Special Proof that golden ratio is irrational How can I proof that the golden ratio is irrational only by using the fact that if a number $n$ is not a square number, then its root $\sqrt n$ must be irrational.
Using this fact, I only would have to show that the golden ratio squared is not a square number. But how can one do this? I only know the typical contradiction proof for irrationality
Can anybody help? :)
Thanks!
 A: The following attempts to show, not that $\phi^2$ is not a square number, but that $\phi$ is irrational, without relying on the irrationality of $\sqrt 5$.
Geometrically, Euclid Elements XIII, 5 proves that if $AB$, in the figure below, is cut in extreme and mean ratio at $C$, and $AB$ is extended so that $AD=AC$ the greater segment, then $DB$ is cut in extreme and mean ratio at $A$ and $AB$ is the greater segment.

Doing this construction in reverse, in the next figure, yields a sequence of smaller and smaller lines cut in extreme and mean ratio. I.e, laying off $CD=CB$, the lesser segment, we get $AC$ cut in extreme and mean ratio at $D$, with $AD$ now the lesser segment. Again, laying off $DE=DA$ gives $DC$ cut in extreme and mean ratio at $E$, and so on.
Since$$BC<CA<2BC$$and hence also$$AD<DC<2AD$$and$$CE<ED<2CE$$then $BC$ does not divide $CA$, $AD$ does not divide $DC$, $CE$ does not divide $ED$, and so on indefinitely.

And since each new greater segment is more than half of the line being cut, then by Euclid X, 2 the original segments $AC$, $CB$ are incommensurable: ”If when the less of two unequal magnitudes is continually subtracted in turn from the greater, that which is left never measures the one before it, the magnitudes will be incommensurable.”
Since the lesser segment continually subtracted from the greater never leaves a remainder that divides it, then the original two magnitudes are incommensurable, i.e. $\frac{AC}{CB}$ is not a ratio of two integers, and $\phi$ is irrational.
