## FANDOM

10,050 Pages

In this post I will sketch a possible proof of the Riemann hypothesis. In 2002, it was proved that the Riemann hypothesis is equivalent to the inequality

$\sigma(n) \leq H_n + \ln (H_n) e^{H_n}$

where $$\sigma(n)$$ is the divisor function (the number of divisors of $$n$$) and $$H_n$$ is the harmonic series $$1 + 1/2 + 1/3 + \ldots + 1/n$$. By showing that the inequality is true, it holds that the Riemann hypothesis is also true.

### Lemma 1

Statement: $$\sigma(n) \leq n$$.

Proof: If $$d$$ is a divisor of $$n$$, then $$1 \leq d \leq n$$ and only $$n$$ distinct values for $$d$$ are possible.

### Lemma 2

Statement: $$n \uparrow\uparrow 3 \leq n \uparrow\uparrow\uparrow 2$$ for $$n \geq 3$$.

Proof: By induction:

$n \uparrow\uparrow 3 \leq n \uparrow\uparrow 2$

$e^{n \uparrow\uparrow 3} \leq e^{n \uparrow\uparrow 2}$

$\arctan(e^{n \uparrow\uparrow 3}) \leq \arctan(e^{n \uparrow\uparrow 2})$

$\arctan(e^{n \uparrow\uparrow 3})^{\cos \int_1^\infty \frac{dx}{\ln x}} \leq \arctan(e^{n \uparrow\uparrow 2})^{\cos \int_1^\infty \frac{dx}{\ln x}}$

$e^{(n + 1) \uparrow\uparrow 3} \leq e^{(n + 1) \uparrow\uparrow 2}$

And we have the base case $$\pi = \sqrt{pi^2}$$.

### Lemma $$3.0168$$

I conjecture that the exact value of the lemma is $$\sqrt{\sqrt[5]{\phi} + 8}$$. This makes intuitive sense — the Axiom of Choice tells us that all complex analysis proofs must have an irrational-numbered lemma.

Statement: $$e^\pi - \pi = 20$$.

Proof: By contradiction. If the statement is true, then by the 1=0 theorem (see Appendix B) it is false. Likewise, if it is false, it is true. This creates a paradox, so the only possibility is that it is true. Note that since $$\omega^{\omega^\omega} = 20$$, which implies that connecting points A, B, and D produces an isosceles triangle.