We’ll stop taking long division for granted in a minute, but first, let’s take a general look at division. What is division? Division is the shortcut for subtracting one value from another value repeatedly until we reach zero. We say that “90 divided by 10 is equal to 9″ because 90-10-10-10-10-10-10-10-10-10 = 0. Using the distributive property, we can rewrite this as 90 – 9(10) = 0.
If we then made steps to solve this equation, we’d get:
90 – 10(9) = 0
+10(9) + 10(9)
90 = 10(9) 90 is “ten, nine times”. Division is the inverse of multiplication.
But division also works in situations where we would not eventually subtract to end with zero. For example, we may end with the number 5. “95 divided by 10 is equal to 9.5″ because 90-10-10-10-10-10-10-10-10-10-5, and that last 5 is one-half (.5) of the 10 we had been subtracting from 90. This 5 is also known as a “remainder”. The result of “95 divided by 10” can then be written as “nine with a remainder of 5”. This remainder is a smaller number than the number being repeatedly subtracted, so does not quite fit in (“remains” outside).
95 – 10(9) = 5
+10(9) + 10(9)
95 = 5 + 10(9)
95 = 10(9) + 5 [commutative property]
95 is “ten nine times with five left over”
However, subtracting repeatedly can be tedious for large numbers, such as “950 divided by 10”, so the long division algorithm was developed. Formatting “950 divided by 10” to the algorithm, we would write:
The sideways “L” is referred to as the “division bar” in this paper.
Here, the decimal can be expanded to 950.00 because of the definition of our number system. This also makes room for any possible “remainders”, or numbers less than 10 that are left over once we subtract all the 10’s from 950 that we can.
To work this algorithm, we want to first ask ourselves, “how many 10’s can be subtracted from 9?” We want to remember both: that our definition of division is “repeated subtraction” and that “multiplication is division’s inverse”. 10×9 is 90, not “9”. We’d answer with “zero” and imagine a “0” above the 9 in 950. Here, we’ll put in a zero, but usually we wouldn’t.
Usually, when the number above the division bar is not zero, we would move to the second step of the algorithm. But since our “0” is imaginary, we won’t move to step two quite yet.
Starting the algorithm over, we’d then move in from the “9” and look at “95” and ask ourselves, “How many 10s cam be subtracted from 95?” We’d answer with “9” and place it above the tens spot in 950.
Our placement of the “9” actually represents a “90” as it is vertically aligned with the 950’s tens placement. Had we placed this “9” over the 9 in 950, it would have represented a 900. By placing our “9” over the 5 in 950, we’re saying that “ninety 10s can be subtracted from 950”, which we know to be true. If you notice the placement of this “90”, you will see that its 9 is vertically aligned with the 950’s hundreds placement. Does this make sense? (remember the definition of division as multiplication’s inverse!)
Continuing to step two of the algorithm, we write this “90” below the “95” to find the next remainder, remembering that this “90” actually represents a “900” because of its placement.
We are left with “5”, which is actually a “50” because of its placement below the tens spot of the 950. The “5” in the original 950 is in the tens spot, therefore our remainder of 5 is actually a remainder of 5×10, or 50.
The algorithm then tells us to “bring down” the next digit in the original 950 and repeat the algorithm. But what is the purpose of this “bringing down”? We always need to keep place value intact and our “5” is really a “50”. By bringing down the “0”, we accomplish this. Also, remembering that our “9” is a 90 because of place value and that division is repeated subtraction, we have:
950 – 10(90), which yields 50 (not 5)
We bring down the “0” from 950 and place it next to our 5 remainder to make a “50”, creating the 50 we already knew to be true.
If that remainder of “5” had been a “0”, we’d be moving into the ones place. Because it is a “5” and aligned in the tens spot of “950”, we’re still working within the tens. We then ask ourselves, “How many 10s can be subtracted from 50?”, and continue on with the algorithm.
We end here by saying that “950 divided by 10 is 95”. In the division algorithm, when we end with a “0”, we stop the algorithm because it means that the divisor (10) can be subtracted from the divisee “950” an even amount of times with no remainders. But we can use this same algorithm when dividing any number by any other number.
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