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How do I give a counter-example of the following logic statement (I think the statement is false):

There exists $x$ $\geq$ 0 s.t. (For All real $y$, $x$ = $y$$^2$)

Since the statement has a "There Exists" I need a general counter-example. Correct me if I am wrong.

2 Answers 2

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A general remark: One may prove $\exists x$ by an example and disprove $\forall x$ by a counter-example. But disproving $\exists x$ or proving $\forall x$ requires other methods.


Assume the statement $\tag{1}\exists x\ge 0\colon \forall y\in\mathbb R\colon x=y^2$ is true. Then for such $x$ we have $x\ge 0 \land \forall y\in\mathbb R\colon x=y^2$ and specifically $\forall y\in\mathbb R\colon x=y^2.$ We specialize for the cases $y=0$ and $y=1$ (which we are allowed to do, because $0\in \mathbb R$ and $1\in \mathbb R$) to find $x=0^2$ and $x=1^2$. By multiplying and the rules of equality we find $1=1^2=x=0^2=0$, i.e. $1=0$. This is a false statement, therefore the assumption $(1)$ must be wrong.

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You need to provide a counter example; you should be able to find one easily. Start by assuming such an $x$ exists. Can you find a $y$ whose square is not x? The trick is to come up with such a $y$ in terms of $x$.