# Pi is transcendental

Pi is transcendental is a famous claim regarding the nature of the number $\displaystyle{ \pi }$. It was first proven by Ferdinand von Lindemann in 1882, buildung on methods developed by Charles Hermite, who was able to prove that the number $\displaystyle{ e }$ is transcendental roughly 10 years earlier in 1873. With the help of Karl Weierstrass their findings could be generalized to the Lindemann-Weierstrass theorem in 1885.

## Proof

The proof shown here is not the original proof by Lindemann but a later proof by David Hilbert which is a bit more accessible.

#### Preparation

Suppose $\displaystyle{ \pi }$ is algebraic. Set $\displaystyle{ \alpha_{1}=i\pi }$ where $\displaystyle{ i }$ is the imaginary unit . Then $\displaystyle{ \alpha_{1} }$ is also algebraic, so it is the root of an $\displaystyle{ n }$-th degree polynomial with integer coefficients. Let $\displaystyle{ \alpha_{2}\text{, . . . , }\alpha_{n} }$ be the other roots of this polynomial.

Since $\displaystyle{ 1+e^{i\pi}=0 }$  we have $\displaystyle{ (1+e^{\alpha_{1}})(1+e^{\alpha_{2}})\text{ . . . }(1+e^{\alpha_{n}})=1+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{N}}=0 }$ for some $\displaystyle{ \beta_{1}\text{, . . . , }\beta_{N} }$. These $\displaystyle{ \beta_i }$ are also algebraic, since they are the sums of other algebraic numbers (the $\displaystyle{ \alpha_i }$). Disregarding the $\displaystyle{ \beta_{i} }$ that are equal to $\displaystyle{ 0 }$, we end up with $\displaystyle{ a+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{M}}=0 }$ for some positive integer $\displaystyle{ a }$ (since $\displaystyle{ e^0=1 }$). Because the $\displaystyle{ \beta_1\text{, . . . , }\beta_M }$ are algebraic, they are the roots of a polynomial $\displaystyle{ f(z)=b z^M+ b_1 z^{M-1}+\text{ . . . }+b_{M-1} z + b_M }$ with integer coefficients $\displaystyle{ b\text{, }b_1\text{, . . . , }b_M }$, and because no $\displaystyle{ \beta_i }$ is equal to zero $\displaystyle{ b_M\neq 0 }$ also.

#### Split

First we will multiply our equation with the integral $\displaystyle{ \int_0^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz }$ , where $\displaystyle{ \rho }$ is some positive integer and $\displaystyle{ g(z)=f(z)\cdot b^M }$. We will write $\displaystyle{ \int_0^{\infty} }$ as a shorthand. This allows us to split the sum into two Parts: 

$\displaystyle{ P_1 = a\int_0^{\infty} +\text{ } e^{\beta_1}\int_{\beta_1}^{\infty} + \text{ . . . } + e^{\beta_M}\int_{\beta_M}^{\infty} }$
$\displaystyle{ P_2 = 0 + e^{\beta_1}\int_0^{\beta_1} + \text{ . . . } + e^{\beta_M}\int_0^{\beta_M} }$

#### First Sum, First Term

Then, one can show that $\displaystyle{ \int_0^{\infty} =\int_0^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz }$ is an integer that is divisible by $\displaystyle{ \rho! }$ by expanding $\displaystyle{ \left[g(z)\right]^{\rho+1}=\left[f(z)b^M\right]^{\rho + 1} }$ to an (integer) polynomial and then using the fact that $\displaystyle{ n!=\int_0^{\infty}z^{n}e^{-z}dz }$ for positive integers $\displaystyle{ n }$ to get rid of the $\displaystyle{ e }$-terms and integrals. 

More specifically, if divided by $\displaystyle{ (\rho+1) }$, one gets the remainder $\displaystyle{ b^{\left(M\rho+M\right)}\cdot b_M^{\left(\rho+1\right)}\cdot\rho! }$. 

#### First Sum, Other Terms

Something slightly weaker can be shown for the terms $\displaystyle{ e^{\beta_i}\int_{\beta_i}^{\infty} =e^{\beta_i}\int_{\beta_i}^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz }$. Using the substitution $\displaystyle{ \omega = z+\beta_i }$ we can set the bounds of integration to $\displaystyle{ 0 }$ and $\displaystyle{ \infty }$,  and then perform some similar algebraic manipulations so that we can use $\displaystyle{ n!=\int_0^{\infty}z^{n}e^{-z}dz }$ again,  which tells us that the integral is equal to $\displaystyle{ \left(\rho+1\right)! G\left(\beta_i\right) }$, where $\displaystyle{ G(\beta_i) }$ is a polynomial in $\displaystyle{ \beta_i }$ with integer coefficients.

Even though the $\displaystyle{ \beta_i }$ are themselves no integers, since they are the roots of the integer polynomial $\displaystyle{ f }$ we get an integer again if we sum them all up. Likewise, the sums of the squares, third powers etc. are all integers aswell.  So, even though the $\displaystyle{ G(\beta_i) }$ are no integers, the sum $\displaystyle{ G(\beta_1)+G(\beta_2)+\text{ . . . }+G(\beta_M) }$ is.

So we can bundle all terms of the first sum, except the first, together to get an integer, that is furthermore divisible by $\displaystyle{ \left(\rho +1\right)! }$.

On the whole we have thus shown that $\displaystyle{ P_1 }$ is an integer that is divisible by $\displaystyle{ \rho ! }$.

That means $\displaystyle{ \frac{P_1}{\rho !} }$ is also an integer, and dividing it by $\displaystyle{ \rho + 1 }$ would give one the remainder $\displaystyle{ b^{\left(M\rho+M\right)}\cdot b_M^{\left(\rho+1\right)} }$. (This is just important to make sure that $\displaystyle{ P_1 }$ is not always equal to zero.)

#### Second Sum

For the terms in the second sum we just calculate upper bounds. Let $\displaystyle{ K }$ be the maximum (absolute) value of $\displaystyle{ z\cdot g(z) }$ on all the line segments $\displaystyle{ \left[0;\beta_i\right] }$ and $\displaystyle{ k }$ the maximum (absolute) value of $\displaystyle{ g(z)\cdot e^{-z} }$ also on these line segments. Then:

$\displaystyle{ \left|\int_0^{\beta_i}\right|\leq \left|\beta_i\right|\cdot k \cdot K^{\rho} }$

So, if we abbreviate $\displaystyle{ \kappa =\left(\left|\beta_1\cdot e^{\beta_1}\right| +\left|\beta_2\cdot e^{\beta_2}\right| +\text{ . . . } +\left|\beta_M\cdot e^{\beta_M}\right| \right)\cdot k }$ we simply get

$\displaystyle{ \left|P_2\right|\leq\kappa\cdot K^{\rho} }$.

Note that neither $\displaystyle{ \kappa }$ nor $\displaystyle{ K }$ depend on $\displaystyle{ \rho }$.

#### Conclusion

Now we come back to our original equation $\displaystyle{ P_1+P_2=0 }$. By dividing with $\displaystyle{ \rho ! }$ on both sides we get

$\displaystyle{ \frac{P_1}{\rho !}+\frac{P_2}{\rho !}=0 }$

Remember that $\displaystyle{ \rho }$ was an arbitrary positive integer. According to the inequality we proved for $\displaystyle{ P_2 }$ we can get $\displaystyle{ \frac{P_2}{\rho !} }$ as close to zero as we want if we choose $\displaystyle{ \rho }$ large enough. But we also proved that $\displaystyle{ \frac{P_1}{\rho !} }$ is always an integer, and there are infinitely many choices  for $\displaystyle{ \rho }$ such that $\displaystyle{ \frac{P_1}{\rho !} }$ is not divisible by $\displaystyle{ \rho + 1 }$ so it can't be zero. One can easily see that this equation can't hold. Therefore our initial premise was faulty and $\displaystyle{ \pi }$ is transcendent.

 Statement of the claim Pi is transcendental Level of certainty Proven Nature Theoretical Counterclaim Pi is algebraic Dependent on Dependency of