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# Archimedes and Numerical Roots

This problem builds on Approximating Pi. This brilliant man Archimedes managed to establish that $3 1/10 < \pi < 3 1/7$.

The problem is how did he calculate the lengths of the sides of the polygons, which he needed to be able to calculate square roots? He didn't have a calculator but needed to work to an appropriate degree of accuracy. To do this he used what we now call numerical roots.

How might he have calculated $\sqrt{3}$?

This must be somewhere between 1 and 2. How do I know this?

Now calculate the average of $3/2$ and $2$ (which is 1.75) - this is a second approximation to $\sqrt{3}$. i.e. we are saying that a better approximation to $\sqrt{3}$ is $(3/n + n)/2$ where $n$ is an approximation to $\sqrt{3}$.

We then repeat the process to find the new (third) approximation to $\sqrt{3}$. $$\sqrt{3} \approx {(3 / 1.75 + 1.75) \over {2}} = 1.73214\dots$$ to find a fourth approximation repeat this process using 1.73214 and so on...

How many approximations do I have to make before I can find $\sqrt{3}$ correct to five decimal places?

Why do you think it works?

Will it always work no matter what I take as my first approximation and does the same apply to finding other roots?

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Age 14 to 16

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This problem builds on Approximating Pi. This brilliant man Archimedes managed to establish that $3 1/10 < \pi < 3 1/7$.

The problem is how did he calculate the lengths of the sides of the polygons, which he needed to be able to calculate square roots? He didn't have a calculator but needed to work to an appropriate degree of accuracy. To do this he used what we now call numerical roots.

How might he have calculated $\sqrt{3}$?

This must be somewhere between 1 and 2. How do I know this?

Now calculate the average of $3/2$ and $2$ (which is 1.75) - this is a second approximation to $\sqrt{3}$. i.e. we are saying that a better approximation to $\sqrt{3}$ is $(3/n + n)/2$ where $n$ is an approximation to $\sqrt{3}$.

We then repeat the process to find the new (third) approximation to $\sqrt{3}$. $$\sqrt{3} \approx {(3 / 1.75 + 1.75) \over {2}} = 1.73214\dots$$ to find a fourth approximation repeat this process using 1.73214 and so on...

How many approximations do I have to make before I can find $\sqrt{3}$ correct to five decimal places?

Why do you think it works?

Will it always work no matter what I take as my first approximation and does the same apply to finding other roots?

The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?

Last year, on the television programme Antiques Roadshow... work out the approximate profit.