The Absolute Value of a Real Number

The Absolute Value of a Real Number

Definition: If $a$ is a real number, then we define the absolute value of the number $a$ denoted $\mid a \mid$ or $\mathrm{abs}(a)$ as: $\mid a \mid = \left\{\begin{matrix} a & \mathrm{if\:a>0,}\\ 0 & \mathrm{if\:a = 0,}\\ -a & \mathrm{if\:a <0.} \end{matrix}\right.$.

For example, suppose we want to find the absolute value of $5$. Well since $5 > 0$, we note that $\mid 5 \mid = 5$. If we wanted to find the absolute value of $-5$ then since $-5 < 0$ we note that $\mid -5 \mid = -(-5) = 5$. We will now look at some important properties of the absolute values of real numbers utilizing the The Order Properties of Real Numbers.

Theorem 1: If $a$ is a real number then $\mid a \mid = \mid -a \mid$.
  • Proof: We will split this proof up into three cases.
  • Case 1: Suppose that $a > 0$. Then $-a < 0$. Therefore by the definition of the absolute value of a number, $\mid a \mid = a$, and $\mid -a \mid = -(-a) = a$, and so $\mid a \mid = \mid -a \mid$.
  • Case 2: Now suppose that $a = 0$. Therefore $-a = 0$ and clearly $\mid a \mid = 0$ and $\mid -a \mid = 0$, and so $\mid a \mid = \mid -a \mid$.
  • Case 3: Lastly suppose that $a < 0$. Then $-a > 0$. We obtain that $\mid a \mid = -a$ and $\mid -a \mid = -a$, and so $\mid a \mid = \mid -a \mid$.
  • In all three cases we get that $\mid a \mid = \mid a \mid$. $\blacksquare$
Theorem 2: If $a$ and $b$ are real numbers then $\mid ab \mid = \mid a \mid \mid b \mid$.
  • Proof: We will split this proof up into three cases.
  • Case 1: Suppose that $a = 0$ or $b = 0$ or both $a, b = 0$. Then $a \cdot b = 0$, and so $\mid a b \mid = 0$. Similarly $\mid a \mid \mid b \mid$ is either $0 \cdot \mid b \mid$ or $\mid a \mid \cdot 0$ or $0 \cdot 0$, all of which equal $0$, so $\mid ab \mid = \mid a \mid \mid b \mid$.
  • Case 2: Suppose that $a, b > 0$. Then $ab > 0$ and so $\mid ab \mid = ab$ and $\mid a \mid \mid b \mid = ab$, so $\mid ab \mid = \mid a \mid \mid b \mid$.
  • Case 3: Suppose that one of $a > 0$ and $b < 0$. Then $ab < 0$. So $\mid ab \mid = -ab$, and $\mid a \mid \mid b \mid = a \cdot -b = -ab$, so $\mid ab \mid = \mid a \mid \mid b \mid$.
  • Case 4: Suppose that $a, b < 0$. Then $ab > 0$ and so $\mid ab \mid = ab$ and $\mid a \mid \mid b \mid = -a \cdot -b = (-1)(-1)ab = ab$. So $\mid ab \mid = \mid a \mid \mid b \mid$.
  • In all four cases we get that $\mid ab \mid = \mid a \mid \mid b \mid$. $\blacksquare$
Theorem 3: If $a$ is a real number then $\mid a \mid ^2 = a^2$.
  • Proof: We know that $a^2 > 0$ and there by applying Theorem 2 we get that $a^2 = \mid a^2 \mid = \mid a\cdot a \mid = \mid a \mid \mid a \mid = \mid a \mid ^2$. $\blacksquare$
Theorem 4: If $c ≥ 0$ then $\mid a \mid ≤ c$ if and only if $-c ≤ a ≤ c$.
  • Proof: $\Rightarrow$ If $\mid a \mid ≤ c$ then we have that both $a ≤ c$ and $-a ≤ c$ or rather $a ≥ -c$ which is equivalent to saying that $-c ≤ a ≤ c$.
  • $\Leftarrow$ Suppose that $-c ≤ a ≤ c$. Then $a ≤ c$ and $-c ≤ a \Leftrightarrow c ≥ -a$ so then $\mid a \mid ≤ c$. $\blacksquare$
Theorem 5: If $a$ is a real number then $-\mid a \mid ≤ a ≤ \mid a \mid$.
  • Proof: We note that $\mid a \mid ≥ 0$ and so by Theorem 5 we get that $\mid a \mid ≤ \mid a$ and so $-\mid a \mid ≤ a ≤ \mid a \mid$. $\blacksquare$
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