Compound inequalities — AND, OR: Question bank

Before the traps, three tiny pieces of vocabulary you will meet on every line. Say each one out loud once so it stops looking like a squiggle:

Definition Three symbols and one word we use throughout

Ray — on a number line, a ray is the whole half-line you get from one inequality: $x > 4$ is every point stretching right forever from (but not including) 4; $x \leq 1$ is every point stretching left forever, including 1. Picture an arrow shooting off one side of a marked point. The parent's number-line picture draws these.
$\emptyset$ — read "the empty set": a solution set containing no numbers at all. It's what you write when nothing works, like a shaded region that turned out to be blank.
$R$ — read "the real numbers": the set of every number on the whole number line (positives, negatives, zero, fractions, decimals). When a solution is $R$ , the entire line is shaded.
overlap ( $\cap$ ) vs combined coverage ( $\cup$ ) — " $\cap$ " is the intersection (the shared middle where both rays land) and " $\cup$ " is the union (everything either ray reaches). These are the exact set operations from Set operations — union and intersection.

The whole topic then rests on three ideas to hold in your head:

AND = intersection ( $\cap$ ) = the overlap where both conditions hold at once.
OR = union ( $\cup$ ) = everything covered by at least one condition.
The compact form $a < x < b$ is secretly two ANDed inequalities, $(a < x)$ AND $(x < b)$ — never an OR.

Keep two coloured rays in mind: AND keeps only where the colours stack; OR keeps every spot that has any colour.

True or false — justify

Two spotlights on a stage: think about where you'd stand for each item.

x > 3

AND

x > 5

simplifies to

x > 5

True. Both must hold, so you need the stricter one; every number bigger than 5 is automatically bigger than 3, so the overlap is

(5, \infty)

x > 3

x > 5

simplifies to

x > 5

False. OR needs only one, so the looser condition wins; the union is

(3, \infty)

because

x = 4

already satisfies "

x > 3

x < 2

x > 2

covers every real number.

False. It covers all reals except

x = 2

itself, which satisfies neither strict inequality; the answer is

(- \infty, 2) \cup (2, \infty)

x \leq 2

x \geq 2

covers every real number.

True. The endpoint

x = 2

is now included on both sides (the

\leq

and

\geq

each catch it), so the union is all of

R

a < x < b

is a valid way to write an OR statement.

False. The chained form means

(a < x)

AND

(x < b)

; there is no chained notation for OR, which must stay as two separate inequalities.

5 > x > - 3

and

- 3 < x < 5

describe the same set.

True. Both say "

x

is between

- 3

and

5

"; the first is just written right-to-left, so rewriting in increasing order gives the identical interval

(- 3, 5)

If two rays point away from each other, their AND is always empty.

True. AND is the overlap; rays like

x < 1

and

x > 4

never share a point, so their intersection is

\emptyset

If two rays point away from each other, their OR is always all of

R

False.

x < 1

x > 4

leaves the gap

[1, 4]

uncovered; a union covers everything only if the rays overlap or touch.

x

is between

2

and

8

" is a compound inequality even though it uses no AND word.

True. "Between" hides an AND: it means

2 < x

AND

x < 8

; English words like between, within, from…to are AND in disguise.

x > 4

AND

x < 1

has the same solution as

x < 1

AND

x > 4

True. AND is order-independent (intersection is commutative); both are empty because no number is simultaneously below 1 and above 4.

Spot the error

Each line contains a wrong step. Say what broke and why it felt right.

Student writes:

- 4 < - 2 x < 6 \Rightarrow 2 < x < - 3

(divided by

- 2

, flipped nothing).

Error: dividing by a negative must flip both relation signs, and "

2 < x < - 3

" is impossible (nothing is above 2 yet below −3). Correct:

2 > x > - 3

, i.e.

(- 3, 2)

Student writes:

- 4 < - 2 x < 6 \Rightarrow 2 > x < - 3

(flipped only the left sign).

Error: you can't flip one sign in a chain — that produces a self-contradictory mixed statement. Both signs flip together, giving

2 > x > - 3

Student solves

x > 4

x < 1

and reports

\emptyset

Error: they used the AND (overlap) reasoning on an OR. OR wants the union, which keeps both rays:

(- \infty, 1) \cup (4, \infty)

Student solves

x > 4

AND

x > 1

and reports

(1, \infty)

Error: they took the union by keeping the looser ray. AND needs both, so only

x > 4

survives; the answer is

(4, \infty)

Student writes the AND-solution of

x \geq 2

and

x \leq 5

[2, 5)

Error: the right endpoint came from

x \leq 5

, which includes 5, so it must be a closed bracket. Correct:

[2, 5]

Student "simplifies"

x < - 2

x \geq 3

- 2 \leq x < 3

Error: they wrote the gap between the rays instead of the rays themselves, and even flipped inclusion. The union is

(- \infty, - 2) \cup [3, \infty)

— the two outer pieces, not the middle.

Student solves

0 \leq 2 x - 1 \leq 7

by adding 1 only to the middle:

0 \leq 2 x \leq 7

Error: an operation must hit all three parts or the chain no longer holds. Adding 1 everywhere gives

1 \leq 2 x \leq 8

, then

\frac{1}{2} \leq x \leq 4

Student claims

x < 5

AND

x > 2

has no solution "because the rays point apart".

Error: these rays point toward each other and overlap on

(2, 5)

; that misread of direction wrongly triggered the empty-AND reflex.

Why questions

The "understand it, don't memorise it" core.

Why is AND the intersection and not just something we're told to memorise?

Because intersection is defined as "belongs to both sets," which is exactly what "condition 1 AND condition 2 both hold" means — the set operation is the logic, not a separate rule. See Set operations — union and intersection.

Why does

x > 3

AND

x > 5

collapse to the stricter inequality but their OR collapses to the looser one?

AND demands both, so it keeps only the smaller overlap (stricter bound); OR demands just one, so the weaker bound already satisfies it and the larger region wins.

Why is there a compact chained form for AND but never for OR?

The AND overlap is a single connected middle band, so one chain

a < x < b

captures it; an OR of separated rays is two disconnected pieces that no single chain can express.

Why does dividing a three-part inequality by a negative flip both signs at once?

Each of the two inequalities in the chain reverses when multiplied by a negative (same rule as Linear inequalities in one variable); since both flip, the chain stays consistent — flipping only one would break it.

Why does "temperature between 20°C and 25°C" translate to AND, not OR?

"Between" requires the value to clear the lower bound and stay under the upper one simultaneously, so both conditions must hold — that's the definition of AND.

Why can an OR of two rays give all of

R

while an AND of the same two rays gives a bounded interval?

OR keeps every point covered by either ray (their combined reach), while AND keeps only the shared overlap; overlapping rays can jointly cover the whole line yet share only a finite band.

Why does the empty set

\emptyset

show up for AND but the whole line

R

show up for OR in "away-pointing" cases?

Rays pointing apart share nothing (empty overlap → AND fails everywhere), yet if they also overlap or meet they can cover everything (full union → OR succeeds everywhere) — opposite operations, opposite extremes.

Why does absolute value turn

∣ x ∣ < 3

into an AND but

∣ x ∣ > 3

into an OR?

∣ x ∣ < 3

means the distance from 0 is small, trapping

x

between

- 3

and

3

(both bounds → AND);

∣ x ∣ > 3

means the distance is large, sending

x

out past either end (one side or the other → OR). More in Absolute value inequalities.

Edge cases

The boundary and degenerate situations that trip people up.

Solve

x > 4

AND

x < 4

No solution,

\emptyset

. No number can be both strictly above and strictly below 4; the strict inequalities exclude the meeting point 4 itself.

Solve

x \geq 4

AND

x \leq 4

Exactly

{4}

— a single point. The overlap shrinks to the shared endpoint because both inequalities now include 4.

Solve

x < 4

x > 4

Everything except 4:

(- \infty, 4) \cup (4, \infty)

. The single point 4 satisfies neither strict condition, so the union punches a hole there.

Solve

x \leq 4

x \geq 4

All reals

R

. The endpoint 4 is caught by both closed pieces, so the union has no gap.

Solve

x > 0

AND

x > 0

(same condition twice).

Just

(0, \infty)

. Intersecting a set with itself changes nothing; duplicate conditions are harmless.

Solve

x < - 100

x > - 100

(- \infty, - 100) \cup (- 100, \infty)

— all reals except

- 100

; the two open rays leave only that one excluded point.

What is the solution set of a "compound inequality" whose two parts are identical under OR, e.g.

x > 2

x > 2

(2, \infty)

. Union of a set with itself is the same set, so the OR adds nothing new.

a < x < b

, what happens if

a = b

, say

3 < x < 3

\emptyset

. The band has zero width — no number is both greater than 3 and less than 3 at once.

a \leq x \leq b

, what happens if

a = b

, say

3 \leq x \leq 3

{3}

. With closed endpoints the degenerate band collapses to the single value 3.

a < x < b

, what happens if

a > b

, say

5 < x < 2

\emptyset

. The lower bound sits above the upper one, so no value can satisfy both; always check that

a < b

before trusting a chain.

Recall One-line summary of every trap here

AND = overlap (can be $\emptyset$ or a single point); OR = combined coverage (can be all of $R$ minus a hole). Strict vs closed endpoints decide whether boundary points join in, and a chain $a < x < b$ is only legal when it's an AND with $a < b$ .

Connections

Compound inequalities — AND, OR — the parent note these traps drill.
Linear inequalities in one variable — the sign-flip rule each half obeys.
Set operations — union and intersection — why AND/OR are $\cap$ / $\cup$ .
Interval notation — writing $\emptyset$ , single points, and unions correctly.
Absolute value inequalities — where AND/OR quietly appear inside $∣ x ∣$ .
Number line representation — the picture behind every item above.