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September 13th, 2017, 12:37 PM   #1
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Algebraic explanation of Division

As an elementary student, I've always struggled with arithmetic, in particular, division and multiplication. This had to do with my lack of interest, as I didn't know why we multiplied and divided the way that we did. (why it worked). Having learned algebra, I now fully understand multiplication, and why we multiplied the numbers the way that we did and then added them up. I still can't seem to find an algebraic explanation of division. I obviously understand the process itself, but can't grasp why the method I used as a kid works. Could anyone convert it all too comfortable and intuitive algebra? Thanks

This is the method I was taught:

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September 15th, 2017, 07:00 PM   #2
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x/y = x*(1/y). Does that help? If 3x = 9, x = 3. You could think about it as dividing both sides by 3 (the normal point of view) or multiplying both sides by 1/3. If (1/3)x = 9, x = 27. You could think about it as multiplying both sides by 3 (the normal point of view) or dividing both sides by 1/3.
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September 16th, 2017, 06:15 AM   #3
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I think that the OP is asking why the algorithm used in manual long division works. If so, I am not sure whether a formal proof is wanted from start to finish or whether a more intuitive explanation is wanted.
Thanks from Antoniomathgini

Last edited by JeffM1; September 16th, 2017 at 06:18 AM.
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September 16th, 2017, 07:22 PM   #4
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Thank you very much for replying. I would prefer an intuitive explanation for why each step is important, and how it links back to algebra. No need for formal proofs.
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Originally Posted by JeffM1 View Post
I think that the OP is asking why the algorithm used in manual long division works. If so, I am not sure whether a formal proof is wanted from start to finish or whether a more intuitive explanation is wanted.
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September 16th, 2017, 10:08 PM   #5
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Let's start with something that we sort of forget.

The mechanics of manual division can be understood in terms division of a non-negative integer by a positive integer. If we have something like

$\dfrac{0.0027}{.3},$

then we can say

$\dfrac{0.0027}{.3} = \dfrac{27}{3} * \dfrac{10^{-4}}{10^{-1}} = 9 * 10^{-3} = 0.009.$

Basically the stuff they taught you to do in grade school about moving decimal points around when doing division is based on that logic.

So, in terms of mechanics, we can always think about a manual division of two numbers as the division of one non-negative integer by a second positive integer that is NOT smaller than the first integer.

We tend to forget that when we divide integers with the same number of digits by hand, it is essentially a matter of guess and check. Of course, we can make an informed guess, but the basic procedure is to guess and check. Long division comes in when we divide an integer with m significant digits by an integer with n significant digits, where m > n > 0. The simplest case is m - n = 1, meaning m = n + 1. Consider then

$\dfrac{a}{b} \text{, where } a,\ b,\ n \in \mathbb Z,\ 1 \le n,\ 10^n \le a < 10^{(n+1)} \text {, and } 0 < b < 10^n.$

We can decompose a as follows:

$a = c + d * 10 \text {, where } c \in \mathbb Z,\ 0 \le c < 10 \text {, and } \dfrac{a - c}{10} \in \mathbb Z.$

All we are saying here is that, if b = 108, a three-digit number, and
a = 2918, a four-digit number, then n = 3 and we can decompose
2918 = 8 + 2910 = 8 + 291 * 10. In our example, n = 3, c = 8, and
d = 291.

It is necessarily true that $d \in \mathbb Z.$

$a = c + 10d \implies d = \dfrac{a - c}{10} \in \mathbb Z \implies d \in \mathbb Z;$

It is also necessarily true that $10^{(n-1)} \le d < 10^n.$

What is the relationship of that decomposition to the process of long division? In that process, we first divide our n-digit divisor into the first n digits of our (n+1)-digit dividend, treating those first n digits as an independent number. The decomposition gives those digits as d. first thr

$\text {Let } e = \left \lfloor \dfrac{d}{b} \right \rfloor.$ In our example,

$e = \left \lfloor \dfrac{291}{108} \right \rfloor = 2.$

In long division, we write down as the first digit of the quotient the highest digit that, when multiplied by the divisor, does not exceed d. That is e. In the case of our example every digit times 108 is less than 291 until we get to 3 so we write down 2 as the first digit of the quotient.

$\text {Let } f = d - be \implies d = be + f.$

In long division, we subtract from d the product of the first digit of the quotient and the divisor. That is exactly what f equals. In our example,

$f = (291 - 2 * 108 ) = 291 - 216 = 75.$

$\text {Let } g = 10f + c.$

In long division, we stick the last digit of the dividend, which is c, on to the end of f. That is just a way to multiply f by 10 and add c to it. In our example, f is 75, and c = 8 so plopping 8 after 75 gives
758 = 10 * 75 + 8.

$\text {Let } h = \left \lfloor \dfrac{g}{b} \right \rfloor.$

In long division, we write down as the second digit of the quotient the highest digit that, when multiplied by the divisor, does not exceed g. That is h. In the case of our example every digit times 108 is less than 758 until we get to 8 so we write down 7 as the second digit of the quotient.

$\text {Let } i = g - bh \implies g = bh + i.$

In grade school, we finally calculate a remainder. That is i. In high school we get a bit more sophisticated and write a fraction at the end of the quotient of
i / b. In our example

$\text {remainder} = i = 758 - 108 * 7 = 758 - 756 = 2.$

OK. I have put the whole painful process into algebraic notation.

Consider

$b * \left ( 10e + h + \dfrac{i}{b} \right ) = 10be + bh + i =$

$10be + g = 10be + 10f + c = 10(be + f) + c = 10d + c = a.$

$\text {But } b * \left ( 10e + h + \dfrac{i}{b} \right ) = a \implies \dfrac{a}{b} = \left ( 10e + h + \dfrac{i}{b} \right ).$

In short, the process does give the correct answer. You can see why they do not explain it in the third grade. But the Indian mathematicians who discovered it back in the 600's were pretty clever.
Thanks from Antoniomathgini

Last edited by JeffM1; September 16th, 2017 at 10:11 PM.
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