Mathematical-Reasoning · Unit 9 · Video 3 · Interactive Practice

Absolute Value by Cases: Proving |x| = a and |−x| = |x|

IKey Formulas

Formula	Name	What it gives the proof
$\|x\| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$	Definition of absolute value	One formula for each sign of $x$ ; both lines give $\|x\| \ge 0$
$\|x\| = a \iff (x = a \text{ or } x = -a)$ , for $a > 0$	Theorem, Part 1	Turns one absolute-value equation into two ordinary equations
$\|-x\| = \|x\|$ for every $x \in \mathbb{R}$	Theorem, Part 2	Proved with three cases: $x > 0$ , $x = 0$ , $x < 0$
$\big((P \vee Q) \to R\big) \equiv \big((P \to R) \wedge (Q \to R)\big)$	A hypothesis that is an "or"	The converse of Part 1 is proved one case at a time

Key Insight: Whenever an absolute value appears, in a proof or an equation, first ask which sign the inside has, and handle each sign as its own labelled case. Here $-x$ means the opposite of $x$ , not a negative number: when $x < 0$ , $-x > 0$ .

IIVisualization 1 — Which Line of the Definition Applies?

Each line of the definition handles one sign, and $x$ and $-x$ fall on opposite lines unless $x = 0$ .

x −7

|u| = u

u \ge 0

|u| = -u

u < 0

IIIVisualization 2 — Why Part 1 Needs $a > 0$

For $a > 0$ , $|x| = a$ holds exactly when $x = a$ or $x = -a$ ; for $a < 0$ the converse fails.

a 3

IVVisualization 3 — Solving Equations with Part 1

Part 1 splits each equation in two: the inside sits at distance $a$ from $0$ , on one side or the other.

Equation Step 1

Step 1 — Identify $x$ and $a$ . In

|t| = 12

the inside is

t

itself, and

a = 12 > 0

, so Part 1 applies.

Step 2 — Split into two cases. Part 1 with

x = t

gives

t = 12

t = -12

Step 3 — Solve each case. Nothing is left to solve:

t = 12

t = -12

Step 4 — Check.

|12| = 12

, and

|-12| = -(-12) = 12

Step 1 — Identify $x$ and $a$ . In

|t + 3| = 5

the expression

t + 3

plays the role of

x

, and

a = 5 > 0

Step 2 — Split into two cases.

t + 3 = 5

t + 3 = -5

Step 3 — Solve each case. Subtract

3

in each case:

t = 2

t = -8

Step 4 — Check.

|2 + 3| = |5| = 5

, and

|-8 + 3| = |-5| = 5

. Keeping only the positive case would lose

t = -8

Step 1 — Identify $x$ and $a$ . In

|3t - 4| = 8

the expression

3t - 4

plays the role of

x

, and

a = 8 > 0

Step 2 — Split into two cases.

3t - 4 = 8

3t - 4 = -8

Step 3 — Solve each case. Add

4

3t = 12

3t = -4

. Divide by

3

t = 4

t = -\tfrac{4}{3}

Step 4 — Check.

3(4) - 4 = 8

, and

3\left(-\tfrac{4}{3}\right) - 4 = -8

with

|-8| = 8

. The solution

t = -\tfrac{4}{3}

is negative, which is allowed: the theorem needs only

a > 0

VQuiz Questions

Problem 1 · Apply Part 1

Solve $|t - 2| = 7$ for the real number $t$ .

✅ Correct! With

t - 2

in the role of

x

and

a = 7 > 0

, Part 1 gives

t - 2 = 7

t - 2 = -7

, so

t = 9

t = -5

❌ Only half the cases.

t = 9

is a solution, but Part 1 also allows

t - 2 = -7

, which gives

t = -5

, and

|-5 - 2| = |-7| = 7

❌ The minus sign goes on the right side, not on $t$ . The second case is

t - 2 = -7

, not

t = -9

; indeed

|-9 - 2| = 11

❌ Check the arithmetic. To undo

t - 2

, add

2

t - 2 = 7

gives

t = 9

, not

5

, and

|5 - 2| = 3 \ne 7

Show solution

Part 1 holds for each real number $x$ , so it holds for $x = t - 2$ . Since $a = 7 > 0$ ,

|t - 2| = 7 \iff t - 2 = 7 \text{ or } t - 2 = -7.

Case 1: $t - 2 = 7$ , so $t = 9$ . Case 2: $t - 2 = -7$ , so $t = -5$ .

Check: $|9 - 2| = |7| = 7$ and $|-5 - 2| = |-7| = 7$ . So $t = 9$ or $t = -5$ .

Problem 2 · Reading the Second Line

Given: $x$ is a real number with $x < 0$ . Which statement is true?

✅ Correct! Since

x < 0

, the second line gives

|x| = -x

, and the opposite of a negative number is positive.

❌ Wrong line. The first line,

|x| = x

, applies only when

x \ge 0

. Here

x < 0

, so

|x| = x

would make

|x|

negative.

❌ $-x$ is not a negative number. The minus sign means "the opposite of

x

". With

x = -7

-x = 7

, so

|x| = 7 > 0

❌ Apply the definition to $-x$ . Since

x < 0

-x > 0

, so the first line gives

|-x| = -x

, which is positive and not equal to

x

. For example,

|-(-7)| = |7| = 7 \ne -7

Show solution

Because $x < 0$ , the second line of the definition applies: $|x| = -x$ . Multiplying $x < 0$ by $-1$ reverses the inequality, so $-x > 0$ . Statement (a) is true.

(b) fails because $|x| = x < 0$ would contradict $|x| \ge 0$ . (c) misreads $-x$ as a negative number. (d) fails because $-x > 0$ gives $|-x| = -x$ , and $-x \ne x$ when $x \ne 0$ .

Problem 3 · A Negative Coefficient

Solve $|5 - 2t| = 3$ for the real number $t$ .

✅ Correct! Both cases give positive solutions here:

5 - 2t = 3

gives

t = 1

, and

5 - 2t = -3

gives

t = 4

❌ Watch the sign when dividing by $-2$ .

-2t = -2

gives

t = 1

, and

-2t = -8

gives

t = 4

. Check:

|5 - 2(-1)| = 7 \ne 3

❌ Only half the cases. Part 1 also gives

5 - 2t = -3

, so

-2t = -8

and

t = 4

; indeed

|5 - 8| = |-3| = 3

❌ The second case has a sign slip.

5 - 2t = -3

gives

-2t = -8

, so

t = 4

. Check:

|5 - 2(-4)| = 13 \ne 3

Show solution

Here $5 - 2t$ plays the role of $x$ and $a = 3 > 0$ , so Part 1 gives two cases.

Case 1: $5 - 2t = 3$ . Subtract $5$ : $-2t = -2$ . Divide by $-2$ : $t = 1$ .

Case 2: $5 - 2t = -3$ . Subtract $5$ : $-2t = -8$ . Divide by $-2$ : $t = 4$ .

Check: $|5 - 2| = 3$ and $|5 - 8| = |-3| = 3$ . So $t = 1$ or $t = 4$ . The case labels refer to the sign of the inside, not of $t$ : both solutions are positive.

Problem 4 · Where the Hypothesis Matters

Consider the statement "for each real number $x$ , if $x = a$ or $x = -a$ , then $|x| = a$ ." For which value of $a$ is this statement false?

✅ Correct! With

a = -3

, the value

x = -3

satisfies

x = a

, but

|-3| = 3 \ne -3

❌ This is the theorem itself. For

a = 3 > 0

, both

|3| = 3

and

|-3| = 3

, so the statement holds.

❌ The boundary case still works. The theorem assumes

a > 0

, but for

a = 0

the hypothesis forces

x = 0

, and

|0| = 0 = a

. The statement is true.

❌ Try a negative $a$ . An absolute value is never negative, so

|x| = a

cannot hold when

a < 0

, yet

x = a

is still possible.

Show solution

The proof of the converse used $a > 0$ in both cases: it put $a$ in the first line of the definition and $-a$ in the second.

For $a = -3$ , take $x = -3$ . Then $x = a$ is true, but since $-3 < 0$ , $|x| = -(-3) = 3$ , and $3 \ne -3$ . One counterexample makes the "for each" statement false.

For $a = 3$ the statement is Part 1 of the theorem. For $a = 0$ the hypothesis says $x = 0$ , and $|0| = 0$ , so the statement is true even though the theorem does not cover this case.

Solved: 0 / 4

Absolute Value by Cases: Proving |x| = a and |−x| = |x|

IKey Formulas

IIVisualization 1 — Which Line of the Definition Applies?

IIIVisualization 2 — Why Part 1 Needs a>0a > 0a>0

IVVisualization 3 — Solving Equations with Part 1

VQuiz Questions

IIIVisualization 2 — Why Part 1 Needs $a > 0$