> [!summary]
This proof uses the definition of the absolute value in different forms to prove the triangle inequality.
>
**Key Result:**
>
$|a+b| \leq |a| + |b|$
>[!info]+ Read Time
**⏱ 1 min**
# Mathematical Proof
> [!warning] Assumptions
To prove the absolute value multiplication rule by [[Proof by Cases|proof by cases]] assume the following:
> - The definition of the absolute value of a $|a+b|$ is:
> $
> | a+b| =
> \begin{cases}
> a+b, & \text{if } a+b \geq 0 \\
> -(a+b), & \text{if } a+b < 0
> \end{cases}
> $
> - The definition of an absolute value of $|a|$ is:
> $
> |a| =
> \begin{cases}
> a, & \text{if } a \geq 0 \\
> -a, & \text{if } a < 0
> \end{cases}
> $
> - The definition of an absolute value of $|b|$ is:
> $
> |b| =
> \begin{cases}
> b, & \text{if } b \geq 0 \\
> -b, & \text{if } b < 0
> \end{cases}
> $
Prove that $|a+b| \leq |a| +|b|$
$
\begin{array}{c}
\text{Case 1:} \quad (a+b) \geq 0: \\
a+b \leq |a| +|b| \quad \text{By defintion of the abs value of a,b} \\
\\
\text{Case 2:} \quad (a+b)<0: \\
\text{This is saying $-(a+b)$ by definiton of abs value of a+b}\\
-a + -b \leq |a| + |b| \quad \text{By definition of the abs value of a,b}
\end{array}
$
# Resources
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