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Consider the statement: $$(\forall x\in \mathbb{R})(\exists y\in \mathbb{R})(xy < x)$$

Is this statement true for all real numbers?

Is my proof enough or do I need to add more to it?

Let $x=1$. Then $1*y < 1$. We must find some constant $y$ multiple of $x$ which is less than $1$. Let $y = 0.5$. Then $1*0.5 < 1$. Thus the statement is false since $y$ is not a whole number.

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    $\begingroup$ Where does the statement mention whole numbers? $\endgroup$
    – quasi
    Apr 19, 2017 at 4:23
  • $\begingroup$ I thought $\mathbb{R}$ meant whole numbers. So the proof must be true then? $\endgroup$ Apr 19, 2017 at 4:25
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    $\begingroup$ Take $x=0$ for instance. Can you find $y\in\Bbb R$ such that $xy<x$? $\endgroup$
    – Juniven
    Apr 19, 2017 at 4:29
  • $\begingroup$ Oh, I get it, So since there's no $y \equiv \mathbb{R}$ then the proof must be false $\endgroup$ Apr 19, 2017 at 4:31
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    $\begingroup$ Of course, when you want to disprove a given statement, you have to produce a counter example. The number $x=0$ is such one counter example. $\endgroup$
    – Juniven
    Apr 19, 2017 at 4:34

3 Answers 3

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Hint: what if $x=0$? $\,\,\,\,\,$

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No it is not, consider $x=0$ then $\forall y \in \mathbb{R}$ xy=0=x$.

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The statement is false. If $x=0$, then $xy = x$ for all $y \in \mathbb{R}$.

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