Ch 3.1 · Equality and the Ruler

Lesson 1 · Number operations and equality

Page 78

Chapter 3 opens with a cartoon of two scientists staring at "Let K = K" on a blackboard: "That, in itself, is a breakthrough." The joke is the lesson. The most obvious statements in algebra (a = a, a + b = b + a) are exactly the ones that get accepted without proof, and the book now gives them their formal rank: they are postulates of algebra, also called axioms, the same rank page 61 gave "two points determine a line."

Look at the five listed properties and sort them yourself before the text does it for you: which are simply accepted, and which could be proved from the others? The last two (squaring a binomial, factoring a difference of squares) are provable, so they are algebra's theorems. The postulate-versus-theorem distinction from chapter 2 was never specifically geometric, and this chapter runs on both kinds at once.

New here axiom

Lesson 1 · Number operations and equality

Page 79

Six named tools, and you will cite them by name for the rest of the book: the reflexive, substitution, addition, subtraction, multiplication, and division properties of equality. Each is one move you are allowed to make on an equation, and a = b means nothing deeper than "a and b are the same number," which is why every one of them is believable at sight. Note the fine print on division: c cannot be 0, and page 80 makes you feel why.

The worked triangle at the bottom is the template the whole chapter imitates: each line is a statement, each parenthesis names the property that justified it. Cover the reasons with your hand and supply them from the list, because "name the reason" is precisely the skill the exercises grade, and it is the two-column proof of page 86 in embryo.

New here properties of equality

Lesson 1 · Exercises, Set I

Page 80

Naming practice first (1 to 12), and the figures are already geometric: the two crossing lines in questions 8 to 10 are quietly proving that vertical angles are equal, a theorem this chapter will make official. Question 8's move (subtract the shared angle from both sides) is the subtraction property doing real geometric work for the first time.

Then the page earns its keep: questions 13 to 17 show what division by zero would cost. Each fraction is rewritten as a multiplication question, and the multiplication question is the honest one. Work all four before the worked example, and end with Euclid's "common notion," which turns out to be a two-line theorem in our system rather than a new assumption.

YouTube TED-Ed: Why can't you divide by zero?

Worked example · page 80

Questions 13 to 17: what division by zero would mean

Fill in each ?: 6/3 = ? because 3 × ? = 6. 6/0 = ? because 0 × ? = 6. 0/6 = ? because 6 × ? = 0. 0/0 = ? because 0 × ? = 0. What is strange about the last one?

Question 13 sets the pattern: every division is secretly a multiplication question. 6/3 = 2 because 3 × 2 = 6.
Question 14: 6/0 asks for a number with 0 × ? = 6. Zero times anything is 0, never 6, so no number works at all.
Question 15: 0/6 asks for 6 × ? = 0, and ? = 0 works fine. Dividing zero is legal; dividing by zero is not.
Question 16: 0/0 asks for 0 × ? = 0, and now every number works: 0, 7, a million. Question 17's strangeness is the opposite disease: not no answer, but all of them.
Now reread the division property on page 79 and notice the clause "and c ≠ 0" was carrying all of this. One fraction has no value, the other has every value, and either would let you prove 2 × 2 = 5, which is literally what happens on page 83.

Lesson 1 · Exercises, Sets I and II

Page 81

The "correcting mistakes" block is chapter 2 cashing a check: to show x = 13 is wrong, the book assumes it is right, substitutes, and rides to 33 = 19, a contradiction. Checking an answer by substitution is an indirect proof, and question 22 wants that name. It also wants you to see the original error: 2x = 26 came from adding 7 when 7 had to be subtracted.

Set II turns equality's properties loose on areas. The crescents figure gives one equation and asks what subtracting the shared regions leaves; the binomial square makes (a + b)² = a² + 2ab + b² visible as one big square carved into four pieces, page 78's "theorem of algebra" drawn instead of expanded. The 1607 Chinese figure is the Pythagorean Theorem, and page 82 will walk its proof.

Worked example · page 81

Questions 23 and 24: the triangle and the crescents

The five regions satisfy I + II + IV + V = II + III + IV. The light blue regions are I and V (the crescents); the dark blue is III (the triangle). How do their areas compare, and why?

Write what is given: I + II + IV + V = II + III + IV. The left side is the two half-circles on the triangle's legs; the right side is the half-circle on its hypotenuse. (That equality is the Pythagorean Theorem wearing half-circles instead of squares.)
Subtract II from both sides: I + IV + V = III + IV. Reason: the subtraction property.
Subtract IV from both sides: I + V = III. Same reason again.
Read the result in words for question 23: the two crescents together have exactly the area of the triangle. No curve was ever measured; two subtractions did everything, which is question 24's answer.
This result has a name worth knowing, the lunes of Hippocrates, proved about 440 B.C., the first curved region in history whose exact area was found. You just reproved it with the subtraction property.

Worked example · page 82

Questions 44 to 46: Dilcue's pie

Dilcue computed the circumference of a giant pie using the wrong expression, πr², and still got the right answer. How? What was the circumference, and what was the area?

Set the wrong expression equal to the right one. The right expression for circumference is 2πr, so his luck requires πr² = 2πr.
Divide both sides by πr (legal, since r is not 0): r = 2. The pie had a radius of exactly 2 feet, which answers question 44: only that one radius forgives the mistake.
Question 45: circumference = 2πr = 4π ≈ 12.6 feet.
Question 46: area = πr² = 4π square feet, the same number with different units, which is the whole joke: at r = 2 the two formulas coincide numerically.
Self-check with page 21's lesson: feet and square feet measure different things, so "the same answer" is a coincidence of numbers, not of quantities. Dilcue was lucky once; the units would have caught him.

Worked example · page 83

Set III: rescuing Lewis Carroll

With x = y = 1: 2(x² − y²) = 0 and 5(x − y) = 0, so 2(x² − y²) = 5(x − y). Divide both sides by (x − y) to get 2(x + y) = 5, so 4 = 5. Where is the crime?

Audit the opening: with x = y = 1, both products really are 0, and 0 = 0, so setting them equal is the substitution property used correctly. So far, sound.
The factoring is page 78's property 5: x² − y² = (x + y)(x − y). Also sound.
Now the division. The division property (page 79) says: if a = b and c ≠ 0, then a/c = b/c. Here c is (x − y) = 1 − 1 = 0. The property's hypothesis fails, so the step has no justification.
Everything after the illegal step is garbage dressed in good algebra: 2(x + y) = 5 was never established, so 4 = 5 never follows.
Answer Carroll the way page 52 answered Aristotle: the logic of each later step is fine, but a chain is no stronger than its weakest link, and this chain's third link divides by zero. That is why the division property carries its c ≠ 0 clause like a loaded warning.

Lesson 2 · The ruler and distance

Page 84

Sally measures Snoopy's mouth ("one lip is on the six and the other lip is on the nine") and cannot figure out what to write down. The strip is the lesson: a ruler does not hand you a distance, it hands you two numbers, and the distance is something you compute from them. Page 85 will make that official.

The historical note deserves a pause: there are no measurements anywhere in Euclid. No ruler, no protractor, no numbers on segments at all; he only compared things. Our book departs from Euclid here, on purpose, because computer graphics and every applied use of geometry run on coordinates. This lesson is where numbers formally enter the deductive system, and they enter the same way everything enters: by postulate.

Lesson 2 · The ruler and distance

Page 85

Three definitions and a postulate, and the chapter's machinery is live. A straightedge draws the line two points determine; a ruler is a straightedge with numbers. Each point gets a coordinate, the distance between two points is the positive difference of their coordinates, and the Ruler Postulate (Postulate 3) guarantees a line can always be numbered this way. Snoopy's mouth: 9 − 6 = 3 inches.

Then coordinates buy their first definition: betweenness. A-B-C means b is between a and c as numbers, either a < b < c or a > b > c. Note the "on the same line" clause; page 88's bent pole is waiting for anyone who skips it. The U-S-A example at the bottom (US + SA = UA, since 5 + 3 = 8) is the seed of the theorem on the next page.

New here straightedge · coordinate · distance · betweenness

Lesson 2 · Theorem 1, then exercises

Page 86

Here is the book's first formally numbered theorem, and its first two-column proof: statements on the left, a reason for each on the right. The Betweenness of Points Theorem says if A-B-C then AB + BC = AC, and the proof is six lines: unpack the definition, apply the Ruler Postulate twice, add, substitute. Read it slowly and check that every reason is something already in the system (a definition, a postulate, or algebra). This format is what "proof" will look like from now on.

Notice what the theorem buys: an obvious-looking fact about walking through B on the way to C now rests on named foundations, so it can be cited by name, and question 47 on page 89 will make you prove the mirror-image case yourself.

Worked example · page 86

Questions 6 to 8: the triple jump

The jumper takes off at A (coordinate 40) and lands at B (47), C, and D (58). The hop from B to C is 5 meters. Find AD, the coordinate of C, and CD.

Question 6: AD is a distance, so subtract coordinates: 58 − 40 = 18 meters for the whole triple jump.
Question 7: C sits 5 meters past B, so its coordinate is 47 + 5 = 52.
Question 8: CD = 58 − 52 = 6 meters.
Now audit with the new theorem: A-B-C-D in order, so the three phases should sum to the whole. Hop 47 − 40 = 7, step 5, jump 6, and 7 + 5 + 6 = 18 = AD. The Betweenness of Points Theorem is the reason that check is guaranteed to work, which is exactly what questions 9 to 11 have you write out.

Lesson 2 · Exercises, Set I

Page 87

Four scenarios, each testing whether numbers on a thing are honest coordinates. The eclipse questions are betweenness with astronomical data (an eclipse of the sun is sun-moon-earth in a line, and the two figures differ only in the order of M and E). The seven league boots build the expression 4 + 7n, your first coordinate formula. Then two traps: the skyscraper and the football field, where the printed numbers fail the Ruler Postulate in two different ways.

The football field's yard lines run 10, 20, ..., 50, then back down to 10, so two different lines wear the same number and differences stop measuring distance (questions 23 to 25); the renumbering on page 88 repairs it with negatives. The skyscraper's flaw is sneakier, and the worked example is there.

Worked example · page 87

Questions 20 to 22: the skyscraper's missing floor

Using floor numbers as coordinates, find the distance in floors from the 2nd to the 9th floor, and from the 5th to the 15th. Are both answers correct?

Question 20: 9 − 2 = 7 floors. Count the drawn floors in the figure to confirm: correct.
Question 21: 15 − 5 = 10 floors, says the subtraction.
Question 22: now look at the figure's numbers: ... 11, 12, 14, 15. There is no 13th floor; the builders skipped the unlucky number. Climbing from 5 to 15 you pass 9 actual floors, not 10.
Say the failure in the postulate's own words: the Ruler Postulate demands points numbered so that differences measure distances. The floor numbers skip a value, so they are labels, not coordinates, and differences across the gap lie by one.
Keep the two failure modes straight: the football field reuses numbers, the skyscraper skips one. Real rulers may do neither, which is exactly what the postulate asserts and why it had to be a postulate at all.

Lesson 2 · Exercises, Sets I and II

Page 88

The renumbered football field (−50 to 50) fixes yesterday's trap: with A at −20 and B at 20, the positive difference is 40 yards, and negative coordinates cost nothing. That is worth noticing: the Ruler Postulate never said coordinates must be positive, only that differences must measure distances. The SAT number line runs the same idea under time pressure.

Set II's pole vaulter is the page's teacher. The vaulter's pole bends, and the hand positions that look "between" the ends fail the definition on page 85, which required the points to lie on one line. Questions 32 and 34 ask you to reproduce the definition and the theorem from memory first; do that honestly before touching the numbers.

Worked example · page 88

Questions 35 and 36: the bent pole

On the bent pole, AB = 1.4 m, BC = 5.0 m, and AC = 6.2 m. How does AB + BC compare with AC? Could AB + BC ever be less than AC?

Add: AB + BC = 1.4 + 5.0 = 6.4, and AC is 6.2. The sum is larger.
Why no contradiction with Theorem 1? The theorem's hypothesis is A-B-C, betweenness, and betweenness lives on a line. B sits on the curve of a bent pole, so the theorem is silent here. A theorem never owes you anything when its hypothesis fails.
Question 36: imagine straightening the pole. The detour through B shrinks until, on the straight line, 1.4 + 5.0 would land exactly on 6.4 = AC. Bending can only lengthen the trip through B, never shorten it, so AB + BC < AC cannot happen.
You have just met, informally, the triangle inequality: the straight path is never beaten. The book will formalize it later; for now it is your sanity check on every distance figure.

Lesson 2 · Exercises, Set II

Page 89

The extension ladder (38 to 41) is a two-line gem: write AC as AB + BC, write BD as BC + CD, set the two ladder halves equal, subtract the shared middle, and the equal-looking end segments are proved equal. Overlapping segments handled by subtraction is a move you will see in triangle proofs for chapters to come. The rung ladder (42 to 44) is coordinate bookkeeping; watch the first and last gaps of 11 inches against the 12-inch spacing.

The stars proof brings back page 56's skeleton with light-years in it, and the signal problem converts 3 seconds at 44 feet per second into a betweenness statement. Question 47 then hands you the honor of proving Theorem 1's other case (a > b > c) yourself, your first written two-column proof.

Worked example · page 89

Question 45: the stars that only look lined up

XY = 7.2, YZ = 9.8, XZ = 16.6 light-years. Complete the indirect proof that star Y is not between star X and star Z.

Beginning assumption, the opposite of the claim: suppose Y is between X and Z, that is, X-Y-Z.
Next line, from the Betweenness of Points Theorem: if X-Y-Z, then XY + YZ = XZ.
Substitute the measured distances: 7.2 + 9.8 = 16.6.
The contradiction: 7.2 + 9.8 = 17.0, and 17.0 ≠ 16.6.
Therefore the assumption is false: star Y is not between the other two, no matter how the night sky looks from here. Compare Orion's belt on page 16, where "apparent" did the same work by eye; now a theorem and 0.4 light-years settle it deductively.

Worked example · page 90

Set III: sizing the oars

Oarlocks are 42 inches apart (BC), and the handles should overlap 4 inches (AD). How long are AB and CD? If the inboard part AB must be 7/25 of the oar's length, how long is the oar, and what length do you order (6 to 10 feet, in 6-inch steps)?

Question 1: look at the overhead figure's order: C, A, D, B. Each inboard section spans from an oarlock past the middle overlap. The span CB = 42 covers CD and AB, but the overlap AD lies inside both, so AB + CD = 42 + 4 = 46. The two oars match, so AB = CD = 23 inches.
Question 2: the company's rule says AB is 7/25 of the whole oar. With the oar drawn as 25X and the inboard part as 7X: 7X = 23, so X = 23/7 ≈ 3.29 inches.
The oar is 25X ≈ 25 × 3.29 ≈ 82.1 inches, about 6 feet 10 inches.
Question 3: order from the catalog's 6-inch steps: 7 feet (84 inches) is the nearest length, and slightly long beats slightly short in an oar. Betweenness, subtraction, and one proportion turned a rowboat problem into arithmetic; that is the whole lesson in one purchase.