<s>[INST] $a/b < (1 + \sqrt{5})/2 \iff a^2 - ab - b^2 < 0$? For positive integers $a$ and $b$, I want to show that $a/b < (1 + \sqrt{5})/2$ if and only if $a^2 - ab - b^2 < 0$.
I had a loose proof ready to go, but I noticed a fatal flaw. Perhaps there is a way to work around this though.
My tactic was to start from $a^2 - ab - b^2 < 0$ and complete the square on the LHS for $a$. I ended up with
$$
\left(a-\frac{b}{2}\right)^2 - \frac{5b^2}{4} < 0 \iff \left(a - \frac{b}{2} \right)^2 < \left( \frac{\sqrt{5} \, b}{2} \right)^2.
$$
Now the tempting thing to do is to show this is equivalent to saying
$$
\quad \quad \qquad \quad \, \, \iff a-\frac{b}{2} < \frac{\sqrt{5} \, b}{2}
$$
but obviously it is necessary for $a > b / 2$ for this to work.
This is not necessarily the case because if $a = 1$ and $b = 5$, then $a^2 - ba - b^2 = -29 < 0$ and $a/b = 1/5 < (1+\sqrt{5})/2$ so the initial claim is true but $a \leq b/2$.
So is there some kind of assumption I can make to get around this, and without loss of generality? Or should I rethink the entire structure of the proof? Cheers!
 [/INST] If $b=0, a^2<0$ which is impossible
So, $b\ne0, b^2>0$ consequently, $$a^2-ab-b^2<0\iff\left(\dfrac ab\right)^2-\left(\dfrac ab\right)-1<0$$
Now the roots of $x^2-x-1=0$ are $$x=\dfrac{1\pm\sqrt5}2$$
We can prove if $(x-a)(x-b)<0$ with $a<b;$
$$a<x<b$$
</s>