|


Subtraction Using Nine's and Ten's Complements
Date: 05/27/2000 at 01:24:37
From: Katie Artus
Subject: Method of complements
I am trying to figure out why the method below, called the "method of
complements" by my university professor, will give me the correct
answer all of the time. It is used in place of regrouping in
subtraction problems.
623 999
-465 -465
----- -----
158 534
+623
-----
1157
/157 (Cross off the 1 in the thousands column and add
+1 this 1 to the ones column.
-----
158
Nines must always be used, and the order of operations depicted above
must be followed. If these conditions are met, the problem will always
work no matter how many numbers are in the problem. This will also
work with standard subtraction when regrouping is not necessary. I
would like to know why this method works. What mathematics are lurking
behind the scenes? I wish to explain this to a third grade child I am
student teaching.
Thank you,
Katie Artus
Date: 05/29/2000 at 14:44:11
From: Doctor TWE
Subject: Re: Method of complements
Hi Katie - thanks for writing to Dr. Math.
The method of complements is perhaps most famous in its binary
variations (called one's complement and two's complement) because
those are the methods that most computers use to subtract. What your
professor showed you is the decimal equivalent of one's complement,
sometimes called nine's complement.
As to why it works, let's examine what we're doing algebraically.
Let's call the minuend (623 in your example) X and the subtrahend (465
in your example) Y. We wish to find X-Y.
In the first step, we subtract Y from 99...9 (the number of 9's
equaling the number of digits in the larger of X and Y). In your
example, 999. So we have:
999-Y
Next, we add X to that to get:
999-Y+X
As long as X > Y, we can see that the result will be greater than or
equal to 1000. When we cross out the leading digit (the 1000's digit
in this example), we are in fact subtracting 1000 from our result.
Thus we have:
999-Y+X-1000
Finally, we add 1 to the units digit. So we have:
999-Y+X-1000+1
Can you see how algebraically this equals X-Y?
If I wanted to explain this to a third grader, I'd probably explain
the ten's complement method instead. Ten's complement works similar to
nine's complement with two slight differences. First, instead of
subtracting the subtrahend (Y) from 99...9, you subtract it from
100...0 instead. [I write this as 99...(10) so I don't have to
borrow.] The second difference is that you don't have to add the 1 at
the end. Can you see why this method produces the same results as
nine's complement?
Here's how I'd explain why it works:
Imagine you want to make some money by selling widgets (pick the
student's favorite thing-a-ma-bob). First, you have to buy the
materials to make widgets. This costs $465. Then you can sell them for
$623. Now you want to figure out how much money you'll make.
But there's one problem - you have to buy the materials BEFORE you can
sell the widgets. So you borrow $1000 from Mom and Dad. From that
$1000 you spend $465 to buy the materials. So you have:
$1000
- 465
----
$ 535
Now you sell the widgets for $623, you now have:
$1000
- 465
----
$ 535
+ 623
----
$1158
Finally, you have to pay back Mom and Dad the $1000 you borrowed to
start up. So your final profit is:
$1000
- 465
----
$ 535
+ 623
----
$/158
Which is the selling price ($623) less the cost of materials ($465).
A final word of caution: While nine's complement and ten's complement
produce the same results when X > Y, they need to be handled
differently when X <= Y. In the cases where X < Y (and where X = Y for
nine's complement), there will not be a "final carry" to cross out.
For example, to subtract 235-687:
9'S Comp. 10'S Comp.
999 1000
- 687 - 687
--- ----
312 313
+ 235 + 235
--- ----
547 548
How do we interpret this? With nine's complement, we subtract the
result from 99...9 and add a negative sign. With ten's complement, we
subtract the result from 100...0 and add a negative sign, like so:
9'S Comp. 10'S Comp.
999 1000
- 547 - 548
--- ----
(-)452 (-)452
I hope this helps. If you have any more questions, write back.
- Doctor TWE, The Math Forum
http://mathforum.org/dr.math/
|
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]


Ask Dr. MathTM
© 1994-2008 The Math Forum
http://mathforum.org/dr.math/