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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, $EC<CA<OC$, as a result: $$CO-AO<EC$$ 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.

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  • $\begingroup$ 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 $a\cos b$. $\endgroup$ – Michael Hardy Jun 14 '13 at 22:54
  • $\begingroup$ Fair enough, then something is wrong with this geometrical construction, because if $\cos(\alpha) \rightarrow 1$ then $\sin(\alpha) \rightarrow 0$ which means $CA \rightarrow 0$ $\endgroup$ – rtybase Jun 14 '13 at 23:07
  • $\begingroup$ When $n \to \infty $ then $\alpha$ changes, it should have probably been indicated as $\alpha_{n}$ $\endgroup$ – rtybase Jun 14 '13 at 23:12
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    $\begingroup$ @rtybase Even if $\sin\alpha\rightarrow 0$, $CA$ does not necessarily tend to zero since simultaneously $AO,CO\rightarrow\infty$. $\endgroup$ – Start wearing purple Jun 14 '13 at 23:12
  • $\begingroup$ Even so $\lim_{n \to \infty }\frac{\sqrt{p_{n}}}{\sqrt{p_{n+1}}}=1$ and $\cos(\alpha_{n})=\frac{AO}{CO}=\frac{\sqrt{p_{n}}}{\sqrt{p_{n+1}}}$ $\endgroup$ – rtybase Jun 14 '13 at 23:15

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