Is there a unique real number $x$ such that for all $y$, $xy = y$?

Question

Is there a unique real number x such that for all $y$, $xy = y$?

In logical form, the equation would be like this: $\exists!x: \forall y, (xy = y)$

My book states that such an $x$ does not exist because when $y = 0$, any value of $x$ would suffice $xy = y$ and therefore $x$ is not unique.

But I'm a little confused by that part. Since we are looking for an $x$ that should work for all $y$, shouldn't $x = 1$ be the answer here since only when $x = 1, xy = y$ holds true for all values of $y$?

For example, when $x = 1$, $x \cdot 1 = 1, x \cdot 2 = y$ and $\forall y, (xy = y)$. And only when $x = 1$ would $\forall y, (xy = y)$ be true.

Answer 1

The statement that your text refutes is $$\forall y \exists! x: ( xy=y )$$

which is a valid refutation as we only need one $y$ for which $\exists! x: ( xy=y )$ fails. And that $y$ is $0$ (the counterexample is also unique).

While $$\exists! x \forall y: (xy =y)$$

is a true statement, because we can take $x=1$ to prove it.

Answer 2

This is probably a typographical error in the book. It'd be helpful for others visiting this website if you included details such as the name of the book, page number, etc.

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Your reasoning is correct. $x = 1$ suffices and is in fact the only number in $\mathbb R$ such that $xy = y$ for all $y\in\mathbb R$. In other words, $x$ is unique.

I believe what the author is trying to explain is that: for every $y\in\mathbb R$, there may not exist a unique $x\in\mathbb R$, such that $xy = y$. The only case where this fails is when $y = 0$. In that case, infinitely many (in fact, all of them) $x\in\mathbb R$ satisfy $xy = y$.