Absolute Value & Distance

Absolute value is distance, not "make it positive"

Here's the definition that actually means something — I want you to engrave this on the inside of your skull:

$|x| = \text{the distance from } x \text{ to } 0 \text{ on the number line.}$

Distance is never negative, and it doesn't care which side you're on. You're $5$ units from zero whether you stand at $5$ or at $-5$, so $|5| = 5$ and $|-5| = 5$. The number $0$ is zero units from itself, so $|0| = 0$. That's it. "Make it positive" gives the right answers for plain numbers, but it hides the idea, and the idea is what survives contact with algebra. The parlor trick gets you through one homework set; the distance picture gets you through every damn thing after it.

There's also a piecewise definition — the same thing written as a rule:

$|x| = \begin{cases} x & \text{if } x \ge 0, \\ -x & \text{if } x < 0. \end{cases}$

Why "$-x$" for negatives? Because if $x$ is negative, its distance from zero is its opposite, and the opposite of a negative is positive (from Zero & the Negatives). E.g. $|-5| = -(-5) = 5$. The two views are identical: the geometric one ("distance from $0$") and the algebraic one ("flip it if it's negative"). Use the distance picture to think, the piecewise rule to compute.

Drag the point and watch its absolute value — the distance bar from $0$ — update no matter which side you're on:

The formula that becomes "norm" later: $|a-b|$

Now the upgrade that makes absolute value indispensable. The distance from $0$ is nice, but we usually want the distance between two arbitrary points $a$ and $b$. It's:

$|a - b| = \text{the distance between } a \text{ and } b.$

Why subtraction? Because $a - b$ measures the gap (the signed displacement from $b$ to $a$), and the absolute value strips the sign to leave pure distance. And it doesn't matter which you subtract: $|a-b| = |b-a|$, because the distance from your house to mine equals the distance from mine to yours. Example: the distance between $3$ and $11$ is $|3 - 11| = |-8| = 8$. Between $-2$ and $5$: $|-2 - 5| = |-7| = 7$.

Burn this in, because it's coming back with a vengeance. In the Matrices stratum, the "length" of a vector $\|v\|$ and the distance between two vectors $\|u - v\|$ are literally this formula, promoted to higher dimensions. $|a-b|$ is the one-dimensional ancestor of every distance in linear algebra. Same idea, more arrows. The whole edifice is built on this one subtraction under an absolute value, and the Federation of Boring Textbook Authors treats it as a throwaway definition. It is not.

Solving $|x| = c$: two points at distance $c$

Translate to distance and the solutions appear for free. $|x| = c$ (with $c > 0$) says "$x$ is exactly distance $c$ from $0$." On a line, two points sit at distance $c$ from zero — one each side:

$|x| = c \;\iff\; x = c \ \text{ or } \ x = -c.$

So $|x| = 7$ means $x = 7$ or $x = -7$. (If $c = 0$, the only point at distance $0$ is $0$ itself. If $c < 0$, there are no solutions — distance can't be negative. The distance picture tells you all three cases instantly.)

Solving $|x| < c$: a band; $|x| > c$: two rays

This is where students who memorize get absolutely slaughtered and students who actually think win effortlessly.

$|x| < c$ says "$x$ is closer than $c$ to zero." The points within distance $c$ of zero form a single band straddling it:

$|x| < c \;\iff\; -c < x < c.$

So $|x| < 3$ means $-3 < x < 3$, the interval $(-3, 3)$. "Less than" $\to$ a band hugging zero. (Recall intervals from Order & Inequality.)

$|x| > c$ says "$x$ is farther than $c$ from zero." Those points are everything outside the band — two rays shooting off in both directions:

$|x| > c \;\iff\; x < -c \ \text{ or } \ x > c.$

So $|x| > 3$ means $x < -3$ or $x > 3$, the rays $(-\infty, -3) \cup (3, \infty)$. "Greater than" $\to$ two rays fleeing the center.

And for centered-elsewhere versions, the distance reading carries you: $|x - 5| < 2$ says "$x$ is within distance $2$ of $5$," i.e. the band $(3, 7)$. Always translate to distance first — "within," "farther than," "at least this far" — then read off the interval. Never memorize the four cases; derive them from the picture every time.

The triangle inequality: the law that runs the universe

I've saved the crown jewel, and I mean that — this thing is genuinely beautiful. For any two real numbers $a$ and $b$:

$|a + b| \le |a| + |b|.$

This is the triangle inequality. In words: the size of a sum is at most the sum of the sizes. Let me show you it's true by cases, using $a^2 \ge 0$-style reasoning and the meaning of $|\cdot|$.

Case 1: $a$ and $b$ have the same sign (or one is zero). Then they pull the same direction, and $|a+b|$ adds up to exactly $|a| + |b|$ — equality. E.g. $|3 + 5| = 8 = |3| + |5|$.

Case 2: $a$ and $b$ have opposite signs. Then they partly cancel, so the sum is smaller in size than the pieces separately: $|a + b| < |a| + |b|$. E.g. $|3 + (-5)| = |-2| = 2$, while $|3| + |-5| = 8$, and indeed $2 < 8$.

Either way, $|a+b| \le |a| + |b|$, with equality exactly when $a$ and $b$ point the same direction. Done.

Why does this "run the universe"? The name is the tell: it's the algebraic version of "the straight path is the shortest — any detour through a third point is at least as long." Going from $0$ to $a+b$ directly never beats going $0 \to a \to a+b$. That single principle, generalized to vectors as $\|u + v\| \le \|u\| + \|v\|$, is the backbone of every notion of distance in mathematics — it's what makes distance behave like distance. You'll meet it again with vectors, and it'll be the exact same inequality wearing arrows. It's also the engine behind half the convergence proofs in analysis. File it away: it detonates across the entire rest of the course, and I am not exaggerating.

You now own absolute value as distance, not as a sign-stripper. Go run the gauntlet.