How do you find the square root of 203?

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How do you find the square root of 203?

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Answer 1 · 2016-08-31T19:08:04

Use a Newton Raphson method to find:

$\sqrt{203} \approx \frac{6497}{456} \approx 14.247807$ $sqrt(203) ~~ 6497/456 ~~ 14.247807$

Explanation:

$203 = 7 \cdot 29$ $203 = 7 * 29$ has no square factors. So its square root cannot be simplified.

It is an irrational number between $14$ $14$ and $15$ $15$ , since:

$14^{2} = 196 < 203 < 225 = 15^{2}$ $14^2 = 196 < 203 < 225 = 15^2$

As such, it cannot be represented in the form $\frac{p}{q}$ $p/q$ for integers $p, q$ $p, q$ .

We can find rational approximations using a Newton Raphson method:

To approximate the square root of $n$ $n$ , start with an approximation $a_{0}$ $a_0$ then iterate using the formula:

$a_{i + 1} = \frac{a_{i}^{2} + n}{2 a_{i}}$ $a_(i+1) = (a_i^2+n)/(2a_i)$

I prefer to re-formulate this slightly using separate integers $p_{i}, q_{i}$ $p_i, q_i$ where $\frac{p_{i}}{q_{i}} = a_{i}$ $p_i/q_i = a_i$ and formulae:

$p_{i + 1} = p_{i}^{2} + n q_{i}^{2}$ $p_(i+1) = p_i^2+n q_i^2$

$q_{i + 1} = 2 p_{i} q_{i}$ $q_(i+1) = 2 p_i q_i$

If the resulting pair of integers have a common factor, then divide by that before the next iteration.

So for our example, let $n = 203$ $n=203$ , $p_{0} = 29$ $p_0 = 29$ and $q_{0} = 2$ $q_0 = 2$ (i.e. $a_{0} = 14.5$ $a_0 = 14.5$ ) ...

${ (p_0 = 29), (q_0 = 2) :}$

${ (p_1 = p_0^2+ n q_0^2 = 29^2 + 203*2^2 = 841+812 = 1653), (q_1 = 2 p_0 q_0 = 2*29*2 = 116) :}$

Note that both $p_1$ and $q_1$ are divisible by $29$ , so divide both by that:

${ (p_(1a) = 1653/29 = 57), (q_(1a) = 116/29 = 4) :}$

Next iteration:

${ (p_2 = p_(1a)^2 + n q_(1a)^2 = 57^2+203*4^2 = 3249+3248 = 6497), (q_2 = 2 p_(1a) q_(1a) = 2*57*4 = 456) :}$

If we stop here, we get:

$sqrt(203) ~~ 6497/456 ~~ 14.247807$

Each iteration roughly doubles the number of significant figures in the approximation.

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How do you find the square root of 203?

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