Prove that $y+1/x=1+y/x$ iff either $x=1$ or $y=1$

Question

This is an exercise from Velleman's "How To Prove It". I would like some feedback on the proof structure for proving goals that involve disjunctions. Also, is it correct to use "$\longrightarrow$" to chain equations together like I do in the proof below, or should they be explained in words?

Prove that $y+1/x=1+y/x$ iff either $x=1$ or $y=1$

Proof: Suppose $y+ \frac{1}{x}= 1 + \frac{y}{x} $. If $x=1$, then clearly $x=1$ or $y=1$. Now suppose $x\neq1$. Then we have $y+\frac{1}{x} = 1+\frac{y}{x} \longrightarrow xy+1 = x+y \longrightarrow xy-y=x-1 \longrightarrow y(x-1)=x-1 \longrightarrow y=1$. Thus, $x=1$ or $y=1$.

Now suppose $x=1$ or $y=1$. We now consider two cases: Case 1. $x=1$. Then $y+\frac{1}{x} = y + \frac{1}{1} = y+1=\frac{y}{1} + 1=\frac{y}{x}+1$. Case 2. $y=1$. Then $y+\frac{1}{x} = 1 + \frac{1}{x} = 1+\frac{y}{x}$.

In every case, we have that $y+1/x=1+y/x$. $\square$

Answer 1

The structure of your proof is entirely correct. However, as you already suspect, chaining equations together using "$\longrightarrow$" is poor form. It would be better to explain each step (briefly) in words. For example:

...Now suppose $x\neq1$. Then from $y+\tfrac1x=1+\tfrac yx$ it follows that $$xy+1=x+y,$$ by multiplying by $x$. Rearranging terms then shows that $$y(x-1)=x-1,$$ and hence that $y=1$.

As suggested in Matteo's answer, a more direct approach would be to note that $$y+\tfrac1x=1+\tfrac yx,$$ is equivalent to $y-1-(\tfrac yx-\tfrac1x)=0$, which is in turn equivalent to $$(y-1)(1-\tfrac1x)=0,$$ which is in turn equivalent to $x=1\ \vee\ y=1$.

Answer 2

First, you have to have $x\neq0$. Rewriting, you have: $$y-1=\frac{1}{x}(y-1) \iff (y-1)\left(\frac{1}{x}-1\right)=0$$ From here, you deduce that: $y=1 \lor x=1$. So, these are the only solutions and your approach is completely correct.