Existence and uniqueness of $y$ such that $xy=(xy)^2$ for every real $x$

Question

I've been asked to prove this theorem: There exists a unique $y \in \mathbb{R}$ such that for every $x \in \mathbb{R}$, we have $$xy = (xy)^2$$

Now, this may seem rather silly, but I'm wondering if this theorem is even true. Clearly, if $y = 1/x$, then regardless of what $x$ is, we would get 1 = 1 on the left hand side and right hand side. However, the theorem also holds if $y = 0$. Since we have two values for which the theorem holds, namely, $y = 0$ and $y = 1/x$, does this mean that the existence is not unique?

$y$ has to be the same for all $x$, that is, it must be independent of $x$. This is not the case for $y=1/x$. — Julián Aguirre, Nov 03 '15 at 17:08
Yes, I got the answer now. y = 1/x does not hold when x = 0, and so it does not hold $\forall x \in \mathbb{R}$. So xy = 0 is what I'm looking for. Thanks. Please mark as resolved. — ShinyPebble, Nov 03 '15 at 17:13

score 2 · Answer 1 · answered Nov 03 '15 at 17:54

2

There is in fact a unique value that holds for every $x$ and that value is $0$. You cannot choose $x$ first and you cannot have $y$ depend on $x$. The order of the quantifiers is very important in this statement .

answered Nov 03 '15 at 17:54

Jean-François Gagnon

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Existence and uniqueness of $y$ such that $xy=(xy)^2$ for every real $x$

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