Rules behind this type of division?

[zorro]

I know this is a silly question for a 12th grader to ask but I don't know the exact rules.

If we divide 950 by 121, the first digit of quotient is 7, then we place a decimal after 7 and bring one 0 down, which is correct way.

In the division of 21780/704, the first digit of quotient is 3, then we bring the 0 after 2178 down so that the remainder (I know its not the correct term here) is 660. Now after this, I always put a decimal after 3 and bring one more 0 after 660 which is wrong. The correct way is to place one 0 after 3 and then bring one 0 down.

Can anyone explain me what are the rules behind this type of division?

 

[arildno]

Well, efficient algorithms will, precisely because they are efficient, mask some of the logic that justifies the use of the algorithm in the first place.

To take a more "unmasked" approach, we have:
A)
\frac{950}{121}=\frac{7*121}{121}+\frac{103}{121}=7+\frac{1}{10}\frac{1030}{121}=7+\frac{1}{10}(\frac{8*121}{121}+\frac{62}{121}=7+\frac{1}{10}*8++++
Note that "adding the zero" (i.e, multiplying by 10!) is coupled to multiplying the whole with 1/10, i.e, so that the net effect is to multiply with 10*1/10=1, i.e, not changing the value of the expression.

B)
\frac{21780}{704}=10*\frac{2178}{704}=10*(\frac{3*704}{704}+\frac{66}{704})=10*(3+\frac{1}{10}*\frac{660}{704})=10*(3+\frac{1}{10}*(\frac{0*704}{704}+\frac{1}{10}*\frac{6600}{704}))=3*10+0*1+\frac{1}{10}*\frac{6600}{704}

Thus, because 66*10=660<704, the coefficient at that decimal place becomes 0.

 

[zorro]

Thanks arildno, you have explained it in a very detailed manner.
But how do you explain it to a kid in 5th or 6th who is learning division.
There are many numbers to which 2nd approach is applied. I just need a rule to remember when to add 0/decimal in the quotient.

 

[symbolipoint]

The real obstacles in understanding Long Division are the young student not yet knowing how to use signed numbers and (for some students) not yet fully understanding fractions. Try writing the divisor and dividend in expanded form and then try to see how many of the divisor you could subtract all at once (being a whole number between 0 and 9). You perform a multiplication to see what number you subtract, and then continue the same process. Actually, polynomial division is just as easy or easier than regular Long Division. The main difference is that your variable is used in place of "10".

 

[zorro]

Try writing the divisor and dividend in expanded form and then try to see how many of the divisor you could subtract all at once (being a whole number between 0 and 9). You perform a multiplication to see what number you subtract, and then continue the same process.

Can you explain with an example?

 

[Grep]

You might want to check out this article I read on BBC a while ago that describes the way some schools are now teaching division and such. It's less efficient that traditional long division, but much more intuitive. Perhaps it's worth trying if you want to explain it to younger children, at least before hitting long division.

http://www.bbc.co.uk/news/magazine-11258175

Just an idea, anyways.

 

[symbolipoint]

Can you explain with an example?

EDIT: The forum system is NOT accepting parentheses to correctly display the two polynomials, so excuse the lacking parentheses. Can someone help with the TEX here?
EDIT: x to the zero power is not showing correctly either, even after I edited for superscript.

Although I prefer not to, I will start one for you.

8045\div 65.

That can be done in expanded form, if you wanted, but would involve much writing.
In expanded form, that expression is:
(8\times 10^{3}+0\times 10^{2}+4\times 10^{1}+5\times 10{0})\div (6\times 10^{1}+5\times 10^{0})

Do you see that the division expression can be just like a polynomial division expression replacing 10 with a variable, such as x ?

Try performing this polynomial division:
(8\times x^{3}+0\times x^{2}+4\times x^{1}+5\times x^{0}) \div (6\times x^{1}+5\times x^{0})

 

[symbolipoint]

That bothers me. Let me try again:

(8\times x^{3}+0\times x^{2}+4\times x^{1}+5\times x^{0})\div(6\times x^{1}+5\times x^{0})

It should look better without the multiplication symbols:
(8x^{3}+0x^{2}+4x^{1}+5x^{0})\div(6x^{1}+5x^{0})

 

[zorro]

Thanks alot!