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I came upon this question,

$\frac1x<\frac12$

thinking that the answer was $x>2$. But the answer turned out to be (-∞, 0) ∪ (2, +∞). Why is the answer in this form, and how do you get it?

suomynonA
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  • Try naming a number in the "surprising" interval $(-\infty,0)$, and plugging into the inequality to see if indeed it holds. This will give you the hint you need. – vadim123 Aug 14 '16 at 01:58
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    Have you ever seen a graph of $y=\frac{1}{x}$? Notice that when $x$ is negative, the left hand side is negative and is therefore indeed less than $\frac{1}{2}$. As for a longer explanation, if you tried to multiply both sides by two and multiply both sides by $x$, you need to take into account the fact that the direction of the inequality changes when multiplying by a negative number, so you must split into cases and consider each individually. – JMoravitz Aug 14 '16 at 02:00
  • The first option definitely – Asinomás Aug 14 '16 at 02:11
  • Your approach is correct, but notice that the inequality is preserved upon taking $x>0$. However, we must also consider the case where $x<0$, right? If we are to consider all real values of $x$, this is a logical progression. – J. Dunivin Aug 14 '16 at 02:15
  • oops, can't believe that i didn't notice values for which x<0. Thanks for the help – suomynonA Aug 14 '16 at 16:35

2 Answers2

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Because $x$ can be negative as well as positive.

If $x > 0$, then you get your $x > 2$.

If $x < 0$, then $\frac1{x} < 0 < \frac12$ is true for all negative $x$.

marty cohen
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1

One is bound to have to split into cases for this. If we are working only over positive $x$, then it is correct to take $$\frac{1}x<\frac{1}2$$ and take the reciprocal of both sides to get $$x>2$$ since the reciprocal reverses the order on positive x (or more generally, when both sides are of the same sign). This is probably what you did. If you'd noted the restriction to positive $x$, you could check out the case for negative $x$ and note that $\frac{1}x<0<2$ for any $x<0$, thus any $x<0$ suffices. This gives the answer.

Another way to go about it is to consider multiplying $$\frac{1}x<\frac{1}2$$ by $2x$. If $x$ is positive, the order is preserved giving $2<x$. If $x$ is negative, the order is reversed, giving $2>x$, which is is satisfied for all negative $x$, thus we know that a positive $x$ greater than $2$ satisfies the equation, as well as all negative $x$. This is, again, the given answer.

Milo Brandt
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