Session 03 · Postulates and Famous Theorems

Lesson 5 · A deductive system

Page 60

The B.C. cartoon is the whole argument in six panels: ecology is the cycle of life, a cycle moves in a circle, and a circle is to end up where you began, which is exactly what just happened to the reader. Every dictionary is circular, because every definition uses words that need definitions of their own. Page 48's vulgar fraction was a first taste of this; here it becomes structural.

Mathematics escapes the circle by refusing to define a few words at all. "Point," "line," and "plane" are the book's undefined terms: starting capital, spent but never earned. Everything session 1 defined (collinear, concurrent) was built from them, and the page restates those definitions in the iff form page 47 promised every definition deserves.

New here undefined term

Lesson 5 · A deductive system

Page 61

What circular definitions are to words, circular proofs are to statements, so a few statements must go unproved too. Those are the postulates, and the book starts with two: two points determine a line, and three noncollinear points determine a plane. The load-bearing word is "determine," which means exactly one. Two points do not merely allow a line; they leave no choice about which line.

The last paragraph assembles the whole machine, and it is the most important paragraph in the chapter: undefined terms feed definitions, definitions and postulates feed proofs, proofs yield theorems. That four-story structure is a deductive system, and it is what the word "geometry" will mean from now on. Euclid built his this way 2,300 years ago; the book is rebuilding it slowly, with you inside.

New here postulate · deductive system

Worked example · page 62

Questions 11 to 15: the duck pond table

Three dipping ducks support a glass table top. What postulate does the table illustrate, what plays each role, and what goes wrong if the ducks line up?

Question 11: a flat surface resting on three supports is Postulate 2: three noncollinear points determine a plane.
Questions 12 and 13: match the parts. The ducks' tail tips are the three points; the glass top is the plane they determine.
Question 14: slide the ducks into a row in your head. The glass now balances on a single line and can tip to either side of it; the top has lost its one fixed position. That wobble is what "determine" rules out.
Question 15: the word that fails is "noncollinear." Three points in a row are collinear, and the postulate deliberately excludes them.
Check yourself on question 6 the same way: the plumb line hangs through two points (the suspension point and the weight's pull), and Postulate 1 is why it has exactly one position. One postulate per object, and each object shows why the postulate's wording earns its keep.

Lesson 5 · Exercises, Sets I and II

Page 63

The intersecting cubes (19 to 25) drill the difference between "are" and "determine." Points B and C determine a line whether or not a third point joins them; F, E, and J are coplanar, but only noncollinear points determine a plane. Read each statement twice before judging it. The Japanese weights replay the dictionary problem with units: picul, kwan, tael, momme, each defined by the next, and question 31 asks why the chain must end somewhere real.

Obtuse Ollie's two curved "lines" meeting twice look silly, but Alice's objection has to cite a postulate, not a definition, and question 39 explains why: the book never defined "line" at all. Line is an undefined term; the postulate is all we have, and it is enough.

Worked example · page 63

Questions 32 to 36: the credit card agreement

Five statements from a credit card agreement. Which are definitions? Which are postulates? Which two combine into a syllogism, and what theorem does it prove?

Question 32: a definition names a word and gives its meaning. Statement (1) defines "transaction finance charge" and statement (3) defines "supercheck." Those two; question 33's answer is those two terms.
Question 34: the rest are rules laid down without proof, the agreement's postulates: (2), (4), and (5).
Question 35: hunt for a conclusion that feeds a hypothesis, the page 51 splice. Statement (2) ends "you will be charged a fee"; statement (4) begins "If you are charged a fee." They chain.
Question 36: run the syllogism. If you go over your credit limit, you will be charged a fee; if you are charged a fee, it will be added to your balance; therefore, if you go over your credit limit, a fee will be added to your balance. A theorem, proved from the bank's own postulates.
Notice what just happened: an entire deductive system (undefined terms, definitions, postulates, one theorem) was hiding in a credit card agreement. Page 61's structure is not special to geometry.

Worked example · page 64

Set III: how many lines do 20 points determine?

Twenty points sit on a circle, no three collinear. Sneak up on the number of lines through the small cases: 3, 4, 5, 6 points.

Count the small cases by drawing: 2 points give 1 line, 3 give 3, 4 give 6, 5 give 10, 6 give 15. The gaps between answers are 2, 3, 4, 5: each new point adds a line to every point already there.
So the seventh point will add 6 lines, the eighth 7, and the pattern is forced by Postulate 1 itself: one new line per old point, no more (each pair determines a line) and no fewer (exactly one).
Shortcut to 20 without adding eighteen numbers: each of the 20 points pairs with 19 others, giving 20 × 19 = 380 ordered pairs, but the pair AB and the pair BA name the same line. Divide by 2: 190 lines.
Check the formula against a case you drew: 5 points give 5 × 4 ÷ 2 = 10. It matches. A counting argument you can verify on small cases and then trust at 20 is deduction doing the work your pencil cannot.

Lesson 6 · Some famous theorems of geometry

Page 65

Theorema, Artmann says, is Greek kin to "theater": something seen. This lesson is a preview of coming attractions, four theorems every educated person meets, starting with the most famous of all. The Pythagorean Theorem is stated here the way the Greeks thought it, in words and areas: the square of the hypotenuse of a right triangle equals the sum of the squares of the other two sides. Pythagoras lived two centuries before Euclid; the theorem was known, as fact, to builders long before either.

You met c² = a² + b² in algebra as a formula to apply. Notice what this book does differently: it will not let the equation into the deductive system until it is proved, and that debt comes due in a later chapter. Until then the theorem has the same status Raoul's paper triangle will have on page 67: believed, used, not yet earned.

Khan Academy The Pythagorean theorem intro

Lesson 6 · Some famous theorems of geometry

Page 66

Three theorems on one page, each stated with a picture that suggests but does not prove. The Pythagorean figures actually draw the squares on the three sides: red area equals blue plus green, and the middle and right figures hint at why, by cutting and rearranging, the same move as Dudeney's puzzle back on page 23. The hypotenuse gets its formal introduction: the side opposite the right angle, always labeled c.

The Triangle Angle Sum Theorem says the three angles of any triangle total 180°, drawn as the three angles lining up along a straight edge. The two circle theorems (circumference πd, area πr²) complete the set. All four are claims you can test with scissors and a protractor tonight, and none of that testing is proof; hold that distinction, because page 67 asks about it point blank.

New here hypotenuse

YouTube The squares-on-sides figure, rebuilt with water

Lesson 6 · Exercises, Set I

Page 67

Pi gets its paragraph: about 3.14, exact value unknowable, decimal without pattern or end. Because π cannot be drawn, the circle theorems cannot even be suggested by pictures, which is the page's quiet concession that some truths outrun scissors entirely. The book then says plainly what it is doing: using these four theorems now, proving them later, and question 6's Greek stamp (a 3-4-5 right triangle in checkerboard squares) shows the Pythagorean claim in countable form.

The 1025 A.D. contest between Raoul and Raimbeau is the page's real lesson, and the worked example takes it up. Before opening it, answer for yourself: what would cutting the corners off a paper triangle demonstrate, and what would it fail to demonstrate?

New here pi · circumference

Khan Academy Circles: radius, diameter and circumference

Worked example · page 67

Questions 8 to 10: Raoul's paper triangle

Challenged to prove a theorem, Raoul drew a triangle, cut off its corners, and fit them together. What theorem was he after? Does his method prove it? Would a thousand triangles prove it?

Question 8: the three corners, reassembled, line up along a straight edge, the second figure on page 66 exactly. He was demonstrating the Triangle Angle Sum Theorem: the angles total 180°.
Question 9: no. The cutting shows that this one triangle's angles total a straight angle, as closely as scissors can show anything. It says nothing about the triangle Raoul did not draw.
Question 10: still no. A thousand triangles are a thousand instances, and the theorem claims every triangle, an infinite family. Piling up instances is induction; page 51 defined proof as deduction from accepted statements, and no pile of examples becomes a deduction.
Self-check: page 12's colored-pencil experiment already taught this moral from the other side. Seeing a pattern persist is the invitation to prove; it is never the proof. Keep Raoul in mind whenever a measurement agrees with a theorem suspiciously well.

Lesson 6 · Exercises, Set I

Page 68

Every problem on this page runs one of the four theorems, and the squares figures (16 and 17) are the Pythagorean Theorem with no algebra visible: the areas themselves add, 36 + 64 giving the hypotenuse square, 169 − 25 giving a missing leg square. The gliding bird and the stereo speakers run the angle sum with one angle known to be 90° or with two angles equal, and both end by asking for the answer "in terms of n," your first taste of proving a little formula instead of computing one number.

Questions 24 and 25 close the loop with page 47: stating the theorem as a conditional invites the converse question, and the converse (if a² + b² = c², the triangle is right) does not follow for free. It happens to be true, and it too will be proved separately, much later.

Worked example · page 68

Questions 16 and 17: the third square

Squares are drawn on the sides of a right triangle. In the first figure the two smaller squares have areas 36 and 64; in the second, the hypotenuse square is 169 and one leg square is 25. Find each missing area.

Read the theorem as an area statement: leg square plus leg square equals hypotenuse square. No side lengths needed; the areas do the arithmetic.
Figure 16: the missing square sits on the hypotenuse, so it is 36 + 64 = 100.
Figure 17: the missing square sits on a leg, so subtract: 169 − 25 = 144.
Check against hidden triangles you already know: areas 36, 64, 100 mean sides 6, 8, 10; areas 25, 144, 169 mean sides 5, 12, 13. Both are famous right triangles, scaled cousins of the 3-4-5 on the Greek stamp one page back.
Notice the direction of each move: add to reach the hypotenuse, subtract to reach a leg. Page 73's ladder question will want the adding direction with real feet.

Lesson 6 · Exercises, Set II

Page 69

The eye pupil question is the circle area theorem earning its keep, and the punchline is the factor of sixteen: the diameter grows four times, the area sixteen, because area runs on the square of the radius. It is the same length-versus-area lesson as page 21's square inches, now drawn on your own eye. The stretched cord (34 to 37) is the Pythagorean Theorem as ancient India wrote it, for the special case of a square's diagonal, and the two light-green squares literally fill the dark-green one.

The pencil experiment at the bottom is Raoul's scissors done with rotation: pivot a pencil through each angle of a triangle and it ends up pointing the opposite way, half a full turn, 180°. A demonstration to enjoy, and by now you can say precisely why it is not a proof.

Khan Academy Area of a circle

Worked example · page 69

Questions 26 to 28: the pupil of the eye

The pupil dilates from 2 mm to 8 mm in diameter. What is its approximate area at each size, and how many times as much light does the wide pupil admit?

The theorem wants the radius, and the questions give diameters. Halve first: 8 mm across is r = 4; 2 mm across is r = 1.
Question 26: area = πr² = π × 16 ≈ 50 square millimeters.
Question 27: area = π × 1 ≈ 3.14 square millimeters.
Question 28: divide, and π cancels: 16π ÷ π = 16 times the light. You could have skipped the areas entirely: the diameter ratio is 4, and area scales by its square, 4² = 16.
Self-check: the ratio question never needed π's value, which is why "approximately" only appears in 26 and 27. Exact ratios from an unknowable number; that is the algebra and the geometry splitting the work cleanly.

Lesson 6 · Exercises, Sets II and III

Page 70

Aryabhata, writing in sixth-century India, states the circle area as half the circumference times half the diameter, and question 41 asks you to check him against the book's theorems; do the substitution and watch πr² fall out. His second rule hides a value of π good to four decimal places, a thousand years before decimal notation reached Europe. Dilcue, the "backward geometry student," is a double joke: read his name backward, then read his answer backward.

Set III's hinged pieces swing between two arrangements, and saying precisely what the two pictures together demonstrate (and, after Raoul, precisely what they do not) is the whole exercise. The chapter's two big ideas, deduction and demonstration, end the lesson sitting side by side on hinges.

Worked example · page 70

Questions 41 and 42: checking Aryabhata

Aryabhata: the area of a circle is half the circumference times half the diameter. Is he right? And if 8 × (100 + 4) + 62,000 were exactly the circumference of a circle of diameter 20,000, what would π be?

Question 41: write his recipe in the book's symbols. Half the circumference is ½(πd) and half the diameter is r, so his area is ½πd × r.
Replace d with 2r: ½π(2r) × r = πr × r = πr². Exactly the page 66 theorem; Aryabhata is correct.
Question 42: follow his arithmetic. 100 + 4 = 104, times 8 is 832, plus 62,000 is 62,832. So π would equal circumference over diameter: 62,832 ÷ 20,000 = 3.1416.
Compare: π is 3.14159..., so his rule is off by less than one part in 100,000, from the sixth century. And note which theorem his recipe leans on: circumference = πd is doing the work, the same theorem you used, because c ÷ d is what π means.

Chapter review · Summary and exercises

Page 71

The summary column is a self-test disguised as a list. Read each of the fourteen Basic Ideas and say its meaning aloud before believing you remember it; every one carries a page number for the ones that hesitate. Then notice the architecture of the list itself: two postulates, then four theorems, the exact page 61 hierarchy, with the theorems still marked by the page where their proof-debt was acknowledged.

The exercises begin immediately with translation practice ("all limericks have five lines" is page 44's koala move) and a Giamatti syllogism to splice. The three birds on the power line are Postulate 1 dressed as a photograph; question 11 asks you to name the kind of statement your answer rests on, which is the review's real question: not what is true, but what kind of true.

Worked example · page 71

Questions 4 to 6: Giamatti's syllogism

"If we have known freedom, then we love it. If we love freedom, then we fear its loss." What is the pattern of a syllogism, what is this one's conclusion, and what might allow a syllogism's conclusion to be false?

Question 4: the pattern, from page 51: a → b, b → c, therefore a → c.
Check the splice before using it: premise one ends "we love it (freedom)" and premise two begins "if we love freedom." The conclusion of one is the hypothesis of the other, so the chain is legal.
Question 5: jump from first hypothesis to last conclusion. If we have known freedom, then we fear its loss.
Question 6: only one thing can betray a syllogism whose form is right: a false premise. The logic never fails; the inputs can. That was Aristotle's statue lesson on page 52, and it is the answer the review wants in one sentence.

Chapter review · Exercises, Set I

Page 72

Morton Salt's slogan gets the full Lesson 1 treatment: name the figure, read it as "if a, then b," rewrite it, take its converse, and say whether the converse rides free. It does not; that is the one-way lesson of page 47 checked one last time. The Spanish geometry sentence is friendlier than it looks: cuadrado, hipotenusa, and igual give themselves away, and the todo-entero-completo dictionary circle is B.C.'s ecology cartoon in another language, which is questions 25 to 27's point.

The SAT triangle in the middle of the page is the review's best computation, three applications of the angle sum in a row, and the worked example takes it slowly. The courtroom transcript beginning at the bottom carries onto page 73; read it straight through once before answering anything.

Worked example · page 72

Questions 17 to 19: the SAT triangle

In triangle ABC, angle A = 48° and angle C = 32°. Segment BM splits angle B into two equal parts marked x°. Find angle ABC, x, and angle BMC.

Question 17: angle sum in the big triangle. ∠ABC = 180 − 48 − 32 = 100°.
Question 18: the two marked angles are equal and total 100°, so x = 50.
Question 19: work inside triangle ABM. Its angles are 48°, x = 50°, and ∠AMB, so ∠AMB = 180 − 48 − 50 = 82°.
∠BMC sits beside ∠AMB on the straight line AC, and angles along a straight line total 180°: ∠BMC = 180 − 82 = 98°.
Check with the other triangle: BMC's angles should be 50, 32, and 98, and 50 + 32 + 98 = 180. It closes. One theorem, used three times, each answer feeding the next; that is what "in terms of the figure" questions reward.

Chapter review · Exercises, Sets I and II

Page 73

The courtroom wrap-up (28 to 30) is an error hunt, and everyone in the transcript deserves a look, including the judge, whose "3.1214" scrambles 3.1416. Lillian Lieber's verse about Euclid is the chapter's structure poem, and questions 31 to 33 ask you to name its parts: postulates, theorems, direct and indirect proof. If any of the four names hesitates, its page number is one column to the left, on 71.

Then the Pythagorean finale: squares of 144 and 256 summing to 400, Heron's too-short ladders (the engineer's version of the same complaint), and the Rhind papyrus, where Egypt's circle-area recipe quietly carries its own value of π for question 38 to extract, some three thousand years before Aryabhata's 3.1416 on page 70.

Worked example · page 73

Question 37: Heron's ladder

A ladder just reaches the top of a wall 16 feet high, and the closest its base can stand to the foot of the wall is 12 feet. How long must the ladder be?

Draw what the figure draws: wall vertical, ground horizontal, the ladder as hypotenuse. The right angle is between wall and ground, so the ladder is c and the legs are 12 and 16.
Add the leg squares, page 68's adding direction: 12² + 16² = 144 + 256 = 400.
The ladder's square is 400, so the ladder is 20 feet, since 20 × 20 = 400.
Check against the family: 12-16-20 is 3-4-5 scaled by four, the stamp triangle yet again. Heron's soldiers needed exactly this computation before an assault, which is as applied as geometry gets.
Compare with the sliding ladder on page 17: there you measured angles as the ladder slid; here one theorem replaces the measuring. That trade, measurement for deduction, is the chapter's whole argument.

Chapter review · Exercises, Set II

Page 74

Two proof skeletons to finish, one of each kind, and naming which is which (questions 41 and 43) matters as much as filling the blanks. The panda proof chains forward: direct. The Clue proof opens by supposing Colonel Mustard didn't do it and rides to a contradiction with the 4 P.M. fact: indirect, the page 56 form exactly. Notice the panda theorem is a pun on a grammar joke; the logic grades out the same either way.

The circle in the square delivers the review's prettiest fact: the circle claims the same fraction of the square's perimeter as of its area, π/4 of each, about 78 percent. Lewis Carroll's tiny planet then runs Eratosthenes backward: from a known circumference to a diameter, dividing by π where page 15 multiplied.

Worked example · page 74

Questions 44 to 49: the circle in the square

A circle of radius r is drawn inside a square, touching all four sides. Find the circumference, the square's perimeter, and what fraction one is of the other; then the same for the areas.

The circle touches all four sides, so the square's side is the circle's diameter, 2r.
Questions 44 and 45: circumference = 2πr; perimeter = 4 × 2r = 8r.
Question 46: 2πr ÷ 8r = π/4. The r cancels, so the answer holds for every such circle, not just one.
Questions 47 and 48: area of circle = πr²; area of square = (2r)² = 4r². Watch the exponent: the whole side gets squared, a page 21 trap in new clothes.
Question 49: πr² ÷ 4r² = π/4 again, about 0.785. Same fraction twice is a coincidence worth staring at; it says the circle is exactly as efficient with the square's boundary as with its interior. Carry π/4 ≈ 78% as a number worth knowing.

Algebra review · Solving linear equations

Page 75

The chapter closes with an algebra tune-up, and its placement is a promise about chapter 3: the geometry ahead will hand you equations (an angle in terms of n on page 68 was the first hint), and the book wants the solving to be automatic before then. The distributive rule and the four-step recipe are the entire page: clear parentheses, simplify each side, gather the variable, divide it free.

Read Example 2 the way you now read proofs, because it is one: each line follows from the line before by a stated reason, written in the right margin like page 56's margin notes. Algebra done with reasons is deduction with numbers, and that is precisely why the review lives in this chapter and not the back of the book.

Worked example · page 76

Question 25: the four steps in one equation

Solve 4(x − 11) = 3x + 16.

Step 1, distribute: 4x − 44 = 3x + 16.
Step 2, simplify each side: nothing to combine; both sides are already tidy.
Step 3, gather the variable: subtract 3x from each side to get x − 44 = 16, then add 44 to each side: x = 60.
Step 4 (divide the variable free) was not needed; the coefficient was already 1. Not every equation uses all four steps, but the steps always come in this order.
Check the way the book checks: substitute back. 4(60 − 11) = 4 × 49 = 196, and 3(60) + 16 = 196. Equal, as Recorde's parallel lines insist. Now try question 26 and watch the x² terms erase each other before your eyes.