Mathematical-Reasoning · Unit 9 · Video 4 · Interactive Practice

The Triangle Inequality by Cases: |x + y| ≤ |x| + |y|

IKey Formulas

Statement	Name	Cases split on
For $a > 0$ : $\ \|x\| < a \iff -a < x < a$	Part (1)	The sign of $x$ : $x \ge 0$ or $x < 0$ (2 cases)
$\|xy\| = \|x\|\,\|y\|$	Part (2)	The signs of $x$ and $y$ together (4 cases)
$z \le \|z\|$ and $-z \le \|z\|$ for every real $z$	Fact	The sign of $z$ (2 cases); used with $z = x$ and $z = y$
$\|x + y\| \le \|x\| + \|y\|$	Part (3), the Triangle Inequality	The sign of $x + y$ (2 cases)

Key Insight: The definition $|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x < 0$ needs the sign of whatever is inside the bars, so the inside sets the split; since $(P \vee Q) \to R \equiv (P \to R) \wedge (Q \to R)$ , proving each case separately proves the statement, provided the cases cover every possibility.

IIPart 1 · |x| < a by Cases on the Sign of x

For $a > 0$ , the statements $|x| < a$ and $-a < x < a$ are true for exactly the same $x$ .

a 3 x 2

IIIThe Triangle Inequality on the Number Line

In each case the Fact bounds $|x + y|$ by $|x| + |y|$ , and the gap shows which Fact inequality is strict.

x 3 y −5

IVThe Inside of the Bars Sets the Split

$|x + y|$ needs only the sign of $x + y$ , two cases; $|xy|$ is handled by the signs of $x$ and $y$ , four cases.

Statement x 3 y −5

VQuiz Questions

Problem 1 · Which Case, Which Values

Given: $x = -4$ and $y = 7$ . Which case of the Triangle Inequality proof applies, and what does it give?

✅ Correct! The case is chosen by the sign of

x + y = 3

, so

|x + y| = 3

, while

|x| + |y| = 4 + 7 = 11

. The inequality is strict because

x

and

y

have opposite signs.

❌ Not quite. The Triangle Inequality proof splits on the sign of

x + y

, not of

x

. Here

x + y = 3 \ge 0

, and the result

-3

cannot be an absolute value.

❌ Not quite.

|x + y| = |3| = 3

, not

11

. The sum

|x| + |y| = 11

is the right-hand side; part of

x + y

cancels because the signs are opposite.

❌ Not quite. The case and

|x + y| = 3

are right, but

|x| = |-4| = 4

, not

-4

, so

|x| + |y| = 4 + 7 = 11

Show solution

The proof splits on the sign of the quantity inside the bars on the left, $x + y = -4 + 7 = 3$ . Since $3 \ge 0$ , this is Case 1, and $|x + y| = x + y = 3$ .

On the right, $|x| + |y| = |-4| + |7| = 4 + 7 = 11$ .

The Fact gives $x \le |x|$ , that is $-4 \le 4$ , and $y \le |y|$ , that is $7 \le 7$ . Adding them gives $x + y \le |x| + |y|$ , that is $3 \le 11$ . So $|x + y| = 3 \le 11 = |x| + |y|$ .

Problem 2 · Choosing the Split

Given: A proof of $|x + y| \le |x| + |y|$ for all real $x$ and $y$ . Which choice of cases covers every pair $x, y$ and lets $|x + y|$ be written without bars inside each case?

✅ Correct! Every real number

x + y

is either

\ge 0

< 0

, and in each case the definition gives

|x + y|

directly:

x + y

-(x + y)

❌ Not quite. These cases cover every

x

, but the sign of

x

does not decide the sign of

x + y

: with

x = 3

y = -5

gives

x + y = -2

while

y = 1

gives

x + y = 4

❌ Not quite. These cases are not exhaustive: they leave out every pair with

x + y = 0

, such as

x = 3

y = -3

❌ Not quite. These four cases cover everything, but in the case

x \ge 0

y < 0

the sum can have either sign (

3 + (-1) = 2

3 + (-5) = -2

), so

|x + y|

still cannot be written without bars there.

Show solution

To remove the bars from $|x + y|$ , the definition of absolute value needs the sign of $x + y$ . The split $x + y \ge 0$ or $x + y < 0$ covers every real number $x + y$ , so it covers every pair $x, y$ , and it gives $|x + y| = x + y$ in the first case and $|x + y| = -(x + y)$ in the second.

Splitting on $x$ alone does not fix the sign of $x + y$ . Using strict inequalities on both sides drops $x + y = 0$ . The four sign cases on $x$ and $y$ settle the sign of $x + y$ only when $x$ and $y$ do not have opposite signs; the two mixed cases would each need a further split on $x + y$ .

Problem 3 · Finishing Case 2

Given: In Case 2 ( $x + y < 0$ ) the proof has $|x + y| = -(x + y) = (-x) + (-y)$ . Which two inequalities from the Fact, added together, finish this case?

✅ Correct! Adding gives

(-x) + (-y) \le |x| + |y|

, and the left side is exactly

|x + y|

in this case.

❌ Not quite. These add to

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

, which is true, but in Case 2 the left side

x + y

is negative and is not

|x + y|

; this is the Case 1 step.

❌ Not quite. These add to

(-x) + y \le |x| + |y|

, and

(-x) + y

is not

|x + y|

in Case 2.

❌ Not quite. These are not part of the Fact and are false in general:

x = 3

would need

3 \le -3

Show solution

Assume $x + y < 0$ . Then $|x + y| = -(x + y) = (-x) + (-y)$ .

The Fact, with $z = x$ and with $z = y$ , gives $-x \le |x|$ and $-y \le |y|$ . Adding these two inequalities:

(-x) + (-y) \le |x| + |y|.

The left side equals $|x + y|$ , so $|x + y| \le |x| + |y|$ in Case 2. The inequalities $x \le |x|$ and $y \le |y|$ are the ones Case 1 uses, where $|x + y| = x + y$ .

Problem 4 · When Is It an Equality?

Given: Real numbers $x$ and $y$ . For exactly which pairs does $|x + y| = |x| + |y|$ hold?

✅ Correct! Equality needs both Fact inequalities used in the case to be equalities, which happens exactly when

x

and

y

do not have opposite signs, that is

xy \ge 0

❌ Not quite. This misses the boundary where one number is

0

x = 0

y = 5

gives

|0 + 5| = 5 = 0 + 5

, yet

xy = 0

❌ Not quite. Two negatives also give equality:

x = -2

y = -3

gives

|-5| = 5 = 2 + 3

❌ Not quite.

x = 3

y = -1

has

x + y = 2 \ge 0

, but

|x + y| = 2 < 4 = |x| + |y|

Show solution

Use the same two cases as the proof and look at the gap $|x| + |y| - |x + y|$ .

Case 1: $x + y \ge 0$ . Then $|x + y| = x + y$ , so the gap is $(|x| - x) + (|y| - y)$ . By the Fact both terms are $\ge 0$ , so the gap is $0$ exactly when $|x| = x$ and $|y| = y$ . Now $|x| = x$ holds when $x \ge 0$ and fails when $x < 0$ (then $|x| = -x > x$ ), so the gap is $0$ exactly when $x \ge 0$ and $y \ge 0$ .

Case 2: $x + y < 0$ . Then $|x + y| = (-x) + (-y)$ , so the gap is $(|x| + x) + (|y| + y)$ , again a sum of two terms that are $\ge 0$ . It is $0$ exactly when $|x| = -x$ and $|y| = -y$ , that is, when $x \le 0$ and $y \le 0$ .

So equality holds exactly when $x \ge 0$ and $y \ge 0$ , or $x \le 0$ and $y \le 0$ (both numbers $\ge 0$ forces Case 1; both $\le 0$ with $x + y \ge 0$ forces $x = y = 0$ , which is also covered). That is exactly the condition $xy \ge 0$ .

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