Prove that the golden ratio is irrational by contradiction I am struggling to see where the contradiction lies in my proof. In a previous example, $1/\phi = \phi-1$ where $\phi$ is the golden ratio $\frac{\sqrt{5} + 1}{2}$. 
Since I am proving by contradiction, I started out by assuming that $ϕ$ is rational. Then, by definition, there exists $a,b$ such that $\phi = a/b$. After some simple calculations and using the result shown from my previous example, I found that $\phi= b/(a-b)$. I also know that $b < a$ from directly calculating the ratio. 
I know there is a contradiction in the result $ϕ = b/(a-b)$ but I cannot see it. Any help would be appreciated. 

Another way:
Let´s assume that φ is rational.
a/b=φ is completely reduced (we can do this when it is rational)
b<a (by definition)
a/b =  (1+ √5)/2  < ( 1 + √9) / 2 = 2   →   a < 2 b →  a-b < b
From  1/φ = φ - 1
→   b/a = a/b -1 = (a - b)/b
b/a is completely reduced but is equal to another fraction with both a smaller denominator and a smaller numerator.
This is a contradiction, so φ is not rational.
I explained a bit more in :
https://www.valgetal.com/physics/Mathematics/Golden%20Ratio/Golden%20Ratio.htm
