0.2 Integers
SLO 6.NS
The Number System. Divide fractions by fractions; compute fluently with
SLO 6.EE
Expressions and Equations. Write, read and evaluate expressions in which
SLO 7.NS
The Number System. Add, subtract, multiply and divide rational numbers,
SLO 7.EE
Expressions and Equations. Apply properties of operations to add, subtract,
SLO 8.NS
The Number System. Know that numbers which are not rational are irrational;
Learning Objectives
By the end of this section, you will be able to:
- locate positive and negative numbers on a number line and compare any two of them;
- find the opposite of a number and explain what its absolute value measures;
- add and subtract signed numbers, and explain why every subtraction can be rewritten as an addition;
- multiply and divide signed numbers and predict the sign of a result before computing it;
- apply the order of operations to expressions that contain negative numbers.
Whole numbers describe amounts you can count. They cannot describe a temperature below zero, a bank balance after an overdraft, an elevation below sea level, or a loss on a spreadsheet. Extending the number system to include negatives fixes that, and it is the single most error-prone topic in prerequisite arithmetic.
The errors are almost never conceptual. Students understand perfectly well that owing $40 is worse than owing $10. The mistakes come from sign bookkeeping in expressions like \(-3 - (-7)\), where three different meanings of the minus sign collide. This section is about making that bookkeeping automatic, because every later chapter assumes it.
Here is a promise worth making early: none of what follows is a list of arbitrary rules to memorize. Every sign rule in this section can be seen on a number line, felt in a money situation, or forced by a pattern that already holds for positive numbers. When a rule feels arbitrary, that is a signal you have the rule but not the reason, and the reason is what survives the year. So each rule below arrives with the argument that produces it.
0.2.1 Negative Numbers and the Number Line
Counting stops at zero. Measuring does not. A thermometer keeps reading past the freezing mark, a bank balance keeps dropping past empty, and an elevation keeps descending past the shoreline. In every one of those situations we need a number that says both how much and which direction, and a whole number cannot do that.
| Situation | Described in words | Written as an integer |
|---|---|---|
| Lowest point in Death Valley | 282 feet below sea level | \(-282\) |
| Home freezer setting | 4 degrees below zero | \(-4\) |
| Overdrawn checking account | 65 dollars owed to the bank | \(-65\) |
| Golf score for the round | 3 strokes under par | \(-3\) |
| Parking garage level | 2 floors below the lobby | \(-2\) |
Notice what the third column is doing. Each situation already had a zero built into it — sea level, the freezing mark, an empty account, par, the lobby — and the negative sign records that we are on the far side of that zero. The number system we need is the whole numbers plus a mirror image of themselves.
The integers are the whole numbers together with their negatives:
$$\dots, -4, -3, -2, -1, 0, 1, 2, 3, 4, \dots$$Every integer is either negative, zero, or positive, and no integer is more than one of those.
A number line is a line with one point chosen as zero and a fixed unit distance marked off in both directions. The positive integers 1, 2, 3, … sit to the right of zero. The negative integers \(-1\), \(-2\), \(-3\), … sit the same distances to the left.
The spacing matters. Each step of one unit is the same size everywhere on the line, which is what makes the picture trustworthy. The gap from \(-7\) to \(-6\) is exactly the gap from 40 to 41.
Fold the number line at zero and the two halves land on each other: 3 covers \(-3\), 8 covers \(-8\). Every fact about negatives in this section is really a fact about that fold. Zero is the crease, which is why it is neither positive nor negative.
Definition 0.2.2 — A number line: one point chosen as zero, one unit distance, repeated.
Order
On a number line, a number is greater than every number to its left. That single rule settles every comparison, including the ones that feel backwards.
- \(5 > 2\), because 5 sits to the right of 2.
- \(-2 > -5\), because \(-2\) sits to the right of \(-5\).
The second one is where intuition misfires. Five is bigger than two, so it feels like \(-5\) should be bigger than \(-2\). It is not. Think of temperature: \(-2^\circ\) is a warmer day than \(-5^\circ\). Or debt: owing $2 leaves you better off than owing $5.
| Comparison | True statement | Why |
|---|---|---|
| 3 and 8 | \(3 < 8\) | 3 is left of 8 |
| \(-3\) and 8 | \(-3 < 8\) | every negative is left of every positive |
| \(-3\) and \(-8\) | \(-3 > -8\) | \(-3\) is right of \(-8\) |
| \(-3\) and 0 | \(-3 < 0\) | negatives are left of zero |
| \(-100\) and \(-1\) | \(-100 < -1\) | \(-100\) is far to the left of \(-1\) |
Zero is neither positive nor negative. It is the boundary, and it is greater than every negative integer and less than every positive one.
That last row is the one to sit with. Among negatives, the number that looks biggest is smallest. A drop to \(-100^\circ\) is colder than a drop to \(-1^\circ\), and an account at \(-100\) dollars is in worse shape than one at \(-1\) dollar. Reading a size and reading a position are two different jobs, and the negative sign is what separates them.
Arrange \(-9,\; 4,\; 0,\; -1,\; 7,\; -12\) from least to greatest, and state which two are farthest apart on the number line.
Solution
Step 1 — Separate by sign. Every negative is left of zero, and zero is left of every positive, so the negatives come first as a block, then 0, then the positives.
Negatives: \(-9,\; -1,\; -12\). Positives: \(4,\; 7\).
Step 2 — Order the negatives. Among negatives, the one with the larger digit sits farther left, so it is smaller. Sorting \(-12,\; -9,\; -1\) left to right gives that order.
Step 3 — Order the positives. These behave normally: \(4\) then \(7\).
Step 4 — Assemble the full list.
$$-12,\; -9,\; -1,\; 0,\; 4,\; 7$$Step 5 — Farthest apart. The extremes of the list are the leftmost and rightmost values, \(-12\) and \(7\). Counting units from \(-12\) up to 0 is 12 steps, and from 0 up to 7 is 7 more, for 19 units in all.
Answer: \(-12,\; -9,\; -1,\; 0,\; 4,\; 7\); the pair \(-12\) and \(7\) are 19 units apart.
Opposites
Fold the line at zero again and every number lands on a partner.
The opposite of a number is the number the same distance from zero on the other side of zero. The opposite of 6 is \(-6\); the opposite of \(-6\) is 6. Zero is its own opposite, since it sits at distance zero on both sides.
A pair of opposites always sums to zero, which is why the opposite of \(a\) is also called the additive inverse of \(a\):
$$6 + (-6) = 0 \qquad -14 + 14 = 0 \qquad a + (-a) = 0$$That fact is the whole engine behind subtraction, and Section 0.2.3 spends its time on it.
We write the opposite of \(a\) as \(-a\), and this is where the notation earns its reputation. The symbol \(-\) does three different jobs.
| Use | Example | Read as |
|---|---|---|
| A negative number | \(-7\) | "negative seven" |
| The opposite of | \(-a\) | "the opposite of \(a\)" |
| Subtraction | \(9 - 4\) | "nine minus four" |
The middle row is the one that trips people up: \(-a\) is not necessarily a negative number. If \(a = -3\), then \(-a = 3\), which is positive. The expression \(-a\) means "flip the sign of whatever \(a\) is," and flipping the sign of a negative gives a positive. Taking the opposite twice returns you to the start: \(-(-6) = 6\).
That double-flip is worth a second look, because it is the exact move that later makes \(-3 - (-7)\) come out positive. Flipping once takes you across zero; flipping again takes you straight back. Reading \(-(-6)\) out loud as "the opposite of negative six" usually settles it faster than any rule, because the opposite of a number six units to the left of zero has to be the number six units to the right.
Definition 0.2.3 — A number and its opposite sit the same distance from zero, on opposite sides.
Evaluate \(-a\) when \(a = 11\) and again when \(a = -11\). Then simplify \(-(-(-4))\).
Solution
Step 1 — Evaluate \(-a\) at \(a = 11\). Substituting gives the opposite of 11.
$$-a = -(11) = -11$$Step 2 — Evaluate \(-a\) at \(a = -11\). Now \(a\) is itself negative, so its opposite lies on the positive side.
$$-a = -(-11) = 11$$Both answers came from the same expression \(-a\). The sign of the result depends entirely on the sign of \(a\), which is exactly why \(-a\) cannot be read as "a negative number."
Step 3 — Simplify \(-(-(-4))\). Work from the inside out. Here three opposite signs are applied to 4.
$$-(-(-4)) = -(\,-(-4)\,) = -(4) = -4$$Three flips starting from \(4\) leave you on the opposite side from where you began; an odd number of flips always does.
Answer: \(-11\), then \(11\), then \(-4\).
Order \(-6,\; 2,\; -15,\; 0,\; -1\) from least to greatest, then evaluate \(-x\) when \(x = -9\) and simplify \(-(-(-(-2)))\).
Solution
Part 1 — Order the list. Negatives first, ordered by how far left they sit; then zero; then positives.
$$-15,\; -6,\; -1,\; 0,\; 2$$Part 2 — Evaluate \(-x\) at \(x = -9\). Substitute with parentheses so the two signs stay separate.
$$-x = -(-9) = 9$$Part 3 — Simplify \(-(-(-(-2)))\). There are four opposite signs applied to 2. Flipping an even number of times returns to the start.
$$-(-(-(-2))) = 2$$Answer: \(-15,\; -6,\; -1,\; 0,\; 2\); then \(9\); then \(2\).
0.2.2 Absolute Value
Sometimes the direction is the whole point, and sometimes it is beside the point. If a hiker is 400 feet below the trailhead and a drone is 400 feet above it, the two are in opposite directions but the same distance away. A tool that reports distance only, ignoring direction, is exactly what absolute value is.
The absolute value of a number is its distance from zero on the number line, written with vertical bars. Distance is never negative, so for every number \(a\), the value \(|a|\) is positive or zero, and \(|a| = 0\) only when \(a = 0\).
$$|6| = 6 \qquad |-6| = 6 \qquad |0| = 0$$Because distance is never negative, absolute value is never negative. It strips the sign and reports only magnitude.
A quality report might ask how far a part is from spec, not whether it is oversized or undersized. Absolute value answers the first question and deliberately throws away the second, which is why \(|-0.04| = |0.04| = 0.04\) treats both misses as the same size error.
That is exactly what you want when the question is "how far" rather than "which way." A balance of \(-250\) dollars and a balance of \(250\) dollars are opposite in direction but equal in size; \(|-250| = |250| = 250\) captures that.
The pairing with opposites is worth naming outright. Opposites always have equal absolute values, since they sit the same distance from zero on either side, so \(|a| = |-a|\) for every number \(a\). And an absolute value can never distinguish a number from its opposite: knowing that \(|n| = 12\) tells you \(n\) is 12 units from zero but leaves two candidates, \(12\) and \(-12\). That ambiguity is not a flaw. It is the correct answer to a question that only asked about distance.
Definition 0.2.4 — Absolute value reports the distance from zero, so it is never negative.
Absolute value as a grouping symbol
The bars group, like parentheses. Simplify what is inside first, then take the absolute value.
$$|3 - 10| = |-7| = 7$$Do not distribute across the bars. The expression \(|3 - 10|\) is 7, not \(|3| - |10| = -7\).
A minus sign outside the bars stays outside and applies afterward:
$$-|{-9}| = -(9) = -9$$That expression is negative, and correctly so — you take the absolute value first, then negate it. The bars only protect what is inside them, and a sign parked outside is untouched by that protection.
| Expression | Value | Note |
|---|---|---|
| \(\lvert -4 \rvert\) | 4 | distance from zero |
| \(-\lvert -4 \rvert\) | \(-4\) | negate after taking absolute value |
| \(\lvert 5 - 9 \rvert\) | 4 | simplify inside first |
| \(\lvert 5 \rvert - \lvert 9 \rvert\) | \(-4\) | different expression entirely |
| \(\lvert -3 \rvert + \lvert -8 \rvert\) | 11 | two separate distances, then add |
The middle two rows differ only in where the bars sit, and they produce opposite answers. Placement is not cosmetic.
Absolute value also gives a clean way to measure the gap between two numbers. The distance between 4 and 9 on the number line is \(|4 - 9| = |-5| = 5\), and subtracting in the other order gives \(|9 - 4| = |5| = 5\) as well. Order does not matter, which is exactly what you want from a distance. In Example 0.2.1 we counted from \(-12\) to 7 by hand and got 19 units; the bars produce the same number without counting, since \(|-12 - 7| = |-19| = 19\).
Simplify each expression: (a) \(\lvert 6 - 14 \rvert\); (b) \(-\lvert 6 - 14 \rvert\); (c) \(\lvert 6 \rvert - \lvert 14 \rvert\); (d) \(8 - \lvert 2 - 7 \rvert\).
Solution
(a) Simplify inside the bars first, then take the distance.
$$\lvert 6 - 14 \rvert = \lvert -8 \rvert = 8$$(b) Same interior work, then apply the outside minus sign. The bars do not protect a sign sitting outside them.
$$-\lvert 6 - 14 \rvert = -\lvert -8 \rvert = -(8) = -8$$(c) Here the bars enclose single numbers, so each is evaluated on its own before subtracting.
$$\lvert 6 \rvert - \lvert 14 \rvert = 6 - 14 = -8$$Parts (a) and (c) use the same three symbols in a different arrangement and disagree by a sign. The bars are grouping symbols, so where they open and close changes the problem.
(d) Grouping first, then subtraction from left to right.
$$8 - \lvert 2 - 7 \rvert = 8 - \lvert -5 \rvert = 8 - 5 = 3$$Answer: 8, \(-8\), \(-8\), and 3.
Find every integer \(n\) with \(\lvert n \rvert = 7\). Then explain why no integer satisfies \(\lvert n \rvert = -7\).
Solution
Step 1 — Translate the first equation into distance. The statement \(\lvert n \rvert = 7\) says that \(n\) sits exactly 7 units from zero.
Step 2 — Count in both directions. Going 7 units right of zero lands on 7. Going 7 units left of zero lands on \(-7\). Nothing else is 7 units away.
$$n = 7 \quad \text{or} \quad n = -7$$Step 3 — Test the second equation. The statement \(\lvert n \rvert = -7\) asks for a number whose distance from zero is negative. Distance counts units traveled, and traveling a negative number of units is not a thing you can do, so no value of \(n\) works.
Answer: \(n = 7\) or \(n = -7\); the equation \(\lvert n \rvert = -7\) has no solution because an absolute value is never negative.
Simplify \(\;\lvert 3 - 11 \rvert\), \(\;-\lvert 3 - 11 \rvert\), and \(\;\lvert -5 \rvert + \lvert -2 \rvert\), then find every integer \(n\) with \(\lvert n \rvert = 4\).
Solution
Part 1 — \(\lvert 3 - 11 \rvert\). Simplify inside the bars, then take the distance.
$$\lvert 3 - 11 \rvert = \lvert -8 \rvert = 8$$Part 2 — \(-\lvert 3 - 11 \rvert\). Same interior, then negate the result.
$$-\lvert 3 - 11 \rvert = -(8) = -8$$Part 3 — \(\lvert -5 \rvert + \lvert -2 \rvert\). Each bar pair encloses a single number, so evaluate both distances and add them.
$$\lvert -5 \rvert + \lvert -2 \rvert = 5 + 2 = 7$$Part 4 — Solve \(\lvert n \rvert = 4\). Two integers sit 4 units from zero, one on each side.
$$n = 4 \quad \text{or} \quad n = -4$$Answer: 8, \(-8\), 7, and \(n = 4\) or \(n = -4\).
0.2.3 Adding and Subtracting Integers
Addition
Adding is movement along the number line. Start at the first number; a positive addend moves right, a negative addend moves left.
Two cases cover everything.
Same signs — add the magnitudes, keep the sign. Both movements go the same direction, so they accumulate.
$$4 + 7 = 11 \qquad -4 + (-7) = -11$$Different signs — subtract the smaller magnitude from the larger, and take the sign of the larger. The movements partly cancel; whichever is bigger wins.
$$-9 + 4 = -5 \qquad 9 + (-4) = 5$$A money model makes this concrete: a debt of $9 combined with a credit of $4 leaves you $5 in debt. The debt was larger, so the result is negative.
Picture the two numbers as teams pulling in opposite directions. Same signs means both teams pull the same way, so the pulls add. Different signs means they pull against each other, so the smaller pull cancels part of the larger, and the winner's direction is the sign of the answer.
Why does the second rule take the sign of the larger magnitude, rather than the sign of the number written first? Because the sign of a sum records which direction you ended up from zero, not which direction you started. Beginning at \(-9\) and walking 4 units right leaves you at \(-5\), still on the negative side, because 4 steps were not enough to cross the 9-unit gap back to zero. Beginning at 9 and walking 4 units left leaves you at 5, still positive, for the mirror-image reason. The magnitudes decide who crosses zero, and crossing zero is what decides the sign.
| Problem | Signs | Work | Result |
|---|---|---|---|
| \(-6 + (-8)\) | same | \(6 + 8 = 14\), keep negative | \(-14\) |
| \(-6 + 8\) | different | \(8 - 6 = 2\), take sign of 8 | \(2\) |
| \(6 + (-8)\) | different | \(8 - 6 = 2\), take sign of \(-8\) | \(-2\) |
| \(6 + 8\) | same | \(6 + 8 = 14\), keep positive | \(14\) |
| \(-7 + 7\) | different | equal magnitudes cancel | \(0\) |
That last row is the additive-inverse fact from Section 0.2.1 showing up inside the addition rule. Whenever the two magnitudes tie, the walk right and the walk left cover the same ground and you land back at zero.
An account starts the week at $120. On Monday a $200 rent payment clears, on Wednesday a $65 paycheck deposit arrives, and on Friday a $30 fee is charged. What is the balance at the end of the week?
Solution
Step 1 — Write each event as a signed number. Money coming in is positive, money going out is negative.
$$120 + (-200) + 65 + (-30)$$Step 2 — Combine the first two. Different signs, so subtract magnitudes and keep the sign of the larger: \(200 - 120 = 80\), and 200 was the negative one.
$$120 + (-200) = -80$$Step 3 — Add Wednesday's deposit. Different signs again: \(80 - 65 = 15\), and 80 was the negative one.
$$-80 + 65 = -15$$Step 4 — Add Friday's fee. Same signs now, so add magnitudes and keep the negative.
$$-15 + (-30) = -45$$Step 5 — Interpret the result. A balance of \(-45\) means the account is overdrawn by $45.
Answer: \(-\$45\), an overdraft of $45.
Subtraction
Every subtraction can be rewritten as an addition.
For any numbers \(a\) and \(b\), the difference \(a - b\) is defined as the sum of \(a\) and the opposite of \(b\):
$$a - b = a + (-b)$$To subtract, add the opposite. This is not a trick to memorize alongside the addition rules — it replaces the need for separate subtraction rules. Convert every subtraction to an addition, then apply the two addition cases above.
$$5 - 8 = 5 + (-8) = -3$$ $$-3 - 7 = -3 + (-7) = -10$$ $$-3 - (-7) = -3 + 7 = 4$$That third line is the one worth slowing down on. Subtracting a negative adds. The two minus signs in \(-(-7)\) are "the opposite of negative seven," which is \(+7\).
If that feels arbitrary, read it as removing a debt — and read it as a change rather than a final total. You are $3 in the hole. Someone cancels a $7 debt of yours, which moves you $7 in your favor. Going up $7 from $3 in the hole leaves you $4 to the good. Taking away a negative leaves you better off, by exactly the size of what was taken away.
There is a second argument for the same fact, and it does not rely on any story. Subtraction asks a missing-addend question: \(9 - 4\) is the number you add to 4 to get 9. Apply that reading to \(-3 - (-7)\) and it asks what you add to \(-7\) to reach \(-3\). Starting at \(-7\) you must move 4 units to the right, so the answer is \(+4\). The number line and the debt story agree, which is a good sign that the rule is forced rather than chosen.
| Expression | Rewritten | Result |
|---|---|---|
| \(12 - 20\) | \(12 + (-20)\) | \(-8\) |
| \(-12 - 20\) | \(-12 + (-20)\) | \(-32\) |
| \(-12 - (-20)\) | \(-12 + 20\) | \(8\) |
| \(12 - (-20)\) | \(12 + 20\) | \(32\) |
| \(0 - (-6)\) | \(0 + 6\) | \(6\) |
Rewriting before computing costs one extra line and eliminates most sign errors. It is worth the line.
Definition 0.2.5 — Rewriting a subtraction as adding the opposite turns it into one walk.
At 6 a.m. the temperature was \(-11^\circ\)F. By 2 p.m. it had reached \(9^\circ\)F. How many degrees did it rise? That evening it fell back to \(-4^\circ\)F. How many degrees did it fall from the afternoon high?
Solution
Step 1 — Set up the morning-to-afternoon change. A change is the ending value minus the starting value.
$$9 - (-11)$$Step 2 — Rewrite the subtraction as an addition. The opposite of \(-11\) is 11.
$$9 - (-11) = 9 + 11 = 20$$The temperature rose 20 degrees. That matches the number line: 11 units to climb from \(-11\) up to zero, then 9 more to reach 9.
Step 3 — Set up the afternoon-to-evening change.
$$-4 - 9 = -4 + (-9) = -13$$Step 4 — Interpret the sign. A change of \(-13\) means the temperature dropped, and the size of the drop is \(\lvert -13 \rvert = 13\) degrees.
Answer: A rise of 20 degrees, then a fall of 13 degrees.
Chains of additions and subtractions
Longer expressions are handled the same way: rewrite every subtraction as an addition first, then work left to right. Rewriting first is what keeps a stray minus sign from attaching itself to the wrong number partway through.
$$-8 - (-3) + (-5) - 2 = -8 + 3 + (-5) + (-2)$$Working left to right: \(-8 + 3 = -5\), then \(-5 + (-5) = -10\), then \(-10 + (-2) = -12\).
Simplify \(\;7 - 15 - (-4) + (-6) - (-9)\).
Solution
Step 1 — Rewrite every subtraction as an addition. There are three subtraction signs; each becomes an addition of the opposite.
$$7 + (-15) + 4 + (-6) + 9$$Step 2 — Work left to right. Different signs, so subtract magnitudes and take the sign of the larger.
$$7 + (-15) = -8$$Step 3 — Continue. Different signs again.
$$-8 + 4 = -4$$Step 4 — Continue. Same signs now, so magnitudes add and the sign stays negative.
$$-4 + (-6) = -10$$Step 5 — Finish. Different signs, and 10 is the larger magnitude.
$$-10 + 9 = -1$$Step 6 — Check by grouping instead. The positives total \(7 + 4 + 9 = 20\) and the negatives total \(15 + 6 = 21\), so the sum is 1 unit onto the negative side. That agrees.
Answer: \(-1\).
Simplify \(\;-14 + 6\), \(\;-14 - 6\), \(\;-14 - (-6)\), and \(\;5 - 12 - (-8) + (-3)\).
Solution
Part 1 — \(-14 + 6\). Different signs: subtract magnitudes, \(14 - 6 = 8\), and take the sign of the larger magnitude, which is negative.
$$-14 + 6 = -8$$Part 2 — \(-14 - 6\). Rewrite as an addition first.
$$-14 + (-6) = -20$$Same signs, so the magnitudes add and the sign stays negative.
Part 3 — \(-14 - (-6)\). The opposite of \(-6\) is 6.
$$-14 + 6 = -8$$Part 4 — \(5 - 12 - (-8) + (-3)\). Rewrite every subtraction, then work left to right.
$$5 + (-12) + 8 + (-3)$$ $$5 + (-12) = -7, \qquad -7 + 8 = 1, \qquad 1 + (-3) = -2$$Answer: \(-8\), \(-20\), \(-8\), and \(-2\).
0.2.4 Multiplying and Dividing Integers
The sign rules
Multiplication and division follow the same, simpler pattern:
- Like signs give a positive result.
- Unlike signs give a negative result.
The magnitude never depends on the signs. In every line above the digits are the same; only the sign changes. That means you can compute the size and the sign as two separate jobs, and doing them separately is faster and safer than trying to track both at once.
| First factor | Second factor | Sign of product | Example |
|---|---|---|---|
| positive | positive | positive | \(4 \cdot 6 = 24\) |
| negative | negative | positive | \((-4)(-6) = 24\) |
| positive | negative | negative | \(4(-6) = -24\) |
| negative | positive | negative | \((-4)(6) = -24\) |
Division obeys the identical table, and that is not a coincidence. Every division statement can be checked by multiplying: \(\frac{-20}{4} = -5\) is correct precisely because \((-5)(4) = -20\). Since the sign rules for multiplication force which product lands where, they force the division answers along with them.
Why two negatives make a positive
Multiplying by a negative reverses direction. Start with a pattern of products and step down.
| Product | Value | Change from the row above |
|---|---|---|
| \((-3)(2)\) | \(-6\) | — |
| \((-3)(1)\) | \(-3\) | up 3 |
| \((-3)(0)\) | \(0\) | up 3 |
| \((-3)(-1)\) | \(3\) | up 3 |
| \((-3)(-2)\) | \(6\) | up 3 |
Each time the second factor decreases by 1, the product increases by 3. Continuing that consistent pattern past zero forces \((-3)(-1)\) to be \(+3\). The rule is not a convention chosen for convenience; it is the only value that keeps arithmetic consistent.
There is a second argument that uses no pattern at all, only the distributive property from Section 0.5 and the fact that anything times zero is zero. Watch what \((-3)(-5)\) is forced to be:
$$0 = (-3)(0) = (-3)\bigl(5 + (-5)\bigr) = (-3)(5) + (-3)(-5) = -15 + (-3)(-5)$$The whole line starts at 0 and ends at \(-15 + (-3)(-5)\), so \((-3)(-5)\) has to be whatever cancels the \(-15\). Only \(+15\) does that. Any other value would break the distributive property, and breaking it would wreck arithmetic everywhere else, so the sign rule is the price of keeping the rest of the system intact.
Multiplying by a negative flips direction on the number line, the way playing a video in reverse flips the action. Do it twice and you are running forward again. That is the whole content of "a negative times a negative is a positive."
Reading the products as repeated groups gives one more angle on the same fact, and it is often the one that finally lands. The product \((-3)(5)\) can be read as five groups of a $3 debt, which is $15 of debt, or \(-15\). Now read \((-3)(-5)\) as removing five groups of a $3 debt. Wiping out $15 of what you owe leaves you $15 better off than before, so the answer is \(+15\). The negative in the second factor is doing the same job the negative did in subtraction: it means take away rather than add on.
Counting the negative factors
For a longer product, count how many negative factors there are:
- An even number of negative factors gives a positive result.
- An odd number of negative factors gives a negative result.
Three negative factors is an odd count, so the product is negative.
The counting rule is just the pairing rule applied repeatedly. Each pair of negatives cancels to a positive, so an even count pairs off completely and leaves nothing negative behind, while an odd count leaves one unmatched negative to set the sign.
This extends to exponents, and connects back to §0.1.4:
$$(-2)^4 = 16 \qquad (-2)^3 = -8$$A negative base raised to an even power is positive; to an odd power, negative. And the placement caution still applies: \((-2)^4 = 16\) but \(-2^4 = -16\), because in the second the exponent attaches only to the 2.
Without finding the numeric value first, predict the sign of each result, then compute it: (a) \((-2)(3)(-5)\); (b) \((-1)(-2)(-3)(-4)(-5)\); (c) \(\dfrac{-72}{-9}\); (d) \((-3)^4\); (e) \(-3^4\).
Solution
(a) Count the negatives in \((-2)(3)(-5)\). There are two, an even count, so the product is positive. The magnitudes give \(2 \cdot 3 \cdot 5 = 30\).
$$(-2)(3)(-5) = 30$$(b) Count the negatives in \((-1)(-2)(-3)(-4)(-5)\). There are five, an odd count, so the product is negative. The magnitudes give \(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120\).
$$(-1)(-2)(-3)(-4)(-5) = -120$$(c) Look at the two signs in the quotient. Both are negative, which is a like pair, so the quotient is positive. The magnitudes give \(72 \div 9 = 8\).
$$\frac{-72}{-9} = 8$$(d) Read the base of \((-3)^4\). The parentheses make the base \(-3\), so four negative factors multiply, an even count, giving a positive result.
$$(-3)^4 = (-3)(-3)(-3)(-3) = 81$$(e) Read the base of \(-3^4\). With no parentheses, the exponent attaches only to the 3, and the minus sign is applied at the end.
$$-3^4 = -(3^4) = -81$$Parts (d) and (e) use the same digits and disagree by a sign, exactly as \(-4^2\) and \((-4)^2\) did in Section 0.1.
Answer: 30, \(-120\), 8, 81, and \(-81\).
Division by zero
Every division statement is a multiplication question in disguise. The quotient \(\frac{20}{4}\) is 5 because \(5 \cdot 4 = 20\). Put a zero in the denominator and that question stops having a good answer.
For every number \(a\), the quotient \(\dfrac{a}{0}\) is undefined — it names no number at all. An expression is undefined when no value satisfies what it asks for, or when more than one value does.
Asking for \(\frac{5}{0}\) means asking what number times 0 gives 5 — and every number times 0 gives 0, so no such number exists.
Zero divided by a nonzero number is fine: \(\frac{0}{5} = 0\), since \(0 \cdot 5 = 0\).
The case \(\frac{0}{0}\) fails for the opposite reason. It asks what number times 0 gives 0, and every number does. A quotient is supposed to name one specific value, so an answer that could be 3 or \(-17\) or 1,000 with equal justification is no answer at all. One case has no candidates and the other has too many, and both are undefined.
| Expression | Value | Why |
|---|---|---|
| \(\dfrac{0}{7}\) | \(0\) | \(0 \cdot 7 = 0\), and only 0 works |
| \(\dfrac{7}{0}\) | undefined | no number times 0 gives 7 |
| \(\dfrac{0}{0}\) | undefined | every number times 0 gives 0 |
| \(\dfrac{-7}{0}\) | undefined | the sign changes nothing; the denominator is still 0 |
This is not a rule to be worked around. An expression that requires dividing by zero has no value, and recognizing when a variable expression could produce a zero denominator matters in Chapter 3.
Definition 0.2.6 — Both 5/0 and 0/0 are undefined, for opposite reasons.
Order of operations with negatives
Nothing about the order of operations changes; only the arithmetic gets more careful. Evaluate \(\;-2(5 - 8)^2 + 3\).
| Step | Expression | Reason |
|---|---|---|
| 1 | \(-2(-3)^2 + 3\) | grouping first: \(5 - 8 = -3\) |
| 2 | \(-2(9) + 3\) | exponent: \((-3)^2 = 9\) |
| 3 | \(-18 + 3\) | multiplication |
| 4 | \(-15\) | addition |
Substituting values keeps the parentheses habit from §0.1.4. Evaluate \(\;x^2 - 4x\;\) when \(x = -5\):
$$(-5)^2 - 4(-5) = 25 - (-20) = 25 + 20 = 45$$There were two places to lose a sign here: the squaring, and the subtraction of a negative. Writing the substituted value in parentheses protects both.
Simplify \(\;\dfrac{-4(3 - 7)}{-2} - (-5)\).
Solution
Step 1 — The fraction bar groups. Simplify the numerator and denominator separately before dividing. Start inside the parentheses in the numerator.
$$3 - 7 = -4$$Step 2 — Finish the numerator. Multiply, using like signs for a positive product.
$$-4(-4) = 16$$Step 3 — Divide. Unlike signs give a negative quotient.
$$\frac{16}{-2} = -8$$Step 4 — Handle the remaining subtraction. Subtracting a negative adds.
$$-8 - (-5) = -8 + 5 = -3$$Answer: \(-3\).
Evaluate \(\;-x^2 + 3x - 8\) when \(x = -4\).
Solution
Step 1 — Substitute with parentheses. Every occurrence of \(x\) becomes \((-4)\), and the leading minus sign stays exactly where it was.
$$-(-4)^2 + 3(-4) - 8$$Step 2 — Exponent first. The parentheses make the base \(-4\), and an even power of a negative is positive.
$$(-4)^2 = 16$$So the first term is \(-(16) = -16\). The minus sign in front was never part of the base, so it is applied after squaring.
Step 3 — Multiply. Unlike signs give a negative product.
$$3(-4) = -12$$Step 4 — Assemble and work left to right.
$$-16 + (-12) - 8 = -28 - 8 = -28 + (-8) = -36$$Answer: \(-36\).
Simplify \(\;(-4)(-3)(-2)\), \(\;\dfrac{-45}{9}\), and \(\;-3(2 - 6)^2 + 10\), then evaluate \(\;x^2 - 5x\) when \(x = -3\).
Solution
Part 1 — \((-4)(-3)(-2)\). Three negative factors is an odd count, so the product is negative. The magnitudes give \(4 \cdot 3 \cdot 2 = 24\).
$$(-4)(-3)(-2) = -24$$Part 2 — \(\dfrac{-45}{9}\). Unlike signs give a negative quotient, and \(45 \div 9 = 5\).
$$\frac{-45}{9} = -5$$Part 3 — \(-3(2 - 6)^2 + 10\). Grouping first, then the exponent, then multiplication, then addition.
$$-3(-4)^2 + 10 = -3(16) + 10 = -48 + 10 = -38$$Part 4 — Evaluate \(x^2 - 5x\) at \(x = -3\). Substitute with parentheses around the \(-3\).
$$(-3)^2 - 5(-3) = 9 - (-15) = 9 + 15 = 24$$Answer: \(-24\), \(-5\), \(-38\), and 24.
Problem Set 0.2
Problem 1. Write an integer for each situation: (a) a debt of $150; (b) an elevation of 240 feet above sea level; (c) a loss of 12 yards on a football play.
Solution
Step 1 — Locate the built-in zero in each situation: As Table 0.2.1 shows, every one of these situations already has a natural zero — no debt, sea level, no yardage lost — and the integer records which side of that zero you land on.
(a) A debt of $150 is $150 below "owing nothing," so it is \(-150\).
(b) An elevation 240 feet above sea level is above that zero, so it is \(240\).
(c) A loss of 12 yards moves the ball backward from where the play started, so it is \(-12\).
Answer: (a) \(-150\); (b) \(240\); (c) \(-12\).
Problem 2. Order from least to greatest: \(-8,\; 3,\; -15,\; 0,\; -2,\; 11\).
Solution
Step 1 — Separate by sign: Negatives: \(-8,\; -15,\; -2\). Positive: \(3, 11\). Zero sits between the two groups.
Step 2 — Order the negatives: The negative farthest left (largest digit) is smallest, so \(-15\) comes before \(-8\), which comes before \(-2\).
Step 3 — Assemble the full list, negatives first, then zero, then positives in normal order:
$$-15,\; -8,\; -2,\; 0,\; 3,\; 11$$Answer: \(-15,\; -8,\; -2,\; 0,\; 3,\; 11\).
Problem 3. Order from greatest to least: \(-1,\; -19,\; -7,\; 5,\; -100\).
Solution
Step 1 — Separate by sign: Positive: \(5\). Negatives: \(-1,\; -19,\; -7,\; -100\).
Step 2 — Order the negatives from greatest to least: The negative closest to zero is greatest, so \(-1\) leads, then \(-7\), then \(-19\), then \(-100\) trails farthest left.
Step 3 — Put the positive first, since it beats every negative, then the ordered negatives:
$$5,\; -1,\; -7,\; -19,\; -100$$Answer: \(5,\; -1,\; -7,\; -19,\; -100\).
Problem 4. Fill in \(<\) or \(>\): \(-6 \;\underline{\phantom{xx}}\; -13\).
Solution
Step 1 — Locate both on the number line: \(-6\) sits six units left of zero; \(-13\) sits thirteen units left of zero, farther out.
Step 2 — Compare positions: A number is greater than every number to its left, and \(-6\) is to the right of \(-13\).
Answer: \(-6 > -13\).
Problem 5. Fill in \(<\) or \(>\): \(-40 \;\underline{\phantom{xx}}\; 1\).
Solution
Step 1 — Locate both: \(-40\) sits on the negative side of zero; \(1\) sits on the positive side.
Step 2 — Compare: Every negative is left of every positive, so \(-40\) is left of \(1\).
Answer: \(-40 < 1\).
Problem 6. Explain in one sentence why \(-2\) is greater than \(-9\).
Solution
Answer: \(-2\) sits closer to zero — and therefore farther right on the number line — than \(-9\) does, and a number is greater than anything to its left, so \(-2 > -9\) the same way a $2 debt beats a $9 debt.
Problem 7. Write the opposite of each number: 14, \(-23\), 0.
Solution
Step 1 — Fold the number line at zero for each value: The opposite of a number is its mirror image across zero, the same distance out on the other side.
$$\text{opposite of } 14 = -14 \qquad \text{opposite of } -23 = 23 \qquad \text{opposite of } 0 = 0$$Zero is its own mirror image, since it sits at distance zero either way.
Answer: \(-14\), \(23\), and \(0\).
Problem 8. Evaluate \(-a\) when \(a = 17\), and again when \(a = -17\).
Solution
Step 1 — Evaluate \(-a\) at \(a = 17\): Substituting gives the opposite of 17.
$$-a = -(17) = -17$$Step 2 — Evaluate \(-a\) at \(a = -17\): Now \(a\) is already negative, so its opposite lands on the positive side.
$$-a = -(-17) = 17$$Answer: \(-17\), then \(17\).
Problem 9. Simplify \(-(-31)\).
Solution
Step 1 — Read \(-(-31)\) as "the opposite of negative 31": Negative 31 sits 31 units left of zero, so its mirror image sits 31 units right of zero.
$$-(-31) = 31$$Answer: \(31\).
Problem 10. Simplify \(-(-(-8))\).
Solution
Step 1 — Work from the inside out: The innermost value is \(-8\), and the first opposite sign flips it back across zero.
$$-(-8) = 8$$Step 2 — Apply the outermost opposite sign:
$$-(-(-8)) = -(8) = -8$$Counting from 8, that is three flips in all, and an odd number of flips always leaves you on the opposite side from where you started.
Answer: \(-8\).
Problem 11. Explain in one sentence why \(-a\) is not always a negative number.
Solution
Answer: The expression \(-a\) means "flip the sign of \(a\)," and when \(a\) is already negative that flip lands on the positive side — for instance \(a = -5\) gives \(-a = 5\) — so \(-a\) is negative only when \(a\) itself is positive.
Problem 12. Simplify \(\lvert -19 \rvert\).
Solution
Step 1 — Read the bars as a distance: \(\lvert -19 \rvert\) asks how far \(-19\) sits from zero on the number line.
\(-19\) is 19 units to the left of zero, so the distance is 19.
Answer: \(\lvert -19 \rvert = 19\)
Problem 13. Simplify \(-\lvert -19 \rvert\).
Solution
Step 1 — Take the absolute value first: the bars only group what is inside them, so \(\lvert -19 \rvert\) is evaluated before anything else touches it.
$$\lvert -19 \rvert = 19$$Step 2 — Apply the outside minus sign: the negative sign sits outside the bars, so it is untouched by them and applies after.
$$-\lvert -19 \rvert = -(19) = -19$$Answer: \(-\lvert -19 \rvert = -19\)
Problem 14. Simplify \(\lvert 4 - 13 \rvert\).
Solution
Step 1 — Simplify inside the bars first: the bars are a grouping symbol, so \(4 - 13\) is computed before taking any distance.
$$4 - 13 = -9$$Step 2 — Take the absolute value of the result: \(-9\) sits 9 units from zero.
$$\lvert 4 - 13 \rvert = \lvert -9 \rvert = 9$$Answer: \(\lvert 4 - 13 \rvert = 9\)
Problem 15. Simplify \(\lvert 4 \rvert - \lvert 13 \rvert\).
Solution
Step 1 — Notice the bars enclose single numbers, not the whole expression: here each of \(4\) and \(13\) gets its own distance from zero before any subtracting happens.
$$\lvert 4 \rvert - \lvert 13 \rvert = 4 - 13$$Step 2 — Subtract:
$$4 - 13 = -9$$This is the same three symbols as problem 0.2.14, only with the bars in a different place, and the two problems land on opposite signs — proof that where the bars open and close is not cosmetic.
Answer: \(\lvert 4 \rvert - \lvert 13 \rvert = -9\)
Problem 16. Simplify \(\lvert -6 \rvert + \lvert -10 \rvert\).
Solution
Step 1 — Evaluate each absolute value separately: each bar pair encloses a single number, so each is its own distance from zero.
$$\lvert -6 \rvert = 6 \qquad \lvert -10 \rvert = 10$$Step 2 — Add the two distances:
$$6 + 10 = 16$$Answer: \(\lvert -6 \rvert + \lvert -10 \rvert = 16\)
Problem 17. Find every integer \(n\) with \(\lvert n \rvert = 11\).
Solution
Step 1 — Translate into distance: \(\lvert n \rvert = 11\) says \(n\) is exactly 11 units from zero.
Step 2 — Count in both directions: 11 units to the right of zero lands on 11, and 11 units to the left lands on \(-11\). No other integer is 11 units away.
$$n = 11 \quad \text{or} \quad n = -11$$Answer: \(n = 11\) or \(n = -11\)
Problem 18. Explain in one sentence why no number satisfies \(\lvert n \rvert = -2\).
Solution
Answer: Absolute value measures a distance, and a distance can never be negative, so no number \(n\) can sit \(-2\) units from zero, which is what \(\lvert n \rvert = -2\) would require.
Problem 19. Use absolute value to find the distance between \(-14\) and 5 on the number line.
Solution
Step 1 — Subtract the two numbers inside the bars: distance between two points on the number line is the absolute value of their difference.
$$-14 - 5 = -19$$Step 2 — Take the absolute value:
$$\lvert -14 - 5 \rvert = \lvert -19 \rvert = 19$$Subtracting in the other order gives the same distance, \(\lvert 5 - (-14) \rvert = \lvert 19 \rvert = 19\), which is exactly what a distance should do.
Answer: The distance between \(-14\) and 5 is 19 units.
Problem 20. Simplify \(-17 + 9\).
Solution
Step 1 — Compare the signs: \(-17\) and \(9\) have different signs, so subtract the smaller magnitude from the larger: \(17 - 9 = 8\). The result takes the sign of the larger magnitude, \(-17\), which is negative.
Answer: \(-17 + 9 = -8\)
Problem 21. Simplify \(-17 + (-9)\).
Solution
Step 1 — Same signs: Both \(-17\) and \(-9\) are negative, so add the magnitudes and keep the sign. \(17 + 9 = 26\), and the sign stays negative.
Answer: \(-17 + (-9) = -26\)
Problem 22. Simplify \(23 + (-31)\).
Solution
Step 1 — Different signs: \(23\) and \(-31\) point opposite ways, so subtract the smaller magnitude from the larger: \(31 - 23 = 8\). The larger magnitude belongs to \(-31\), so the result is negative.
Answer: \(23 + (-31) = -8\)
Problem 23. Rewrite \(6 - 19\) as an addition, then simplify it.
Solution
Step 1 — Rewrite as addition: Subtracting 19 is the same as adding its opposite.
$$6 - 19 = 6 + (-19)$$Step 2 — Combine: The addends have different signs, so subtract magnitudes: \(19 - 6 = 13\), and take the sign of the larger magnitude, \(-19\).
Answer: \(6 - 19 = 6 + (-19) = -13\)
Problem 24. Rewrite \(-6 - (-19)\) as an addition, then simplify it.
Solution
Step 1 — Rewrite as addition: The opposite of \(-19\) is \(19\), so subtracting \(-19\) becomes adding \(19\).
$$-6 - (-19) = -6 + 19$$Step 2 — Combine: Different signs, so subtract magnitudes: \(19 - 6 = 13\). The larger magnitude, \(19\), is positive, so the result is positive.
Answer: \(-6 - (-19) = -6 + 19 = 13\)
Problem 25. Simplify \(-10 - 4 - (-7)\).
Solution
Step 1 — Rewrite the subtraction as an addition: Only one minus sign in the expression is a true subtraction — of \(-7\) — so it becomes adding \(7\); the rest were already additions of negatives.
$$-10 - 4 - (-7) = -10 + (-4) + 7$$Step 2 — Combine the first two terms: Same signs, so add the magnitudes and keep the sign.
$$-10 + (-4) = -14$$Step 3 — Add the last term: Different signs, so subtract magnitudes: \(14 - 7 = 7\), taking the sign of the larger magnitude, \(-14\).
$$-14 + 7 = -7$$Answer: \(-10 - 4 - (-7) = -7\)
Problem 26. Simplify \(8 - 15 - (-3) + (-11)\).
Solution
Step 1 — Rewrite every subtraction as an addition: There are two subtraction signs to convert; the \(+(-11)\) is already an addition.
$$8 - 15 - (-3) + (-11) = 8 + (-15) + 3 + (-11)$$Step 2 — Work left to right. First, \(8 + (-15)\): different signs, \(15 - 8 = 7\), sign of the larger magnitude \(-15\).
$$8 + (-15) = -7$$Step 3 — Continue: \(-7 + 3\): different signs, \(7 - 3 = 4\), sign of the larger magnitude \(-7\).
$$-7 + 3 = -4$$Step 4 — Finish: \(-4 + (-11)\): same signs, add magnitudes and keep negative.
$$-4 + (-11) = -15$$Step 5 — Check by grouping: The positive terms total \(8 + 3 = 11\); the negative terms total \(15 + 11 = 26\). Since \(26 - 11 = 15\) with the negatives larger, the sum is \(-15\), which agrees.
Answer: \(8 - 15 - (-3) + (-11) = -15\)
Problem 27. At 5 a.m. the temperature was \(-13^\circ\)F, and by noon it was \(6^\circ\)F. How many degrees did it rise?
Solution
Step 1 — Set up the change: A rise is the ending temperature minus the starting temperature.
$$6 - (-13)$$Step 2 — Rewrite as an addition: The opposite of \(-13\) is \(13\).
$$6 - (-13) = 6 + 13 = 19$$Answer: The temperature rose \(19^\circ\)F.
Problem 28. An account holding $85 is charged a $120 payment and then receives a $40 deposit. Write the balance as a signed number.
Solution
Step 1 — Write each event as a signed number. The starting balance is positive; the payment charged is subtracted, and the deposit is added.
$$85 - 120 + 40$$Step 2 — Rewrite the subtraction as an addition, then combine the first two terms: Different signs, so subtract magnitudes: \(120 - 85 = 35\), taking the sign of the larger magnitude, \(-120\).
$$85 + (-120) = -35$$Step 3 — Add the deposit: Different signs again: \(40 - 35 = 5\), and this time the larger magnitude, \(40\), is positive.
$$-35 + 40 = 5$$Step 4 — Interpret the sign: A positive balance means the account holds money rather than owing it.
Answer: The balance is \(\$5\) (i.e., \(+5\)) — the account is $5 in the black, not overdrawn.
Problem 29. Simplify \((-7)(4)\).
Solution
Step 1 — Identify the signs. One factor is negative and one is positive — unlike signs, so the product is negative.
Step 2 — Multiply the magnitudes. \(7 \cdot 4 = 28\).
$$(-7)(4) = -28$$Answer: \(-28\).
Problem 30. Simplify \((-7)(-4)\).
Solution
Step 1 — Both factors are negative. Like signs, so the product is positive.
Step 2 — Multiply the magnitudes. \(7 \cdot 4 = 28\).
$$(-7)(-4) = 28$$Answer: 28.
Problem 31. Simplify \((-2)(5)(-3)\).
Solution
Step 1 — Count the negative factors. Two of the three factors, \(-2\) and \(-3\), are negative — an even count, so the product is positive.
Step 2 — Multiply the magnitudes. \(2 \cdot 5 \cdot 3 = 30\).
$$(-2)(5)(-3) = 30$$Answer: 30.
Problem 32. Simplify \((-1)(-2)(-3)(-4)\).
Solution
Step 1 — Count the negative factors. All four factors are negative — an even count, so the product is positive.
Step 2 — Multiply the magnitudes. \(1 \cdot 2 \cdot 3 \cdot 4 = 24\).
$$(-1)(-2)(-3)(-4) = 24$$Answer: 24.
Problem 33. Simplify \(\dfrac{-56}{-8}\).
Solution
Step 1 — Numerator and denominator are both negative. Like signs, so the quotient is positive.
Step 2 — Divide the magnitudes. \(56 \div 8 = 7\).
$$\frac{-56}{-8} = 7$$Answer: 7.
Problem 34. Simplify \(\dfrac{36}{-6}\).
Solution
Step 1 — The numerator is positive and the denominator is negative. Unlike signs, so the quotient is negative.
Step 2 — Divide the magnitudes. \(36 \div 6 = 6\).
$$\frac{36}{-6} = -6$$Answer: \(-6\).
Problem 35. State whether each is defined, and give its value when it is: (a) \(\dfrac{0}{9}\); (b) \(\dfrac{9}{0}\).
Solution
Step 1 — Part (a), zero divided by a nonzero number. The question \(\dfrac{0}{9}\) asks what number times 9 gives 0. Only 0 does, so the quotient is defined and equals that number.
$$\frac{0}{9} = 0$$Step 2 — Part (b), a nonzero number divided by zero. The question \(\dfrac{9}{0}\) asks what number times 0 gives 9. Every number times 0 gives 0, never 9, so no number satisfies it.
Answer: (a) defined, \(\dfrac{0}{9} = 0\); (b) undefined.
Problem 36. Explain in one sentence why \((-5)^2\) and \(-5^2\) have different values, and give both values.
Solution
Step 1 — Explain the difference. In \((-5)^2\) the parentheses make \(-5\) the base that gets squared, but in \(-5^2\) there are no parentheses around the \(-5\), so the exponent attaches only to the 5 and the negative sign is applied after squaring.
Step 2 — Evaluate \((-5)^2\). The base is \(-5\), and an even power of a negative number is positive.
$$(-5)^2 = (-5)(-5) = 25$$Step 3 — Evaluate \(-5^2\). Square 5 first, then apply the minus sign.
$$-5^2 = -(5^2) = -(25) = -25$$Answer: \((-5)^2 = 25\) and \(-5^2 = -25\); they differ because only the first has the negative sign inside the squared base.
Problem 37. Simplify \((-2)^5\).
Solution
Step 1 — Read the base and exponent. The parentheses make the base \(-2\), raised to the 5th power. Five is odd, so the result is negative.
Step 2 — Compute the magnitude. \(2^5 = 32\).
$$(-2)^5 = -32$$Answer: \(-32\).
Problem 38. Without computing the value, state the sign of \((-1)(-3)(-5)(-7)(-9)(-11)\) and explain how you know.
Solution
Step 1 — Count the negative factors. All six factors, \(-1,\, -3,\, -5,\, -7,\, -9,\, -11\), are negative.
Step 2 — Apply the counting rule. Six is an even count, and an even number of negative factors always gives a positive result, so the sign is settled without ever finding the magnitude.
Answer: The product is positive, because it has an even number (six) of negative factors.
Problem 39. Simplify \(-4(3 - 9)^2 + 20\).
Solution
Step 1 — Grouping first. Simplify inside the parentheses.
$$3 - 9 = -6$$Step 2 — Exponent next. An even power of a negative number is positive.
$$(-6)^2 = 36$$Step 3 — Multiply. Unlike signs give a negative product.
$$-4(36) = -144$$Step 4 — Add.
$$-144 + 20 = -124$$Answer: \(-124\).
Problem 40. Evaluate \(x^2 - 6x\) when \(x = -2\).
Solution
Step 1 — Substitute with parentheses. Every occurrence of \(x\) becomes \((-2)\).
$$(-2)^2 - 6(-2)$$Step 2 — Exponent first. An even power of a negative number is positive.
$$(-2)^2 = 4$$Step 3 — Multiply. Unlike signs give a negative product.
$$6(-2) = -12$$Step 4 — Subtract, rewriting as an addition. Subtracting a negative adds.
$$4 - (-12) = 4 + 12 = 16$$Answer: 16.
Problem 41. Evaluate \(-x^2 + 4x - 1\) when \(x = -3\).
Solution
Step 1 — Substitute with parentheses. Every \(x\) becomes \((-3)\). The leading minus sign in \(-x^2\) is not part of the base, so it stays exactly where it was.
$$-(-3)^2 + 4(-3) - 1$$Step 2 — Exponent first. The parentheses around \(-3\) make it the base being squared; the minus sign out front is applied afterward.
$$(-3)^2 = 9, \qquad -(-3)^2 = -9$$Step 3 — Multiply. Unlike signs give a negative product.
$$4(-3) = -12$$Step 4 — Assemble and work left to right.
$$-9 + (-12) - 1 = -21 - 1 = -22$$Answer: \(-22\).
Problem 42. Simplify \(\dfrac{-3(8 - 12)}{-6} - (-4)\).
Solution
Step 1 — The fraction bar groups the numerator. Simplify inside the parentheses first.
$$8 - 12 = -4$$Step 2 — Finish the numerator. Like signs give a positive product.
$$-3(-4) = 12$$Step 3 — Divide. Unlike signs give a negative quotient.
$$\frac{12}{-6} = -2$$Step 4 — Handle the remaining subtraction. Subtracting a negative adds.
$$-2 - (-4) = -2 + 4 = 2$$Answer: 2.
Key Terms
integer — a whole number or the negative of a whole number: …, \(-2\), \(-1\), 0, 1, 2, ….
number line — a line with a marked zero and a fixed unit distance, positives to the right and negatives to the left.
opposite — the number the same distance from zero as a given number but on the other side of zero.
additive inverse — another name for the opposite; a number and its additive inverse sum to zero.
absolute value — the distance a number sits from zero on the number line, written \(\lvert a \rvert\), and never negative.
magnitude — the size of a number with its sign ignored; the value its absolute value reports.
undefined — describes an expression that names no number, as division by zero does.