Question

What is the best way of finding `sqrt(13)` without using a calculator?

Answer 1

I'd suggest Newton's Method, though I'm not prepared to claim that is is easier than guess and check, then adjust the guess.

Explanation:

Newton's Method is an iterative method of approximation. (It works because of calculus, but this question is posted in Algebra, so let's leave that alone.)

Make a first approximation. In your example, say `x_1 = 3`

The next approximation is: `x_2 = 1/2(13/x_1+x_1)`

In other words, divide `13` by the current approximation and average that with your last approximation.

Knowing `x_n`, we find `x_(n+1)` by:

`x_(n+1) = 1/2( 13/x_n + x_n)`

So we get: `x_1 = 3`

To find `x_2`:
`13/3 = 4.33`
The average of our current approximation, `3` and the quotient `4.33` is `3.67`
So `x_2 = 3.67`

To find `x_3`:
`13/3.67 = 3.54`
The average of our current approximation, `3.67` and the quotient `3.54` is `3.61`
So `x_3 = 3.61`

Yes, it used to be tedious doing calculations.

Answer 2

There is a (perhaps not well known) method for finding the square root of a number which I have attempted to demonstrate below.

Explanation:

Start as if you were setting up a long division (but note the absence of a divisor). The number is divided into blocks of 2 digits with as many pairs of zeros after the decimal point as you care to write. The decimal point should be written directly above the decimal point of the number for which you are trying to find the square root (I seem to have lost mine).

Decide on the largest digit whose square is not larger than the first digit pair of the value you are working with and enter them as indicated below

Multiply the number above the line by the number to the left of the vertical line and subtract this product from the value above it.

Copy the next pair of digits down as a suffix to the previous remainder.
Double the value above the line and allow for a suffix digit (so, in this case 3 becomes something between 60 and 69; yet to be determined).

Determine the largest digit which when used as the suffix digit on the left and then used to multiply the resultant value is not greater than the working value (in this case not greater than 400).

Multiply, subtract, bring down the next digit pair.
Double the value from the top and write with space for a suffix digit to the left of the work area.

Continue the process as indicated below:

Please; if anyone can provide a simpler explanation of how to work this process, please do so.

Answer 3

Rather than write a long comment to Jim's, here's 'another' answer.

To find `sqrt(n)`, iterate your approximations using:

`a_(i+1) = a_i + (n - a_i^2)/(2a_i)`

Explanation:

I usually use this with 'improper' fractions to derive a sequence of approximations, stopping when I think I have enough significant digits, then long dividing the resulting integers.

Alternatively, if I only want the square root to 4 significant digits or so, I start with a reasonable 2 digit approximation and perform one or two steps.

I try to memorise the squares of `2` digit numbers too. So in the case of `13` I should remember that `36^2 = 1296` is rather close to `1300`, so `36` makes a good approximation to `sqrt(1300)`.

The next approximation would be `36 + 4/72 = 36 + 1/18 ~= 36.056`

Hence `sqrt(13) ~= 3.6056`