Pi is transcendental
Pi is transcendental is a famous claim regarding the nature of the number [math]\displaystyle{ \pi }[/math]. It was first proven by Lindemann in 1882, buildung on methods developed by Hermite, who was able to prove that the number [math]\displaystyle{ e }[/math] is transcendental roughly 10 years earlier in 1873. With the help of Weierstrass their findings could be generalized to the Lindemann-Weierstrass theorem in 1885.
Proof
The proof sketched out here is not the original proof by Lindemann but a later proof by David Hilbert which is more accessible.
Suppose [math]\displaystyle{ \pi }[/math] is algebraic. Then [math]\displaystyle{ \alpha_{1}:=i\pi }[/math] is also algebraic, so it is the root of an [math]\displaystyle{ n }[/math]-th degree polynomial with integer coefficients. Let [math]\displaystyle{ \alpha_{2}\text{, . . . , }\alpha_{n} }[/math] be the other roots of this polynomial. Since [math]\displaystyle{ 1+e^{i\pi}=0 }[/math] we have [math]\displaystyle{ (1+e^{\alpha_{1}})(1+e^{\alpha_{2}})\text{ . . . }(1+e^{\alpha_{n}})=1+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{N}}=0 }[/math] with some algebraic [math]\displaystyle{ \beta_{1}\text{, . . . , }\beta_{N} }[/math]. Disregarding the [math]\displaystyle{ \beta_{i} }[/math] that are equal to [math]\displaystyle{ 0 }[/math], we end up with [math]\displaystyle{ a+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{M}}=0 }[/math] for some positive integer [math]\displaystyle{ a }[/math].
We now have to multiply both sides with an improper integral which I will just call [math]\displaystyle{ I_{\rho} }[/math], with some positive integer [math]\displaystyle{ \rho }[/math] as parameter. This will allow us to split the sum into two parts [math]\displaystyle{ P_1+P_2=0 }[/math], where [math]\displaystyle{ P_1 }[/math] and [math]\displaystyle{ P_2 }[/math] have the following properties:
- [math]\displaystyle{ P_1 }[/math] is an integer that can be divided by [math]\displaystyle{ \rho! }[/math] and [math]\displaystyle{ \frac{P_1}{\rho!}\equiv a b^{\rho M+M}b_{M}^{\rho +1}\text{ mod }(\rho+1) }[/math]
- where [math]\displaystyle{ b }[/math], [math]\displaystyle{ b_M }[/math] are the first and last coefficient of the polynomial with integer coefficients that has the [math]\displaystyle{ \beta_{1}\text{, . . . , }\beta_{M} }[/math] as it's roots (so neither [math]\displaystyle{ a }[/math], [math]\displaystyle{ b }[/math] nor [math]\displaystyle{ b_M }[/math] is zero).
- [math]\displaystyle{ |P_2|\lt\chi K^{\rho} }[/math] for some positive constants [math]\displaystyle{ \chi\text{ and }K }[/math].
So, if we choose [math]\displaystyle{ \rho }[/math] big enough [math]\displaystyle{ \frac{P_2}{\rho!} }[/math] tends towards zero while [math]\displaystyle{ \frac{P_1}{\rho!} }[/math] always stays an integer. Furthermore the complicated congruence guarantees that [math]\displaystyle{ \frac{P_1}{\rho!} }[/math] is not zero if [math]\displaystyle{ \rho }[/math] is a multiple of [math]\displaystyle{ a b b_M }[/math]. Since [math]\displaystyle{ P_1+P_2=0 }[/math] it should also be the case that [math]\displaystyle{ \frac{P_1}{\rho!}+\frac{P_2}{\rho!}=0 }[/math]. But that can't be when the first part get's arbitrarily small while the second part stays a non-zero integer.
| Statement of the claim | Pi is transcendental |
| Level of certainty | Proven |
| Nature | Theoretical |
| Counterclaim | Pi is algebraic |
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