Nib asked
Why is pi an irrational number?
Pi is irrational because assuming it equals a fraction a over b lets you build an area that would have to be a whole number and also be strictly between zero and one.
7 steps, drawn by Nib. Read it here, or watch him draw it. Updated .
1never ends
People often say pi is irrational because its decimal never ends. That is not the reason: one third is 0.3333 forever and is a perfectly good fraction.
A number is irrational when no fraction of two whole numbers equals it. Its decimal then never ends and never repeats.
Irrational means no whole-number fraction equals it, not just endless digits.
a common mix-upPi is not irrational just because its digits go on forever: one third's digits go on forever too, and it is a fraction.
2stand-ins
The fraction 22/7 equals 3.142857..., while pi is 3.141592.... They differ by about 0.0013. The fraction 355/113 is much closer, off by under three ten-millionths, but it is still not pi.
Every fraction near pi is only an approximation.
3suppose a over b
Pi is the circumference of a circle divided by its diameter, the same for every circle.
We use proof by contradiction. Suppose pi equals a over b, where a and b are whole numbers. If that leads to nonsense, the supposition must be false.
Assume pi = a/b and hunt for a contradiction.
4a tiny hump
Define f(x) = x^n (a - b x)^n / n!. Since pi = a/b, the factor (a - b x) is zero at x = pi, so f is zero at both ends of zero to pi. Inside it is a hump, and as n grows the factorial in the bottom shrinks it.
f is a hump on 0 to pi that shrinks as n grows. (Plotted here with a = pi, b = 1 only to show the shape.)
5a whole number
Integrate f(x) sin x from 0 to pi by parts repeatedly. The derivatives of f at 0 and at pi are all whole numbers, because the factorial cancels, so the area equals a sum of whole numbers: it is itself a whole number.
Under the assumption, the area is a whole number.
6trapped
On 0 to pi, x is at most pi and a - b x is between 0 and a, so f is at most (pi a)^n / n!. The area is therefore at most pi (pi a)^n / n!, which goes to zero as n grows, because the factorial eventually outruns any power.
So the area is strictly between 0 and 1 for large n. No whole number lives there, which contradicts the previous step.
The same area would be a whole number and strictly between 0 and 1: impossible.
Why does n! win against (pi a)^n for large n?
Show the answer
Each step multiplies the factorial by a bigger number, but the power only by the same fixed pi a, so eventually the factorial grows faster.
7the whole proof
The proof in one line: assuming pi is a fraction forces one number to be both a whole number and strictly between zero and one.
Pi is irrational: no fraction a/b equals it.
Pi isn't irrational because its digits roam forever; it is irrational because no fraction can fit, and a clever area proves it.