Long Division Algorithm (3)

Note: Visitors, , the LaTeX system seems to not work perfectly here, so  images are now used here instead, to help illustrate the process.  Further editing may still be needed.

How does long division really work?

For best understanding, a person should understand the meaning of fractions and should understand signed numbers, but we will not worry too much about those right now.    How about an example,   ?

We usually do not know our multiples of 15, so we may like to know what factors of 15 to use to subtract from the 850; but how may we decide what factors?

Try expression the dividend and divisor in expanded form. (I’ll show that here but without use of exponents):

latex$ 8×100+5×10+0×1\div 1×10+5×1$   and please excuse the use of ‘x’ as the multiplication symbol.   We can arrange into the “long division” expression form:

 

Long Division Expanded Form Before Start

We want to ask, what is the biggest amount of times that we can subtract the divisor, based upon the first term of the dividend, which is 8×100?  We focus on the “1×10″ and on the “8×100″, for now.  It either goes in 0 times or something between 1 and 9.  Just figure,  8×100 divided by 1×10 is  8×10.   Good! Now what do we do?

Report the first part of the division, perform the multiplication, and perform the first subtraction based on this:

first partial division, multiplication, resulting subtraction

  

                              (Notice this is a negative number, -350)

We will now bring down the next term and repeat the process:

 

finished basic algorithm, remainder showing

                        

How many times is 1×10 contained in  -35×10 ?   Answer is  -35.  -350/10=-35.     Going through the process, the subtraction gives us  +175.   For our purposes, we are almost done.   +175 seems like the REMAINDER.   We really had to let in negative numbers.

Notice again, the remainder means, since the divisor was 15, the fraction latex$ \fract{175}{15}$.

The quotient then can be initially shown as 8×10 – 35 + latex$\frac{175}{15}$.   The fraction simplifies to (yes, let me skip showing the steps here) 11 latex$\frac{2}{3}$,

So the quotient then is 80 – 35 + 11latex$\frac{2}{3}$ = 56 latex$\frac{2}{3}$

 

Again, as a reminder, some of this may be better understood if a student is well introduced to adding and subtracting of signed numbers, not very difficult to learn, but often enough not taught in the lower elementary school levels.

Long Division Algorithm (2)

The problem with subtraction of the divisor quantity one at a time until no more can be subtacted is obviously too time-consuming for most long divisions.  How about subtracting MULTIPLES of the divisor!  Maybe, 2 times the divisor or 3 times the divisor?

Let’s try that and see what it could look like:

,  maybe try 2×15=30 and see how this goes.

90 – 30 = 60   this was 2 times
60 – 30 = 30   this was 2 times
30 – 30 = 0     this was 2 times and finally finishes this division.

So count the 15’s that were subtracted from the starting 90.
2 + 2 + 2 = 6.
Six of WHAT?  6 of the 15’s.   This again means

Still, this method is inefficient for many long division computations.  We really want something either closer to or identical to the standard long division algorithm.  Let’s jump to that.  See part 3 next.

Long Division Algorithm (1)

The long division algorithm is complicated to learn.  First, students need to understand and know the basic multiplication facts.    We can understand how multiplication works. 

15+15+15+15+15+15 = 15 x 6 = 90.   No problem.  Fifteen is summed and it appears in the sum a total count of six times. 

Students are soon taught the typical long division method and to use various trials and what to do if a step works or what to do if a step does not work; and use trial divisions using 10 or rounding up to use 20.  Students must do various subtractions and bring down a number and if the student has not yet understood how all this works, they (some, at least) become confused.   Eventually, nearly EVERYONE can learn long division, but for some, the process stays confusing and seems impossible to follow.  No matter to worry about too soon.

Try this thought:  You know that repeated addition of the same quantity gave you a multiplication result.  Division is the OPPOSITE PROCESS.   Subtract a given number from a starter number, and count these subtractions, changing the starter number each time, until what remains is too small to subtract any more,   like this:

90 —– you start with,

15 —–the number to subtract.

90 – 15 = 75  

75 – 15 =60

60 – 15  =45

45 – 15 = 30

30 – 15 = 15

15 – 15 = 0

How many subtractions of 15 from starter of 90 until no more to subtract?  SIX.    This means  .  This is good because we know 6×15=90.  Division is just the reverse of multiplication.

Keep in mind the positions of the numbers in the equation are important.

, and “divisor” is always “the quantity to be subtracted from” the dividend.

You can do this with any long division with whole numbers when dividend is larger than divisor. 

There is one main problem with the method that was just shown:  for divisions other than common basic multiplication-division fact, it is too slow, not efficient.   See part 2 for an improvement.

Improper Fraction: Denominator crawls into the Numerator location

Here is how a beginner or any confused student may understand improper fractions.  Just like a reflection of the division operation, answer the question;  “How does the Denominator behave to the Numerator”?

Imagine that the part above the fraction bar is a space which can hold something, and that the thing under the fraction bar is something that wants to crawl into the space above the fraction bar.  How many of the denominators will fit in the space in the numerator location? 

Can you imagine seeing the “3″ crawling upward to fill the “18″?  How many times can this happen until all of the “18″ is filled?  (Note you have a “3″ replenished every time one of them crawls upward and fits.  )  In this example, obviously there is room upstairs for 6 of the “3’s” to fit.

    Now, how is THAT for 18 over 3 equals 6 ?

Strange – Algebra 1 harder than Basic Arithmetic?

So many students who were able to learn enough of Arithmetic and much of Basic Math in elementary school and junior high suddenly struggle with and hate Algebra in eighth grade and in high school.  They find that their success is not as high, not nearly as easy.  What happened?   Once the emphasis goes to looking at the rules by which Arithmetic works, students are confused.  This almost makes you wonder why they were able to do well enough before starting Algebra.  The same rules are there.