User:Johnny phi/sandbox

$$if$$, $$\tau$$ = $$\frac{\phi}{\sqrt{2}}$$, then: $$\tau$$ * $$n$$ = $$\sqrt{2}$$, where $$n$$ = $$\frac{1}{2}$$ $$\phi$$ $$\phi$$ = $$\frac{1}{m}$$ = $$1+m$$

$$\Sampi$$ = $$m^2$$, which equals -2+  $$\phi$$. This is the beauty of $$\phi$$. And so we derive the following definition: $$\tau * m$$ = $$\frac{1}\sqrt{2}$$

Now for $$\Sampi$$ we also can state that:

$$if$$: $$\sqrt\frac{\Sampi}$$ = $$\sqrt{3}$$. Than, $$\frac{\Sampi}{\beta}$$ = $$k$$  ⟺ $$\sqrt{2}$$....  $$\frac{\beta}$$ = $$Q$$ ⟺ $$\sqrt{10}$$. And so we see, $$\frac{\sampi}$$ = $$y$$ ⟺ $$\sqrt{5}$$

Now lets annalyze a little more on $$\alpha$$..When we say annalyze in this connection we revere to what is $$\alpha$$ $$to$$ $$\varphi$$

$$4+\alpha$$ = $$3\varphi$$ $$3+\alpha$$ = $$(2+\sampi)*\varphi$$ $$2+\alpha$$ = $$(1+ 2\sampi)*\varphi$$ $$1+\alpha$$ = $$3\sampi*\varphi$$..or $$(1+\sampi^2)*\varphi$$

Lets annalyze $$\beta$$ in connection to $$\sampi$$ $$2\beta^2 = \sampi^2$$

So what I discoverd from this was that if we do:

$$1/\beta = 2\tau*\varphi$$

Which gave afterwards a beatifull determination of $$\sqrt{2}$$ in the form of: $$\beta + \tau$$

We can conclude in this way that the next definition for $$\sqrt{2}$$ is very beatifull:

$$\beta + \tau$$ = $$\frac{\sampi}{\beta}$$