# Set Proof: xy <x

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.

• Where does the statement mention whole numbers? Apr 19, 2017 at 4:23
• I thought $\mathbb{R}$ meant whole numbers. So the proof must be true then? Apr 19, 2017 at 4:25
• Take $x=0$ for instance. Can you find $y\in\Bbb R$ such that $xy<x$? Apr 19, 2017 at 4:29
• Oh, I get it, So since there's no $y \equiv \mathbb{R}$ then the proof must be false Apr 19, 2017 at 4:31
• 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. Apr 19, 2017 at 4:34

Hint: what if $x=0$? $\,\,\,\,\,$
No it is not, consider $x=0$ then $\forall y \in \mathbb{R}$ xy=0=x$. The statement is false. If$x=0$, then$xy = x$for all$y \in \mathbb{R}\$.