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

algebra-precalculusproof-writing

Here is my proof:

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How can I improve it?
In the bottom one, I decided to use proof by contradiction because I think it's not legitimate to assume that the conclusion's equation is true.

Would it be invalid to do this instead?:

Case 1. $x=1$. Evaluating the equation we get $y + 1/1 = 1 + y/1$, which is true, as required.

Case 2. $y=1$. Evaluating the equation we get $1 + 1/x = 1+1/x$, which is true, as required.

Best Answer

Regarding your first proof, note that when writing a proof, you shouldn't be too stingy with displaying your steps; in particular, claiming that "solving for $y$ we get the required equation", although correct, doesn't make for a persuasive proof (and persuading the reader really is the whole point of a proof)!

It also needs a small correction in its fourth line: \begin{align}&y+\frac1x=1+\frac yx\\\implies{}& xy+1=x+y\\\implies{}& y(x-1)-(x-1)=0\\\implies{}&(y-1)(x-1)=0\\\implies{}& x=1 \quad\text{ or }\quad y=1,\end{align} as required. (Alternatively: $\cdots\implies y(x-1)=x-1\implies x=1\quad\text{or}\quad y=\dfrac{x-1}{x-1}\cdots,$ where the left disjunct protects against division by zero, which, with practice, one knows to look out for.)

I think it's not legitimate to assume that the conclusion's equation is true.

There's no such rule. (In two of your previous questions, you were proving assertions by first assuming them to be true, which is circular reasoning. The other point is not that you cannot make particular assumptions about equations or whatever, but that assumptions should be treated as provisional.)

Would it be invalid to do this instead?:

Case 1. $x=1$. Evaluating the equation we get $y + 1/1 = 1 + y/1$, which is true, as required.

Case 2. $y=1$. Evaluating the equation we get $1 + 1/x = 1+1/x$, which is true, as required.

Well what's actually being required is $$\color\red{ y+\frac1x=}\color\violet{1+\frac yx},$$ so a better presentation is (just showing Case 1 as example):

  • Case 1. $x=1:$ $$\color\red{y+\frac1x=}y+\frac11=y+1=1+\frac y1=\color\violet{1+\frac yx},$$ as required.

I was actually secretly working separately from each side of the equation and meeting in the middle but, for a more succinct presentation, I'm sewing the two pieces together as one line.

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