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we all know the basic rules for operations of addition, subtraction, multiplication and division.

but what i don't know is why these rules (of addition, subtraction, multiplication and division) works.

as if we have been given algorithm to do these operations but not explained how they were derived.

if anyone can explain please expain .

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Can you give some context regarding how much math do you know? For instance, do you know about elementary set theory and Dedekind Peano Axioms? I'm assuming you don't. So what you want is an intuitive motivation to define $+$ and $\cdot$ the way we do and a justification of the standard algorithms? –  Git Gud May 19 '13 at 19:22
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This is one of those things one learns in back alleys because it's considered indecent to mention it in school. –  Michael Hardy May 19 '13 at 19:22
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Your last paragraph mentions algorithms. Does your question mean you want to know how algorithms for doing arithmetic are justified, or does it mean you want to know such things as why multiplication is commutative and why it distributes over addition? Some clarification of the question would help. –  Michael Hardy May 19 '13 at 19:26
    
i know elementary set theory but i don't know Dedekind Peano Axioms. –  pratik May 19 '13 at 19:27
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@Michael: The OP responded to your query. –  Cameron Buie May 19 '13 at 19:31
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2 Answers

One thing you'll want to understand here is that you are only talking about 2 operations: addition and multiplication.

Subtraction is a way of abstracting the concept of additive inverses (e.g. $3 + (-3) = 0$), and division is really multiplicative inverses ($3 \times \frac{1}{3} = 1$).

I know this doesn't fully answer your question, but it does simplify it.

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yes you are right –  pratik May 19 '13 at 19:32
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$$ \begin{array}{cccccc} & & 4 & 3 \\ & \times & 7 & 9 \\ \hline & 3 & 8 & 7 \\ 3 & 0 & 1 \\ \hline 3 & 3 & 9 & 7 \end{array} $$

Why is this algorithm, taught in elementary school, justified?

It's because multiplication distributes over addition: $$ 9\cdot(43) = 9\cdot(4\cdot10 \quad+\quad 3) = (9\cdot4)(10)\quad+\quad(9\cdot3) $$ Hence $$ \begin{array}{ccc} & 2 & 7 \\ 3 & 6 \\ \hline 3 & 8 & 7 \end{array} $$

Similarly one multiplies $43$ by $7$, getting $301$, but the $301$ is moved one place to the left because it is $70$ rather than $7$ by which $43$ was to be multiplies; thus one multiplies by $10$ by moving one place to the left.

A think and explanation of long division might be more involved. Or might not$\ldots$

Later edit: $1352/43$. There's a quotient and a remainder: $1352 = 31\cdot 43 + 19$. The quotient is $31$ and the remainder is $19$. So $\dfrac{1352}{43} = 31 + \dfrac{19}{43}$. The remainder must be less than $43$ so the fraction must be less than $1$. So if we move the decimal point over to get a one-digit quotient, we have $$ \frac{135.2}{43} = 3.1 +\text{less than $0.1$} = 3 + \text{less than $1$} $$ Since $43$ goes $3+\text{fraction}$ times into $135.2$, and $43\cdot3$ must be an integer, we have $43$ goes $3+\text{some fraction}$ times into $135$. In particular, $$ 135 = 43\cdot3 + 6.\quad \text{The quotient is $3$ and the remainder is $6$.} $$ Now look at the usual algorithm:

$$ \begin{array}{cccccccccc} & & & & 3 \\ \hline 43 & ) & 1 & 3 & 5 & 2 \\ & & 1 & 2 & 9 \\ \hline & & & & 6 \end{array} $$ Why should we now bring down the $2$ and then divide the resulting $62$ by $43$?

$$ \frac{135.2}{43} = \frac{135}{43} + \frac{0.2}{43} = 3 + \underbrace{\frac{6}{43} + \frac{0.2}{43}} = 3 + \frac{6.2}{43}. $$

So $$ \frac{1352}{43} = 30 + \frac{62}{43}. $$ So that's why.

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