Main Page
Welcome to $\mathsf{Pr} \infty \mathsf{fWiki}$
 $\mathsf{Pr} \infty \mathsf{fWiki}$ is an online compendium of mathematical proofs! Our goal is the collection, collaboration and classification of mathematical proofs. If you are interested in helping create an online resource for math proofs feel free to register for an account. Thanks and enjoy! If you have any questions, comments, or suggestions please post on the discussion page, or contact one of the administrators. Also, feel free to take a look at the frequently asked questions because you may not be the first with your idea. To see what's currently happening in the community, visit the community portal. |
Featured Proof
Theorem
Pi ($\pi$) is irrational.
Proof
Aiming for a contradiction, suppose $\pi$ is rational.
Then from Existence of Canonical Form of Rational Number:
- $\exists a \in \Z, b \in \Z_{>0}: \pi = \dfrac a b$
Let $n \in \Z_{>0}$.
We define the polynomial function:
- $\forall x \in \R: \map f x = \dfrac {x^n \paren {a - b x}^n} {n!}$
We differentiate this $2 n$ times, and then we build:
- $\ds \map F x = \sum_{j \mathop = 0}^n \paren {-1}^j \map {f^{\paren {2 j} } } x = \map f x + \cdots + \paren {-1}^j \map {f^{\paren {2 j} } } x + \cdots + \paren {-1}^n \map {f^{\paren {2 n} } } x$
That is, $\map F x$ is the alternating sum of $f$ and its first $n$ even derivatives.
First we show that:
- $(1): \quad \map F 0 = \map F \pi$
From the definition of $\map f x$, and our supposition that $\pi = \dfrac a b$, we have that:
- $\forall x \in \R: \map f x = b^n \dfrac {x^n \paren {\pi - x}^n} {n!} = \map f {\pi - x}$
Using the Chain Rule for Derivatives, we can apply the Principle of Mathematical Induction to show that, for all the above derivatives:
- $\forall x \in \R: \map {f^{\paren j} } x = \paren {-1}^j \map {f^{\paren j} } {\pi - x}$
In particular, we have:
- $\forall j \in \set {1, 2, \ldots, n}: \map {f^{\paren {2 j} } } 0 = \map {f^{\paren {2 j} } } \pi$
From the definition of $F$, it follows that:
- $\map F 0 = \map F \pi$
Next we show that:
- $(2): \quad \map F 0$ is an integer.
We use the Binomial Theorem to expand $\paren {a - b x}^n$:
- $\ds \paren {a - b x}^n = \sum_{k \mathop = 0}^n \binom n k a^{n - k} \paren {-b}^k x^k$
By substituting $j = k + n$, we obtain the following expression for $f$:
- $\ds \map f x = \frac 1 {n!} \sum_{j \mathop = n}^{2 n} \binom n {j - n} a^{2 n - j} \paren {-b}^{j - n} x^j$
Note the following:
- The coefficients of $x^0, x^1, \ldots, x^{n - 1}$ are all zero
- The degree of the polynomial $f$ is at most $2 n$.
So we have:
- $\forall j < n: \map {f^{\paren j} } 0 = 0$
- $\forall j > 2 n: \map {f^{\paren j} } 0 = 0$
But for $n \le j \le 2 n$, we have:
- $\map {f^{\paren j} } 0 = \dfrac {j!} {n!} \dbinom n {j - n} a^{2 n - j} \paren {-b}^{j - n}$
Because $j \ge n$, it follows that $\dfrac {j!} {n!}$ is an integer.
So is the binomial coefficient $\dbinom n {j - n}$ by its very nature.
As $a$ and $b$ are both integers, then so are $a^{2 n - j}$ and $\paren {-b}^{j - n}$.
So $\map {f^{\paren j} } 0$ is an integer for all $j$, and hence so is $\map F 0$.
Next we show that:
- $(3): \quad \ds \dfrac 1 2 \int_0^\pi \map f x \sin x \rd x = \map F 0$
As $\map f x$ is a polynomial function of degree $n$, it follows that $f^{\paren {2 n + 2} }$ is the null polynomial.
This means:
- $F + F = f$
Using the Product Rule for Derivatives and the derivatives of sine and cosine, we get:
- $\paren {\map {F'} x \sin x - \map F x \cos x}' = \map f x \sin x$
By the Fundamental Theorem of Calculus, this leads us to:
- $\ds \frac 1 2 \int_0^\pi \map f x \sin x \rd x = \frac 1 2 \bigintlimits {\paren {\map {F'} x \sin x - \map F x \cos x} } {x \mathop = 0} {x \mathop = \pi}$
From Sine and Cosine are Periodic on Reals, we have that $\sin 0 = \sin \pi = 0$ and $\cos 0 = - \cos \pi = 1$.
So, from $\map F 0 = \map F \pi$ (see $(1)$ above), we have:
- $\ds \frac 1 2 \int_0^\pi \map f x \sin x \rd x = \map F 0$
The final step:
On the interval $\openint 0 \pi$, we have from Sine and Cosine are Periodic on Reals that $\sin x > 0$.
So from $(2)$ and $(3)$ above, we have that $\map F 0$ is a positive integer.
Now, we have that:
- $\paren {x - \dfrac \pi 2}^2 = x^2 - \pi x + \paren {\dfrac \pi 2}^2$
and so:
- $x \paren {\pi - x} = \paren {\dfrac \pi 2}^2 - \paren {x - \dfrac \pi 2}^2$
Hence:
- $\forall x \in \R: x \paren {\pi - x} \le \paren {\dfrac \pi 2}^2$
Also, from Real Sine Function is Bounded, $0 \le \sin x \le 1$ on the interval $\openint 0 \pi$.
So, by the definition of $f$:
- $\ds \frac 1 2 \int_0^\pi \map f x \sin x \rd x \le \frac {b^n} {n!} \paren {\frac \pi 2}^{2 n + 1}$
But this is smaller than $1$ for large $n$, from Radius of Convergence of Power Series over Factorial.
Hence, for these large $n$, we have $\map F 0 < 1$, by $(3)$.
This is impossible for the (strictly) positive integer $\map F 0$.
So our assumption that $\pi$ is rational must have been false.
$\blacksquare$
Historical Note
This proof was first published by Ivan Morton Niven in $1947$, and is considered a classic.
The underlying method was developed by Charles Hermite in $1873$, but it was Niven who provided the final details.
Sources
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.17$: More About Irrational Numbers. $\pi$ is Irrational