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Proof that Pi is an Irrational Number 📂Analysis

Proof that Pi is an Irrational Number

Theorem

π∉Q\pi \notin \mathbb{Q}

Q\mathbb{Q} represents the set of rational numbers.

Proof

Strategy: Utilize the fact that integers do not have the property of being dense. Define the functions ff, FF very cleverly, applying various tricks. This method was devised by Ivan Niven and is among the easiest to prove that π\pi is irrational, but unfortunately, because it employs the epsilon-delta argument, it cannot be proven within the scope of high school curriculum without leaps.

N\mathbb{N} represents the set of natural numbers, Z\mathbb{Z} represents the set of integers.

  • Part 1. f(j)(0),f(j)(π)∈Zf^{(j)} (0), f^{(j)} (\pi) \in \mathbb{Z}

    Assuming that π∈Q\pi \in \mathbb{Q}, the pi π\pi should be expressible in terms of some relatively prime a,b∈Na,b \in \mathbb{N} as π=ab\pi = {{ a } \over {b}}. Regarding this, let’s define the function ff as follows.

    f(x):=xn(a−bx)nn! f(x) := {{ x^{n} ( a - bx )^{n} } \over { n! }} Then, f(x)f(x) does not have terms of lower order than nn, and since a,b,n∈Na, b, n \in \mathbb{N}, by the binomial theorem, all coefficients of terms in n!f(x)n! f(x) are integers. Therefore, if x=0x = 0, for all j=0,1,2,⋯j = 0, 1, 2, \cdots

    f(j)(0)∈Z f^{(j)} (0) \in \mathbb{Z}

    Meanwhile, since f(x)=f(ab−x)=f(π−x)f(x) = f \left( {{ a } \over { b }} - x \right) = f( \pi - x)

    f(j)(π)∈Z f^{(j)} (\pi) \in \mathbb{Z}

  • Part 2. ∫0πf(x)sin⁡xdx∈Z\int_{0}^{\pi} f(x) \sin x dx \in \mathbb{Z}

    F(x):=∑j=0n(−1)jf(2j)(x)=f(x)−f′′(x)+f(4)(x)−⋯+(−1)nf(2n)(x) F(x) := \sum_{j=0}^{n} ( -1 )^{j} f^{(2j)} (x) = f(x) - f ''(x) + f^{(4)} (x) - \cdots + (-1)^{n} f^{(2n)} (x)

    Defining a new function FF for ff as above

    ddx[F′(x)sin⁡x−F(x)cos⁡x]=F’’(x)sin⁡x+F(x)sin⁡x=[−∑j=1n(−1)jf(2j)(x)+∑j=0n(−1)jf(2j)(x)]sin⁡x=f(x)sin⁡x \begin{align*} {{ d } \over { dx }} \left[ F ' (x) \sin x - F(x) \cos x \right] =& F’’(x) \sin x + F(x) \sin x \\ =& \left[ - \sum_{j=1}^{n} ( -1 )^{j} f^{(2j)} (x) + \sum_{j=0}^{n} ( -1 )^{j} f^{(2j)} (x) \right] \sin x \\ =& f(x) \sin x \end{align*}

    Conversely, integrating f(x)sin⁡xf(x) \sin x from [0,π][0,\pi] gives

    ∫0πf(x)sin⁡xdx=[F′(x)sin⁡x−F(x)cos⁡x]0π=F(π)+F(0) \begin{align*} \int_{0}^{\pi} f(x) \sin x dx =& \left[ F ' (x) \sin x - F(x) \cos x \right]_{0}^{\pi} \\ =& F( \pi ) + F( 0 ) \end{align*}

    But since it was f(j)(0),f(j)(π)∈Zf^{(j)} (0), f^{(j)} (\pi) \in \mathbb{Z} in Part 1, F(0),F(π)∈ZF(0), F(\pi) \in \mathbb{Z} hence

    (F(0)+F(π))∈Z \left( F( 0 ) + F(\pi) \right) \in \mathbb{Z}

  • Part 3. ∫0πf(x)sin⁡xdx∈N\int_{0}^{\pi} f(x) \sin x dx \in \mathbb{N}

    If we say 0<x<π0 < x < \pi then

    f(x)=xn(a−bx)nn!=bnxn(ab−x)nn!=bnxn(π−x)nn! \begin{align*} f(x) =& {{ x^{n} ( a - bx )^{n} } \over { n! }} \\ =& {{ b^{n} x^{n} \left( {{ a } \over { b }} - x \right)^{n} } \over { n! }} \\ =& {{ b^{n} x^{n} \left( \pi - x \right)^{n} } \over { n! }} \end{align*}

    Hence f(x)>0f(x) > 0 and since sin⁡x>0\sin x > 0

    0<f(x)sin⁡x 0 < f(x) \sin x

    Therefore, ∫0πf(x)sin⁡xdx∈Z\int_{0}^{\pi} f(x) \sin x dx \in \mathbb{Z} of Part 2. is actually

    ∫0πf(x)sin⁡xdx∈N \int_{0}^{\pi} f(x) \sin x dx \in \mathbb{N}

  • Part 4. 0<f(x)sin⁡x<πnann!0 < f(x) \sin x < {{ \pi^{n} a^{n} } \over { n! }}
    Differentiating ff gives f′(x)=nxn−1(a−bx)nn!−bnxn(a−bx)n−1n!=xn−1(a−bx)n−1(n−1)![(a−bx)−bx]=2bxn−1(a−bx)n−1(n−1)![π2−x] \begin{align*} f '(x) =& {{ n x^{n-1} ( a - bx )^{n} } \over { n! }} - {{ b n x^{n} ( a - bx )^{n-1} } \over { n! }} \\ =& {{ x^{n-1} ( a - bx )^{n-1} } \over { (n-1) ! }} \left[ (a-bx) - bx \right] \\ =& 2b {{ x^{n-1} ( a - bx )^{n-1} } \over { (n-1) ! }} \left[ {{ \pi } \over {2}} - x \right] \end{align*} Therefore, ff has its maximum value at x=π2x = {{\pi} \over {2}}. Directly substituting and calculating gives f(x)≤bnn!(π2)n(π2)n=ann!(12)n(π2)n=ann!(π4)n<anπnn! \begin{align*} f(x) \le & {{ b^{n} } \over { n! }} \left( {{\pi} \over {2}} \right)^{n} \left( {{\pi} \over {2}} \right)^{n} \\ =& {{ a^{n} } \over { n! }} \left( {{1} \over {2}} \right)^{n} \left( {{\pi} \over {2}} \right)^{n} \\ =& {{ a^{n} } \over { n! }} \left( {{\pi} \over {4}} \right)^{n} \\ <& {{ a^{n} \pi^{n} } \over { n! }} \end{align*} Of course, since from 0<x<π0 < x < \pi, ∣sin⁡x∣≤1| \sin x | \le 1, for all n∈Nn \in \mathbb{N} 0<f(x)sin⁡x<(aπ)nn!=lim⁡n→∞(aπ)nn!=0 0 < f(x) \sin x < {{ (a \pi)^{n} } \over { n! }} = \lim_{n \to \infty} {{ (a \pi)^{n} } \over { n! }} = 0 Therefore n≥N  ⟹  ∫0πf(x)sin⁡xdx<1 n \ge N \implies \int_{0}^{\pi} f(x) \sin x dx < 1 A N∈NN \in \mathbb{N} satisfying this exists. However, this contradicts ∫0πf(x)sin⁡xdx∈N\int_{0}^{\pi} f(x) \sin x dx \in \mathbb{N}, leading to the following conclusion. π∉Q \pi \notin \mathbb{Q}

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