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Subtraction: Decomposition (Regrouping/Borrowing)
Date: 06/06/2003 at 07:41:52
From: Hala
Subject: Subtraction
How do you subtract using the decomposition method?
I think it is like this:
82
-63
and then you have to split the units or something - I don't remember!
Date: 06/06/2003 at 12:25:55 From: Doctor Peterson Subject: Re: Subtraction Hi, Hala. I'm not familiar with the term "decomposition method," but it seems to be the method I'm used to calling either "borrowing" or "regrouping." You can read one description of the method here: Borrowing in Subtraction http://mathforum.org/library/drmath/view/59035.html Let's take your problem slowly. We want to subtract 63 from 82. We can think of that as taking 6 dimes and three pennies from 8 dimes and 2 pennies: DDDDDDDD PP - DDDDDD PPP -------------- But we don't have enough pennies to take three away, so we have to take one of the dimes and change it into ten pennies: PPPPPPPPPP DDDDDDD PP - DDDDDD PPP -------------- Now there are plenty of pennies; we take 3 away from the 12 we have now, and there are 9 left. Then we take 6 dimes from the 7 we have left, and there is one remaining: PPPPPPPPPP DDDDDDD PP - DDDDDD PPP -------------- D PPPPPPPPP In numbers, we can think of it this way: 82 = 80 + 2 = 70 + 12 - 63 = 60 + 3 = 60 + 3 ---- ------ ------- 10 + 9 = 19 That is, we think of 82 as 80+2 (that's the dimes and pennies); then we rewrite it as 70+12 by taking 10 away from the dimes and putting it with the pennies; then we can subtract. The short way to write this is 7 1 8 2 - 6 3 ------ 1 9 I can't show the crossing out I would really do. When I see that I need more "ones" to subtract from, I cross out the 8 and write a 7 above it to show that I only have 7 "tens" of the 8 I started with; and I put a little 1 next to the 2 to change the 2 to a 12. Then I subtract 12-3 and 7-6. Does that help? If you have any further questions, feel free to write back. You might want to show me your work on a problem so I can see where you have trouble. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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