The real number line and its subsets, ordering negatives, the five properties of addition and multiplication, subtraction and division as definitions, and unit analysis as a self-check.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 1 — Equations and Inequalities
Apply Properties of Real Numbers
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 2-9 — the lesson these objectives are drawn from
Warm-up
You have been using these properties since middle school without naming them. Naming them is the whole point of this lesson.
Discussion prompt
Rearrange 17 plus 48 plus 3 in your head to make it easy. Which two moves did you just make, and could you name either one?
Hint: You almost certainly moved the 3 next to the 17. Why were you allowed to?
Answer:
\[ 17 + 48 + 3 = 17 + 3 + 48 = 20 + 48 = 68 \]
Moving the 3 past the 48 is the commutative property of addition. Regrouping so the 17 and the 3 get added first is the associative property. Both are so familiar they feel like nothing — but they are exactly what fails in matrix multiplication later this year, so it is worth knowing their names.
Concept
Algebra 2 rests on two things: every real number is a point on one line, and there is a fixed, short list of moves you are allowed to make. Everything else in this course is those moves applied in a longer chain.
real number line — A line on which every real number is exactly one point and every point is exactly one real number, arranged so that numbers increase from left to right.
That is why a wrong answer can always be traced: some step used a move that is not on the list.
Figure (svg): A real number line from -6 to 6 with negative five fourths marked between -2 and -1 and the square root of three marked between 1 and 2
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 2-3
Section
Section 1
Concept
A real number either can be written as a quotient of two integers or it cannot. That single question splits the whole line in two, with nothing left over.
irrational number — A real number that cannot be written as a quotient of two integers; as a decimal it neither terminates nor repeats.
Figure (svg): Nested boxes showing whole numbers inside integers inside rational numbers, with irrational numbers in a separate box, both inside the real numbers
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 2-2 — the Subsets of the Real Numbers key-concept box
Picture it
The picture is not decoration — it is the answer to almost every classification question you will be asked.
Figure (svg): Nested boxes showing whole numbers inside integers inside rational numbers, with irrational numbers in a separate box, both inside the real numbers
Read it inward: every whole number is an integer, every integer is rational (put it over one), and every rational is real. Read it outward and none of those reverse.
Worked example
This is Example 1 from the textbook. The move that matters is the first one.
\[ \text{Graph } -\tfrac{5}{4} \text{ and } \sqrt{3} \text{ on a number line.} \]
Turn the fraction into a decimal
Why: A fraction bar is a division sign, and you cannot place a number you cannot say out loud as a decimal.
\[ -\frac{5}{4} = -1.25 \]
Approximate the radical to the nearest tenth
Why: The square root of three is irrational, so it has no exact decimal. One decimal place is enough to place it between two ticks.
\[ \sqrt{3} = 1.7\text{ approx} \]
Place -1.25 between -2 and -1, closer to -1
Why: One and a quarter units to the left of zero, so a quarter of the way past -1.
\[ \text{point between } -2\text{ and } -1 \]
Place 1.7 between 1 and 2, closer to 2
Why: Seven tenths of the way from 1 to 2.
\[ \text{point between } 1\text{ and } 2 \]
Figure (svg): The solution to Worked example graph a fraction and a radical shown as a ladder of expressions, one row per algebraic move
\[ -\tfrac{5}{4} \approx -1.25 \qquad \sqrt{3} \approx 1.7 \]
Verify: square the approximation
Why: 1.7 squared is 2.89, just under 3, so 1.7 is slightly low but correct to the nearest tenth. And -1.25 really is between -2 and -1, which is where a number a bit past -1 belongs.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 2-2
Sorting
One question decides every one of these: can it be written as one integer over another?
Sort into buckets
Drop each number into the right column.
Notice that every integer landed on the rational side. That is the point of the nested boxes.
Worked example
Guided Practice 1. Do them all as decimals first, then place them in one pass.
\[ \text{Graph } -0.2, \; \tfrac{7}{10}, \; -1, \; \sqrt{2}, \; -4. \]
Convert every entry to a decimal
Why: Mixed notation is the only thing making this hard. Seven tenths is 0.7; the square root of two is about 1.4; the rest already are decimals.
\[ -0.2, 0.7, -1, 1.4, -4 \]
Sort the decimals mentally from least to greatest
Why: Sorting before plotting means you plot left to right and never have to erase.
\[ -4, -1, -0.2, 0.7, 1.4 \]
Plot the two integers first
Why: Negative four and negative one land exactly on ticks, so they anchor the rest.
\[ -4\text{ and } -1\text{ on ticks} \]
Fit the three non-integers between the ticks
Why: Negative two tenths sits just left of zero; 0.7 is most of the way to 1; 1.4 is a bit under one and a half.
\[ -0.2, 0.7, 1.4\text{ between ticks} \]
Figure (svg): The solution to Worked example five numbers at once shown as a ladder of expressions, one row per algebraic move
\[ -4 < -1 < -0.2 < \tfrac{7}{10} < \sqrt{2} \]
Verify: re-read the line left to right
Why: Reading the plotted points left to right gives -4, -1, -0.2, 0.7, 1.4 — the same order the sort produced, so nothing was misplaced.
Trap
\[ \sqrt{9}, \; \sqrt{3}, \; \sqrt{16} \]
Call all three irrational because each one has a radical sign
Why: The radical sign is treated as if it were the definition of irrational.
This misses that a radical is just a question — and sometimes the answer is a plain integer.
\[ \sqrt{9}, \; \sqrt{3}, \; \sqrt{16} \]
Evaluate each one first, then classify
Why: Irrational is a fact about the number, not about how it was written.
\[ \sqrt{9} = 3 \quad \sqrt{16} = 4 \quad \sqrt{3} = 1.7320508\ldots \]
Two of them are whole numbers. Only the square root of three refuses to terminate or repeat, so only it is irrational.
Elimination
Every option here is a real number. Only one of them is irrational.
Eliminate the wrong options
Which of these is irrational?
Survives elimination: B
Why: Seven is not a perfect square, so its square root cannot be written as a quotient of integers and its decimal never terminates or repeats. Compare 22/7, which looks like pi but is rational precisely because it is written as a quotient of integers.
Notation
This one sentence contains four separate claims. Take them apart.
Annotate
On: \( \mathbb{W} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \)
Each symbol says the left set sits strictly inside the right one, so the reverse claim fails every time: some integers are not whole, some rationals are not integers, some reals are not rational.
Prediction
Both of these are close to negative three and a bit. Commit before you compute.
Predict first
Which is further left on the number line: negative 22 over 7, or negative pi?
Correct: -22/7 is further left.
\[ \tfrac{22}{7} \approx 3.142857 \qquad \pi \approx 3.141593 \]
\[ -\tfrac{22}{7} \approx -3.142857 < -\pi \approx -3.141593 \]
Why: 22/7 is about 3.142857, and pi is about 3.141593, so 22/7 is the larger of the two positives. Negating flips the order: the larger positive becomes the smaller negative, so -22/7 sits to the left of -pi. This flip is the single most common source of ordering mistakes with negatives.
Section
Section 2
Concept
Ordering is not a separate skill from graphing. Put the numbers on the line and read them left to right; the ordering falls out. The only reason it feels hard is that negatives invert the digit-size instinct you built in elementary school.
\[ -80 < -70 < -61 < -52 < -23 < -2 \]
Eighty is bigger than seventy. Negative eighty is smaller than negative seventy, because it is further from zero on the losing side.
Figure (svg): A number line from -90 to 10 with the six record-low state temperatures plotted, showing that -80 sits furthest left
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 3-3 — the record-low temperature table
Picture it
Alaska, Colorado, Florida, Montana, New York and Rhode Island, in degrees Fahrenheit.
Figure (svg): A number line from -90 to 10 with the six record-low state temperatures plotted, showing that -80 sits furthest left
Nobody argues about which is coldest once the numbers are plotted. Plotting is the cure for the instinct, not willpower.
Worked example
Example 2 from the textbook, written as a multiple-choice question — which is exactly how it will appear on a test.
\[ \text{Order } -80, -61, -2, -70, -52, -23 \text{ from lowest to highest.} \]
Notice that every value is negative
Why: When all values share a sign the comparison is purely about distance from zero, which simplifies the sort.
Sort by size ignoring the signs, largest size first
Why: Among negatives, the furthest from zero is the smallest, so the biggest-looking number comes first.
\[ 80, 70, 61, 52, 23, 2 \]
Put the minus signs back in that same order
Why: Reattaching the signs without re-sorting is the whole trick.
\[ -80, -70, -61, -52, -23, -2 \]
Read the answer as a chain of inequalities
Why: A chain is easier to check than a list, because each neighbouring pair can be checked on its own.
\[ -80 < -70 < -61 < -52 < -23 < -2 \]
Figure (svg): The solution to Worked example order six negative temperatures shown as a ladder of expressions, one row per algebraic move
\[ -80 < -70 < -61 < -52 < -23 < -2 \]
Verify: scan the chain for a rising pair
Why: Walking the chain, every step moves right on the number line: -80 to -70 rises, -70 to -61 rises, and so on to -2. No pair is out of order, so the sort is complete.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 3-3
Ranking
Highest elevations of five states, in feet. Least to greatest.
Put in order
Why: All five are positive, so ordinary digit-size reasoning works and the only real risk is misreading 14,494 as smaller than it is because of the comma. Count digits first: 535 has three, 2407 and 4145 and 6643 have four, and 14,494 has five, which settles the extremes before you compare anything.
Worked example
Exercise 10, using the state-elevation table. This one adds two complications at once: mixed signs, and a reversed direction.
\[ \text{Order } 0, \; -282, \; 257, \; -8, \; 178 \text{ from greatest to least.} \]
Split the list by sign before sorting anything
Why: Every positive beats every negative and zero sits between them, so this one split does most of the work.
\[ \text{positives } 257, 178 |\text{ zero } |\text{ negatives } -8, -282 \]
Sort the positives greatest first
Why: Among positives, bigger digits really do mean bigger number.
\[ 257, 178 \]
Place zero next
Why: Zero is greater than every negative and less than every positive.
\[ 257, 178, 0 \]
Sort the negatives, closest to zero first
Why: Going down from zero, the nearer negative is the greater one, so -8 comes before -282.
\[ -8\text{ then } -282 \]
Join the three groups in order
Why: The direction asked for was greatest to least, so positives lead and the most negative value ends the list.
\[ 257, 178, 0, -8, -282 \]
Figure (svg): The solution to Worked example mixed signs, greatest to least shown as a ladder of expressions, one row per algebraic move
\[ 257 > 178 > 0 > -8 > -282 \]
Verify: reverse it and check it reads least to greatest
Why: Reversed, the list is -282, -8, 0, 178, 257, which climbs steadily. A list that is correct in one direction is correct in the other, so this is a genuine check and not a restatement.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 6-6
Trap
\[ \text{Order } -80, -2, -61 \text{ from lowest to highest.} \]
Say -2 is lowest because 2 is the smallest digit
Why: The size of the digits is being read as the size of the number.
\[ -2, -61, -80 \quad \text{(wrong)} \]
\[ \text{Order } -80, -2, -61 \text{ from lowest to highest.} \]
Place all three on a number line first
Why: Lowest means furthest left, and the line is the only place left and right are defined.
\[ -80 < -61 < -2 \]
Negative two is the warmest of the three, so it belongs at the end of a lowest-to-highest list, not the start.
Two truths and a lie
All three statements are about the same three numbers. One of them is false.
Eliminate the wrong options
Two of these are true. Knock those two out and keep the false one.
Survives elimination: C
Why: The survivor is the broken statement. It compares the digits after the decimal point as if they were whole numbers and then ignores the minus signs entirely. Negative three quarters is further from zero on the losing side than negative one half, so it is the smaller of the two — the opposite of what the sentence claims.
Discrimination
Do not order these. Just say which ones your instinct will get wrong if you rush.
Sort into buckets
Sort each comparison by whether digit-size instinct gives the right answer.
Prediction
Commit to an answer before you reason it through.
Predict first
If a is less than b and both are negative, what is true about the opposites, negative a and negative b?
Correct: -a is greater than -b.
\[ a < b \quad \Longrightarrow \quad -a > -b \]
\[ -5 < -3 \quad \Longrightarrow \quad 5 > 3 \]
Why: Negating reflects both numbers across zero, and a reflection reverses left and right. So whichever one was on the left becomes the one on the right. Concretely, -5 is less than -3, and their opposites are 5 and 3, where 5 is now the greater. The distance between them is irrelevant; only the reflection matters.
Section
Section 3
Concept
Closure, commutative, associative, identity and inverse each have an addition version and a multiplication version. The distributive property is the only one that mixes the two operations, which is exactly why it is the one that does real work.
associative property — A property about regrouping: which pair you combine first does not change the result. It is about parentheses, never about order.
Learn the pairs together. Later this year matrix multiplication keeps the associative property and loses the commutative one, and that distinction only means something if you kept them apart now.
Figure (svg): A two-column table of the properties of addition and multiplication: closure, commutative, associative, identity and inverse, plus the distributive property beneath
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 3-3 — Properties of Addition and Multiplication
Picture it
The two are confused constantly, and the confusion is visual: one changes order, the other changes brackets.
Figure (svg): Two columns contrasting the commutative property, which changes the order of the terms, with the associative property, which changes only the parentheses
Read the two columns bottom-up: the tell is whether the letters moved, not whether the answer stayed the same.
Worked example
Example 3. Each of these is one move, so each has exactly one name.
\[ \text{a) } 7 + 4 = 4 + 7 \qquad \text{b) } 13 \cdot \tfrac{1}{13} = 1 \]
For (a), check whether the letters moved or the brackets moved
Why: There are no brackets at all, and the 7 and the 4 swapped seats, so the order changed.
\[ 7 + 4 = 4 + 7 \]
Name it: commutative property of addition
Why: Order changed, operation is addition.
For (b), notice the result is 1, not 0
Why: The identity you land on tells you which operation's inverse you are looking at: 0 for addition, 1 for multiplication.
\[ 13 \times 1 / 13 = 1 \]
Name it: inverse property of multiplication
Why: A number times its reciprocal gives the multiplicative identity.
Figure (svg): The solution to Worked example name the property shown as a ladder of expressions, one row per algebraic move
\[ \text{a) commutative property of addition} \qquad \text{b) inverse property of multiplication} \]
Verify: check each against the property table
Why: The table's commutative row reads a plus b equals b plus a, which is line (a) with 7 and 4 substituted. The multiplication inverse row reads a times one over a equals 1, which is line (b) with a equal to 13.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 4-4
Matching
Each statement uses exactly one property.
Match the pairs
Why: Two questions sort all five. Did the letters move, or only the brackets? And did the line land on 0 or on 1? Landing on 0 means an additive inverse, landing on 1 means a multiplicative inverse, and a sum appearing inside a product means the distributive property.
Worked example
Guided Practice 3 through 6. Speed comes from asking the same two questions every time.
\[ \begin{aligned} &(2 \cdot 3) \cdot 9 = 2 \cdot (3 \cdot 9) \qquad && 15 + 0 = 15 \\ &4(5 + 25) = 4(5) + 4(25) && 1 \cdot 500 = 500 \end{aligned} \]
First statement: the 2, 3 and 9 never moved, only the brackets
Why: Same order, different grouping — that is the definition of associative.
Second statement: adding zero changed nothing
Why: Zero is the additive identity, the number that leaves everything alone under addition.
Third statement: one factor was shared out over a sum
Why: This is the only property that turns one operation into the other, so it is unmistakable once you look for a sum inside a product.
Fourth statement: multiplying by one changed nothing
Why: One is the multiplicative identity, the multiplication counterpart of zero.
Figure (svg): The solution to Worked example four in a row shown as a ladder of expressions, one row per algebraic move
\[ \text{associative (mult.)}, \; \text{identity (add.)}, \; \text{distributive}, \; \text{identity (mult.)} \]
Verify: evaluate both sides of each line
Why: Left to right: 54 equals 54, 15 equals 15, 120 equals 120, and 500 equals 500. Every line is a true numerical statement, which is the minimum a property has to deliver.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 4-4
Trap
\[ 3 + (5 + 8) = 3 + (8 + 5) \]
Call this the associative property because there are brackets
Why: The presence of brackets is being read as the property, rather than what happened to them.
The brackets are in exactly the same place on both sides. Nothing was regrouped.
\[ 3 + (5 + 8) = 3 + (8 + 5) \]
Compare the two sides position by position
Why: The 5 and the 8 swapped seats; the brackets are unchanged.
Order changed, grouping did not, so this is the commutative property of addition — applied inside the bracket.
\[ (3 + 5) + 8 = 3 + (5 + 8) \quad \text{is the associative one} \]
Sorting
Same five property names, two operations. Sorting them is how you stop mixing the identities up.
Sort into buckets
File each statement under the operation whose property it is.
Only one entry in the entire property table carries a condition, and it is the multiplicative inverse. That is worth remembering on its own.
Fill the middle
The algebra is done. The reason for one line has been rubbed out.
\[ \text{Fill the blank with the name of the property used.} \]
Fill in the blanks
4(5 + 25) \;=\; 4(5) + 4(25) \qquad \Longleftarrow \; distributive property
Why: A single factor outside a bracket was handed to each term inside it, turning one multiplication into two multiplications joined by addition. That mixing of the two operations happens in exactly one property, so no other name fits. Every other property in the table keeps you inside a single operation.
Counterexample
Here is a statement that sounds like it belongs on the property list.
\[ a - b = b - a \]
Discussion prompt
Subtraction looks like it should be commutative — after all, addition is. Find one pair of numbers that proves it is not, and then say what that failure tells you about why subtraction is defined as adding the opposite.
Hint: Any two numbers that are not equal will do. Try the smallest pair you can think of.
Answer:
\[ 7 - 3 = 4 \qquad 3 - 7 = -4 \]
One counterexample is enough to kill a general claim, and this is why the property table lists only addition and multiplication. Subtraction and division get no properties of their own — they inherit whatever they inherit by being rewritten as addition and multiplication, which is the subject of the next section.
Section
Section 4
Concept
Subtraction is defined as adding the opposite, and division is defined as multiplying by the reciprocal. They are not extra operations with extra rules — they are shorthand, and rewriting them is how you get access to the property list.
reciprocal — The multiplicative inverse of a nonzero number b, written as one over b; multiplying b by it gives 1.
\[ a - b = a + (-b) \qquad a \div b = a \cdot \tfrac{1}{b}, \; b \neq 0 \]
This is why the property table has only addition and multiplication columns. Anything true of a subtraction is true because of what it becomes when you rewrite it.
Figure (svg): Two rewrite arrows: a minus b becomes a plus the opposite of b, and a divided by b becomes a times the reciprocal of b
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 4-4 — Defining Subtraction and Division
Picture it
Each arrow converts an operation you have no properties for into one you do.
Figure (svg): Two rewrite arrows: a minus b becomes a plus the opposite of b, and a divided by b becomes a times the reciprocal of b
The opposite of b is the number that adds to zero with it. The reciprocal of b is the number that multiplies to one with it. Those two sentences are the entire inverse row of the table.
Worked example
Example 4. The answer is not the point; the column of reasons is the point.
\[ \text{Show that } a + (2 - a) = 2, \text{ justifying each step.} \]
Rewrite the subtraction as addition of the opposite
Why: Definition of subtraction. Nothing in the property table applies to a minus sign, so this rewrite has to come first.
\[ a + [2 + (-a)] \]
Swap the two terms inside the bracket
Why: Commutative property of addition. This puts the negative a next to the a that will cancel it.
\[ a + [(-a) + 2] \]
Regroup so that a and negative a are bracketed together
Why: Associative property of addition. The terms have not moved; only the brackets have.
\[ [a + (-a)] + 2 \]
Replace the bracketed pair with zero
Why: Inverse property of addition: a number plus its opposite is the additive identity.
\[ 0 + 2 \]
Drop the zero
Why: Identity property of addition.
\[ 2 \]
Figure (svg): The solution to Worked example justify every step shown as a ladder of expressions, one row per algebraic move
\[ a + (2 - a) = 2 \]
Verify: substitute a number for a
Why: Take a equal to 9. The left side is 9 plus the quantity 2 minus 9, which is 9 plus negative 7, which is 2. Take a equal to -4: negative 4 plus the quantity 2 plus 4, which is negative 4 plus 6, which is 2. The identity holds for both, as a proof using only universal properties must.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 4-4
Ranking
These five lines prove that the quantity c minus 3, plus 3, equals c. They are shuffled.
Put in order
Why: The rewrite always comes first, because none of the properties apply to a minus sign. Regrouping then puts the two threes together, the inverse property collapses them to zero, and the identity property drops the zero. That order — rewrite, rearrange, cancel, simplify — is the shape of nearly every justification question in this chapter.
Worked example
Exercise 17. Same shape, other operation — which is the point of learning the properties in pairs.
\[ \text{Show that } 6 \cdot (a \div 3) = 2a, \text{ justifying each step.} \]
Rewrite the division as multiplication by the reciprocal
Why: Definition of division. Again the rewrite comes first, because there are no properties of division to appeal to.
\[ 6 \times(a \times 1 / 3) \]
Swap the two factors inside the bracket
Why: Commutative property of multiplication, moving the one third next to the 6.
\[ 6 \times(1 / 3 \times a) \]
Regroup so that 6 and one third are bracketed together
Why: Associative property of multiplication: the factors kept their order, only the brackets moved.
\[ (6 \times 1 / 3) \times a \]
Multiply the bracketed pair
Why: Six times one third is 2, an ordinary arithmetic step rather than a property.
\[ 2 \times a \]
Figure (svg): The solution to Worked example the division version shown as a ladder of expressions, one row per algebraic move
\[ 6 \cdot (a \div 3) = 2a \]
Verify: substitute a number for a
Why: Take a equal to 12. The left side is 6 times the quantity 12 divided by 3, which is 6 times 4, which is 24. The right side is 2 times 12, which is 24 as well. Trying a equal to 0 gives 0 on both sides, so the identity survives the awkward case too.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 6-6
Error analysis
The algebra below arrives at the right answer. One of the four reasons is the wrong name, which on a justify-each-step question costs the mark even though the arithmetic is fine.
Annotate
On: \( \begin{aligned} 3x + (6 + 4x) &= 3x + (4x + 6) && \text{associative property of addition} \\ &= (3x + 4x) + 6 && \text{associative property of addition} \\ &= 7x + 6 && \text{distributive property} \end{aligned} \)
The tell never changes: if the letters changed seats it was commutative, and if only the brackets moved it was associative.
Fill the middle
The first and last lines are given. The middle one is where the property actually gets used.
Fill in the blanks
15 \cdot (3 \div b) \;=\; 15 \cdot \left(3 \cdot \frac{1}{b}\right) \;=\; \tfrac______
Why: Division has no properties of its own, so the first move on any line containing a division sign is to convert it into multiplication by the reciprocal. Once it is 3 times one over b, the associative property lets you regroup as 15 times 3, all over b, which is 45 over b. Skipping the rewrite is what makes this kind of problem feel impossible.
Explain it to yourself
One line from the worked example, on its own.
\[ a + [(-a) + 2] = [a + (-a)] + 2 \]
Discussion prompt
In your own words: what changed between these two expressions, what did not change, and which property permits it? Then say why the same move would be illegal if the operations were mixed, as in a plus the quantity negative a times 2.
Hint: Look at the order of the symbols before you look at the brackets.
Answer:
Nothing moved. The symbols read a, negative a, 2 on both sides, in that order. Only the brackets shifted, which is precisely the associative property of addition.
\[ a + [(-a) \cdot 2] \neq [a + (-a)] \cdot 2 \]
The associative property is stated for one operation at a time. With a mix of addition and multiplication there is no regrouping property at all, which is why order of operations exists as a separate convention.
Definition probe
The multiplicative inverse rule is the only line in the table with a condition attached. Test its edges.
Sort into buckets
Sort each case by whether it satisfies the multiplicative inverse property.
Section
Section 5
Concept
Carry the units through the arithmetic and they behave like factors: the same unit above and below cancels, and whatever survives is the unit of your answer. If the surviving unit is wrong, the setup was wrong, and you know it before you check the number.
unit analysis — Carrying units through a calculation as algebraic factors so that the units of the result confirm the setup was correct.
\[ \frac{36 \text{ dollars}}{4 \text{ hours}} = 9 \text{ dollars per hour} \]
A conversion factor is always a fraction equal to one, which is why multiplying by it changes the units but never the quantity.
Figure (svg): A chain of four fractions converting 45 miles per hour into feet per second, with the cancelling units struck through
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 5-5 — Use unit analysis with operations
Picture it
Converting 45 miles per hour into feet per second, one unit at a time.
Figure (svg): A chain of four fractions converting 45 miles per hour into feet per second, with the cancelling units struck through
Every fraction in that chain equals one: an hour really is sixty minutes, a mile really is 5280 feet. Multiplying by one is why the speed is unchanged even though the number is.
Worked example
Example 5c. Four conversion factors, each one chosen so that a unit cancels.
\[ \text{Convert } 45 \text{ miles per hour into feet per second.} \]
Write the given rate as a fraction with its units attached
Why: Units cannot cancel if they are not written down, so this step is not optional bookkeeping.
\[ 45 \text{mi} / 1 h \]
Multiply by one hour over sixty minutes
Why: Hours are upstairs, so the conversion factor must have hours downstairs for them to cancel.
\[ 45 \text{mi} / 60 \min \]
Multiply by one minute over sixty seconds
Why: Same reasoning one level down: minutes are now upstairs, so put minutes downstairs.
\[ 45 \text{mi} / 3600 s \]
Multiply by 5280 feet over one mile
Why: Miles are upstairs and the answer needs feet, so miles go downstairs in the last factor.
\[ 45 \times 5280 \text{ft} / 3600 s \]
Do the arithmetic that is left
Why: Every unit except feet and seconds has cancelled, so the surviving unit confirms the setup before the number is even computed.
\[ \frac{237600}{3600} = 66 \]
Figure (svg): The solution to Worked example miles per hour into feet per second shown as a ladder of expressions, one row per algebraic move
\[ 45 \; \frac{\text{mi}}{\text{h}} = 66 \; \frac{\text{ft}}{\text{s}} \]
Verify: run the check backwards
Why: Sixty-six feet per second times 3600 seconds per hour is 237,600 feet per hour, and dividing by 5280 feet per mile gives exactly 45 miles per hour. Coming back to the starting value confirms both the arithmetic and the direction of every factor.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 5-5
Prediction
You are converting 4 gallons into pints, and there are 8 pints in a gallon.
Predict first
Which setup cancels correctly?
Correct: 4 gal times (8 pt / 1 gal), giving 32 pints.
\[ 4 \; \text{gal} \cdot \frac{8 \text{ pt}}{1 \text{ gal}} = 32 \text{ pt} \]
A second check: a pint is smaller than a gallon, so the count of pints must be larger than the count of gallons. Thirty-two is larger than four, so the answer is at least the right size.
Why: Gallons are upstairs in the given quantity, so the factor must carry gallons downstairs for them to cancel, leaving pints. The second option leaves gallons squared over pints, which is not a volume. The direction of a conversion factor is never a matter of taste; it is forced by which unit you need to kill.
Worked example
Example 6a. One factor only, but the direction of that factor is the whole question.
\[ \text{Montpelier to Montreal is about } 132 \text{ miles. Convert to kilometres, using } 1 \text{ mile} \approx 1.61 \text{ km.} \]
Write the quantity with its unit
Why: One hundred and thirty-two miles, not just 132.
\[ 132 \text{mi} \]
Choose the factor that puts miles downstairs
Why: Miles are upstairs in the given quantity, so they must appear downstairs in the factor to cancel. That fixes the direction before any arithmetic.
\[ 132 \text{mi} \times(1.61 \text{km} / 1 \text{mi}) \]
Cancel miles and multiply
Why: Only kilometres are left, which is the unit that was asked for.
\[ 132 \times 1.61 = 212.52 \]
Round sensibly
Why: The conversion factor was given to three digits, so an answer to three digits is honest and 213 is the textbook's figure.
\[ 213 \text{km} \]
Figure (svg): The solution to Worked example miles into kilometres shown as a ladder of expressions, one row per algebraic move
\[ 132 \text{ mi} \approx 213 \text{ km} \]
Verify: convert the answer back
Why: 213 kilometres divided by 1.61 kilometres per mile is about 132.3 miles, which returns the starting distance to within rounding. Had the factor been used upside down the result would have been about 82, which is smaller than the mileage — impossible, since a kilometre is shorter than a mile so the count of them must be larger.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 5-5
Error analysis
A student converts 253 kilometres into miles. The arithmetic is done correctly; the setup is not.
Annotate
On: \( 253 \text{ km} \cdot \frac{1.61 \text{ km}}{1 \text{ mile}} = 407.33 \text{ km}^2 \text{ per mile} \)
Unit analysis is not extra work. It is the only check in this lesson that catches an error before you have committed to an answer.
Matching
Match each everyday question to the arrangement that answers it.
Match the pairs
Why: In every case the units decide the operation, not the wording. Dollars over hours gives a rate. Miles over miles-per-hour leaves hours, because the miles cancel. Hours times miles-per-hour leaves miles. And yards times feet-per-yard leaves feet. If you cannot remember whether to multiply or divide, write the units and let them tell you.
Estimation
Sixteen years, converted into seconds.
Predict first
Roughly how many seconds are in 16 years?
Correct: About 500 million seconds.
\[ 16 \cdot 365 \cdot 24 \cdot 60 \cdot 60 = 504\,576\,000 \]
Estimating first is not a shortcut around the arithmetic — it is a trap for the arithmetic's mistakes.
Why: A year is roughly 31.5 million seconds, a number worth carrying around, so sixteen years is about sixteen times that, which is a bit over 500 million. Doing it exactly with 365 days gives 504,576,000. Getting the order of magnitude first means a slipped decimal point in the long multiplication cannot survive unnoticed.
Constraint
The table of measures has been taken away. You remember only that there are 5280 feet in a mile.
Discussion prompt
You need to convert 90 kilometres per hour into miles per hour and you cannot look up the kilometre-to-mile factor. What could you use instead, and how would you know your answer is in the right range even if your remembered factor is slightly off?
Hint: You do not need the exact factor to bound the answer. What do you know about which unit is longer?
Answer:
A kilometre is shorter than a mile, so the count of miles must be smaller than 90. That bound holds no matter what factor you use, and it kills the most common error, which is multiplying when you should divide.
\[ 90 \; \frac{\text{km}}{\text{h}} \cdot \frac{1 \text{ mile}}{1.61 \text{ km}} \approx 56 \; \frac{\text{mi}}{\text{h}} \]
Even a rough remembered factor of 1.6 gives about 56, and a rough factor of 1.5 gives 60. Both are comfortably under 90, so the bound confirms the direction while the factor only affects the last digit.
Comparison
Fill the blanks from memory before you scroll back. The pairing is the thing worth memorising, not the individual lines.
Comparison matrix
| Property | Addition version | Multiplication version |
|---|---|---|
| Commutative | a + b = b + a | ab = ba |
| Associative | (a + b) + c = a + (b + c) | (ab)c = a(bc) |
| Identity | a + 0 = a | a times 1 = a |
| Inverse | a + (-a) = 0 | a times 1/a = 1, a not 0 |
| Closure | a + b is a real number | ab is a real number |
The distributive property is deliberately missing from this table: it is the only one that does not come in an addition-and-multiplication pair, because it already is one.
Pattern
Whether the question says graph, order, name the property or justify each step, the same five moves cover it.
Steps three and four are the pair that most students skip, and skipping them is why justify-each-step questions feel arbitrary.
OpenStax Algebra and Trigonometry 2e, §1.1 Real Numbers: Algebra Essentials §1.1
Check
Classification. Solve it before you click.
Check your understanding
Which of these numbers is rational?
Answer: B
Why: The square root of 25 is exactly 5, and 5 is an integer, so it is rational — the radical sign is doing no work. A radical only signals irrationality when the number under it is not a perfect square.
Check
Naming a property. Look at what moved.
Check your understanding
Which property does this statement illustrate? (6 times 5) times 7 = 6 times (5 times 7)
Answer: B
Why: The factors appear in the same order on both sides — 6, then 5, then 7 — and only the brackets changed position. Regrouping without reordering is exactly the associative property, and the operation throughout is multiplication.
Check
Unit analysis. Set it up before you compute.
Check your understanding
A car travels 60 kilometres per hour. Using 1 mile approximately 1.61 kilometres, what is its speed in miles per hour?
Answer: A
Why: Kilometres are upstairs, so the factor must carry kilometres downstairs: 60 divided by 1.61 is about 37.3. A mile is longer than a kilometre, so the count of miles must come out smaller than 60, which confirms the direction before the arithmetic.
Real world
A recipe calls for 2 cups of stock per serving, you are cooking for 7 people, and stock comes in 32 fluid ounce cartons. There are 8 fluid ounces in a cup.
Discussion prompt
How many cartons do you buy, and which step of the calculation is the one where unit analysis saves you? Say what unit each intermediate number carries.
Hint: Work in fluid ounces in the middle, and notice what happens at the very end when you have to round.
Answer:
\[ 7 \text{ servings} \cdot \frac{2 \text{ cups}}{1 \text{ serving}} = 14 \text{ cups} \]
\[ 14 \text{ cups} \cdot \frac{8 \text{ fl oz}}{1 \text{ cup}} = 112 \text{ fl oz} \]
\[ 112 \text{ fl oz} \cdot \frac{1 \text{ carton}}{32 \text{ fl oz}} = 3.5 \text{ cartons} \]
Unit analysis fixes the direction of all three factors. The rounding at the end is the one judgement it cannot make for you: 3.5 cartons means buying 4, because a carton is not divisible at the shop. That is the difference between a number and an answer.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Is every integer a rational number, and is every rational number an integer?
Correct: Every integer is rational, but not every rational is an integer.
\[ -27 = \tfrac{-27}{1} \in \mathbb{Q} \qquad \tfrac{1}{2} \in \mathbb{Q}, \; \tfrac{1}{2} \notin \mathbb{Z} \]
Why: Any integer n can be written as n over 1, which is a quotient of integers, so every integer is rational. The reverse fails immediately: one half is a quotient of integers but is not an integer. That asymmetry is exactly what the nested boxes picture is drawing, and it is the reason those arrows point one way only.
Explain it
They know how to add and subtract negatives, but they have never heard the word associative.
Discussion prompt
In no more than four sentences, and using no symbols, explain the difference between the commutative and the associative property. Then give them one test they can apply to any statement to decide which it is.
Hint: The test is one question about what changed between the two sides.
Answer:
A usable answer: commutative means you may change the order of the things being added or multiplied. Associative means you may change which pair you combine first. The test is a single question — did the items change seats, or did only the brackets move?
If they can apply that test, they can name every property in this lesson without memorising the table.
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: Classification is fixed by evaluating radicals before judging them. Ordering is fixed by drawing the line rather than reasoning about digits. Naming properties is fixed by the two-question test: did the letters move, or only the brackets. Conversion direction is fixed by writing the units down and asking which one you need to cancel. 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
Draw one long horizontal number line across the middle of the page. Above it, sketch the nested boxes — whole inside integer inside rational, with irrational beside them — and put one example number from this lesson in each region. Below the line, write the five property names in a column, and beside each write its addition version and its multiplication version from memory. Then, in the bottom corner, write the two rewrite arrows: a minus b becomes what, and a divided by b becomes what. Finally, draw a box around the one entry in the whole page that carries a condition, and write the condition next to it.
The boxed entry should be the multiplicative inverse, and the condition should be that the number is not zero. If you boxed something else, go back to Section 4.
Recap
Five things, and the fifth one is the one that keeps checking your work for the rest of the year.
| If the question says | Your first move is |
|---|---|
| Graph these numbers | Convert every one to a decimal |
| Order from least to greatest | Split by sign, then sort each group |
| Identify the property | Ask whether letters or brackets moved |
| Justify each step | Rewrite the subtraction or division first |
| Convert the units | Write the units, then pick the factor that cancels |
Chapter 1 continues by putting these moves to work on expressions and then on equations. Nothing new gets added to the list of legal moves — the chains just get longer.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.1 Apply Properties of Real Numbers §1.1, pp. 2-9 — everything on these slides traces back here
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