# Evaluating a simple proof about rational numbers

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?

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

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.