The first rigorous proof that π is irrational is from Johann Heinri...
That is, assume $\pi$ is rational.
To prove that $f(x)$ and its derivatives $f^{(j)}(x)$ have integral...
Shouldn't that be $F_n(x)$ ?
\begin{eqnarray*} & F''(x) &=f^{2}(x)-f^{4}(x)+f^{6}(x) - ... \\ &...
The idea now is to show that the integral of $f(x) sin(x)$ cannot a...
We can see that $f(x)$ is positive over $(0, \pi)$, and since $\pi$...
A SIMPLE PROOF THAT
TT
IS IRRATIONAL
IVAN NIVEN
Let
7T
= a/6, the quotient of positive integers. We define the poly-
nomials
x
n
(a
bx)
n
F(x) = ƒ<» - fW(x) +/
(4)
0) + (- l)
n
/
(2n)
0)>
the positive integer n being specified later. Since nlf(x) has integral
coefficients and terms in x of degree not less than n, f(x) and its
derivatives ƒ
{j)
(x) have integral values for # = 0; also for x*=ir~a/b,
since ƒ(x) =f(a/b—x). By elementary calculus we have
d
\F'(x) sin x F(x) cos x\ = F"(x) sin x + F(x) sin x = f(x) sin #
dx
and
(1) ! ƒ(» sin xdx = [F'(a) sin x - i?(» cos
#]
0
*
= F(v) + F(0).
J
o
Now
F(TT)-{-F(fi)
is aninteger, since ƒ
(
#(?r) and/
(j,)
(0) are integers. But
for 0<x<7T,
7r
n
a
n
0 < ƒ(#) sin x < j
n\
so that the integral in (1) is positive, but arbitrarily small for n suffi-
ciently large. Thus (1) is false, and so is our assumption that
TT
is
rational.
PURDUE UNIVERSITY
Received by the editors November 26, 1946, and, in revised form, December 20,
1946.
509

Discussion

Shouldn't that be $F_n(x)$ ? \begin{eqnarray*} & F''(x) &=f^{2}(x)-f^{4}(x)+f^{6}(x) - ... \\ & F''(x) & + F(x) = f^{2}(x)-f^{4}(x)+f^{6}(x) - ... \\ &+& f(x)-f^{2}(x)+f^{4}(x)-f^{6}(x)+... \\ & = & f(x) \end{eqnarray*} No; $n$ is specified later in the proof to be a *sufficiently large* constant; the definition of $F(x)$ is simply in terms of this constant $n$. That is, assume $\pi$ is rational. Starting with $0<x<\pi$, we can raise to the $n$th power to get $0<x^n<\pi^n$. Starting again with $0<x<\pi$, multiply across by $-b$, add $a$, then raise to the $n$th power to get $a^n > (a-bx)^n>(a-b\pi)^n=0$. Multiplying both equations and dividing by $n!$ gets us $0 < \frac{x^n(a-bx)^n}{n!} < \frac{\pi^na^n}{n!}$. In particular, this will work for all $n$ large enough such that $\pi^na^n<n!$. (So $f$ and $F$ in fact depend on $a$, but there's no harm in that since $a$ has already been chosen to be arbitrary and fixed) The first rigorous proof that π is irrational is from Johann Heinrich Lambert in 1761. He proved that if $x \not= 0$ is rational, then $tan(x)$ must be irrational. Since $tan(π /4) = 1$ is rational, then π must be irrational. The simpler proof given here is due to Ivan Niven in 1947, at the time a young mathematician at the University of Oregon. It only assumes a knowledge of basic Calculus. nice clarification. The idea now is to show that the integral of $f(x) sin(x)$ cannot always be an integer for all values of n, which will be in contradiction with $(1)$, thus proving that $\pi$ must be irrational. Consider a value of x that is between 0 and $\pi$. It's not difficult to see that $f(x)$ in this interval is always less than $\frac{\pi^{n}a^{n}}{n!}$. At the same time $sin(x)$ will always be less than 1 in the same interval and so $f(x)sin(x)<\frac{\pi^{n}a^{n}}{n!}$. As a consequence $\int_{0}^{\pi} f(x)sin(x)<\frac{\pi^{n+1}a^{n}}{n!}$. Now if we recall that $e^{\pi a}=\sum \frac{\pi^n a^n }{n!} $, and any term in a convergent series will have to tend to 0 as $n \rightarrow \infty $, we conclude that $\int_{0}^{\pi} f(x)sin(x)$ will have to tend to 0 for a sufficient large $n$, which contradicts the previous result stating that the integral must be an integer. By contradiction we can now conclude that $\pi$ needs to be irrational. We can see that $f(x)$ is positive over $(0, \pi)$, and since $\pi$ is the smallest positive zero of the sine function, $\sin x$ is positive there as well; thus their product also is. This is the first half of the inequality: $0 < f(x) \sin x$. For the other half, note that the zeroes of $f$ are at $0$ and $\pi$, and that the derivative of $f(x)$ is: $$f'(x) = \frac{n (a-2bx) \cdot x^{n-1} (a-bx)^{n-1}}{n!}$$ It has zeroes at $0$, $\pi/2$, and $\pi$. Since $f$ is positive over $(0, \pi)$, we can deduce that over the interval $[0, \pi]$, the graph of $f$ is a Bell curve: it goes up from $0$ at $x=0$, to $f(\pi/2)$ halfway, and back down from there to $0$ at $x=\pi$. Thus over the interval $(0, \pi)$, we conclude that $$f(x) < f(\pi/2) = \frac{\pi^n a^n}{4n!} < \frac{\pi^na^n}{n!},$$ completing the inequality. (Is there an easier way to show this half?) To prove that $f(x)$ and its derivatives $f^{(j)}(x)$ have integral values for $x=0$ we should first note that $f(x)$ is a polynomial with terms in $x$ from $x^n...x^{2n}$. To do that we just need to expand the polynomial $(a-bx)^{n}$ using the binomial expansion and note that we end up with a polynomial with terms from $x^0...x^{n}$. Finally, when multiplied by $x^{n}$ we end up with terms ranging from $x^n...x^{2n}$. Now let's divide the derivatives in 3 groups: \begin{eqnarray*} & \text{1. } & j<n \\ & \text{2. } & n\leq j \leq 2n \\ & \text{3. } & j>2n \end{eqnarray*} For $j<n$ the terms in $f^{(j)}(x)$ will range from $x^{n-j}...x^{2n-j}$ and so $f^{(j)}(0) = 0 $. For $j>2n$ all the terms will vanish and again $f^{(j)}(0) = 0$. Finally for $n\leq j \leq 2n$, there will be at least a term that doesn't vanish for $x=0$ - the term in $x^{j}$ - and so $f^{(j)}(0) = C_{j}$, where $C_j$ is a binomial coefficient multiplied by $\frac{j!}{n!}$. The binomial coefficient is an integer as well as $\frac{j!}{n!}$, because $j>n$ and so their product ($C_j$) will also be an integer. We finally proved that $f(x)$ and its derivatives $f^{(j)}(x)$ have integral values for $x=0$, for all $j$. Now if we observe that: \begin{eqnarray*} f(a/b-x) & =& (a/b-x)^{n}(a-b(a/b-x))^{n}/n! \\ & = & (a/b-x)^{n}(bx)^{n}/n! \\ & = & (a-bx)^{n}x^{n}/n! \\ & = & f(x) \end{eqnarray*} We verify that the results derived for $x=0$ are also applicable for $x=a/b=\pi$.