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As part of an assignment I was asked to perform an evaluation of the following proof:

$5.46$) Evaluate the proposed proof of the following result.

Result: If $x$ is an irrational number and $y$ is a rational number, then $z = x - y$ is irrational.

Proof: Assume, to the contrary, that $z = x - y$ is rational. Then $z = \dfrac{a}{b}$, where $a, b \in\Bbb Z$ and $b\neq0$. Since $\sqrt2$ is irrational, we let $x = \sqrt2$. Since $y$ is rational, $y = \dfrac{c}{d}$, where $c, d\in\Bbb Z$ and $d\neq0$. Therefore, $sqrt(2) = x$ and $x = y+z$ and $y+z = \dfrac{c}{d}+\dfrac{a}{b}$ and $\dfrac{c}{d}+\dfrac{a}{b} = \dfrac{ad+bc}{bd}$; Since $ad + bc$ and $bd$ are integers, where $bd\neq0$, it follows that $\sqrt2$ is rational, producing a contradiction.

I took issue with the proof using $\sqrt2$, stating that it does not prove the statement for all irrational numbers. I basically suggested that we let $x$ be an arbitrary irrational number, otherwise proceeding as was done. The professor has said that the proof as written is correct.

She went on to explain, and I'm copy/pasting the email directly,"The statement “If $x$ is an irrational number and $y$ is an irrational number, then $z = x – y$ is irrational.” is the same as “For any $x$ irrational number and for any $y$ a rational number, $z = x – y$ is an irrational number.” Similarly, the statement “Assume, to the contrary, that $z = x – y$ is rational.” is the same as “For any $x$ irrational number and for any $y$ a rational number, $z = x – y$ is a rational number.” So, if there exist $x$ and $y$ such that the statement ($z = x – y$ is rational) does not hold, then we get contradiction. Why? Because the statement should hold for all $x$ and $y$ irrational numbers, and if does not hold for a particular $x$ and for a particular $y$, then we cannot say that the statement holds for all $x$ and $y$ irrational numbers."

I don't understand the bolded part. Shouldn't the contrary position be, or imply, an existence statement? i.e., if the statement is

For any irrational number $x$ and rational number $y$, $z = x - y$ is irrational.

Shouldn't the contrary be, or imply,

There exists an irrational number $x$ and rational number $y$ such that $z = x + y$ is rational

and the proof should then proceed the way I described? By saying that in all cases, a contradiction is reached?

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    $\begingroup$ I would agree with you. The professor is torturing language, although she may be right. $\endgroup$ Nov 15, 2019 at 4:14

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You are correct and your professor is wrong. Regardless of the language being used, the result in question cannot be proved by assuming $x$ is a fixed irrational number. For the proof by contradiction to be valid, the statement in question should more formally read "Assume that there exist an irrational $x$ and a rational $y$ such that $z = x + y$ is rational" and a contradiction would have to be derived for any irrational $x$ and rational $y$. Just because the statement wasn't explicit doesn't mean it can be interpreted differently.

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    $\begingroup$ Thanks for the insight. I was sure of this but wanted a second opinion before I brought it up in person. This problem was worth 9/100 points and I was given 0 for it. $\endgroup$ Nov 15, 2019 at 13:39

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