Integers — operations, number line, absolute value: Worked examples

The scenario matrix

Every integer question this topic can throw at you falls into one of these classes. "Sign" means positive $(+)$ , negative $(-)$ , or the special value zero $(0)$ . Read $Z$ as "the integers" — the set ${\dots, - 2, - 1, 0, 1, 2, \dots}$ .

Cell	Case class	What makes it tricky
A	Addition, same signs $(+, +)$ or $(-, -)$	Directions accumulate
B	Addition, opposite signs $(+, -)$	One movement cancels the other
C	The tie: $a + (- a) = 0$ / degenerate zero	Result lands exactly on origin
D	Subtraction, including minus a negative	Rewrite $a - b = a + (- b)$ first
E	Multiplication all four sign combos + zero	$(-) \times (-) = (+)$ ; anything $\times 0 = 0$
F	Division, including $0 \div n$ , $n \div 0$ , $0 \div 0$	Some cells have no answer
G	Absolute value of a messy expression	Evaluate inside first, then measure distance
H	Triangle inequality — equal / strict / zero	Same sign → equal; opposite → strict; zero → equal
I	Word problem (debt / temperature)	Translate words → signed integers
J	Exam twist (order of operations + signs)	Bracket-first rules meet sign rules

We now walk ten examples, one per cell, A through J.

Cell A — Addition, same signs

Worked example

$(- 6) + (- 7)$ Forecast: guess the sign and size before reading on. Same direction twice — will the answer be bigger or smaller than each piece?

Step 1. Both numbers are negative, so both are moves to the left on the number line. Why this step? Adding a negative means "move left." Two left-moves point the same way, so they pile up.

Step 2. Add the magnitudes (distances from zero): $6 + 7 = 13$ . Why this step? When two movements share a direction, total distance is just their sum — nothing cancels.

Step 3. Keep the shared sign: the answer is $- 13$ . Why this step? Starting left of zero and moving further left keeps you left of zero. $(- 6) + (- 7) = - 13$ Verify: Reverse it — from $- 13$ , undo the last move by going $7$ right → $- 6$ . Undo the first by going $6$ right → $0$ . Back home, so the sum is consistent. ✓

The figure below draws exactly this walk. Read it left to right along the number line: the blue arrow is the first move (6 units left, ending at $- 6$ ), the orange arrow stacked above is the second move (7 more units left), and the red dot marks where you finally stop, $- 13$ . Notice both arrows point the same way — that is why the distances simply add.

Figure — Integers — operations, number line, absolute value

Cell B — Addition, opposite signs

In the figure, compare the two arrows' directions: the long blue arrow goes 9 units left, then the orange arrow doubles back 4 units right (it points the opposite way). The red dot at $- 5$ shows the leftover: the orange backtrack ate into the blue move, so only $9 - 4 = 5$ units of "left" survive.

Cell C — The tie / degenerate zero

Cell D — Subtraction, including minus-a-negative

Cell E — Multiplication, all four sign combos + zero

Worked example Evaluate

$4 \times 5$ , $4 \times (- 5)$ , $(- 4) \times 5$ , $(- 4) \times (- 5)$ , and $0 \times (- 5)$ Forecast: predict all five signs. Which one surprises people? And what does multiplying by zero do?

Step 1. $(+) \times (+)$ : read $4 \times 5$ as "four copies of $5$ " — repeated addition of a positive stays positive. $4 \times 5 = 20$ . Why this step? $5 + 5 + 5 + 5$ — four moves right of size $5$ , all positive. (Order won't matter for the value, but keeping "first factor = how many copies of the second" avoids confusion.)

Step 2. $(+) \times (-)$ : read $4 \times (- 5)$ as "four copies of $- 5$ " — four moves left. $4 \times (- 5) = - 20$ . Why this step? Repeating a leftward move stays leftward.

Step 3. $(-) \times (+) = (-)$ by commutativity: $(- 4) \times 5 = - 20$ , same value as Step 2. Why this step? Order doesn't change a product, so this must match the previous cell.

Step 4. $(-) \times (-) = (+)$ . Use the forced-by-algebra argument: $(- 4) \times (- 5) + (- 4) \times 5 = (- 4) \times ((- 5) + 5) = (- 4) \times 0 = 0.$ Since $(- 4) \times 5 = - 20$ , we need $(- 4) \times (- 5) - 20 = 0$ , so $(- 4) \times (- 5) = 20$ . Why this step? The distributive law forces the negative-times-negative to be $+ 20$ , otherwise the sum couldn't be zero. It's not a convention — it's the only value that keeps arithmetic consistent.

Step 5 (the zero edge case). $0 \times (- 5) = 0$ , and likewise $(- 5) \times 0 = 0$ . Why this step? $0 \times (- 5)$ means "zero copies of $- 5$ " — you take no steps at all, so you never leave the origin. Any integer times zero is zero, with no sign to worry about. $0 \times n = n \times 0 = 0 for every integer n$ Verify: Signs alternate for the four nonzero cases: $20, - 20, - 20, 20$ , magnitudes all $20$ . The zero case gives $0$ . ✓

Cell F — Division: $0 \div n$ , $n \div 0$ , and $0 \div 0$

Worked example

$(- 20) \div 5$ , $(- 20) \div (- 4)$ , $0 \div 7$ , $7 \div 0$ , and $0 \div 0$ Forecast: two have clean signed answers, one is a "quiet zero," and two do not exist. Which two, and — importantly — for different reasons?

Step 1. $(- 20) \div 5$ : opposite signs → negative. $20 \div 5 = 4$ , so $- 4$ . Why this step? Division inherits the exact sign rules of multiplication, because it's the inverse operation.

Step 2. $(- 20) \div (- 4)$ : same signs → positive. $20 \div 4 = 5$ , so $+ 5$ . Why this step? $(- 4) \times 5 = - 20$ , and division asks "what times $- 4$ gives $- 20$ ?" — answer $5$ .

Step 3 ( $0 \div n$ , the quiet zero). $0 \div 7$ : ask "what times $7$ gives $0$ ?" The answer is $0$ , since $0 \times 7 = 0$ . Why this step? Dividing zero by any nonzero number is fine — zero shared into $7$ groups is still zero in each group. $0 \div 7 = 0$

Step 4 ( $n \div 0$ with $n \neq = 0$ → UNDEFINED: no answer exists). $7 \div 0$ : ask "what number $x$ satisfies $0 \times x = 7$ ?" Why this step? Division is the inverse of multiplication. But $0 \times x = 0$ for every $x$ , and $0$ is never $7$ . So no $x$ works — the answer simply does not exist. We call this undefined. $7 \div 0 is undefined (no value exists).$

Step 5 ( $0 \div 0$ → INDETERMINATE: too many answers). $0 \div 0$ : ask "what $x$ satisfies $0 \times x = 0$ ?" Why this step? Here the problem is the opposite of Step 4: every $x$ works — $0 \times 3 = 0$ , $0 \times (- 8) = 0$ , all of them. With no single answer to pick, $0 \div 0$ is called an indeterminate form. This is a different failure from Step 4: there we had zero candidates, here we have infinitely many. Both are left without a value, but for opposite reasons. $0 \div 0 is indeterminate (no unique value).$ Verify: Step 1: $5 \times (- 4) = - 20$ ✓. Step 2: $(- 4) \times 5 = - 20$ ✓. Step 3: $0 \times 7 = 0$ ✓. Step 4 (undefined): no $x$ solves $0 \cdot x = 7$ → zero candidates ✓. Step 5 (indeterminate): $0 \cdot 3 = 0$ and $0 \cdot (- 8) = 0$ → many candidates ✓.

Cell G — Absolute value of a messy expression

The figure makes the last step visual. Read it as a distance measurement: the orange dot sits at the inside value $- 7$ , the gray dot marks the origin $0$ , and the green double-headed arrow between them is what $∣ \cdot ∣$ measures — its length, $7$ . The arrow has no preferred direction, which is exactly why absolute value ignores the minus sign.

Cell H — Triangle inequality: equal / strict / zero

Worked example Compare

$∣ a + b ∣$ with $∣ a ∣ + ∣ b ∣$ for (i) $a = 6, b = 2$ ; (ii) $a = 6, b = - 2$ ; (iii) $a = 6, b = 0$ Forecast: the triangle inequality says $∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$ . When is it equal, when strictly less, and what happens when one term is zero?

Step 1 (same signs). $∣6 + 2∣ = ∣8∣ = 8$ and $∣6∣ + ∣2∣ = 6 + 2 = 8$ . Why this step? Both moves go right, nothing cancels, so total distance = sum of distances → equality. $∣6 + 2∣ = ∣6∣ + ∣2∣ (equal)$

Step 2 (opposite signs). $∣6 + (- 2) ∣ = ∣4∣ = 4$ , but $∣6∣ + ∣ - 2∣ = 6 + 2 = 8$ . Why this step? The leftward $- 2$ backtracks over part of the rightward $6$ , so the net distance $4$ is less than the total walking distance $8$ . $∣6 + (- 2) ∣ = 4 < 8 = ∣6∣ + ∣ - 2∣ (strict)$

Step 3 (zero case). $∣6 + 0∣ = ∣6∣ = 6$ and $∣6∣ + ∣0∣ = 6 + 0 = 6$ . Why this step? Adding zero is no movement at all, so there's nothing to cancel — the inequality is an equality, just like the same-sign case. Zero counts as "same direction as anything." $∣6 + 0∣ = ∣6∣ + ∣0∣ (equal)$ Verify: All three obey $∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$ : $8 \leq 8$ ✓, $4 \leq 8$ ✓, $6 \leq 6$ ✓. Equality holds when $a, b$ share a sign or one is zero; strict only when they have genuinely opposite signs. ✓

The figure stacks all three walks. Compare the three rows: in the top row both arrows point right, so their tips reach the full combined length ( $8$ ) — equality. In the middle row the second arrow reverses, pulling the endpoint back to $4$ — strict inequality. In the bottom row the second term is zero, so the second arrow has no length at all and the endpoint stays at the full $6$ — equality again.

Cell I — Word problem (temperature / debt)

Worked example At dawn the temperature is

$- 9^{\circ}$ C. It rises $4^{\circ}$ by noon, then drops $6^{\circ}$ by night. Final temperature? Forecast: "rise" and "drop" are opposite directions — guess the final sign.

Step 1. Translate words to signed integers: start $- 9$ , rise $+ 4$ , drop $- 6$ . Why this step? A rise is a move right (add positive), a drop is a move left (add negative). This connects the story to the number line.

Step 2. Add left to right: $- 9 + 4 = - 5$ (Cell B). Why this step? Do operations in the order they happen in time — noon before night.

Step 3. Then $- 5 + (- 6) = - 11$ (Cell A, same signs). Why this step? The night drop is another leftward move, accumulating with the current negative value. $- 9 + 4 - 6 = - 1 1^{\circ} C$ Verify: Net change $= + 4 - 6 = - 2$ , so final $= - 9 + (- 2) = - 11$ . Units: degrees Celsius ✓. Below freezing, which matches a cold night. ✓

Cell J — Exam twist (order of operations + signs)

Worked example Evaluate

$- 3 \times (2 - 7) + (- 8) \div (- 2)$ Forecast: brackets first, then $\times$ and $\div$ , then $+$ . Track every sign — predict the final number before reading on.

Step 1. Brackets first. Rewrite the subtraction inside as addition of the opposite: $2 - 7 = 2 + (- 7) = - 5$ (Cell B). Why this step? The priority list demands the innermost bracket before anything else; and we rewrite $2 - 7$ as $2 + (- 7)$ because subtraction is adding the opposite, which lets us reuse the opposite-signs addition rule (larger magnitude $7$ is negative, so the result is $- 5$ ). $- 3 \times (- 5) + (- 8) \div (- 2)$

Step 2. Multiplication: $(- 3) \times (- 5) = 15$ (Cell E, negative × negative = positive). Why this step? Multiply and divide rank above addition; do them before the final sum.

Step 3. Division: $(- 8) \div (- 2) = 4$ (Cell F, same signs → positive). Why this step? Division shares the same rank as multiplication, so it's handled in this same pass, left to right. $15 + 4$

Step 4. Add last: $15 + 4 = 19$ . Why this step? Addition is the lowest priority in the list, so it comes after every multiply/divide is done. $- 3 \times (2 - 7) + (- 8) \div (- 2) = 19$ Verify: Recompute each piece independently: $(- 3) \times (- 5) = 15$ , $(- 8) \div (- 2) = 4$ , $15 + 4 = 19$ ✓.

Recall Self-check (cover the answers)

Which cell is $(- 5) + (- 2)$ ? ::: Cell A (same signs) → $- 7$ Which two division cases have no value, and how do they differ? ::: Cell F: $n \div 0$ ( $n \neq = 0$ ) is undefined — no $x$ solves $0 \times x = n$ ; $0 \div 0$ is indeterminate — every $x$ solves $0 \times x = 0$ What is $0 \div 7$ ? ::: $0$ — dividing zero by a nonzero number is fine When is $∣ a + b ∣ = ∣ a ∣ + ∣ b ∣$ ? ::: When $a, b$ have the same sign, or at least one of them is zero (Cell H) Rewrite $x - (- 6)$ before computing. ::: $x + 6$ (minus a negative = plus) Evaluate $- 3 \times (2 - 7) + (- 8) \div (- 2)$ . ::: $19$ (brackets → multiply/divide → add)

Connections

Parent: Integers — operations, number line, absolute value
Builds on: Natural numbers
Leads to: Rational numbers, Linear equations
Related pictures: Coordinate plane, Vectors in 2D, Inequalities