0.5 Properties of Real Numbers
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:
- use the commutative and associative properties, and explain which operations do not have them;
- rearrange and regroup a calculation on purpose to make the arithmetic easier;
- identify the additive and multiplicative identities, and find the opposite and the reciprocal of a number;
- use the distributive property forwards to expand and backwards to factor;
- distribute a negative number without losing a sign;
- state the properties of zero, including why division by zero has no value.
You have been using the properties in this section for years without naming them. When you add a column of numbers and start with the pair that makes ten, you are using the commutative and associative properties. When you multiply \(7(x + 2)\) and get \(7x + 14\), you are using the distributive property. When you rewrite \(\dfrac{2}{5}\) as \(\dfrac{6}{15}\), you are using the multiplicative identity. None of that is new to you. What is new is the naming.
Naming them matters because algebra eventually asks a harder question than "what is the answer?" It asks "why are you allowed to do that?" An expression can be rewritten in a hundred different-looking ways, and only some of those rewritings keep its value the same. The properties below are the list of moves that are always safe. Every other rule you meet for the rest of this course is built out of them.
This pays off in Chapter 3. Solving an equation is a chain of rewritings, and each link in the chain needs a reason. When you write out a solution with a reason beside every step, the reasons will almost always be the property names from this section. Learning them now turns Chapter 3 into bookkeeping instead of guesswork.
0.5.1 Commutative and Associative Properties
Two questions organize everything here. Does the order of the numbers matter? Does the grouping of the numbers matter? For addition and multiplication the answer to both is no, and that freedom is worth a lot. For subtraction and division the answer to both is yes, and forgetting that is a reliable way to get a wrong answer.
For any real numbers \(a\) and \(b\),
$$a + b = b + a \qquad \text{and} \qquad a \cdot b = b \cdot a$$Changing the order of the numbers does not change the sum or the product.
The word commute means to travel back and forth, which is a decent picture of what is happening: the two numbers swap seats and nothing else changes. Check it once with real numbers.
$$9 + 4 = 13 \qquad \text{and} \qquad 4 + 9 = 13$$ $$6 \cdot 7 = 42 \qquad \text{and} \qquad 7 \cdot 6 = 42$$A property claims something about every pair of numbers, so a single pair where it fails is enough to sink it. That is why the two lines below matter so much: you do not need to test subtraction on a hundred pairs to know it is not commutative. One pair does the job.
Now try the same swap on subtraction and on division. This is the half of the property students skip, and it costs them points for years.
$$11 - 4 = 7 \qquad \text{but} \qquad 4 - 11 = -7$$ $$18 \div 6 = 3 \qquad \text{but} \qquad 6 \div 18 = \dfrac{1}{3}$$Since \(7\) and \(-7\) are different, and \(3\) and \(\dfrac{1}{3}\) are different, order clearly matters for both. Subtraction and division are not commutative.
Which operations let you swap. The table below collects the four tests:
| Operation | Commutative? | The test | What happened |
|---|---|---|---|
| Addition | yes | \(9 + 4 = 13\) and \(4 + 9 = 13\) | same result |
| Multiplication | yes | \(6 \cdot 7 = 42\) and \(7 \cdot 6 = 42\) | same result |
| Subtraction | no | \(11 - 4 = 7\) and \(4 - 11 = -7\) | opposite results |
| Division | no | \(18 \div 6 = 3\) and \(6 \div 18 = \tfrac{1}{3}\) | reciprocal results |
This is the same warning Section 0.1 gave about translating words. "Five less than \(n\)" is \(n - 5\) and not \(5 - n\) exactly because subtraction is not commutative. If it were, the distinction would not exist and nobody would have to be careful.
There is a way to get the freedom back. Section 0.2 showed that any subtraction can be rewritten as an addition, \(a - b = a + (-b)\). Once you have done that, you are adding, so you may reorder as much as you like.
$$11 - 4 = 11 + (-4) = -4 + 11 = 7$$What you may not do is leave the minus sign behind. The term is \(-4\), and its sign travels with it wherever it goes.
For any real numbers \(a\), \(b\), and \(c\),
$$(a + b) + c = a + (b + c) \qquad \text{and} \qquad (ab)c = a(bc)$$The commutative property lets the numbers change seats: \(3 + 8\) becomes \(8 + 3\). The associative property leaves them in the same seats and moves the parentheses around them, like moving a fence: \((3 + 8) + 5\) becomes \(3 + (8 + 5)\). Order versus grouping — two different freedoms.
Definition 0.5.1 — Commutative Property: swapping the two numbers leaves a sum or a product alone, and changes a difference or a quotient.
Changing the grouping of the numbers does not change the sum or the product.
Notice that the order stays exactly the same in both statements. Only the parentheses move. Here it is with numbers:
$$(13 + 27) + 8 = 40 + 8 = 48 \qquad \text{and} \qquad 13 + (27 + 8) = 13 + 35 = 48$$ $$(4 \cdot 25) \cdot 17 = 100 \cdot 17 = 1700 \qquad \text{and} \qquad 4 \cdot (25 \cdot 17) = 4 \cdot 425 = 1700$$Both routes land in the same place, but the first route in each pair is much easier to walk. That is the practical payoff, and we will use it deliberately in a moment.
Subtraction and division fail this one too:
$$(20 - 6) - 5 = 14 - 5 = 9 \qquad \text{but} \qquad 20 - (6 - 5) = 20 - 1 = 19$$ $$(48 \div 8) \div 2 = 6 \div 2 = 3 \qquad \text{but} \qquad 48 \div (8 \div 2) = 48 \div 4 = 12$$Order and grouping side by side. Here is the whole subsection in four rows:
| Property | What is free to move | Addition | Multiplication | Subtraction | Division |
|---|---|---|---|---|---|
| Commutative | the order | holds | holds | fails | fails |
| Associative | the grouping | holds | holds | fails | fails |
Because grouping is free for addition and multiplication, we can write \(a + b + c\) and \(abc\) with no parentheses at all — every way of inserting them gives the same value, so none of them is needed. That is not true of \(a - b - c\), which the order of operations settles left to right by convention, not by property.
Definition 0.5.2 — Associative Property: the fence moves, the numbers stay where they are, and the total holds.
Using both properties on purpose
Put the two properties together and they say something useful: a sum can be shuffled and regrouped any way you like, and so can a product. Good mental arithmetic is mostly the habit of hunting for the convenient arrangement before you compute anything.
Look for a pair that makes a round number:
$$37 + 46 + 63 = 37 + 63 + 46 = (37 + 63) + 46 = 100 + 46 = 146$$Or a pair of factors that makes a power of ten:
$$8 \cdot 19 \cdot 25 = 8 \cdot 25 \cdot 19 = (8 \cdot 25) \cdot 19 = 200 \cdot 19 = 3800$$Rearrangements worth looking for. Each row rewrites the calculation into one you can do in your head:
| As written | Rearranged | Why that helps | Result |
|---|---|---|---|
| \(58 + 27 + 42\) | \((58 + 42) + 27\) | the first pair makes 100 | \(127\) |
| \(5 \cdot 39 \cdot 20\) | \((5 \cdot 20) \cdot 39\) | the first pair makes 100 | \(3900\) |
| \(-16 + 91 + 16\) | \((-16 + 16) + 91\) | the first pair cancels to 0 | \(91\) |
| \(\tfrac{3}{4} \cdot 23 \cdot \tfrac{4}{3}\) | \(\left(\tfrac{3}{4} \cdot \tfrac{4}{3}\right) \cdot 23\) | the first pair multiplies to 1 | \(23\) |
The last two rows are a preview of Section 0.5.2. Pairing a number with its opposite, or with its reciprocal, makes the pair vanish, and that is the most useful rearrangement of all.
The same moves work when letters are involved, which is how simplification actually gets done.
Simplify \(\;12 + (n + 5)\), naming the property behind each step.
Solution
Step 1 — Swap the order inside the parentheses. We want the two numbers next to each other, so move the 5 in front of the \(n\). This is the commutative property of addition.
$$12 + (5 + n)$$Step 2 — Move the parentheses. Regroup so that 12 and 5 are inside the same set. This is the associative property of addition.
$$(12 + 5) + n$$Step 3 — Add the constants. Now the two numbers are together and we can just add them.
$$17 + n$$We could not have added 12 and 5 in the original expression, because the parentheses grouped the 5 with the \(n\) instead. The two properties are what let us take that grouping apart and rebuild it differently.
Answer: \(17 + n\).
Simplify \(\;(y + 9) + 14\), and use the properties to compute \(\;25 \cdot 17 \cdot 4\) mentally.
Solution
Part 1 — Simplify \((y + 9) + 14\).
Regroup so the two numbers sit together. That is the associative property of addition:
$$y + (9 + 14)$$Add the constants:
$$y + 23$$Part 2 — Compute \(25 \cdot 17 \cdot 4\).
Swap the order so 25 and 4 are neighbors. That is the commutative property of multiplication:
$$25 \cdot 4 \cdot 17$$Regroup them together, which is the associative property of multiplication, and multiply the easy pair first:
$$(25 \cdot 4) \cdot 17 = 100 \cdot 17 = 1700$$Answer: \(y + 23\) and \(1700\).
0.5.2 Identity and Inverse Properties
Some numbers do nothing at all. That sounds useless, and it is actually one of the most useful facts in algebra.
The number \(0\) is the additive identity: for any real number \(a\),
$$a + 0 = a \qquad \text{and} \qquad 0 + a = a$$Definition 0.5.3 — Additive Identity: zero is the jump that goes nowhere, and the place a number and its opposite land.
The number \(1\) is the multiplicative identity: for any real number \(a\),
$$a \cdot 1 = a \qquad \text{and} \qquad 1 \cdot a = a$$Adding zero leaves a number alone. Multiplying by one leaves a number alone. Each of these is the one number that does nothing to its own operation.
$$-14 + 0 = -14 \qquad 0 + \dfrac{7}{9} = \dfrac{7}{9} \qquad 62 \cdot 1 = 62 \qquad 1 \cdot (-3.5) = -3.5$$Notice that 0 is not the multiplicative identity. Multiplying by zero does not leave a number alone, it wipes the number out, and that is Section 0.5.4's business.
An identity is a place to land. An inverse is the thing that gets you there.
Definition 0.5.4 — Multiplicative Identity: one is the factor that moves nothing, and where a number meets its reciprocal.
The additive inverse of a real number \(a\) is \(-a\), the number that adds to it to give the additive identity:
$$a + (-a) = 0$$The multiplicative inverse of a real number \(a\), where \(a \neq 0\), is \(\dfrac{1}{a}\), the number that multiplies it to give the multiplicative identity:
$$a \cdot \dfrac{1}{a} = 1$$The additive inverse is the opposite from Section 0.2: the same distance from zero, on the other side. The multiplicative inverse is the reciprocal: flip the fraction over.
$$11 + (-11) = 0 \qquad \dfrac{4}{9} \cdot \dfrac{9}{4} = 1$$To find a reciprocal, write the number as a fraction and turn it upside down. A whole number \(n\) is really \(\dfrac{n}{1}\), so its reciprocal is \(\dfrac{1}{n}\).
Both inverses for the same number. Read across each row and check that the pairs really do give 0 and 1:
| Number | Additive inverse (opposite) | Multiplicative inverse (reciprocal) | Check |
|---|---|---|---|
| \(8\) | \(-8\) | \(\dfrac{1}{8}\) | \(8 + (-8) = 0\); \(8 \cdot \tfrac{1}{8} = 1\) |
| \(-5\) | \(5\) | \(-\dfrac{1}{5}\) | \(-5 + 5 = 0\); \(-5 \cdot \left(-\tfrac{1}{5}\right) = 1\) |
| \(\dfrac{4}{9}\) | \(-\dfrac{4}{9}\) | \(\dfrac{9}{4}\) | \(\tfrac{4}{9} \cdot \tfrac{9}{4} = 1\) |
| \(-\dfrac{2}{7}\) | \(\dfrac{2}{7}\) | \(-\dfrac{7}{2}\) | \(-\tfrac{2}{7} \cdot \left(-\tfrac{7}{2}\right) = 1\) |
| \(1\) | \(-1\) | \(1\) | 1 is its own reciprocal |
| \(0\) | \(0\) | none | see below |
Two of those rows deserve a comment. A reciprocal always keeps the sign of the original number, so the reciprocal of \(-5\) is \(-\dfrac{1}{5}\) and not \(\dfrac{1}{5}\) — a negative times a positive is negative, which could never come out to \(+1\). And 0 is its own opposite, which is another way of saying that zero is neither positive nor negative.
That last row is not an oversight. There is no number you can multiply by 0 to get 1, because every product with 0 is 0. The reciprocal of zero would have to be \(\dfrac{1}{0}\), and Section 0.2 already showed that this is undefined.
Dividing a nonzero number by zero is undefined because no number times 0 gives a nonzero result. Zero has no reciprocal because no number times 0 gives 1. Those are one fact seen from two sides, and it is why the multiplicative inverse carries the condition \(a \neq 0\).
Watch for that condition. The rest of algebra is full of it, and every time it appears it traces back to this one place.
Any fraction with matching top and bottom equals 1: \(\dfrac{5}{5}\), \(\dfrac{7}{7}\), \(\dfrac{12}{12}\). Multiplying by one of those changes how a fraction looks without changing what it is worth, the way trading four quarters for a dollar changes the coins but not the money.
That disguise is the whole reason we can build equivalent fractions. Because \(\dfrac{c}{c} = 1\) for any nonzero \(c\), multiplying by \(\dfrac{c}{c}\) is really multiplying by 1:
$$\dfrac{3}{8} = \dfrac{3}{8} \cdot 1 = \dfrac{3}{8} \cdot \dfrac{5}{5} = \dfrac{15}{40}$$The value never moved. It got multiplied by 1. That single line justifies every common denominator you have ever found.
Choosing the right disguised 1. Pick the fraction that turns the denominator you have into the denominator you want:
| Goal | Disguised 1 to use | Result |
|---|---|---|
| Write \(\dfrac{5}{6}\) with denominator 42 | \(\dfrac{7}{7}\) | \(\dfrac{35}{42}\) |
| Write \(\dfrac{2}{9}\) with denominator 45 | \(\dfrac{5}{5}\) | \(\dfrac{10}{45}\) |
| Write \(7\) as a fraction with denominator 3 | \(\dfrac{3}{3}\) | \(\dfrac{21}{3}\) |
| Add \(\dfrac{1}{4} + \dfrac{1}{6}\) | \(\dfrac{3}{3}\) and \(\dfrac{2}{2}\) | \(\dfrac{3}{12} + \dfrac{2}{12} = \dfrac{5}{12}\) |
Reducing a fraction is the same move run backwards. In \(\dfrac{15}{40} = \dfrac{3 \cdot 5}{8 \cdot 5} = \dfrac{3}{8} \cdot \dfrac{5}{5} = \dfrac{3}{8}\), nothing mysterious is being cancelled. You are peeling off a factor of 1.
Find the additive inverse and the multiplicative inverse of \(-\dfrac{3}{10}\), and verify each one.
Solution
Step 1 — Additive inverse. The additive inverse is the opposite, so flip the sign.
$$-\dfrac{3}{10} \quad \longrightarrow \quad \dfrac{3}{10}$$Verify that the pair adds to the additive identity:
$$-\dfrac{3}{10} + \dfrac{3}{10} = 0$$Step 2 — Multiplicative inverse. The multiplicative inverse is the reciprocal, so turn the fraction over. The sign stays negative, since a negative times a positive could never give \(+1\).
$$-\dfrac{3}{10} \quad \longrightarrow \quad -\dfrac{10}{3}$$Verify that the pair multiplies to the multiplicative identity:
$$-\dfrac{3}{10} \cdot \left(-\dfrac{10}{3}\right) = \dfrac{30}{30} = 1$$Answer: additive inverse \(\dfrac{3}{10}\); multiplicative inverse \(-\dfrac{10}{3}\).
Find the reciprocal of \(-6\), and rewrite \(\dfrac{4}{7}\) with a denominator of 63.
Solution
Part 1 — The reciprocal of \(-6\).
Write \(-6\) as a fraction, then turn it over. The sign stays with it:
$$-6 = -\dfrac{6}{1} \quad \longrightarrow \quad -\dfrac{1}{6}$$Check by multiplying:
$$-6 \cdot \left(-\dfrac{1}{6}\right) = 1$$Part 2 — Rewrite \(\dfrac{4}{7}\) with denominator 63.
Ask what turns 7 into 63. Since \(7 \cdot 9 = 63\), the disguised 1 we need is \(\dfrac{9}{9}\):
$$\dfrac{4}{7} \cdot \dfrac{9}{9} = \dfrac{36}{63}$$Answer: \(-\dfrac{1}{6}\) and \(\dfrac{36}{63}\).
0.5.3 The Distributive Property
The last two subsections handled one operation at a time. This one is the bridge between multiplication and addition, and it is the property you will use more than any other.
For any real numbers \(a\), \(b\), and \(c\),
$$a(b + c) = ab + ac \qquad \text{and} \qquad (b + c)a = ba + ca$$Multiplying a sum by a number gives the same result as multiplying each term of the sum and then adding.
The second form is not a new property. Because multiplication is commutative, the outside factor may sit on either side of the parentheses and the result is the same. That right-hand version is the one you will use when you factor a variable out of a sum such as \(5w + 8w\).
The outside factor gets handed out — distributed — to everything inside the parentheses.
$$6(x + 4) = 6 \cdot x + 6 \cdot 4 = 6x + 24$$Picture a rectangle 6 units tall whose width is split into a piece of length \(x\) and a piece of length 4. You can measure its area as one rectangle, \(6(x + 4)\), or as two rectangles added together, \(6x + 24\). Same rectangle, so the two expressions have to be equal.
Test it on numbers where you can compute both sides. Take \(9(20 + 3)\). By the order of operations that is \(9 \cdot 23 = 207\). Distributing instead gives \(9 \cdot 20 + 9 \cdot 3 = 180 + 27 = 207\). Same answer, different route — and that second route is how most people multiply in their heads, so you have been distributing for years without a name for it.
That picture is worth carrying around, because it also explains the most common way to get this wrong. Writing \(6(x + 4) = 6x + 4\) would mean measuring the second rectangle as if it were only 4 square units instead of \(6 \cdot 4 = 24\). The height applies to both pieces, so every term inside the parentheses gets multiplied. No exceptions.
Subtraction needs no new rule, since subtracting is adding the opposite:
$$a(b - c) = ab - ac$$ $$7(2y - 5) = 7 \cdot 2y - 7 \cdot 5 = 14y - 35$$If you would rather not track a minus sign, rewrite first: \(7(2y - 5) = 7\bigl(2y + (-5)\bigr) = 14y + (-35) = 14y - 35\). Same answer, one extra line, far fewer sign errors while you are learning.
Definition 0.5.7 — Distributive Property: one rectangle cut in two, so the outside factor reaches both terms.
Distributing a negative
This is where the real mistakes live. When the factor outside the parentheses is negative, every sign inside changes.
$$-3(m - 8) = -3 \cdot m + (-3)(-8) = -3m + 24$$The second term comes out positive, because a negative times a negative is positive, exactly as Section 0.2 said. The single most common error in beginning algebra is distributing the 3 and leaving the minus sign behind, which produces \(-3m - 24\).
When you distribute \(-3\), the thing being handed out is \(-3\), not 3. Circle the sign together with the number before you start, and multiply that whole package by each term inside. Most sign errors here are really bookkeeping errors, not arithmetic errors.
A minus sign sitting in front of a parenthesis is a multiplication by \(-1\), even though nobody writes the 1. So \(-(4k - 9) = -1(4k - 9) = -4k + 9\). In practice: a minus sign in front of a parenthesis flips every sign inside.
Four ways this goes wrong. Compare the correct column against what students most often write:
| Expression | Correct | Common error | What went wrong |
|---|---|---|---|
| \(-3(m - 8)\) | \(-3m + 24\) | \(-3m - 24\) | the minus sign was not distributed |
| \(-5(2k + 3)\) | \(-10k - 15\) | \(-10k + 15\) | sign dropped from the second term |
| \(-(p - 6)\) | \(-p + 6\) | \(-p - 6\) | the invisible \(-1\) was not distributed |
| \(4 - 2(x + 1)\) | \(4 - 2x - 2\) | \(2(x + 1)\) | subtracted before distributing |
That last row is a different mistake, and a sneaky one. The \(4 -\) out front does not combine with the 2 first. Multiplication outranks subtraction, so the \(-2\) has to be distributed before anything gets combined.
Reading it backwards
Every property is an equation, and an equation can be read in either direction. Read right to left, the distributive property pulls a shared factor out front:
$$ab + ac = a(b + c)$$ $$12n + 18 = 6 \cdot 2n + 6 \cdot 3 = 6(2n + 3)$$A common factor of an expression is a factor that appears in every term. Rewriting \(ab + ac\) as \(a(b + c)\) is called factoring out the common factor.
Factoring is one of the two directions you will travel constantly, and Chapter 7 builds a whole unit on it.
Pulling out the common factor. Check each row by distributing back and confirming you get the original:
| Expression | Common factor | Factored form |
|---|---|---|
| \(12n + 18\) | \(6\) | \(6(2n + 3)\) |
| \(15t - 25\) | \(5\) | \(5(3t - 5)\) |
| \(8h + 8\) | \(8\) | \(8(h + 1)\) |
| \(-6r - 21\) | \(-3\) | \(-3(2r + 7)\) |
The third row is worth a second look. Factoring 8 out of \(8h + 8\) leaves a 1 behind, not nothing, because that second term is \(8 \cdot 1\). Writing \(8(h)\) would be wrong, and distributing back catches it immediately.
Section 0.1 claimed that \(5w + 8w = 13w\) and promised an explanation later. Here it is:
$$5w + 8w = (5 + 8)w = 13w$$That middle step is the distributive property read right to left, with \(w\) as the common factor. Combining like terms is not a separate rule you have to memorize — it is factoring out the variable part, adding the coefficients, and putting the variable back.
It also explains the restriction. Why will \(9x^2\) and \(8x\) not combine? Because factoring out the variable part leaves \(x(9x + 8)\), and that parenthesis still has a variable in it. Nothing collapses into one term. Like terms combine exactly when pulling out the variable part leaves a plain numerical sum behind.
Definition 0.5.8 — Common Factor: the shared factor lifts out of every term and stands in front.
Simplify \(\;-2(5a - 3) + 7a\), naming the property behind each step.
Solution
Step 1 — Distribute the \(-2\). Hand out the whole factor, sign included, to both terms inside. This is the distributive property.
$$-10a + 6 + 7a$$Step 2 — Move the like terms together. Reorder the sum so the two \(a\) terms are neighbors. This is the commutative property of addition.
$$-10a + 7a + 6$$Step 3 — Factor out the \(a\). This is the distributive property read backwards.
$$(-10 + 7)a + 6$$Step 4 — Add the coefficients.
$$-3a + 6$$Answer: \(-3a + 6\).
Simplify \(\;4(2x + 5) - 3(x - 6)\).
Solution
Step 1 — Distribute the 4. Multiply both terms in the first parentheses by 4.
$$8x + 20 - 3(x - 6)$$Step 2 — Distribute the \(-3\). The factor is \(-3\), not 3, so the second term becomes \(-3 \cdot (-6) = +18\).
$$8x + 20 - 3x + 18$$Step 3 — Group the like terms. Reorder so the \(x\) terms sit together and the constants sit together, by the commutative property of addition.
$$8x - 3x + 20 + 18$$Step 4 — Factor out the \(x\) and add the constants. Pulling out the \(x\) is the distributive property backwards.
$$(8 - 3)x + 38$$Step 5 — Finish the arithmetic.
$$5x + 38$$Every line here had a reason, and every reason was a property from this section. That is the format Chapter 3 will ask you to produce for equations.
Answer: \(5x + 38\).
Simplify \(\;-4(3b - 2) + 5b\), then factor \(\;21c - 14\).
Solution
Part 1 — Simplify \(-4(3b - 2) + 5b\).
Distribute \(-4\) to both terms, keeping the sign attached:
$$-12b + 8 + 5b$$Reorder so the \(b\) terms are together, by the commutative property of addition:
$$-12b + 5b + 8$$Factor out the \(b\), which is the distributive property backwards, and add the coefficients:
$$(-12 + 5)b + 8 = -7b + 8$$Part 2 — Factor \(21c - 14\).
Both terms are divisible by 7, so 7 is the common factor:
$$21c - 14 = 7 \cdot 3c - 7 \cdot 2 = 7(3c - 2)$$Check by distributing back: \(7 \cdot 3c = 21c\) and \(7 \cdot 2 = 14\), which rebuilds the original.
Answer: \(-7b + 8\) and \(7(3c - 2)\).
0.5.4 Properties of Zero
Zero behaves unlike any other number, and it is worth collecting the differences in one place.
Zero is the additive identity, which we already stated: \(a + 0 = a\). Adding zero changes nothing, subtracting zero changes nothing, and subtracting a number from zero gives its opposite, \(0 - a = -a\). Multiplication is where zero stops being polite.
For any real number \(a\),
$$a \cdot 0 = 0 \qquad \text{and} \qquad 0 \cdot a = 0$$Zero times anything is zero, no matter how big the other factor is. Fifteen groups of nothing is nothing, and nothing groups of fifteen is also nothing.
$$47 \cdot 0 = 0 \qquad 0 \cdot (-936) = 0 \qquad 0 \cdot \dfrac{5}{8} = 0$$This holds inside longer products too. A single zero factor collapses the whole thing:
$$(-8)(93)(0)(41) = 0$$There is no need to compute anything else. One zero factor is enough.
If two numbers multiply to 12, you learn almost nothing about them — there are infinitely many pairs. If two numbers multiply to 0, you learn a great deal: at least one of them was 0. There is no other way to reach zero by multiplying.
Zero has a second property that runs in the other direction, and it is a separate fact rather than the first one restated: if \(ab = 0\), then \(a = 0\) or \(b = 0\). The reason is that in the real numbers two nonzero factors always have a nonzero product. That one sentence is the engine behind solving factored equations, which is how quadratics get solved in Chapter 7. If \((x - 4)(x + 9) = 0\), one of the two factors has to be zero, and that narrows the whole search to \(x = 4\) and \(x = -9\). The mechanics come later; for now just notice that the fact making it work is a property of zero.
Division is the other place zero behaves strangely, and two very different situations get confused with each other.
Definition 0.5.9 — Multiplication Property of Zero: one zero factor collapses the whole product.
For any real number \(a\) with \(a \neq 0\),
$$\dfrac{0}{a} = 0 \qquad \text{and} \qquad \dfrac{a}{0} \text{ is undefined}$$Division by zero is undefined for every numerator, including \(\dfrac{0}{0}\).
Every division question is really a multiplication question, and that is the fastest way to see why these two cases split apart.
It does not mean "very large" either. It means there is no such number, so an expression containing it has no value to report. Later, denominators will contain variables, and part of your job will be spotting which values would make a denominator zero and ruling them out first.
Asking for \(\dfrac{0}{13}\) means asking what number times 13 gives 0. The answer is 0, since \(0 \cdot 13 = 0\). Splitting nothing among thirteen people gives each of them nothing, which is a perfectly good answer.
Asking for \(\dfrac{13}{0}\) means asking what number times 0 gives 13. Nothing does. The multiplication property of zero guarantees that every such product is 0 and never 13, so no number qualifies and the expression names nothing at all.
Asking for \(\dfrac{0}{0}\) fails for the opposite reason. What times 0 gives 0? Every number does. An expression has to name exactly one value, and this one names all of them, so it names none.
Every division involving zero. Read the third column as the question the division is really asking:
| Expression | Value | The multiplication question |
|---|---|---|
| \(\dfrac{0}{13}\) | \(0\) | what times 13 gives 0? Only 0. |
| \(\dfrac{13}{0}\) | undefined | what times 0 gives 13? Nothing. |
| \(\dfrac{0}{0}\) | undefined | what times 0 gives 0? Everything. |
| \(\dfrac{13}{13}\) | \(1\) | what times 13 gives 13? Only 1. |
The rule of thumb is short: zero on top is fine, zero on the bottom is not.
Every property in this section, in symbols. Assume \(a\), \(b\), and \(c\) are real numbers:
| Property | Addition | Multiplication |
|---|---|---|
| Commutative | \(a + b = b + a\) | \(ab = ba\) |
| Associative | \((a + b) + c = a + (b + c)\) | \((ab)c = a(bc)\) |
| Identity | \(a + 0 = a\) | \(a \cdot 1 = a\) |
| Inverse | \(a + (-a) = 0\) | \(a \cdot \dfrac{1}{a} = 1\), for \(a \neq 0\) |
The properties that stand alone. These four do not come in an addition-and-multiplication pair:
| Property | Statement |
|---|---|
| Distributive | \(a(b + c) = ab + ac\) |
| Multiplication by zero | \(a \cdot 0 = 0\) |
| Division into zero | \(\dfrac{0}{a} = 0\), for \(a \neq 0\) |
| Division by zero | \(\dfrac{a}{0}\) is undefined |
Definition 0.5.10 — Division Involving Zero: zero on top is fine, zero on the bottom is not.
Evaluate \(\;\dfrac{n - 5}{n + 2}\;\) when \(n = 5\), and then when \(n = -2\).
Solution
Step 1 — Substitute \(n = 5\). Put the value in wherever \(n\) appears.
$$\dfrac{5 - 5}{5 + 2} = \dfrac{0}{7}$$The numerator is zero and the denominator is not, so this is the safe case. What times 7 gives 0? Only 0.
$$\dfrac{0}{7} = 0$$Step 2 — Substitute \(n = -2\). Now do the same with the second value.
$$\dfrac{-2 - 5}{-2 + 2} = \dfrac{-7}{0}$$The denominator is zero, so we are asking what number times 0 gives \(-7\). Nothing does, so this expression has no value.
Answer: \(0\) when \(n = 5\); undefined when \(n = -2\).
Simplify \(\;(14)(-77)(0)(3)\), and state the value of \(\dfrac{0}{9}\) and of \(\dfrac{9}{0}\).
Solution
Part 1 — The product.
One of the factors is 0, and zero times anything is zero, so the entire product collapses without any other arithmetic:
$$(14)(-77)(0)(3) = 0$$Part 2 — The two divisions.
For \(\dfrac{0}{9}\), ask what times 9 gives 0. Only 0 does:
$$\dfrac{0}{9} = 0$$For \(\dfrac{9}{0}\), ask what times 0 gives 9. Every product with 0 is 0, so nothing works and the expression is undefined.
Answer: \(0\); \(0\); undefined.
Up to now we have used these properties to simplify — to take an expression and rewrite it more cleanly. Chapter 3 puts them to a second use. Solving an equation means producing a chain of equivalent equations, each one a small rewriting of the last, ending with the variable by itself. Every link in that chain has to be legal, and this section is the list of legal moves. When you are asked to justify a solution step by step, these are the names you will write: distributive property, commutative property, additive inverse, multiplicative identity. The question "why am I allowed to do that?" now has an answer you can point to.
Problem Set 0.5
Problem 1. Name the property shown by \(\;17 + 32 = 32 + 17\).
Solution
Step 1 — Compare both sides of the equation: The equation \(17 + 32 = 32 + 17\) has the same two numbers added in a different order, and the sum does not change.
Step 2 — Name the property: Swapping the order of the terms in an addition without changing the result is exactly what the commutative property of addition says.
Answer: commutative property of addition
Problem 2. Name the property shown by \(\;(6 \cdot 5) \cdot 9 = 6 \cdot (5 \cdot 9)\).
Solution
Step 1 — Compare both sides of the equation: In \((6 \cdot 5) \cdot 9 = 6 \cdot (5 \cdot 9)\) the three factors stay in the same order, but the grouping (which pair is multiplied first) changes.
Step 2 — Name the property: Changing the grouping of factors without changing their order or the product is the associative property of multiplication.
Answer: associative property of multiplication
Problem 3. Name the property shown by \(\;8(y + 3) = 8y + 24\).
Solution
Step 1 — Look at the structure of the equation: In \(8(y + 3) = 8y + 24\), the factor outside the parentheses, \(8\), is multiplied onto each term inside the parentheses separately.
Step 2 — Check the arithmetic: \(8 \cdot y = 8y\) and \(8 \cdot 3 = 24\), so \(8(y+3) = 8y + 24\), which matches the given equation.
Step 3 — Name the property: Multiplying a factor onto every term inside a sum is the distributive property.
Answer: distributive property
Problem 4. Give a counterexample showing that subtraction is not commutative.
Solution
Step 1 — Recall what "commutative" would require: Subtraction would be commutative only if \(a - b\) always equaled \(b - a\) for every choice of \(a\) and \(b\).
Step 2 — Pick specific numbers and subtract both ways: Let \(a = 5\) and \(b = 3\). Then \(5 - 3 = 2\), but \(3 - 5 = -2\), using the rule that subtracting a larger number from a smaller one gives a negative result.
Step 3 — Compare the two results: \(2 \neq -2\), so switching the order of the numbers changes the answer.
Answer: \(5 - 3 = 2\) but \(3 - 5 = -2\), and since \(2 \neq -2\), subtraction is not commutative.
Problem 5. Give a counterexample showing that division is not associative.
Solution
Step 1 — Recall what "associative" would require: Division would be associative only if \((a \div b) \div c\) always equaled \(a \div (b \div c)\) for every choice of \(a\), \(b\), and \(c\).
Step 2 — Pick specific numbers and group both ways: Let \(a = 8\), \(b = 4\), \(c = 2\). Grouping the first two: \((8 \div 4) \div 2 = 2 \div 2 = 1\).
Step 3 — Try the other grouping: Grouping the last two: \(8 \div (4 \div 2) = 8 \div 2 = 4\), because \(4 \div 2 = 2\) is computed first.
Step 4 — Compare the two results: \(1 \neq 4\), so changing the grouping changes the answer.
Answer: \((8 \div 4) \div 2 = 1\) but \(8 \div (4 \div 2) = 4\), and since \(1 \neq 4\), division is not associative.
Problem 6. Use the commutative and associative properties to compute \(\;48 + 76 + 52\) mentally, and describe the regrouping you used.
Solution
Step 1 — Reorder the addends (commutative property of addition): Rewrite \(48 + 76 + 52\) as \(48 + 52 + 76\), moving \(52\) next to \(48\) because those two numbers are easy to add together.
Step 2 — Regroup the addends (associative property of addition): Group the first two together: \((48 + 52) + 76\), so that "friendly" pair can be added first.
Step 3 — Add the friendly pair: \(48 + 52 = 100\), because the ones digits \(8\) and \(2\) make a ten, which carries cleanly into a round hundred.
Step 4 — Add the remaining number: \(100 + 76 = 176\).
Step 5 — Check by adding in the original order: \(48 + 76 = 124\), and \(124 + 52 = 176\), which matches.
Answer: \(48 + 76 + 52 = 176\); regroup \(48\) with \(52\) to make \(100\) first, then add \(76\).
Problem 7. Use the commutative and associative properties to compute \(\;2 \cdot 87 \cdot 50\) mentally.
Solution
Step 1 — Reorder the factors (commutative property of multiplication): Rewrite \(2 \cdot 87 \cdot 50\) as \(2 \cdot 50 \cdot 87\), moving \(50\) next to \(2\) because those two multiply to a round number.
Step 2 — Regroup the factors (associative property of multiplication): Group the first two together: \((2 \cdot 50) \cdot 87\), so the round-number pair is multiplied first.
Step 3 — Multiply the friendly pair: \(2 \cdot 50 = 100\).
Step 4 — Multiply by the remaining factor: \(100 \cdot 87 = 8700\), since multiplying by \(100\) just appends two zeros to \(87\).
Step 5 — Check by multiplying in the original order: \(2 \cdot 87 = 174\), and \(174 \cdot 50 = 8700\), which matches.
Answer: \(2 \cdot 87 \cdot 50 = 8700\); regroup \(2\) with \(50\) to make \(100\) first, then multiply by \(87\).
Problem 8. Simplify \(\;(w + 6) + 11\).
Solution
Step 1 — Regroup the addends (associative property of addition): Rewrite \((w + 6) + 11\) as \(w + (6 + 11)\), shifting which two numbers are added first without changing their order.
Step 2 — Add the known numbers: \(6 + 11 = 17\).
Step 3 — Write the simplified expression: \(w + (6 + 11) = w + 17\).
Answer: \(w + 17\)
Problem 9. Simplify \(\;9 + (m + 4)\).
Solution
Step 1 — Reorder the terms inside the parentheses (commutative property of addition): Rewrite \(9 + (m + 4)\) as \(9 + (4 + m)\), since \(m + 4\) and \(4 + m\) name the same sum.
Step 2 — Regroup the addends (associative property of addition): Rewrite \(9 + (4 + m)\) as \((9 + 4) + m\), shifting which two numbers are added first.
Step 3 — Add the known numbers: \(9 + 4 = 13\).
Step 4 — Write the simplified expression: \((9 + 4) + m = 13 + m\).
Answer: \(13 + m\) (equivalently \(m + 13\))
Problem 10. What is the additive identity, and what is the multiplicative identity?
Solution
Step 1 — Recall what "identity" means: An identity element is the number that, when combined with any number \(a\) under an operation, leaves \(a\) unchanged.
Step 2 — Find the additive identity: We need a number \(a\) such that \(a + 0 = a\) for every \(a\); that number is \(0\), since adding \(0\) never changes a value.
Step 3 — Find the multiplicative identity: We need a number \(a\) such that \(a \cdot 1 = a\) for every \(a\); that number is \(1\), since multiplying by \(1\) never changes a value.
Answer: The additive identity is \(0\); the multiplicative identity is \(1\).
Problem 11. Find the additive inverse and the multiplicative inverse of \(\;12\).
Solution
Step 1 — Find the additive inverse: The additive inverse of a number is what you add to it to get the additive identity \(0\), so we need \(x\) with \(12 + x = 0\), giving \(x = -12\).
Step 2 — Find the multiplicative inverse: The multiplicative inverse of a number is what you multiply it by to get the multiplicative identity \(1\), so we need \(y\) with \(12 \cdot y = 1\), giving \(y = \dfrac{1}{12}\).
Step 3 — Check both: \(12 + (-12) = 0\) confirms the additive inverse, and \(12 \cdot \dfrac{1}{12} = 1\) confirms the multiplicative inverse.
Answer: The additive inverse of \(12\) is \(-12\); the multiplicative inverse is \(\dfrac{1}{12}\).
Problem 12. Find the additive inverse and the multiplicative inverse of \(\;-\dfrac{5}{8}\).
Solution
Step 1 — Find the additive inverse: We need \(x\) so that \(-\dfrac{5}{8} + x = 0\); flipping the sign gives \(x = \dfrac{5}{8}\).
Step 2 — Find the multiplicative inverse: We need \(y\) so that \(-\dfrac{5}{8} \cdot y = 1\); flipping the fraction (and keeping the sign, since a reciprocal never changes the sign of the original number) gives \(y = -\dfrac{8}{5}\).
Step 3 — Check both: \(-\dfrac{5}{8} + \dfrac{5}{8} = 0\) confirms the additive inverse, and \(-\dfrac{5}{8} \cdot \left(-\dfrac{8}{5}\right) = \dfrac{40}{40} = 1\) confirms the multiplicative inverse.
Answer: The additive inverse of \(-\dfrac{5}{8}\) is \(\dfrac{5}{8}\); the multiplicative inverse is \(-\dfrac{8}{5}\).
Problem 13. Explain in one or two sentences why 0 has no multiplicative inverse.
Solution
Step 1 — Restate what a multiplicative inverse would require: A multiplicative inverse of \(0\) would be a number \(y\) satisfying \(0 \cdot y = 1\).
Step 2 — Apply the property of zero: By the multiplication property of zero, \(0 \cdot y = 0\) for every real number \(y\), so \(0 \cdot y\) can never equal \(1\); no such \(y\) exists.
Answer: \(0\) has no multiplicative inverse because multiplying \(0\) by any number always gives \(0\), never \(1\).
Problem 14. Rewrite \(\dfrac{3}{5}\) with a denominator of 40, and state which disguised 1 you used.
Solution
Step 1 — Find the scale factor: To turn a denominator of \(5\) into \(40\), we ask what \(5\) is multiplied by: \(40 \div 5 = 8\).
Step 2 — Build the disguised 1: Since \(\dfrac{8}{8} = 1\), multiplying by it does not change the value of the fraction, only its form.
Step 3 — Multiply: $$\dfrac{3}{5} \cdot \dfrac{8}{8} = \dfrac{3 \cdot 8}{5 \cdot 8} = \dfrac{24}{40}$$
Step 4 — Check: \(\dfrac{24}{40}\) simplifies back to \(\dfrac{24 \div 8}{40 \div 8} = \dfrac{3}{5}\), so the value is unchanged.
Answer: \(\dfrac{3}{5} = \dfrac{24}{40}\), using the disguised 1 \(\dfrac{8}{8}\).
Problem 15. Distribute: \(\;5(x + 7)\).
Solution
Step 1 — Apply the distributive property: Multiply the \(5\) outside the parentheses by each term inside: \(5(x + 7) = 5 \cdot x + 5 \cdot 7\).
Step 2 — Simplify each product: \(5 \cdot x = 5x\) and \(5 \cdot 7 = 35\).
Step 3 — Check by substitution: Let \(x = 2\): the original gives \(5(2+7) = 5 \cdot 9 = 45\), and the distributed form gives \(5(2) + 35 = 10 + 35 = 45\), so the results match.
Answer: \(5(x + 7) = 5x + 35\).
Problem 16. Distribute: \(\;6(3n - 4)\).
Solution
Step 1 — Apply the distributive property: Multiply the \(6\) outside the parentheses by each term inside: \(6(3n - 4) = 6 \cdot 3n - 6 \cdot 4\).
Step 2 — Simplify each product: \(6 \cdot 3n = 18n\) and \(6 \cdot 4 = 24\).
Step 3 — Check by substitution: Let \(n = 1\): the original gives \(6(3-4) = 6(-1) = -6\), and the distributed form gives \(18(1) - 24 = 18 - 24 = -6\), so the results match.
Answer: \(6(3n - 4) = 18n - 24\).
Problem 17. Distribute: \(\;-2(q + 9)\).
Solution
Step 1 — Apply the distributive property with the negative factor: The entire signed factor \(-2\) is distributed to each term inside, not just its size: \(-2(q + 9) = -2 \cdot q + (-2) \cdot 9\).
Step 2 — Simplify each product, tracking the sign flip: \(-2 \cdot q = -2q\) and \(-2 \cdot 9 = -18\), so the \(+9\) inside becomes \(-18\) outside.
Step 3 — Check by substitution: Let \(q = 1\): the original gives \(-2(1+9) = -2(10) = -20\), and the distributed form gives \(-2(1) - 18 = -2 - 18 = -20\), so the results match.
Answer: \(-2(q + 9) = -2q - 18\).
Problem 18. Distribute: \(\;-7(2v - 5)\).
Solution
Step 1 — Apply the distributive property with the negative factor: The entire signed factor \(-7\) is distributed to each term inside, and the \(-5\) keeps its own sign: \(-7(2v - 5) = -7 \cdot 2v + (-7)(-5)\).
Step 2 — Simplify each product, tracking the sign flip: \(-7 \cdot 2v = -14v\), and \(-7 \cdot (-5) = 35\) since a negative times a negative is positive, so the \(-5\) inside becomes \(+35\) outside.
Step 3 — Check by substitution: Let \(v = 1\): the original gives \(-7(2-5) = -7(-3) = 21\), and the distributed form gives \(-14(1) + 35 = -14 + 35 = 21\), so the results match.
Answer: \(-7(2v - 5) = -14v + 35\).
Problem 19. Simplify: \(\;-(8k - 3)\).
Solution
Step 1 — Read the negative sign as multiplying by \(-1\): The expression \(-(8k - 3)\) means \(-1 \cdot (8k - 3)\), so distribute the \(-1\) across both terms inside the parentheses.
$$-(8k - 3) = -1(8k) + (-1)(-3) = -8k + 3$$Answer: \(-8k + 3\)
Problem 20. Simplify: \(\;10 - 4(z + 2)\).
Solution
Step 1 — Distribute the \(-4\) first: Multiplication outranks subtraction in the order of operations, so \(-4\) must multiply into the parentheses before anything is combined.
$$10 - 4(z + 2) = 10 - 4z - 8$$Step 2 — Combine the like (constant) terms: \(10\) and \(-8\) are both plain numbers, so they combine into one number.
$$10 - 8 - 4z = 2 - 4z$$Answer: \(2 - 4z\)
Problem 21. Simplify: \(\;3(2p + 1) - 5(p - 4)\).
Solution
Step 1 — Distribute \(3\) into the first parentheses: Multiply \(3\) by each term inside \((2p + 1)\).
$$3(2p + 1) = 6p + 3$$Step 2 — Distribute \(-5\) into the second parentheses: Multiply \(-5\) by each term inside \((p - 4)\), being careful with the sign.
$$-5(p - 4) = -5p + 20$$Step 3 — Combine like terms: Add the two results together, grouping the \(p\)-terms and the constants separately.
$$(6p + 3) + (-5p + 20) = (6p - 5p) + (3 + 20) = p + 23$$Answer: \(p + 23\)
Problem 22. Factor out the greatest common factor: \(\;18g + 27\).
Solution
Step 1 — Find the greatest common factor of the coefficients: The factors of \(18\) are \(1, 2, 3, 6, 9, 18\) and the factors of \(27\) are \(1, 3, 9, 27\), so the largest number common to both is \(9\).
Step 2 — Divide each term by the GCF and write it out front: \(18g \div 9 = 2g\) and \(27 \div 9 = 3\), so the GCF comes out front and what's left goes inside parentheses.
$$18g + 27 = 9(2g + 3)$$Step 3 — Check by distributing back: \(9(2g) + 9(3) = 18g + 27\), which matches the original expression.
Answer: \(9(2g + 3)\)
Problem 23. Factor out the greatest common factor: \(\;24d - 16\).
Solution
Step 1 — Find the greatest common factor of the coefficients: The factors of \(24\) are \(1, 2, 3, 4, 6, 8, 12, 24\) and the factors of \(16\) are \(1, 2, 4, 8, 16\), so the largest number common to both is \(8\).
Step 2 — Divide each term by the GCF and write it out front: \(24d \div 8 = 3d\) and \(16 \div 8 = 2\), so the GCF comes out front and what's left goes inside parentheses.
$$24d - 16 = 8(3d - 2)$$Step 3 — Check by distributing back: \(8(3d) - 8(2) = 24d - 16\), which matches the original expression.
Answer: \(8(3d - 2)\)
Problem 24. Explain how combining \(\;9s + 4s\) into \(13s\) uses the distributive property.
Solution
Step 1 — Identify what's common to both terms: In \(9s + 4s\), both terms share the same variable factor \(s\); only the coefficients \(9\) and \(4\) are different.
Step 2 — Apply the distributive property in reverse: The distributive property normally turns \(a(b + c)\) into \(ab + ac\); read backwards, it turns \(ab + ac\) into \(a(b + c)\). Here, factoring \(s\) out of both terms gives
$$9s + 4s = (9 + 4)s$$Step 3 — Add the coefficients: Since \(9 + 4 = 13\), the expression simplifies to \(13s\).
Answer: Combining \(9s + 4s\) into \(13s\) is the distributive property applied in reverse: \(9s + 4s = (9+4)s = 13s\), where \(s\) is the common factor pulled out front.
Problem 25. Simplify: \(\;(-5)(38)(0)\).
Solution
Step 1 — Apply the multiplication property of zero: Any number multiplied by \(0\) equals \(0\), so it doesn't matter what \(-5\) and \(38\) are — as soon as one factor in the product is \(0\), the whole product is \(0\).
$$(-5)(38)(0) = 0$$Answer: \(0\)
Problem 26. State the value of \(\dfrac{0}{21}\) and of \(\dfrac{21}{0}\).
Solution
Step 1 — Evaluate \(\dfrac{0}{21}\): Zero divided by any nonzero number is \(0\), since \(0\) can be split into \(21\) equal groups of size \(0\).
$$\dfrac{0}{21} = 0$$Step 2 — Evaluate \(\dfrac{21}{0}\): Division by zero asks "how many groups of \(0\) make \(21\)?" — no number works, since any number of groups of size \(0\) still totals \(0\), so this division names no number at all.
$$\dfrac{21}{0} \text{ is undefined}$$Answer: \(\dfrac{0}{21} = 0\), and \(\dfrac{21}{0}\) is undefined.
Problem 27. For what value of \(n\) is \(\dfrac{n + 6}{n - 3}\) undefined?
Solution
Step 1 — Recall when a fraction is undefined: A fraction is undefined exactly when its denominator equals \(0\), since division by \(0\) names no number.
Step 2 — Set the denominator equal to zero and solve: The denominator here is \(n - 3\), so find the value of \(n\) that makes it \(0\).
$$n - 3 = 0 \quad\Rightarrow\quad n = 3$$Step 3 — Confirm the numerator doesn't change the conclusion: At \(n = 3\), the numerator is \(3 + 6 = 9\), a nonzero number over a zero denominator, which confirms the fraction is undefined (not equal to \(0\)) at this value.
Answer: The expression is undefined when \(n = 3\).
Key Terms
commutative property — changing the order of the numbers does not change a sum or a product.
associative property — changing the grouping of the numbers does not change a sum or a product.
additive identity — the number 0, which leaves any number unchanged when added to it.
multiplicative identity — the number 1, which leaves any number unchanged when multiplied by it.
additive inverse — the number that adds to a given number to give 0; also called its opposite.
multiplicative inverse — the number that multiplies a given number to give 1; also called its reciprocal.
distributive property — multiplying a sum by a number equals multiplying each term and then adding.
common factor — a factor shared by every term of an expression, which can be pulled out front.
undefined — describes an expression that names no number at all, as division by zero does.