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What is the domain for $$\dfrac{1}{x}\leq\dfrac{1}{2}$$

according to the rules of taking the reciprocals, $A\leq B \Leftrightarrow \dfrac{1}{A}\geq \dfrac{1}{B}$, then the domain should be simply $$x\geq2$$

however negative numbers less than $-2$ also satisfy the original inequality. When am I missing in my understanding?

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up vote 1 down vote accepted

The equivalence $$A\leq B\iff \frac{1}{A}\geq\frac{1}{B}$$ only holds for numbers $A$ and $B$ that have the same sign (i.e., are both positive or both negative). Remember, when $c>0$, we have $$A\leq B \iff c A\leq c B$$ and when $c<0$, we have $$A\leq B\iff cA\geq cB.$$ To go from $A\leq B$ to $$\frac{1}{A} \mathbin{\fbox{$\leq$ or $\geq$}} \frac{1}{B}$$ you'll multiply both sides by $c=\frac{1}{AB}$. This $c$ is greater than $0$ when $AB>0$, which happens when $A$ and $B$ have the same sign, and it is less than $0$ when $AB<0$, which happens when $A$ and $B$ have opposite signs.

Thus, $$\frac{1}{x}\leq\frac{1}{2}\iff\begin{cases} x\geq 2 &\text{ if }x\text{ has the same sign as 2},\\ x\leq 2 &\text{ if }x\text{ has the opposite sign as 2}. \end{cases}$$

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When you convert $A \le B$ to $\cfrac 1 A \ge \cfrac 1 B$, you are in fact dividing by $AB$. This works if $AB$ is positive, but if $AB \lt 0$ you have to reverse the inequality.

The inequality in the question is true whenever $x \lt 0$ because the left hand side is negative and the right-hand side is positive.

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Inequalities with fractions require some care. If you write yours in the form $$ \frac{1}{x}-\frac{1}{2}\le0 $$ you see that it's equivalent to $$ \frac{2-x}{2x}\le0 $$ or to $$ \frac{x(2-x)}{2x^2}\le0 $$ Since $2x^2>0$, we can remove the denominator (but keeping the condition that $x\ne0$). So we have the standard $$ x(2-x)\le0 $$ or $$ x(x-2)\ge0 $$ whose solution set is $(-\infty,0]\cup[2,\infty)$. Remembering we can't accept $0$ as solution, we end up with $$ x<0\text{ or }x\ge2 $$

Of course such a long discussion is too much for this simple case, but it should show how things can go wrong with "hasty simplifications". Here, one can simply observe that when $x<0$ the inequality is clearly satisfied and only for $x>0$ one has to do something more: but for $x>0$ one can indeed reverse the fractions, so this becomes $x\ge2$.

This is analogous to irrational inequalities such as $\sqrt{1-x}\ge x-2$. All numbers satisfying $$ \begin{cases} 1-x\ge0 & \text{(existence of the square root)}\\ x-2<0 & \text{(the RHS is negative)} \end{cases} $$ are solutions of the inequality. When $x-2\ge0$, instead, you can square both sides and proceed as usual: $$ \begin{cases} 1-x\ge(x-2)^2\\ x-2\ge0 \end{cases} $$

Don't be hasty and inequalities will bite you no more.

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No, you are not missing anything, that rule only works for $A; B > 0$, so $\dfrac{1}{x} \le \dfrac{1}{2} \Leftrightarrow \left[ \begin{array}{l} x < 0 \\ x \ge 2 \end{array} \right.$

Of course it's easy to see that for $x < 0$, we'll always have $\dfrac{1}{x} < 0 < \dfrac{1}{2}$.

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