Question

What is the square root of `3` ?

Answer 1

`sqrt3 = 1.732`

Explanation:

3 is not a perfect square, so does not have an exact square root.

`sqrt3` is an irrational number. The answer is an infinite, non-recurring decimal.

Using a calculator gives:

`sqrt3 = 1.7320508.....`

Give your answer rounded to a reasonable number of decimal places.

Answer 2

The square root of `3` is an irrational number which we can approximate as:

`sqrt(3) ~~ 18817/10864 ~~ 1.7320508`

Explanation:

"The" square root of `3` is an irrational number. It is not expressible in the form `p/q` for integers `p, q` and its decimal expansion neither repeats nor terminates.

It can be expressed by a (non terminating) continued fraction:

`sqrt(3) = [1;bar(1,2)] = 1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(2+1/(1+...)))))))`

This positive square root is also known as the principal square root of `3`. The number `-sqrt(3)` is also a square root of `3`.

We can get rational approximations to `sqrt(3)` by truncating the continued fraction early.

For example:

`sqrt(3) ~~ [1;1,2,1] = 1+1/(1+1/(2+1/1)) = 7/4 = 1.75`

`sqrt(3) ~~ [1;1,2,1,2,1] = 1+1/(1+1/(2+1/(1+1/(2+1/1)))) = 26/15 = 1.7bar(3)`

`sqrt(3) ~~ [1;1,2,1,2,1,2,1] = 1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(2+1/1)))))) = 97/56 = 1.732bar(142857)`

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Another method to find approximations to `sqrt(n)` begins with a rational approximation `p_0/q_0` then repeatedly applies the formulae:

`{ (p_(i+1) = p_i^2 + n q_i^2), (q_(i+1) = 2 p_i q_i) :}`

Let us start with `n=3`, `p_0=2`, `q_0=1`.

Then:

`{ (p_1 = p_0^2+n q_0^2 = 2^2+3*1^2 = 4+3=7), (q_1 = 2 p_0 q_0 = 2*2*1 = 4) :}`

`{ (p_2 = p_1^2 + n q_1^2 = 7^2+3*4^2 = 49+48 = 97), (q_2 = 2 p_1 q_1 = 2*7*4 = 56) :}`

`{ (p_3 = p_2^2 + n q_2^2 = 97^2+3*56^2 = 9409+9408 = 18817), (q_3 = 2 p_2 q_2 = 2*97*56 = 10864) :}`

Stopping here, we get:

`sqrt(3) ~~ 18817/10864 ~~ 1.7320508`

As you may notice, this method is exponentially faster than the continued fraction method, effectively doubling the number of terms each time.