Reciprocals and the inverse property of multiplication, the division rule that turns any division into a multiplication, the sign rule for quotients, simplifying complex fractions, why division by zero is undefined, and evaluating expressions and velocities that involve division.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 2 — Properties of Real Numbers
Dividing Real Numbers
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 113-118 — the lesson these objectives are drawn from
Warm-up
Lesson 2.4 made subtraction disappear by rewriting it as addition. The same trick is about to work on division.
Discussion prompt
Work out 1 divided by 3, then work out 1 times one third. Do you get the same answer, and can you say why?
Hint: Think about what a fraction bar means.
Answer:
\[ 1 \div 3 = \tfrac{1}{3} \qquad 1 \cdot \tfrac{1}{3} = \tfrac{1}{3} \]
They agree, and not by accident. Dividing by three and multiplying by one third are the same operation written two ways — exactly as subtracting and adding the opposite were. That single observation turns division into multiplication, and the multiplication rules from Lesson 2.5 then cover it.
Concept
To divide a number by a nonzero number, multiply by its reciprocal. That turns division from a fourth operation with its own rules into a rewriting step followed by the multiplication rules you already have.
reciprocal — One of two numbers whose product is one. The reciprocal of a nonzero number a is one over a.
\[ a \div b = a \cdot \tfrac{1}{b} \qquad \text{Example: } 1 \div 3 = 1 \cdot \tfrac{1}{3} = \tfrac{1}{3} \]
Zero is the one number with no reciprocal, which is exactly why dividing by zero is not allowed.
Figure (svg): The division rule stating that a divided by b equals a times the reciprocal of b
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 113-113
Section
Section 1
Concept
Two numbers whose product is one are called reciprocals. The inverse property of multiplication says that every nonzero number has exactly one such partner.
\[ a \cdot \tfrac{1}{a} = 1 \qquad \text{and} \qquad \tfrac{1}{a} \cdot a = 1 \]
For a fraction, the reciprocal is the fraction turned upside down: two fifths and five halves multiply to one.
Figure (svg): Pairs of reciprocals shown multiplying to one
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 113-113 — the definition of reciprocals and the Inverse Property of Multiplication
Picture it
Every row multiplies to exactly one.
Figure (svg): Pairs of reciprocals shown multiplying to one
Notice that a negative number's reciprocal is also negative, since two negatives are needed to make the product positive one. Zero is missing from the list, and that absence is the subject of the last section.
Worked example
The mixed number is the one worth converting first.
\[ \text{Find the reciprocal of } \; 3, \quad \tfrac{2}{5}, \quad -4, \quad \tfrac{1}{2}, \quad 4\tfrac{1}{3}. \]
The reciprocal of 3 is one third
Why: Write the whole number as three over one, then flip it.
\[ \frac{1}{3} \]
The reciprocal of two fifths is five halves
Why: Turn the fraction upside down.
\[ \frac{5}{2} \]
The reciprocal of negative 4 is negative one quarter
Why: The sign stays, since two negatives are needed to give positive one.
\[ -\frac{1}{4} \]
The reciprocal of one half is 2
Why: Flipping one half gives two over one, which is two.
\[ 2 \]
Convert the mixed number first, then flip
Why: Four and one third is thirteen thirds, so its reciprocal is three thirteenths.
\[ \frac{3}{13} \]
Figure (svg): The solution to Worked example find five reciprocals shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{1}{3}, \quad \tfrac{5}{2}, \quad -\tfrac{1}{4}, \quad 2, \quad \tfrac{3}{13} \]
Verify: multiply each number by its reciprocal
Why: Three times one third is one; two fifths times five halves is ten tenths, which is one; negative four times negative one quarter is positive one. Every pair multiplies to one, which is the definition rather than a consequence of it.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 113-113
Matching
Every pair must multiply to one.
Match the pairs
Why: Each reciprocal is the original number turned upside down, with the sign left alone. The third pair is the one worth checking: negative four and negative one quarter multiply to positive one, and swapping either sign would give negative one instead.
Worked example
The sign question is settled by the definition rather than by a separate rule.
\[ \text{Show that the reciprocal of } -5 \text{ is } -\tfrac{1}{5} \text{ and not } \tfrac{1}{5}. \]
State what a reciprocal has to do
Why: Multiplied by the original number, it must give one.
\[ \text{product must be } 1 \]
Test the positive candidate
Why: Negative five times one fifth has one negative factor, so it is negative one — not one.
\[ -5 \cdot(\frac{1}{5}) = -1 \]
Test the negative candidate
Why: Negative five times negative one fifth has two negative factors, so it is positive one.
\[ -5 \cdot(-\frac{1}{5}) = 1 \]
Conclude
Why: Only the negative candidate satisfies the definition, so a negative number's reciprocal is negative.
\[ -\frac{1}{5} \]
Figure (svg): The solution to Worked example reciprocals of negatives shown as a ladder of expressions, one row per algebraic move
\[ -5 \cdot \left(-\tfrac{1}{5}\right) = 1 \]
Verify: check the general claim on another negative
Why: Negative two thirds times negative three halves is positive one, again requiring both to be negative. A number and its reciprocal always share a sign, which follows directly from the product having to be positive one.
Trap
\[ \text{the reciprocal of } -4 \]
Flip the number and also flip the sign, giving one quarter
Why: Reciprocals and opposites are both operations that undo something, so their effects get merged.
\[ -4 \cdot \tfrac{1}{4} = -1 \neq 1 \]
The product came out negative one rather than one, so the candidate fails the definition.
\[ \text{the reciprocal of } -4 \text{ is } -\tfrac{1}{4} \]
Flip the number and keep the sign exactly as it was
Why: A reciprocal must multiply to positive one, and that needs the two signs to agree.
Opposites and reciprocals are different: opposites add to zero and change sign, reciprocals multiply to one and keep it. Confusing the two is the commonest slip in this section.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Opposite | Reciprocal | |
|---|---|---|
| What the pair does | adds to zero | multiplies to one |
| Applied to 5 | -5 | 1/5 |
| Applied to -5 | 5 | -1/5 |
| Does the sign change? | Yes, always | No, never |
The bottom row is the whole distinction. An opposite always flips the sign and a reciprocal never does, and every confusion between the two comes from forgetting which is which.
Elimination
The number is negative two thirds.
Eliminate the wrong options
Which number is its reciprocal?
Survives elimination: A
Why: Turning negative two thirds upside down while keeping its sign gives negative three halves, and the product of the two is positive one because both factors are negative. Checking the product against the definition settles this immediately without any rule to remember.
Edge cases
The inverse property is stated for every nonzero number. Find out why.
Discussion prompt
Try to find a reciprocal for zero. What would it have to satisfy, and why can no number satisfy it? Then say what that means for division.
Hint: Write down the equation a reciprocal of zero would have to solve.
Answer:
\[ 0 \cdot x = 1 \quad \text{for some } x? \]
By the property of zero from Lesson 2.5, zero times anything is zero, never one. So no number can be zero's reciprocal, and zero is the single exception to the inverse property of multiplication.
Since dividing by a number means multiplying by its reciprocal, dividing by zero would mean multiplying by a number that does not exist. That is why division by zero is undefined rather than merely difficult — there is nothing to multiply by.
Section
Section 2
Concept
To divide a number a by a nonzero number b, multiply a by the reciprocal of b. The result is called the quotient of a and b.
\[ a \div b = a \cdot \tfrac{1}{b} \]
After the rewrite there are no divisions left, so the multiplication rules and properties from Lesson 2.5 apply directly.
Figure (svg): The division rule stating that a divided by b equals a times the reciprocal of b
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 113-113 — the Division Rule box and Example 1
Picture it
One statement, and it converts an operation into one you already have.
Figure (svg): The division rule stating that a divided by b equals a times the reciprocal of b
This is the same move as Lesson 2.4's subtraction rule, one operation up. Subtraction became addition of an opposite; division becomes multiplication by a reciprocal.
Worked example
This is Example 1 from the textbook. The third one has a mixed number.
\[ \text{Find } \; -10 \div (-2), \quad 0 \div 5, \quad -39 \div 4\tfrac{1}{3}. \]
Rewrite the first as a multiplication
Why: Multiply negative ten by the reciprocal of negative two, which is negative one half.
\[ -10 \cdot(-\frac{1}{2}) \]
Apply the multiplication rule
Why: Two negative factors, so positive, and ten halves is five.
\[ 5 \]
Rewrite the second
Why: Zero times one fifth, which the property of zero makes zero.
\[ 0 \]
Convert the mixed number, then rewrite the third
Why: Four and one third is thirteen thirds, whose reciprocal is three thirteenths, so this is negative thirty-nine times three thirteenths.
\[ -39 \cdot(\frac{3}{13}) = -9 \]
Figure (svg): The solution to Worked example three quotients shown as a ladder of expressions, one row per algebraic move
\[ -10 \div (-2) = 5, \quad 0 \div 5 = 0, \quad -39 \div 4\tfrac{1}{3} = -9 \]
Verify: multiply each quotient back by the divisor
Why: Five times negative two is negative ten; zero times five is zero; negative nine times thirteen thirds is negative thirty-nine. A quotient times its divisor must give back the dividend, which is the definition of division and therefore a genuine check.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 113-113
Translation
Four divisions, four rewrites. Watch the signs.
Match the pairs
Why: In every rewrite the divisor is turned upside down and its sign is left alone, while the dividend is untouched. The third one shows why mixed numbers are converted first: two and a half becomes five halves, and flipping that gives two fifths, which cannot be read off the mixed form.
Worked example
Guided Practice 1 to 4. Convert any mixed numbers before rewriting.
\[ \text{Find } \; -8 \div (-4), \quad 5 \div 2\tfrac{1}{2}, \quad -\tfrac{3}{4} \div 3, \quad \tfrac{3}{4} \div (-3). \]
Rewrite the first
Why: Negative eight times negative one quarter, with two negative factors.
\[ 2 \]
Convert and rewrite the second
Why: Two and a half is five halves, whose reciprocal is two fifths, so this is five times two fifths.
\[ 2 \]
Rewrite the third
Why: Negative three quarters times one third.
\[ -\frac{1}{4} \]
Rewrite the fourth
Why: Three quarters times negative one third, which has one negative factor.
\[ -\frac{1}{4} \]
Figure (svg): The solution to Worked example four from guided practice shown as a ladder of expressions, one row per algebraic move
\[ 2, \quad 2, \quad -\tfrac{1}{4}, \quad -\tfrac{1}{4} \]
Verify: compare the last two
Why: The third and fourth have the same absolute values arranged differently and both come out as negative one quarter, since each has exactly one negative factor. Moving a minus sign from the dividend to the divisor does not change a quotient, which the sign rule in the next section states directly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 114-114
Error analysis
The student applied the division rule to three quotients. Two are wrong.
Annotate
On: \( -10 \div (-2) = -10 \cdot \left(\tfrac{1}{2}\right) = -5 \qquad 0 \div 5 = 0 \qquad \tfrac{3}{4} \div 3 = \tfrac{3}{4} \cdot 3 = \tfrac{9}{4} \)
Both errors are caught by the same check: multiply the answer back by the divisor and see whether the dividend returns. Negative five times negative two is ten rather than negative ten, and nine quarters times three is twenty-seven quarters rather than three quarters.
Faded example
The reciprocal is what you need to supply.
Fill in the blanks
-39 \div 4\tfrac3/13-9 = -39 \div \tfrac______ = -39 \cdot ___ = ___
Why: Four and one third is thirteen thirds, whose reciprocal is three thirteenths. Negative thirty-nine times three thirteenths is negative nine, since thirty-nine divided by thirteen is three and three times three is nine. Converting the mixed number first is what makes the flip possible at all.
Prediction
Dividing does not always make a number smaller.
Predict first
Is 6 divided by one half larger or smaller than 6?
Correct: Larger — it is 12.
\[ 6 \div \tfrac{1}{2} = 6 \cdot 2 = 12 \]
Why: Dividing by one half means multiplying by two, since the reciprocal of one half is two. So the answer doubles rather than halves. Dividing makes a number smaller only when the divisor is greater than one, and the rewrite makes that obvious: the reciprocal of a number below one is a number above one.
Socratic
Division is a familiar operation. The rewrite still earns its place.
Discussion prompt
Give two advantages of rewriting a division as a multiplication by a reciprocal, at least one of which is about a property you gain. Then say which earlier lesson made exactly the same kind of move.
Hint: Think about what division cannot do that multiplication can.
Answer:
First, you gain the commutative and associative properties. Division has neither — six divided by three is not three divided by six, and the bracketing of a chain of divisions changes the answer — while multiplication has both, so a rewritten expression can be reordered and regrouped freely.
Second, the sign rules become the ones you already know from Lesson 2.5, so there is nothing new to learn about signs. The same move was made in Lesson 2.4, where subtraction was rewritten as addition of the opposite for exactly the same two reasons.
Section
Section 3
Concept
Because dividing is multiplying by a reciprocal, and a reciprocal has the same sign as its number, the sign rule for quotients is exactly the sign rule for products.
\[ \tfrac{-a}{b} = \tfrac{a}{-b} = -\tfrac{a}{b} \]
Figure (svg): The sign rule for quotients, with same signs giving positive and opposite signs giving negative
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 114-114 — The Sign of a Quotient Rule box
Picture it
Each column holds two arrangements of the same digits.
Figure (svg): The sign rule for quotients, with same signs giving positive and opposite signs giving negative
Twenty and five give four or negative four depending only on whether the two signs agree. The magnitudes are identical in all four cases, exactly as with multiplication.
Worked example
The Sign of a Quotient box gives all four arrangements.
\[ \text{Find } \; -20 \div (-5), \quad 20 \div 5, \quad -20 \div 5, \quad 20 \div (-5). \]
Check the signs of the first
Why: Both negative, so they agree and the quotient is positive.
\[ 4 \]
Check the second
Why: Both positive, so they agree and the quotient is positive.
\[ 4 \]
Check the third
Why: One of each, so the quotient is negative.
\[ -4 \]
Check the fourth
Why: One of each again, so also negative.
\[ -4 \]
Figure (svg): The solution to Worked example four quotients of twenty and five shown as a ladder of expressions, one row per algebraic move
\[ 4, \quad 4, \quad -4, \quad -4 \]
Verify: multiply each quotient by its divisor
Why: Four times negative five is negative twenty; four times five is twenty; negative four times five is negative twenty; negative four times negative five is twenty. Every one returns its own dividend, so all four signs are right.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 114-114
Sorting
Compare the two signs; do not compute the size.
Sort into buckets
Sort each quotient by the sign of its answer.
Notice that the first four use the same two digits and split evenly between the columns. Only the arrangement of the signs changed, and only that decided the answer.
Worked example
Example 3 from the textbook, with a equal to negative 2 and b equal to 3.
\[ \text{Evaluate } \; \frac{2a}{a - b} \; \text{ when } a = -2 \text{ and } b = 3. \]
Substitute both values, in brackets
Why: Two times negative two on top, and negative two minus three underneath.
\[ 2(-2) / (-2 - 3) \]
Simplify the numerator
Why: Two times negative two has one negative factor, so it is negative four.
\[ -4 \]
Simplify the denominator
Why: Negative two minus three rewrites as negative two plus negative three, which is negative five.
\[ -5 \]
Divide, using the sign rule
Why: Both negative, so the signs agree and the quotient is positive four fifths.
\[ \frac{4}{5} \]
Figure (svg): The solution to Worked example evaluate an expression with a fraction bar shown as a ladder of expressions, one row per algebraic move
\[ \frac{2(-2)}{-2 - 3} = \frac{-4}{-5} = \tfrac{4}{5} \]
Verify: check the sign against the rule
Why: Numerator and denominator both came out negative, so the quotient must be positive — and it is. Had only one of them been negative, the answer would have been negative four fifths, which is the commonest wrong answer here.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 114-114
Trap
\[ \frac{-4}{-5} \]
See two minus signs and write a minus sign in the answer
Why: Two visible minus signs suggests something negative is going on.
\[ = -\tfrac{4}{5} \quad \text{(wrong)} \]
Two negatives in a quotient agree, so they cancel. The answer is positive four fifths, exactly as two negative factors in a product give a positive.
\[ \frac{-4}{-5} = \tfrac{4}{5} \]
Ask whether the two signs agree, and let agreement mean positive
Why: The rule is about agreement rather than about counting marks on the page.
Multiplying back confirms it: four fifths times negative five is negative four, which is the numerator we started with.
Elimination
Three of these name the same number.
Eliminate the wrong options
Which one is NOT equal to the others?
Survives elimination: D
Why: The last has two negative signs, which agree, so its quotient is positive three quarters. The other three each contain exactly one negative and are all equal to negative three quarters. A single minus sign may be written in any of three places without changing the value, which is worth knowing when simplifying fractions.
Faded example
The substitution is done. Simplify and divide.
Fill in the blanks
\frac-4-5 = \frac___}___} = \tfrac______
Why: The numerator is negative four and the denominator is negative five, and the two signs agree, so the quotient is positive four fifths. Simplifying the top and bottom completely before dividing is what makes the sign question a single comparison rather than a tangle.
Socratic
It would be reasonable to expect division to need its own rule.
Discussion prompt
Explain why the sign rule for quotients is identical to the sign rule for products, using the division rule and one fact about reciprocals.
Hint: What sign does a reciprocal have?
Answer:
Dividing by b means multiplying by the reciprocal of b, and a reciprocal always has the same sign as the number it came from — since their product has to be positive one. So dividing by a negative number is multiplying by a negative number, and dividing by a positive is multiplying by a positive.
That means the count of negative factors in the rewritten product is exactly the count of negatives among the dividend and divisor. Two negatives give an even count and therefore a positive answer; one negative gives an odd count and a negative answer. The two rules are not merely similar — they are the same rule seen through the rewrite.
Section
Section 4
Concept
A fraction whose numerator or denominator is itself a fraction is called a complex fraction. The long central bar is a division, and the division rule handles it like any other.
Figure (svg): A complex fraction rewritten as a division and then as a multiplication by a reciprocal
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 114-114 — Example 2, Simplify Complex Fractions
Picture it
Two rewrites, and it becomes an ordinary product.
Figure (svg): A complex fraction rewritten as a division and then as a multiplication by a reciprocal
Nothing here is new. The only skill is spotting which bar is the main one, and the answer to that is always the longest bar on the page.
Worked example
This is Example 2 from the textbook. In one the fraction is on top, in the other underneath.
\[ \text{Find } \; \frac{\tfrac{1}{3}}{-4} \; \text{ and } \; \frac{-3}{\tfrac{1}{4}}. \]
Rewrite the first as a division
Why: One third divided by negative four.
\[ (\frac{1}{3}) \div(-4) \]
Apply the division rule
Why: Multiply by the reciprocal of negative four, which is negative one quarter.
\[ (\frac{1}{3}) \cdot(-\frac{1}{4}) \]
Multiply
Why: One negative factor, so negative, and one twelfth in size.
\[ -\frac{1}{12} \]
Do the second the same way
Why: Negative three divided by one quarter is negative three times four, which is negative twelve.
\[ -12 \]
Figure (svg): A complex fraction rewritten as a division and then as a multiplication by a reciprocal
\[ \frac{\tfrac{1}{3}}{-4} = -\tfrac{1}{12} \qquad \frac{-3}{\tfrac{1}{4}} = -12 \]
Verify: multiply each quotient back by its divisor
Why: Negative one twelfth times negative four is one third, which is the first numerator. Negative twelve times one quarter is negative three, which is the second numerator. Both check, and note that the two answers differ enormously in size — dividing by a small fraction makes a number much larger.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 114-114
Translation
Rewrite each as a multiplication by a reciprocal.
Match the pairs
Why: In each case the numerator is left alone and the denominator is flipped, keeping its sign. The second is the one worth studying: flipping one quarter gives four, so what looked like a division turns into a multiplication that makes the number bigger — which is exactly what dividing by a number below one does.
Worked example
Guided Practice 3 and 4. Same numbers, different arrangement.
\[ \text{Find } \; -\tfrac{3}{4} \div 3 \; \text{ and } \; \tfrac{3}{4} \div (-3). \]
Rewrite the first
Why: Negative three quarters times one third.
\[ (-\frac{3}{4}) \cdot(\frac{1}{3}) \]
Multiply
Why: One negative factor, so negative, and three twelfths is one quarter.
\[ -\frac{1}{4} \]
Rewrite the second
Why: Three quarters times negative one third.
\[ (\frac{3}{4}) \cdot(-\frac{1}{3}) \]
Multiply
Why: Again one negative factor, and the same size.
\[ -\frac{1}{4} \]
Figure (svg): The solution to Worked example two more from guided practice shown as a ladder of expressions, one row per algebraic move
\[ -\tfrac{3}{4} \div 3 = -\tfrac{1}{4} \qquad \tfrac{3}{4} \div (-3) = -\tfrac{1}{4} \]
Verify: explain why the two agree
Why: Each has exactly one negative among its dividend and divisor, so both quotients are negative, and the absolute values are identical. Moving the minus sign from the top to the bottom changes nothing, which is what the three-places rule says.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 114-114
Trap
\[ \frac{-3}{\tfrac{1}{4}} \]
Divide the small fraction by the whole number, since that seems the natural direction
Why: One quarter divided by three feels more familiar than three divided by one quarter.
\[ = \tfrac{1}{4} \div (-3) = -\tfrac{1}{12} \quad \text{(wrong)} \]
The correct answer is negative twelve, which is a hundred and forty-four times larger. The main bar says top divided by bottom, and reversing it gives the reciprocal of the right answer.
\[ \frac{-3}{\tfrac{1}{4}} = -3 \div \tfrac{1}{4} = -3 \cdot 4 = -12 \]
Read the main bar as the numerator divided by the denominator, always in that order
Why: The bar's meaning is fixed by position, and division is not commutative.
A size check catches this instantly: dividing by a quarter should make a number four times bigger, so the answer must be larger than three, not smaller.
Estimation
Dividing by a small number gives a large answer.
Predict first
Roughly how big is negative 3 divided by one quarter?
Correct: About -12.
\[ -3 \div \tfrac{1}{4} = -3 \cdot 4 = -12 \]
Why: There are four quarters in every unit, so there are twelve quarters in three, and the sign is negative because exactly one of the two numbers is. Dividing by a number below one always makes the magnitude larger, and noticing that before computing rules out three of these four options immediately.
Elimination
A complex fraction has several bars, and only one of them is the division being asked for.
Eliminate the wrong options
In the expression with one third on top and negative four underneath, which bar is the main one?
Survives elimination: A
Why: The longest bar separates the entire numerator from the entire denominator, and it is the division the expression is asking for. Identifying it first, and rewriting the expression with an explicit division sign, removes the ambiguity that makes complex fractions feel harder than they are.
Socratic
Dividing has always made numbers smaller, until now.
Discussion prompt
Explain, using the division rule, why dividing by one quarter multiplies a number by four. Then say exactly when dividing makes a number smaller and when it makes it larger.
Hint: Look at the size of the reciprocal.
Answer:
Dividing by one quarter means multiplying by its reciprocal, which is four. So the operation really is a multiplication by four, and the answer is four times as large. The intuition that dividing shrinks things came from always dividing by numbers greater than one.
Dividing by a number greater than one shrinks, because its reciprocal is less than one. Dividing by a number between zero and one enlarges, because its reciprocal is greater than one. Dividing by one changes nothing. The reciprocal is what decides, which is another reason the rewrite is worth making automatic.
Section
Section 5
Concept
Zero divided by any nonzero number is zero, since the rewrite gives zero times a reciprocal. Dividing by zero is undefined, because zero has no reciprocal to multiply by.
\[ 0 \div 5 = 0 \cdot \tfrac{1}{5} = 0 \qquad 5 \div 0 \text{ is undefined} \]
The two cases look symmetric on the page and behave completely differently, which is why they have to be kept apart deliberately.
Figure (svg): Zero divided by a number giving zero, contrasted with a number divided by zero being undefined
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 113-113 — Example 1 part b and the nonzero condition in the Division Rule
Picture it
One side is ordinary and the other is impossible.
Figure (svg): Zero divided by a number giving zero, contrasted with a number divided by zero being undefined
The division rule explains the asymmetry directly: the first asks you to multiply zero by something, which is fine, and the second asks for a reciprocal of zero, which does not exist.
Worked example
Both come straight from the division rule, and they behave completely differently.
\[ \text{Evaluate } \; 0 \div 5 \; \text{ and } \; 5 \div 0. \]
Rewrite the first
Why: Zero times the reciprocal of five, which is zero times one fifth.
\[ 0 \cdot(\frac{1}{5}) \]
Apply the property of zero
Why: Zero times anything is zero.
\[ 0 \]
Rewrite the second
Why: Five times the reciprocal of zero — but zero has no reciprocal.
State the conclusion
Why: The expression is undefined, not zero and not infinite.
Figure (svg): Zero divided by a number giving zero, contrasted with a number divided by zero being undefined
\[ 0 \div 5 = 0 \qquad 5 \div 0 \text{ undefined} \]
Verify: check the first by multiplying back
Why: Zero times five is zero, which is the dividend, so the first answer is right. For the second there is nothing to check, because no candidate answer exists: whatever number you propose, multiplying it by zero gives zero rather than five.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 113-113
Sorting
Look at which position the zero occupies.
Sort into buckets
Sort each expression by whether it has a value.
Every expression in the left column equals zero and every one in the right column has no value. Which side of the bar the zero sits on is the entire difference.
Worked example
A hot-air balloon's velocity is its change in position divided by the time taken.
\[ \text{A balloon drops } 90 \text{ feet in } 6 \text{ seconds. Find its velocity.} \]
Write the displacement with its sign
Why: Downward is negative, so the change in position is negative ninety feet.
\[ -90\text{ feet} \]
Write the formula and substitute
Why: Velocity is change in position divided by time.
\[ -90 \div 6 \]
Apply the sign rule
Why: One negative and one positive, so the signs differ and the quotient is negative.
Divide and attach the unit
Why: Ninety over six is fifteen, so the velocity is negative fifteen feet per second.
\[ -15 \text{ft} / s \]
Figure (svg): A hot air balloon's velocity found by dividing a negative displacement by a positive time
\[ \frac{-90 \text{ ft}}{6 \text{ s}} = -15 \text{ ft/s} \]
Verify: check the sign and the unit
Why: Feet divided by seconds gives feet per second, which is a velocity. The negative sign records downward motion, which matches a balloon that is dropping — and the speed, by Lesson 2.2, would be the absolute value, fifteen feet per second.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 117-117
Trap
\[ 5 \div 0 = 0 \]
Reason that anything involving zero gives zero
Why: The property of zero from Lesson 2.5 really does say that any product with a zero factor is zero, and it feels as though division should follow.
Check it: if the answer were zero, then zero times zero would have to give back five. It gives zero, so the answer cannot be zero — or anything else.
\[ 0 \div 5 = 0 \qquad 5 \div 0 \text{ is undefined} \]
Ask which position the zero is in before saying anything
Why: Zero as a dividend is ordinary; zero as a divisor is impossible, and the two look almost identical on the page.
The check that settles it is multiplying back. A quotient must satisfy quotient times divisor equals dividend, and with a divisor of zero that equation has no solution unless the dividend is zero too.
Elimination
Four attempted explanations. Only one is a real argument.
Eliminate the wrong options
Which explanation actually works?
Survives elimination: A
Why: A quotient must satisfy quotient times divisor equals dividend. With a divisor of zero, the left side is always zero by the property of zero, so no quotient exists unless the dividend is zero too. The division rule says the same thing in another way: it needs the reciprocal of zero, and no such number exists.
Missing information
A question can be perfectly well written and still be unanswerable.
Discussion prompt
A balloon drops 90 feet. What is its velocity? Say exactly what is missing, and give two velocities the balloon could have depending on how the gap is filled.
Hint: Velocity is a rate, so it compares a change with something.
Answer:
The time taken is missing. A velocity is a change in position per unit of time, so a displacement on its own cannot give one.
\[ \text{in 6 seconds: } -90 \div 6 = -15 \text{ ft/s} \qquad \text{in 30 seconds: } -90 \div 30 = -3 \text{ ft/s} \]
The same descent is five times faster in the first case. A change alone says nothing about a rate, which is why every rate in this book is a quotient of two quantities rather than a single measurement.
Socratic
Mathematicians define things all the time. This one is left undefined on purpose.
Discussion prompt
Suppose someone decided that 5 divided by 0 should equal 7. Show what breaks, using the relationship between a quotient and its divisor. Then say why leaving it undefined is the better choice.
Hint: A quotient times its divisor must return the dividend.
Answer:
\[ \text{if } 5 \div 0 = 7 \text{ then } 7 \cdot 0 = 5, \text{ but } 7 \cdot 0 = 0 \]
The definition would contradict the property of zero, which says every product with a zero factor is zero. And the same argument would work for any proposed answer, so no choice is consistent — the problem is not that the right value is hard to find but that no value works.
Leaving it undefined keeps the rest of arithmetic consistent. Defining it would force either the property of zero or the relationship between a quotient and its divisor to be abandoned, and both of those are used constantly. An undefined case is a small price for keeping everything else intact.
Comparison
Fill the blanks from memory before you scroll back. Two operations were made to disappear.
Comparison matrix
| Operation | Rewritten as | What it buys |
|---|---|---|
| Subtraction | adding the opposite | commutativity and associativity |
| Division | multiplying by the reciprocal | commutativity and associativity |
| Addition and multiplication | nothing — they already have both | the properties directly |
Chapter 2 has really taught two operations rather than four. Subtraction and division are rewritings, and everything in the rest of the book uses them in that form.
Pattern
Whether the division is between whole numbers, fractions or a complex fraction, the same five moves cover it.
Step three keeps the divisor's sign. A reciprocal flips the number over and never flips its sign, which is the single distinction between reciprocals and opposites.
OpenStax Elementary Algebra 2e, §1.4 Multiply and Divide Integers §1.4
Check
The sign rule. Compare the two signs first.
Check your understanding
What is negative 36 divided by negative 9?
Answer: A
Why: The two signs agree, both being negative, so the quotient is positive. Thirty-six divided by nine is four. Multiplying back confirms it: four times negative nine is negative thirty-six.
Check
The division rule. Flip the divisor.
Check your understanding
What is 8 divided by two thirds?
Answer: A
Why: The reciprocal of two thirds is three halves, so the division becomes eight times three halves, which is twenty-four halves, or twelve. Dividing by a number below one makes the answer larger, which rules out three of these options on size alone.
Check
Zero. Notice which position it is in.
Check your understanding
Which statement is correct?
Answer: A
Why: Zero as a dividend gives zero times a reciprocal, which is zero. Zero as a divisor would require the reciprocal of zero, which does not exist, so that expression is undefined. The two cases are not symmetric even though they look it.
Real world
A submarine changes depth from negative 40 metres to negative 130 metres over 3 minutes. A second submarine changes from negative 130 metres to negative 40 metres over 5 minutes.
Discussion prompt
Compute each submarine's velocity in metres per minute as a change in position divided by a time, and say what the sign of each answer records. Then explain why the second one is positive even though both submarines stayed below the surface the whole time.
Hint: Compute the change first, later value minus earlier, then divide by the time.
Answer:
\[ \text{first: } \frac{-130 - (-40)}{3} = \frac{-90}{3} = -30 \text{ m/min} \]
\[ \text{second: } \frac{-40 - (-130)}{5} = \frac{90}{5} = 18 \text{ m/min} \]
The first submarine descends at thirty metres per minute and the second rises at eighteen. The sign records the direction of the change, not the position: the second is positive because it moved upward, even though it was below the surface throughout.
This is exactly the distinction from Lesson 2.4 between a position and a change of position, now divided by a time to give a rate. Chapter 4 will call a quantity of this shape a rate of change, and it is the central idea of that chapter.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Does dividing a number always make it smaller?
Correct: No — dividing by a number between 0 and 1 makes it larger.
\[ 6 \div \tfrac{1}{2} = 6 \cdot 2 = 12 \qquad 6 \div 2 = 6 \cdot \tfrac{1}{2} = 3 \]
Dividing by a number above one shrinks, by a number below one enlarges, and by one changes nothing.
Why: Dividing by one quarter means multiplying by four, since a reciprocal of a number below one is a number above one. Six divided by a half is twelve, and three divided by a quarter is twelve. The intuition that dividing shrinks comes from years of dividing only by numbers greater than one, and the division rule shows exactly why it fails.
Explain it
They can divide whole numbers and have never divided by a fraction.
Discussion prompt
In no more than four sentences, explain why dividing by one half is the same as multiplying by two, using a concrete situation rather than a rule. Then tell them the one thing they must never do, and why.
Hint: How many halves are there in a whole pizza?
Answer:
A usable answer: dividing by one half asks how many halves fit into the number. There are two halves in every whole, so six wholes contain twelve halves, and six divided by a half is twelve. Dividing by a small piece gives a big count, which is why the answer came out larger.
The thing never to do is divide by zero. Asking how many zeros fit into six has no answer — you could fit any number of them and never reach six — so the expression has no value at all rather than a very large one.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Reciprocals are fixed by remembering they flip the number and never the sign, and by converting mixed numbers first. Quotient signs are fixed by asking whether the two signs agree rather than counting marks. Complex fractions are fixed by finding the longest bar and rewriting with an explicit division sign. Zero is fixed by asking which position it occupies before saying anything. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write four reciprocal pairs, including one negative and one mixed number, and show each pair multiplying to one. Underneath, write the division rule in symbols and beside it the subtraction rule from Lesson 2.4, with a note saying what each one buys you. In the middle, work one complex fraction all the way through, marking the main bar before you start. Near the bottom, draw two boxes headed zero on top and zero underneath, put an example in each, and write the value or the word undefined. Finally, in the margin, write the check that a quotient must pass, and apply it to one of your answers.
The check in your margin should be that the quotient times the divisor returns the dividend. Applying it to the zero-underneath case is what shows why no answer can exist there.
Recap
Five things, and the first one makes the fourth operation disappear as thoroughly as the second one did.
| If the question says | Your first move is |
|---|---|
| Find the quotient | Check the divisor is not zero, then flip it |
| Divide by a mixed number | Convert it to an improper fraction first |
| Simplify the complex fraction | Find the longest bar |
| Evaluate the expression | Simplify top and bottom, then compare signs |
| Find the velocity | Divide the change in position by the time |
That completes Chapter 2. Chapter 3 puts all four operations to work solving equations, and the rewrites from Lessons 2.4 and 2.8 are what make the solving steps reversible — which is the property the whole of equation solving rests on.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.8 Dividing Real Numbers §2.8, pp. 113-118 — everything on these slides traces back here
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