Dec 29, 2021 #2 skeeter Messages 1,103 Reaction score 1 $\bigg[(2+2\sqrt{2})^2 \cdot (6-4\sqrt{2}) \bigg]^{1/6}$ $\bigg[2(6+4\sqrt{2})(6-4\sqrt{2}) \bigg]^{1/6}$ $\bigg[2(36-32) \bigg]^{1/6}$ $\bigg[2^3\bigg]^{1/6} = \sqrt{2}$
$\bigg[(2+2\sqrt{2})^2 \cdot (6-4\sqrt{2}) \bigg]^{1/6}$ $\bigg[2(6+4\sqrt{2})(6-4\sqrt{2}) \bigg]^{1/6}$ $\bigg[2(36-32) \bigg]^{1/6}$ $\bigg[2^3\bigg]^{1/6} = \sqrt{2}$
Thread 'There are only finitely many primes' I just saw this one. If there are finitely many primes, then ##0<\prod_{p}\sin(\frac\pi p)=\prod_p\sin\left(\frac{\pi(1+2\prod_q q)}p\right)=0## Of course it is in a way just a variation of Euclid's idea, but it is a one liner. View full post »
I just saw this one. If there are finitely many primes, then ##0<\prod_{p}\sin(\frac\pi p)=\prod_p\sin\left(\frac{\pi(1+2\prod_q q)}p\right)=0## Of course it is in a way just a variation of Euclid's idea, but it is a one liner.