**Solution 1** The number 0 has no reciprocal, and 1 and -1 are their own reciprocals. This leaves only 2 and -2. The reciprocal of 2 is 1/2, but 2 is not less than 1/2. The reciprocal of -2 is -1/2, and -2 is less than -1/2, so it is $A$. **Solution 2** The statement "a number is less than its reciprocal" can be translated as $x < \frac{1}{x}$. Multiplication by $x$ can be done if you do it in three parts: $x>0$, $x=0$, and $x<0$. You have to be careful about the direction of the inequality, as you do not know the sign of $x$. If $x>0$, the sign of the inequality remains the same. Thus, we have $x^2 < 1$ when $x > 0$. This leads to $0 < x < 1$. If $x=0$, the inequality $x < \frac{1}{x}$ is undefined. If $x<0$, the sign of the inequality must be switched. Thus, we have $x^2 > 1$ when $x < 0$. This leads to $x < -1$. Putting the solutions together, we have $x<-1$ or $0 < x < 1$, or in interval notation, $(-\infty, -1) \cup(0, 1)$. The only answer in that range is -2 **Solution 3** Starting again with $x < \frac{1}{x}$, we avoid multiplication by $x$. Instead, move everything to the left, and find a common denominator: $x < \frac{1}{x}$, $x - \frac{1}{x} < 0$, $\frac{x^2 - 1}{x} < 0$, $\frac{(x+1)(x-1)}{x} < 0$ Divide this expression at $x=-1$, $x=0$, and $x=1$, as those are the three points where the expression on the left will "change sign". If $x<-1$, all three of those terms will be negative, and the inequality is true. Therefore, $(-\infty, -1)$ is part of our solution set. If $-1 < x < 0$, the $(x+1)$ term will become positive, but the other two terms remain negative. Thus, there are no solutions in this region. If $0 < x < 1$, then both $(x+1)$ and $x$ are positive, while $(x-1)$ remains negative. Thus, the entire region (0, 1) is part of the solution set. If $1 < x$, then all three terms are positive, and there are no solutions. At all three "boundary points", the function is either 0 or undefined. Therefore, the entire solution set is $(-\infty, -1) \cup (0, 1)$, and the only option in that region is $x=-2$, leading to $A$. **Solution 4** We can find out all of their reciprocals. Now we compare and see that the answer is $A$ **Solution 5** Look at each number. Notice that the number must be negative. The number cannot be -1, 0, 1,.... -2 is all that is left