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!


3 Answers 3


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 $a<b;$


  • $\begingroup$ Thank you! Perfect solution! $\endgroup$ May 14, 2017 at 14:44
  • $\begingroup$ @TristanBatchler If your question has been fully answered, please accept said answer. $\endgroup$
    – Omry
    May 14, 2017 at 14:46
  • $\begingroup$ @Omry Yep, I definitely will when able. Need to wait another 5 minutes or so to accept this answer. $\endgroup$ May 14, 2017 at 14:47

I want to answer to this question with matrix method. Consider the following matrix that is well-know to $Q$ matrix $$ Q=\left[ \begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array} \right] \, . $$ with the induction on $n$, you can prove that the $n$th power of matrix $Q$, is in the following form $$ Q_2^n= \left[ \begin{array}{cc} F_{n-1} & F_{n} \\ F_{n} & F_{n+1} \end{array} \right] \Longrightarrow F_{n-1}\,F_{n+1}-F_n^2=(-1)^n $$ by definition of Fibonacci sequence ($F_{n+1}=F_n+F_{n-1}$) you can conclude that $$ F_n^2-F_{n-1}\,F_{n+1}-F_{n-1}^2={(-1)}^{n+1} $$ Now, Suppose that $F_n=a$ and $F_{n-1}=b$, then for $n=2k$, we have $$ a^2-ab-b^2=-1<0 $$ In addition, we know that the limit values of Fibonacci number is the golden number as follows $$ \lim_{n \to \infty}\frac{a}{b}=\lim_{n \to \infty} \frac{F_n}{F_{n-1}}= \frac{1+\sqrt{5}}{2} $$

that it completes the discussion.

  • 1
    $\begingroup$ That's awesome! Nice answer with matrices. $\endgroup$ May 15, 2017 at 22:26

Let $g=(1+\sqrt 5\;)/2.$ Let $a/b=g-d.$ Then $$(a/b)^2-(a/b)-1<0\iff (g-d)^2-(g-d)-1<0\iff$$ $$\iff (g^2-g-1)+(d^2-2dg+d)<0\iff d^2-2gd+d<0\iff$$ $$\iff d^2-d(1+\sqrt 5\;)+d<0\iff d(d-\sqrt 5\;)<0.$$ [I]. Now $0<a/b<g \implies 0<d<g<\sqrt 5 \implies d(d-\sqrt 5\;)<0$.

[II]. And $a/b>g \implies d<0$ $ \implies d(d-\sqrt 5\;)=|d|(|d|+\sqrt 5\;)>0\implies$ $$\implies \neg [(a/b)^2-(a/b)-1<0]$$ which implies $(a/b)^2-(a/b)-1>0$, because the only solutions to $x^2-x-1=0$ are $g$ and $(-g^{-1})$, neither of which is rational.


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