Absolute Value and the Triangle Inequality

Real analysis is the part of mathematics that proves why calculus works. Almost every proof in it measures how close two numbers are, and the tool for that is the absolute value. You solved inequalities like $|x - 3| < 2$ in Algebra 1, and in Discrete Math you learned to write a direct proof. This course puts the two together.

Idea clave

$|x - a|$ is the distance from $x$ to $a$, so $|x - a| < \varepsilon$ means $a - \varepsilon < x < a + \varepsilon$. Two inequalities turn distances into bounds: the triangle inequality $|a + b| \le |a| + |b|$, and the reverse triangle inequality $|a| - |b| \le |a - b|$.

Por qué funciona

Since $-|a| \le a \le |a|$ and $-|b| \le b \le |b|$, adding gives $-(|a| + |b|) \le a + b \le |a| + |b|$, which says exactly $|a + b| \le |a| + |b|$. For the reverse version, write $a = (a - b) + b$ and apply the triangle inequality: $|a| \le |a - b| + |b|$, so $|a| - |b| \le |a - b|$. Swapping $a$ and $b$ gives $|b| - |a| \le |a - b|$ too, so in fact $\big||a| - |b|\big| \le |a - b|$.

Analysts use these two facts in two directions. The triangle inequality gives an upper bound for a sum: if two quantities are each small, so is their sum. The reverse inequality gives a lower bound: a number close to $a$ cannot be much smaller in size than $|a|$. We will use the second one to keep denominators away from $0$.

Worked examples

Ejemplo resuelto: Unwrapping an absolute value

Write the set of $x$ with $|x + 2| < 0.1$ as an interval.

$|x + 2| = |x - (-2)|$ is the distance from $x$ to $-2$.
So $-0.1 < x + 2 < 0.1$, that is, $-2.1 < x < -1.9$.

Respuesta: $(-2.1, -1.9)$, an open interval because the inequality is strict.

Ejemplo resuelto: Bounding a factor

If $|x - 3| < 1$, how big can $|x + 3|$ be?

From $|x - 3| < 1$: $2 < x < 4$.
Then $5 < x + 3 < 7$, so $|x + 3| < 7$.
This is the step that makes $x^{2}$ continuous at $3$: $|x^{2} - 9| = |x - 3|\,|x + 3| < 7|x - 3|$, which is small when $|x - 3|$ is.

Respuesta: $|x + 3| < 7$, and no smaller bound works for every such $x$.

Ejemplo resuelto: Keeping away from zero

If $|x - 5| < 2$, show that $|x| > 3$.

Reverse triangle inequality: $|x| \ge |5| - |x - 5|$.
Since $|x - 5| < 2$, we get $|x| > 5 - 2 = 3$.

Respuesta: $|x| > 3$. (So $\frac{1}{|x|} < \frac{1}{3}$: a bound on a reciprocal, the kind the quotient law needs.)

Error común

Running the inequality backward. The tempting "$|a - b| \le |a| - |b|$" is false: with $a = 1$ and $b = 2$, the left side is $1$ and the right side is $-1$. The difference of sizes is a lower bound for the size of the difference. And $|a + b| = |a| + |b|$ holds only when $a$ and $b$ have the same sign (or one is $0$). Check any inequality like these with $a = 1$, $b = -1$ and with $a = 1$, $b = 2$ before you use it.

In a proof you will often see a chain like $|x^{2} - 9| = |x - 3||x + 3| \le |x - 3|(|x - 3| + 6)$. The last step is the triangle inequality: $|x + 3| = |(x - 3) + 6| \le |x - 3| + 6$.

Worked examples

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