Exercises — Integers — operations, number line, absolute value

• Integer $\mathbb{Z}=\{\dots ,-2,-1,0,1,2,\dots \}$ — whole numbers going both ways.
• Number line: a horizontal ruler with $0$ in the middle; right is $+$, left is $-$; one step $=$ one unit.
• Adding a number $=$ a move on that line: add positive $\to$ step right, add negative $\to$ step left.
• Subtracting $=$ adding the opposite: $a-b=a+(-b)$.
• Absolute value $|a|$ $=$ distance from $0$, always $\ge 0$: $|a|=a$ if $a\ge 0$, and $|a|=-a$ if $a<0$.

What we do: "steps left" means add a negative, "steps right" means add a positive.

• $5$ left of $2$: start at the point $2$, walk $5$ chalk-steps toward the negative side. Look at the blue arrow: $2\to 1\to 0\to -1\to -2\to -3$. That is $2+(-5)===-3==$.
• $5$ right of $-3$: start at $-3$, walk $5$ steps toward the positive side (yellow arrow): $-3\to -2\to -1\to 0\to 1\to 2$. That is $(-3)+5===2==$.

Notice they swapped places — that is the whole "left/right symmetry" of the line.

What we do: separate the multiply/divide sign rule from the add rule.

• $(-8)\times (-3)$: multiplication, negative times negative $\Rightarrow$ positive $(+)$.
• $(-8)+(-3)$: addition, both moves go left $\Rightarrow$ negative $(-)$.
• $(-8)÷4$: division, negative over positive $\Rightarrow$ negative $(-)$.

Signs: $==+==,==-==,==-==$.

What we do: turn every subtraction into "add the opposite," then read as movements. $(-6)-(-9)+(-4)=(-6)+9+(-4)$

• Start at $-6$. Move $9$ right: $-6+9=3$.
• From $3$, move $4$ left: $3+(-4)=-1$.

Answer: $==-1==$. Why each step: subtracting $-9$ is adding $+9$ because $-(-9)=9$; adding $-4$ is a leftward step.

(a) Directly: $(-4)\times 7=-28,(-4)\times (-2)=+8$ $-28+8=-20.$ (b) Distributive — factor the shared $(-4)$: $(-4)\times 7+(-4)\times (-2)=(-4)\times (7+(-2))=(-4)\times 5=-20.$ Both give $==-20==$. The distributive law is just "do the inside first, then one multiply" — often faster.

What we do: always evaluate inside the bars before taking absolute value.

• $3-11=-8$, so $|3-11|=|-8|=8$ (distance of $-8$ from $0$).
• $|-4|=4$. $8-4===4==.$

What we do: read $|x|$ as "distance from $0$," then ask which points sit at that distance.

• $|x|=6$: points exactly $6$ steps from $0$. There are two — one each side (pink dots in the figure): $x===6==$ or $x===-6==$.
• $|x|<3$: points closer than $3$ steps from $0$, i.e. strictly between $-3$ and $3$. Since $x$ must be an integer, the endpoints $\pm 3$ are excluded (distance $=3$ is not $<3$): $x\in \{-2,-1,0,1,2\}.$ That is $==5==$ integers (the yellow band in the figure). This ties to 2.1.05-Inequalities — an absolute-value inequality carves out an interval around $0$.

What we do: notice $a-b$ and $b-a$ are opposites, then use property $|-n|=|n|$. Let $n=a-b$. Then $b-a=-(a-b)=-n$. Distance-from-zero doesn't care about side: $|b-a|=|-n|=|n|=|a-b|.$ So it is always true. Check: $a=2,b=9$: $|2-9|=|-7|=7$ and $|9-2|=|7|=7$. ✓ Meaning: on the line, the distance between two points is the same whichever one you start from.

What we do: model "down" as adding negatives, "up" as adding positives; track the running position. Running positions (each is previous $+$ move):

• Start: $0$
• $-18$: $0+(-18)=-18$
• $+7$: $-18+7=-11$
• $-11$: $-11+(-11)=-22$
• $+22$: $-22+22=0$

(a) Final depth $===0==$ m (back at sea level). (b) Greatest depth $=$ the most negative running position $=-22$, so she reached $==22==$ m down. (c) Total distance swum $=$ sum of the magnitudes of each move (distance ignores direction — that is exactly what absolute value is for): $|-18|+|7|+|-11|+|22|=18+7+11+22===58==\text{ m}.$ Why absolute value here: her final position is $0$, but she clearly worked hard — position (a) and path length (c) are different questions, and $|\cdot |$ is the tool that measures path length regardless of sign.

What we do: rearrange to pair each number with its opposite (additive inverse), using commutativity.

• $15\times 1=15$ (multiplicative identity).
• Reorder (commutative + associative for addition): $(7+(-7))+(15+(-15))+(-3).$
• $7+(-7)=0$ and $15+(-15)=0$ (additive inverse). $0+0+(-3)=-3.$ Answer: $==-3==$. Spotting inverse pairs turns a messy sum into one number.

What we do: bootstrap from the ordinary triangle inequality $|x+y|\le |x|+|y|$ (proven in the parent note).

Step 1 — one direction. Write $a$ as $(a-b)+b$. Apply the triangle inequality with $x=a-b,y=b$: $|a|=|(a-b)+b|\le |a-b|+|b|.$ Subtract $|b|$ from both sides: |a| - |b| \le |a-b|. \tag{i}

Step 2 — the other direction. By symmetry, swap the roles of $a$ and $b$. Write $b=(b-a)+a$: $|b|\le |b-a|+|a|\Rightarrow |b|-|a|\le |b-a|=|a-b|.$ The last equality uses Exercise 3.2 ($|b-a|=|a-b|$). So: -\bigl(|a| - |b|\bigr) \le |a-b|. \tag{ii}

Step 3 — combine. Lines (i) and (ii) say the number $|a|-|b|$ is $\le |a-b|$ and its negative is $\le |a-b|$. "A number and its negative are both $\le M$" is exactly "$|\text{that number}|\le M$." Hence: $||a|-|b||\le |a-b|.■$

Verification with $a=-2,b=7$:

• Left: $||-2|-|7||=|2-7|=|-5|=5.$
• Right: $|-2-7|=|-9|=9.$
• $5\le 9$ ✓.

What we do: read $|x+4|\le 3$ as "$x+4$ is at most $3$ away from $0$," i.e. squeezed between $-3$ and $3$. $-3\le x+4\le 3.$ Subtract $4$ from all three parts (this is the 2.1.05-Inequalities move — same operation on every part keeps the chain valid): $-7\le x\le -1.$ Integer solutions: $\{-7,-6,-5,-4,-3,-2,-1\}$ — that is $==7==$ integers. Geometric read: $|x+4|$ is the distance from $x$ to the point $-4$. So we want all integers within $3$ steps of $-4$ — exactly $-4$ and three neighbours on each side.