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Distance on the line, absolute-value inequalities as intervals, and the inequality every later estimate uses.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to read $|x - c|$ as a distance, convert any absolute-value inequality into the interval or pair of rays it describes with the right open and closed ends, apply the triangle inequality and its reverse to bound a quantity you cannot compute directly, and say when each bound is tight. This is the algebra the epsilon-delta definition is written in, so an hour spent here is an hour not lost in the next three lessons.
$|x|$ is $x$ when $x \ge 0$ and $-x$ when $x < 0$. That definition is worth reading twice, because $-x$ is positive when $x$ is negative: the minus sign is not making anything negative, it is undoing a negativity that is already there.
The useful reading is geometric. $|x|$ is the distance from $x$ to $0$, and $|x - c|$ is the distance from $x$ to $c$. Every statement in this lesson is a statement about distance once it is read that way:
| Written | Read as | The set |
|---|---|---|
| $\|x - c\| < r$ | closer to $c$ than $r$ | the open interval $(c - r,\ c + r)$ |
| $\|x - c\| \le r$ | at most $r$ from $c$ | the closed interval $[c - r,\ c + r]$ |
| $\|x - c\| > r$ | further from $c$ than $r$ | two rays, everything outside |
| $0 < \|x - c\| < \delta$ | near $c$ but not equal to it | the interval with the centre punctured |
The last row is the one the definition of a limit uses, and the $0 <$ is not decoration: it is what lets a limit ignore the value of the function at the point.
The triangle inequality. $$|a + b| \le |a| + |b|$$ with equality exactly when $a$ and $b$ have the same sign (or one is zero). Going somewhere via a detour is never shorter than going directly. Its companion, the reverse triangle inequality $\bigl||a| - |b|\bigr| \le |a - b|$, says that numbers close together have sizes close together.
Another way: picture
A number line with $c$ marked and a band of half-width $r$ drawn round it. $|x - c| < r$ is the inside of the band; $|x - c| > r$ is everything outside it; and $0 < |x - c| < \delta$ is the band with the single point $c$ lifted out, leaving a hole you can see through.
Another way: steps
To turn an absolute-value condition into intervals:
Almost every estimate in analysis has the same shape: you want to bound something you cannot compute, so you write it as a sum of things you can bound, and add the bounds. The triangle inequality is the licence to do that.
The three-term version, used constantly, is $$|a - c| \le |a - b| + |b - c|,$$ obtained by writing $a - c = (a - b) + (b - c)$. In words: to show $a$ is near $c$, find a $b$ near both. In the proofs of this course that $b$ is usually a limit, a partial sum or a value of an approximating function, and the argument is called an $\varepsilon/2$ argument: bound each of the two pieces by $\varepsilon/2$ and the whole is below $\varepsilon$.
The reverse inequality comes from the same move. From $a = (a - b) + b$, $$|a| \le |a - b| + |b| \quad\Longrightarrow\quad |a| - |b| \le |a - b|,$$ and swapping the letters gives $|b| - |a| \le |a - b|$ as well; together they are $\bigl||a| - |b|\bigr| \le |a - b|$.
Treating $|x|$ as "drop the minus sign". That works on numbers and fails on expressions: $|x - 5|$ is not $x + 5$, and $|{-x}|$ is $|x|$ rather than $x$.
Squaring to remove the absolute value without care. $|a| = |b|$ does follow from $a^2 = b^2$, but $|a| < b$ is equivalent to $a^2 < b^2$ only when $b \ge 0$ — and if $b$ is an unknown, that is a hypothesis you have to have.
*Joining the outside cases with and. $|x - 1| > 3$ gives $x > 4$ or $x < -2$. Writing and* produces the empty set, which is the most common wrong answer to this kind of question and is wrong for a reason you can say out loud: no point is far from $1$ in both directions.
Forgetting that the triangle inequality is an inequality. $|a + b|$ can be much smaller than $|a| + |b|$ — it is $0$ when $b = -a$. The bound is a ceiling, never an estimate of the size.
Learners reach for the triangle inequality as if it were an identity. It is an inequality, and the gap can be as large as the numbers themselves: $|3 + (-3)| = 0$ while $|3| + |{-3}| = 6$. Equality holds exactly when $a$ and $b$ point the same way. Knowing when it is tight is what tells you whether a bound you have derived is the best available or merely true — and the practice item that asks for the "smallest bound" is asking precisely that.
Suppose $|a - 4| < 0.01$ and $|b - 4| < 0.01$. How far apart can $a$ and $b$ be?
Two points near one place.
$|a - b| = |(a - 4) + (4 - b)| \le |a - 4| + |b - 4| < 0.02$.
Insert the common point, then split.
Run in reverse: to force $|a - b| < \varepsilon$ it is enough to put both within $\varepsilon/2$ of a common value. That is the whole technique.
Halve the target, hit each piece.
Solve $|2x - 6| \le 4$. First factor out the coefficient: $|2x - 6| = 2|x - 3|$.
Pull the $2$ out; $|2u| = 2|u|$ because $2 > 0$.
So $2|x - 3| \le 4$, that is $|x - 3| \le 2$: at most $2$ from $3$.
Now it is a distance statement.
The solution is $[1, 5]$, closed because the inequality is not strict.
$|3x + 6| = 3|x + 2| = 3|x - (-2)|$, so the condition is $|x - (-2)| < 3$.
Factor, then read the centre.
Centre $-2$, radius $3$, strict: the open interval $(-5, 1)$.
Match each interval or point condition to its distance notation.
| $|x-1|<3$ | $|x-1|\le3$ | $|x-1|=3$ | $|x-1|>3$ | |
|---|---|---|---|---|
| $(-2,4)$ | ||||
| $[-2,4]$ | ||||
| The points $-2$ and $4$ | ||||
| $(-\infty,-2)\cup(4,\infty)$ |
Match each distance claim to the bound or equality condition that justifies it.
| Is at most $|a|+|b|$ | Is at most $|a-b|$ | Holds when $a$ and $b$ have the same sign (or one is zero) | Is at most $|x-y|$ | |
|---|---|---|---|---|
| $|a+b|$ | ||||
| $||a|-|b||$ | ||||
| $|a+b|=|a|+|b|$ | ||||
| $||x|-|y||$ |
Match each absolute-value condition to its geometric solution set.
| The open interval $(-1,5)$ | The closed interval $[-5,3]$ | Two open rays, $x<4$ or $x>6$ | Exactly the two points $-6$ and $6$ | |
|---|---|---|---|---|
| $|x-3|<2$ | ||||
| $|x+1|\leq4$ | ||||
| $|x-5|>1$ | ||||
| $|x|=6$ |
If $|a| \le 7$ and $|b| \le 4$, what is the smallest bound the triangle inequality gives for $|a + b|$?
Answer:
Write $\{\, x : |x - 3| < 5 \,\}$ as an interval.
This task has no paper form; do it on a device.
$f(x) = 2x + 5$ and $|x - c| \le \frac{1}{4}$. What is the best bound on $|f(x) - f(c)|$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Write $\{\, x : |x - 3| \ge 6 \,\}$ as a set of intervals.
This task has no paper form; do it on a device.
You can turn an absolute-value inequality into intervals, and bound a sum or a difference with the triangle inequality. Say in your own words what $0 < |x - a| < \delta$ excludes, and why a limit needs it excluded. Next: functions, domains and composition.
8. Your turn: write $\{x : |3x + 6| < 9\}$ as an interval, step 2