Chapter 3.2 — Exercise 3.1 — Basic Concepts

Five Quick-Recall Facts

None of these five need working out — they're facts from the introduction that later questions in this exercise, and later chapters entirely, keep assuming are already fluent.

True or False, With the Exact Axiom Named

Question 2 tests whether a familiar-sounding statement actually matches Euclid's wording, or only sounds like it does.

The first row is the one most often answered wrong, and it's worth seeing exactly why: Postulate 1 is about two distinct points, not one. Through a single point alone, nothing stops a line from being drawn in any direction at all — rotate a ruler around a pin stuck in a page, and every angle it passes through gives a different valid line through that same point.

A Whole Made of Seven Parts

Question 3 gives eight points A through H, in that order, lying on one straight line, and asks for AH > AB + BC + CD to be shown.

Since the points lie in order along the line, AH = AB + BC + CD + DE + EF + FG + GH — the full length is the sum of every small segment between consecutive points. AB + BC + CD is only three of those seven pieces, meaning it's a genuine part of the whole. By the same "whole is greater than a part" axiom used already in Question 2, AH > AB + BC + CD follows immediately, with no measurement needed at all.

The Midpoint Proof, by Substitution

Question 4 places Q between P and R with PQ = QR, and asks for PQ = ½PR to be proved.

From the figure, PR = PQ + QR. Since PQ and QR are given as equal, QR can be replaced by PQ, turning the equation into PR = PQ + PQ = 2PQ. Dividing both sides by 2 gives PQ = ½PR — a one-line substitution once QR has been swapped out.

Constructing an Equilateral Triangle

Question 5 asks for an equilateral triangle with every side 5.2 cm, built with compass and straightedge rather than measured out freehand.

• Draw line segment AB of length 5.2 cm.
• Draw a circle of radius 5.2 cm centred at A.
• Draw a second circle of radius 5.2 cm centred at B, intersecting the first at C.
• Join A to C and B to C.

Since C sits on both circles, AC and BC are each a radius of one of the two equal circles — 5.2 cm each — while AB was drawn to that length directly. All three sides equal, so triangle ABC is equilateral, and the whole construction rests on nothing more than Postulates 1 and 3: a unique line through two points, and a circle from any centre and radius.

A Conjecture, By Definition and by Example

Question 6 asks for a conjecture to be defined and illustrated. A conjecture is a statement neither proved nor disproved — an educated guess that has held up in every case checked so far, without a general argument establishing it for every case. The Goldbach Conjecture is the standard example: every even number greater than 4 can be written as the sum of two primes (6 = 3+3, 8 = 3+5, 10 = 3+7 or 5+5, and so on for every even number checked, without exception found yet — and without proof either).

Infinitely Many Parallels, and When Two Lines Must Meet

Question 7 marks two points P and Q, draws the line through them, and asks how many lines can be drawn parallel to it. The answer is infinitely many — a parallel line exists at every possible distance from PQ, on either side, and distance itself can take infinitely many values.

Question 8 gives a transversal n crossing two lines l and m, with interior angles ∠1 and ∠2 on one side summing to less than 180°. By Euclid's fifth postulate, that's exactly the condition under which l and m, extended far enough, meet — and they meet specifically on the side where the angle sum falls short of 180°.

Chaining Equalities Through a Shared Third Value

Question 9 gives ∠1 = ∠3, ∠2 = ∠4, and ∠3 = ∠4, and asks for the relationship between ∠1 and ∠2. Since ∠1 equals ∠3, and ∠3 equals ∠4, and ∠2 also equals ∠4, both ∠1 and ∠2 are tied to the same value (∠4) — so ∠1 = ∠2, by the axiom that things equal to the same thing are equal to one another.

Halves of Equal Segments

Question 10 places X on AB and Y on BC of a triangle, with BX = ½AB, BY = ½BC, and AB = BC given, then asks for BX = BY.

Since AB = BC, taking half of each side keeps them equal: ½AB = ½BC (halves of equal things are equal). But BX and BY were defined to equal exactly ½AB and ½BC respectively, so both are tied to the same value — giving BX = BY, by the same "equal to the same thing" axiom used in Question 9.

Where This Reasoning Style Reappears

Chaining equalities through a shared third value (Questions 9 and 10) is exactly the reasoning behind every congruence proof in the Triangles chapter, just applied there to whole triangles instead of single segments. The construction technique from Question 5 — two circles locating a third point — reappears in more elaborate form throughout Construction of Quadrilaterals, and the fifth-postulate reasoning behind Question 8 is the direct ancestor of every parallel-line angle relationship studied later in Class 9 and 10.