I am reading Laczkovich's proof ([can be found here:](https://en.m.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational#Laczkovich's_proof)) 


Let $f\_k(x) = 1 - \frac{x^2}k+\frac{x^4}{2! k(k+1)}-\frac{x^6}{3! k(k+1)(k+2)} + \cdots \qquad (k\notin\{0,-1,-2,\ldots\})$


Claim 3: If $x\neq 0$ and if $x^2$ is rational, then $(\forall k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):f\_k(x)\neq0\text{ and }\frac{f\_{k+1}(x)}{f\_k(x)}\notin\mathbb{Q}.$


And there is this part that I didn't understand:


When he proves this claim he starts with..
Otherwise, there would be a number $y\neq 0$ and integers $a$ and $b$ such that $f\_{k}(x)=ay$ and $f\_{k+1}(x)=by$. In order to see why, take $y=f\_{k+1}(x)$, $a=0$ and $b=1$ if $f\_{k}(x)=0$; otherwise, choose integers $a$ and $b$ such that ${f\_{k+1}(x)\over f\_{k}(x)}={b\over a}$ and define $y= {f\_{k}(x)\over a}={f\_{k+1}(x)\over b}$. $(\*)$


first of all does he assume here that $f\_k(x)=0\text{ and }\frac{f\_{k+1}(x)}{f\_k(x)}\in\mathbb{Q}?$
When he writes this "there would be a number $y\neq 0$ and integers $a$ and $b$ such that $f\_{k}(x)=ay$ and $f\_{k+1}(x)=by$ "is it because he assumes what if 
$\frac{f\_{k+1}(x)}{f\_k(x)}\in\mathbb{Q}?$ but does he also assume that what if $f\_k(x)=0\text ?$


I mean what he is doing here is the use of contrapositive of the claim3?
I just want to know how is the process of thinking made here. Can someone explain/paraphrase the reasoning in $(\*)?$ I am literally stuck there. Thanks in advance



Clarification of the claim 3 in the proof of the irrationality of $\pi ^2$
Arnaud is correct, and here is a fleshed-out version of the start of the proof of the claim.


Assume the claim is false.


Then for some $x\neq0$ with $x^2$ rational, there exists $k\in\mathbb Q\setminus \{0,-1,-2,\dots\}$ for which either $f\_k(x)=0$ or $\dfrac{f\_{k+1}(x)}{f\_k(x)}\in\mathbb Q$. Fix such $x$ and $k$.


Case 1: ($f\_k(x)=0$): Taking $y=f\_{k+1}$, $a=0$, and $b=1$, we have $f\_k(x)=ay$ and $f\_{k+1}(x)=by$ for some integers $a$ and $b$ and real $y$.


Case 2: ($f\_k(x)\neq0$): Then $\dfrac{f\_{k+1}(x)}{f\_k(x)}\in\mathbb Q$ (because the claim is false), so choose integers $a$ and $b$ so that $\dfrac{f\_{k+1}(x)}{f\_k(x)}=\dfrac{b}{a}$, and let $y=\dfrac{f\_{k}(x)}{a}$. Again, we have $f\_k(x)=ay$ and $f\_{k+1}(x)=by$ for some integers $a$ and $b$ and real $y$.


In case 2, $y\neq0$ is clear, but in either case, $y=\dfrac{f\_{k+1}(x)}b$, and if $y=0$, it would follow from claim 1 that $f\_{k+n}(x)=0$ for all integers $n\geq1$, in which case the limit in claim 2 would not hold.


Thus $y\neq0$, and summarizing so far: if claim 3 is false (as witnessed by fixed $k$ and $x$), there are integers $a$ and $b$ and real $y\neq0$ for which $f\_k(x)=ay$ and $f\_{k+1}(x)=by$.


Now recall that $x^2$ is rational, and (remembering that $k$ is rational and $a$ and $b$ are integers) choose $c\in\mathbb N$ such that $bc/k$, $ck/x^2$, and $c/x^2$ are all integers. Then consider the sequence $g\_n$ [and so on…]


