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The Elements, Book I
MONT 104Q – Mathematical
Journeys: Known to Unknown
Preliminaries, 1
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Following the Common Notions (Axioms) and
Postulates comes a sequence of Propositions and
their proofs.
Two traditional types of Propositions:
The “problems” show how to construct something
(and show the construction works)
The “theorems” claim something (and prove those
claims)
Copies of the Elements also contained extensive
scholia, or commentaries, added by later scholars.
We won't read them, but they were effectively part
of the Elements for later readers.
Preliminaries, 2
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Want to follow Euclid's logical structure, without
necessarily sticking that closely to the specific ways
he said things.
So in particular, we'll freely use modern symbolic
forms of geometric statements.
Example: Congruence of figures (line segments,
angles, triangles, etc.) We'll write something like
ΔAEB ≅ ΔBDA to say triangles are congruent. Euclid
doesn't do this – just says line segments, angles,
triangles are “equal.”
So, this might be a bit different from the online
translations of Euclid you're reading.
Proposition 1
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To construct on a given line segment AB an
equilateral triangle.
The Proof
1.
AC = AB, since radii of a circle are all equal (this
was stipulated in the definition of a circle).
2.
Similarly BC = AB.
3.
Therefore, AC = BC (Common Notion 1)
4.
Hence ∆ABC is equilateral (again, this was given
as a definition previously). QEF
A Question
Do we actually know that such a point C exists?
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The answer: strictly speaking, we don't (!) It does not
follow from any of the Postulates that Euclid set out at
the start – we would need additional postulates to assure
this.
 Euclid is pretty clearly appealing to our intuition about
physical circles here and either does not appreciate that
there is something missing, and/or does not want to
address that point at this stage.
 This is definitely a flaw of a sort, but perhaps justified
from educational/pedagogical orientation of the text.
(He doesn't come back to it later either(!))
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Propositions 2 and 3
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These are somewhat technical constructions aimed at
showing that the straightedge and “collapsing” compass are
sufficient for routine tasks such as measuring off a given length
on a given line.
Proposition 2. Given a line segment AB and a point P,
construct a point X such that PX = AB.
The construction uses Proposition 1 and Postulate 3
Proposition 3. Given two unequal line segments, lay off on the
greater a line segment equal to the smaller.
This construction uses Proposition 2 and Postulate 3 again, but
without using the compass to “transfer” the length
A natural Question
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It’s natural to ask: Why did Euclid go to the trouble of making
these somewhat involved constructions for relatively simple
tasks that would be easy if we had an implement like the
modern compass that can be used to measure and transfer
lengths in a construction?
The answer seems to be that his goal was to show that a very
small set of simple starting assumptions was sufficient to
develop the basic facts of geometry.
So some technical stuff would be acceptable at the start to
establish “routines” for those tasks under the more restrictive
working conditions or hypotheses.
The “SAS” Congruence Criterion
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Proposition 4. Two triangles are congruent if two sides of one
triangle are congruent with two sides of the other triangle, and
the included angles are also congruent.
The proof given amounts to saying: move the first triangle until
the sides bounding the two equal angles coincide, then the
third sides must coincide too.
This idea gives a valid proof, of course, but it again raises a
question: What in the Postulates says we can move any
figure? Again, there is physical intuition and/or an unstated
assumption being used here(!)
The “Pons Asinorum”
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Proposition 5. In any isosceles triangle, the base
angles are equal; also the angles formed by the
extensions of the sides and the base are equal.
Start of proof of Proposition 5
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By the construction CD = CE, and CA = CB by
hypothesis. Moreover ᐸ ACE = ᐸ BCD.
Hence ΔDCB ≅ ΔECA (Proposition 4)
It follows that BD = AE, ᐸ CBD = ᐸCAE, and
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CDB = ᐸ CEA (corresponding parts of congruent
triangles)
Then since CE = CD and CA = CB, we also have AD =
BE (Common Notion 3)
Therefore, ΔAEB ≅ ΔBDA (Proposition 4 – note the
angle at D is the same as the angle at E from the third
line above.)
Conclusion of proof of Proposition 5
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This shows that ᐸ DAB = ᐸ EBA, which is the second
part of the statement of the Proposition.
To finish the proof we must show that the base angles in
ΔCAB are equal.
ᐸ EAB = ᐸ DBA from the congruence of the small
triangles at the bottom of the figure.
But also ᐸ EAC = ᐸ DBC by the first congruence
established before.
Hence ᐸ BAC = ᐸ EAC - ᐸ EAB = ᐸ DBC - ᐸ DBA = ᐸ
ABC (Common Notion 3). QED
Propositions 6 and 7
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First, a converse of Proposition 5:
Proposition 6. If two angles of a triangle are equal,
then the sides opposite those angles are equal.
Euclid gives a proof by contradiction, using Propositions
3 and 4. Next:
Proposition 7. If in the triangles ΔABC and ΔABD,
with C and D on the same side of AB, we have AC =
AD and BC = BD, then C = D.
Another proof by contradiction, using Proposition 5; Euclid
gives only one case out of several.
Proposition 8
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Proposition 8. If the three sides of one triangle are
equal to the three sides of another triangle, then the
triangles are congruent.
This is the “SSS” congruence criterion
Proof is based on Proposition 7; in fact can almost see
that Euclid wanted to present the reasoning “broken
down” into easier steps by doing it this way.
Modern mathematicians call a result used primarily to
prove something else a “lemma.”
Next, a sequence of “bread-and-butter”
constructions
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Proposition 9. To bisect a given angle.
Angle Bisection – The Proof
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Proof: AB = AC by construction.
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BD = CD since the triangle ΔBCD is equilateral
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AD is common to the two triangles ΔABD and
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ΔACD
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Therefore, ΔABD and ΔACD are congruent (Proposition
8).
Hence <ADB = <ADC, and we have done what we set
out to do – the angle at A is bisected. QEF
Line Segment Bisection
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Proposition 10. To bisect a given line segment.
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Construction: Let AB be the given line segment
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Using Proposition 1, construct the equilateral triangle
ΔABC
Using Proposition 9, bisect the angle at C
Let D be the intersection of the angle bisector and AB.
Then D bisects AB.
Line Segment Bisection – Proof
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Proof:
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AC = BC since ΔABC is equilateral
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<ACD = <BCD since CD bisects <ACB
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The side CD is in both triangles ΔACD and ΔBCD
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Therefore, ΔACD and ΔBCD are congruent by
Proposition 4.
Hence AD = BD since the corresponding parts of
congruent triangles are equal. QEF