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How do I solve this inequality:

$\frac{1}{x} < 0 $

Its deceptively tricky. I've spent some time thinking about it, but came up with nothing. The answer is obviously $x < 0$, but how do I derive that algebraically?

Can the result be derived by performing algebraic operations on each side of the equation instead of reasoning about it?

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  • $\begingroup$ Multiply $x^2$ on both sides, which does not change signs since $x^2$ is always non-negative $\endgroup$ – user27126 Nov 8 '13 at 3:06
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$\frac{1}{x} < 0 \iff x^2\frac{1}{x} < x^2\times 0 =0 \iff x < 0$.

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The numerator is always positive. The denominator switches signs at zero. A positive numberline "divided" by a numberline that changes sign at zero equals (so to speak) a numberline that changes sign at zero. Since the left side of zero is negative, the fraction is negative, so $x<0$

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  • $\begingroup$ Can the result be derived by performing algebraic operations on each side of the equation instead of reasoning about it? $\endgroup$ – user106310 Nov 8 '13 at 3:00
  • $\begingroup$ You got to be careful with that, because if you would multiply by, say $x$, and $x$ is negative, your inequality sign reverses. I did not draw the numberlines, but my method works for any fraction compared with zero. The signs on the numberlines of the numerator and denominator divided over on-and -another, gives you the final signs for each interval where either the numerator or denominator is zero. For $<0$ you pick the negative intervals. Algebraically perfectly sound $\endgroup$ – imranfat Nov 8 '13 at 3:03

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