real mechanism behind addition, subtraction, multiplication, division 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 .
 A: 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.
A: $$
\begin{array}{cccccc}
& & 4 & 3 \\
& \times & 7 & 9 \\
\hline & 3 & 8 & 7 \\
3 & 0  & 1 \\
\hline 3 & 3 & 9 & 7 
\end{array}
$$
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.
