Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is Pi? Why is Pi the only number able to calculate the circumference of a circle? Also, how was Pi discovered?

share|cite|improve this question

marked as duplicate by vadim123, amWhy, joriki, Clayton, Micah May 23 '13 at 16:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

It is the ratio of the circumference of a circle and its diameter. It was observed that this is constant, i.e. independent of the circle – M Turgeon May 23 '13 at 16:32
It can be shown that the complex exponential function $f(z)=e^z$ is periodic. If $P$ is its period, we can define $\pi$ as $$\pi:=\frac{P}{2i}$$ – Jared May 23 '13 at 16:32

1 Answer 1

Usually, Pi ($\pi$) is defined to the number that is the ratio of the circumference of a circle to its diameters. That is, if you measure a circle perfectly and find the circumference equals $C$ and the diameter equals $d$, then you will have $$\pi = \frac{C}{d}.$$

Now, there are many other good numbers out there for determining the circumference of a circle. For example, the number $\tau$ is the ratio of the circumference of a circle, $C$, to its radius, $r$. That is $$\tau = \frac{C}{r}.$$ However, since the diameter is twice the radius ($d = 2r$) this all comes down to the same idea and we have $$\pi = \frac{C}{d} = \frac{C}{2r}.$$ Therefore we have $\tau = 2\pi$.

The value of $\pi$ (or at least approximations) were most likely discovered through measuring the circumference and diameter very precisely. As these measurements were carried out, someone probably noticed a pattern and that the ratios were very close. He or she then began to measure more precisely and noticed that the ratio was in fact constant.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.