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Whether $\dfrac{1}{|x|} \lt y \implies |x|\gt \dfrac{1}{y}$

My attempt

$\dfrac{1}{|x|} \lt y \implies -y \lt \dfrac{1}{x} \lt y$

I know $a>b$ can be written as $\dfrac{1}{a} \lt\dfrac{1}{b}$ only if both $a$ and $b$ are positive or negative.

Therefore, I am not able to take inequality $\dfrac{1}{x} \lt y$ and convert it to $x \gt \dfrac{1}{y}$ because $x$ and $y$ can be of different signs.

My idea was to take $\dfrac{1}{x} \lt y$ and $-y \lt \dfrac{1}{x}$ separately, take reciprocals on both sides and combine.

Please help with this.

Note: I am a beginner with inequality related concepts

3 Answers 3

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if $x=0$ we get $$y>0$$ thus we have $$\frac{1}{y}>0$$ if $$x\ne 0$$ then we get $$\frac{1}{|x|}>\frac{1}{y}$$

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    when $x=0, ~~\dfrac{1}{x} = \infty$, and then how $y \gt 0$ – 2017-02-01
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Yes, of course! Because from given we have $y>0$.

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$\forall x\in\mathbb{R},x=0$ or $0 <\frac{1}{|x|}$, since $|x|$ is positive definite! Therefore you do in fact know that $y>0$, and that they have the same (positive) sign.

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    ok got it, so i went in a completely wrong direction with my try, right? – 2017-02-01
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    @Kiran You can get it that way, but it's a little more confusing because the casework depends on the sign of $x$ Specifically, when you consider $-x$, that value of $x$ is negative and so $-x$ is positive. It should be clearer if you just let $n=|x|$ and then look at $\frac{1}{n} – 2017-02-01
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    right, it is indeed a simple solution then. – 2017-02-01
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    one doubt, when $x=0, ~~\dfrac{1}{x} = \infty$, and then this will not be a valid case, right? – 2017-02-01
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    $\frac{1}{|x|}$ isn't defined for $x=0$, so I assumed that $x$ would not be zero in the context of the problem you're working on. I would have to see more of the context to give a more complete response. – 2017-02-01