Absolute Value of a Number
For any $a \in \mathbb{R}$, if $a = 0$, the absolute value is $0$. If $a \neq 0$, the absolute value of $a$ is defined as the positive value between $a$ and $-a$. The absolute value of $a$ is denoted by $|a|$ and is algebraically expressed as a piecewise function:
$$\left| a \right| =
\begin{cases}
-a, & \text{if } a < 0 \\ 0, & \text{if } a = 0 \\a, & \text{if } a > 0
\end{cases}$$
- $\left| -2\right|=-(-2)=2, \quad\left| 3\right|= 3, \quad\left|0\right|=0$
- $\left| -\frac{3}{2} \right|= -(-\frac{3}{2})=\frac{3}{2}, \quad \left| \frac{4}{7} \right| =\frac{4}{7}$
- If $x<y$, then $x-y<0$, which implies: $\left| x-y\right|=-(x-y)=y-x$
- If $x<0<y$ and $\left| x\right|>y$, then $x+y<0$, which yields: $\left|x+y\right|=-x-y$
- If $x<0$, then $x^7<0$, which yields: $\left| x^7\right|= -x^7$
- $\left|x^4+1\right|=x^4+1$
Example:
Given $x < y < z < 0$, let us determine the value of the expression $\frac{\left| x\right|}{x}-\frac{\left| y\right|}{y}+\frac{\left| z\right|}{z}$. Since $x < 0$, $y < 0$, and $z < 0$, their arguments are negative:
$$\left|x\right| = -x ,\quad \left|y\right| = -y ,\quad \left|z\right| = -z$$ Substituting these evaluations into the expression yields:
$$\frac{\left| x\right|}{x}-\frac{\left| y\right|}{y}+\frac{\left| z\right|}{z}=\frac{-x}{x}-\frac{-y}{y}+\frac{-z}{z}$$
$$=-1-(-1)+(-1)=-1$$
QUESTION 6
Given $x, y, z \in \mathbb{R}$ and $x \cdot y \cdot z \neq 0$,
$$\frac{|x|}{x} + \frac{|y|}{y} + \frac{|z|}{z}$$
how many distinct values can this sum take?
$\text{A) } 8 \quad \text{B) } 7 \quad \text{C) } 6 \quad \text{D) } 5 \quad \text{E) } 4$
Solution:
Let the given expression be represented by $S$:
$$S = \frac{|x|}{x} + \frac{|y|}{y} + \frac{|z|}{z}$$
The component ratio $\frac{|a|}{a}$ evaluates to $+1$ if $a > 0$, and $-1$ if $a < 0$. Therefore, to evaluate $S$, we analyze the signs of $x$, $y$, and $z$, which yield 8 potential sign combinations:
$(x > 0, y > 0, z > 0 \implies S = 1 + 1 + 1 = 3)$
$(x > 0, y > 0, z < 0 \implies S = 1 + 1 – 1 = 1) $
$(x > 0, y < 0, z > 0 \implies S = 1 – 1 + 1 = 1)$
$(x > 0, y < 0, z < 0 \implies S = 1 – 1 – 1 = -1) $
$(x < 0, y > 0, z > 0 \implies S = -1 + 1 + 1 = 1)$
$(x < 0, y > 0, z < 0 \implies S = -1 + 1 – 1 = -1) $
$(x < 0, y < 0, z > 0 \implies S = -1 – 1 + 1 = -1)$
$(x < 0, y < 0, z < 0 \implies S = -1 – 1 – 1 = -3)$
Collecting these outcomes, the set of unique values that the sum $S$ can take is:
$S \in \{3, 1, -1, -3\}$
Thus, there are exactly 4 distinct values that the sum can take.
Correct Answer:
$\text{E) } 4$