# While finding root of a square number in dividing method why we make double of the first root number and why we take the numbers in pair?

Mar 10, 2017

#### Explanation:

Let a number be $k p q r s t m$. Observe that square of a single digit number can have up to two digits, square of a two digit number can have up to four digits, square of a three digit number can have up to six digits and square of a four digit number can have up to eight digits. You may have already got a hint now that why we take the numbers in pairs.

As the number has seven digits, so square root will have four digits. And making them in pairs we get $\underline{k} \text{ "ul(pq)" "ul(rs)" } \underline{t m}$ and as$k$ is single digit, square root could start from $3 , 2$ or $1$.

The numerical value of number is

$k \times 1000000 + p \times 100000 + q \times 10000 + r \times 1000 + s \times 100 + t \times 10 + m$

we an also write it the following way, which we say (A)

$k \times 1000000 + \left(10 p + q\right) \times 10000 + \left(10 r + s\right) \times 100 + \left(10 t + m\right)$

Let us consider a two digit number $a b c$ and let its square root be $f g$. Actually numerical value of these numbers is $100 a + 10 b + c$ and $10 f + g$ and hence we must have

$100 a + 10 b + c = {\left(10 f + g\right)}^{2} = 100 {f}^{2} + 20 f g + {g}^{2}$

or $100 a + 10 b + c = 100 {f}^{2} + \underline{2 \left(10 f + g\right)} g$

Hence, in division method we first search for some $f$, whose square is equal or just less than $a$. Naturally $f$ comes in the place for quotient and remainder would be $\left(a - {f}^{2}\right)$, with place value $100 \left(a - {f}^{2}\right)$.

For next digit, we choose divisor as double of $f$ (note that its place value is $10 f$ and choose a $g$, which makes it $10 f + g$.

I hope this makes this clear. Would have gone for a bigger number like $k p q r s t m$, but things get too complicated.