Absolute Value: Distance from 0

How do we measure the size of a number without its sign? That measure is the absolute value.

On the number line it is the distance from 0. Positive numbers stay as they are; negative numbers shed their sign. The rules of distance come with it. A length is never negative, and a combined walk never exceeds the sum of its parts. That is the triangle inequality.

For example, 🌡 −3 degrees and +3 degrees are the same distance from 0 degrees. The sign is the direction; the absolute value is the distance.

Examples & Exercises

Prove completely that for all real $x, y$ , the triangle inequality $|x + y| \le |x| + |y|$ holds.

|x + y| \le |x| + |y|

Opposite signs — the sum cancels in part, leaving slack.

Show proof

Both sides are non-negative, so compare their squares. $|x + y|^2 = (x + y)^2 = x^2 + 2xy + y^2$ .
Trap the middle term: $xy \le |xy| = |x| \, |y|$ , so $|x + y|^2 \le |x|^2 + 2|x| \, |y| + |y|^2 = (|x| + |y|)^2$ .
On non-negative numbers the square root preserves order. Taking square roots of both sides gives $|x + y| \le |x| + |y|$ . The proof is complete.

Prove completely that for all real $x, y$ , $\bigl| |x| - |y| \bigr| \le |x - y|$ holds.

\bigl| |x| - |y| \bigr| \le |x - y|

Two points in the same direction — the difference of lengths never exceeds the distance.

Show proof

Since $x = (x - y) + y$ , the triangle inequality gives $|x| \le |x - y| + |y|$ . Rearranging, $|x| - |y| \le |x - y|$ .
Swapping the roles of $x$ and $y$ gives $|y| - |x| \le |y - x|$ . But $|y - x| = |x - y|$ , so $|y| - |x| \le |x - y|$ holds too.
Combining the two, $\bigl| |x| - |y| \bigr| = \max(|x| - |y|, |y| - |x|) \le |x - y|$ . The proof is complete.

Find all real $x$ satisfying $|x - 1| < 2$ , and prove the equivalence completely.

|x - 1| < 2 \iff -1 < x < 3

x = 1

— at the center the distance is 0.

Show proof

By the definition, for $r > 0$ , $|a| < r$ is equivalent to $-r < a < r$ . Check by sign: if $a \ge 0$ then $|a| = a$ , so $a < r$ and $-r < 0 \le a$ ; if $a < 0$ then $|a| = -a$ , so $-a < r$ , i.e. $-r < a$ , and $a < 0 < r$ .
Applying this with $a = x - 1$ and $r = 2$ , $|x - 1| < 2$ is equivalent to $-2 < x - 1 < 2$ . Adding 1 throughout gives $-1 < x < 3$ .
So the solution set is the open interval $(-1, 3)$ : exactly the points whose distance from 1 is less than 2.