subtract by Adding … really!

Posted by mark schwartz on March 31, 2016

__Introduction__

Take the problem 30105 – 5908. Typically, it would be set up in a vertical format in order to do it:

30105

– __ 5908__

Doing this problem using the standard procedure, a student would have to invoke “borrowing” or “bundling and unbundling”. However, this problem has a bunch of zeroes in it and has been intentionally crafted to point out some of the issues with “borrowing” by including zeroes. It just gets fussy and I’ve seen the work of many students where they’ve struck out numbers but indicated uncertainty about what to do by noting a “10” where there should be a “9” and conversely. This led me to consider an alternative which turns out to totally bypass the need for knowing how to borrow.

__The Story__

Below is a demonstration (in slow-motion math) of how the alternative is derived, followed by a discussion and examples. You need not do every step as done in this demonstration and I’ll show you examples below how to use this idea.

Begin with the above problem:

30105 – 5908

To this, add ‘zero’, in the form of 10000 − 10000, placing it as:

10000 + 30105 – 5908 − 10000

Adding zero doesn’t change the value of the outcome of the subtraction.

Then replace the first 10000 with 9999 + 1, giving:

9999 + 1 + 30105 – 5908 − 10000

Now, using the commutative property rearrange the terms, giving:

30105 + 1 + 9999 – 5908 − 10000

Using the associatve property gives:

(30105 + 1) + (9999 – 5908) − 10000

And doing the operations inside the parentheses gives:

30106 + 4091 – 10000

Adding the first two terms gives, 34917 – 10000

And doing the subtraction gives 24197.

So, basically the problem 30105 – 5908 could be done in two steps, seen as 30106 + 4091 (** got this by adding 1 to 30105 and subtracting 5908 from 9999**). Breathe slowly and look at the stuff in the parentheses again before going on. Add these two terms. When the addition is done, any leading ‘1’ can be dropped because it has already been added to 30105. Sounds strange but mathematically this is what happens.

Consider these.

If the original problem were 723 – 58, it could be rewritten as 724 + 941 which gives 1665, and dropping the leading ‘1’ gives the final answer of 665. Please note that 58 has to be understood to be 058.

If the original problem were 6502 ̶ 3409, it could be rewritten as 6503 + 6590 which gives 13093 and dropping the leading ‘1’ gives the final answer of 3093.

And, the problem 123 ̶ 56 could be rewritten as 124 + 943, giving 67 (dropped the leading 10).

Please note that the leading ‘1’ need not be written if you see that a ‘1’ would result from the addition.

In essence, the procedure IS …

- ADD ONE TO THE UNITS COLUMN IN THE FIRST TERM (the minuend if that term is still used).
- SUBTRACT EACH NUMBER IN THE SECOND TERM FROM 9.
- ADD THESE TWO NUMBERS.
- IF THERE IS A LEADING “1”DROP IT (OR JUST DON’T WRITE IT).

With practice the subtraction can be done mentally, one number at a time. You just have to work slowly.

Try this one, doing it one number at a time: 624 − 57.

The first addition would be 5 + 2 (the 5 coming from 4 + 1 and the 2 coming from 9 – 7: (The 4 from the first term; the 7 from the second term). There is no carry.

The next addition would be 2 + 4 (the 2 from the 10s column of the first term) and the 4 from 9 – 5: (the 5 from the 10s column of the second term). Again, there is no carry.

The last addition would be 6 + 9 (the 6 from the 100s column of the first term) and the 9 from 9 – 0; (he 0 from the 100s column of the second term).

The result would be 1567, but drop the leading 1, giving 567.

Note that in that example, when the addition was done, there was never a carry, but if the addition does result in a carry, please do so.

One advantage to this algorithm is not having to borrow. Another advantage is that addition is more comfortable to most people than the standard subtraction algorithm. Make up some problems and try it.

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