# What is the catch in this geometrical/number theory question?

Here is a simple geometrical construction: $$CA \perp x$$ and $$DE \perp OC$$

As a result: $$\bigtriangleup CDE \cong \bigtriangleup CAO$$, because $$\angle CDE=\angle CAO=\frac{\pi }{2}$$ and $$\angle ECD$$ is common. This means: $$\frac{ED}{AO}=\frac{CD}{CA}=\frac{EC}{OC}$$ Or $$\frac{ED}{AO}=\frac{CO-AO}{CA}=\frac{EC}{OC}$$ because $$AO=DO$$. Obviously, $$EC, as a result: $$CO-AO Also, if $$\alpha \to 0$$, where $$\cos(\alpha)=\frac{AO}{CO}$$, then $$CA \to 0$$ and $$EC \to 0$$ as well.

Now, if we take $$CO=BO=\sqrt{p_{n+1}}$$ and $$DO=AO=\sqrt{p_{n}}$$, square roots of the two consecutive primes, we obtain: $$\sqrt{p_{n+1}} - \sqrt{p_{n}} < EC$$

Because $$\lim_{n \to \infty }\frac{\sqrt{p_{n}}}{\sqrt{p_{n+1}}}=1$$ which seems to be true from (Limit inferior of the quotient of two consecutive primes), we have $$\cos\left ( \alpha \right )=\frac{\sqrt{p_{n}}}{\sqrt{p_{n+1}}}\rightarrow 1, n \to \infty$$ which means $$\alpha \to 0$$ and as a result $$EC \to 0$$. So $$\sqrt{p_{n+1}} - \sqrt{p_{n}} \to 0$$ which is inline with my previous invstigations http://rtybase.blogspot.co.uk/2013/05/tackling-andricas-conjecture-part-3.html.

In a way, this proves Andrica's conjecture (https://en.wikipedia.org/wiki/Andrica%27s_conjecture), unless there is a catch.

The problem arises when substituting $$CO=BO=p_{n+1}$$ and $$DO=AO=p_{n}$$ $$p_{n+1} - p_{n} < EC$$ $$\lim_{n \to \infty }\frac{p_{n}}{p_{n+1}}=1$$ But then $$p_{n+1} - p_{n} \to 0$$ is simply false. Where is the catch?

Remark: because $$p_{n+1}<2 \cdot p_{n}$$, $$\frac{p_{n}}{p_{n+1}} > \frac{1}{2}$$ and $$\frac{\sqrt{p_{n}}}{\sqrt{p_{n+1}}}>\frac{\sqrt{2}}{2}$$ these impose some bounds on $$\alpha$$, like $$0< \alpha <\frac{\pi }{3}$$ for the first case and $$0< \alpha <\frac{\pi }{4}$$ for the second.

• I changed two instances of $cos(\alpha)$ to $\cos(\alpha)$. That is standard. This not only prevents italicization, but also provides proper spacing in expressions like $a\cos b$. – Michael Hardy Jun 14 '13 at 22:54
• Fair enough, then something is wrong with this geometrical construction, because if $\cos(\alpha) \rightarrow 1$ then $\sin(\alpha) \rightarrow 0$ which means $CA \rightarrow 0$ – rtybase Jun 14 '13 at 23:07
• When $n \to \infty$ then $\alpha$ changes, it should have probably been indicated as $\alpha_{n}$ – rtybase Jun 14 '13 at 23:12
• @rtybase Even if $\sin\alpha\rightarrow 0$, $CA$ does not necessarily tend to zero since simultaneously $AO,CO\rightarrow\infty$. – Start wearing purple Jun 14 '13 at 23:12
• Even so $\lim_{n \to \infty }\frac{\sqrt{p_{n}}}{\sqrt{p_{n+1}}}=1$ and $\cos(\alpha_{n})=\frac{AO}{CO}=\frac{\sqrt{p_{n}}}{\sqrt{p_{n+1}}}$ – rtybase Jun 14 '13 at 23:15