Download Slides on Elements, Book I

The Elements, Book I
MATH 392 – Geometry Through
History
Feb. 1 and 3, 2016
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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 on the course homepage.
The 5 Common Notions
1. Things that are equal to the same thing are
equal to one another.
2. If equals be added to equals, the wholes are
equal.
3. If equals be subtracted from equals, the
remainders are equal.
4. Things that coincide with one another are
equal to one another.
5. The whole is greater than the part.
The First Four Postulates
1. [It is possible] to draw a straight line from
any point to any point.
2. [It is possible] to produce any finite straight
line continuously in a straight line.
3. [It is possible] to describe a circle with any
center and given distance as radius.
4. All right angles are equal (i.e. congruent)
to one another.
The Fifth Postulate
5. If a straight line falling on two straight lines
makes the angles on the same side less than
two right angles, the two straight lines, if
produced indefinitely, meet on that side on
which are the angles less than the two right
angles.
The situation in Postulate 5
After Producing the lines sufficiently far,
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Comments
1.
2.
3.
Postulates 1, 2, and 3 describe the constructions
possible with an (unmarked) straightedge and a
“collapsing” compass – that is the compass can
be used to draw circles but not to measure or
transfer distances
Postulate 4 is a statement about the
homogeneous nature of the plane – every right
angle at one point is congruent to a right angle at
any other point
Postulate 5 is both more complicated than, and
less “obvious” than the others(!)
Content of Euclid's Text
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Following the Common Notions (Axioms) and
Postulates comes a sequence of Propositions and
their proofs – 48 in all in Book I.
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)
Manuscript copies of the Elements typically also
contained extensive scholia, or commentaries,
added by later scholars. They were effectively part
of the Elements for later readers.
Proposition 1
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To construct, on a given line segment AB, an
equilateral triangle.
Proof: AC = AB (this was stipulated in the definition of
a circle). Similarly BC = AB. Therefore, AC = BC
(Common Notion 1) Hence ∆ABC is equilateral
(again, a definition). QEF
A Possible Criticism
Strictly speaking, we don't know the circles must
intersect somewhere. That does not follow from the
Common Notions and Postulates.
 Euclid is 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 (he doesn't come
back to it later either(!))
 This is a flaw, but perhaps justified because of the
pedagogical orientation of the text.
 Quite a few similar issues elsewhere.
 A reworking of Euclid's foundations done by D.
Hilbert (early 20th century – see Chapter 2 in
McCleary) – adding quite a few additional
postulates – was designed to overcome all of these,
and it was entirely successful.
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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 equal angles coincide, then the third
sides must coincide too (since there's just one line joining two
distinct point – another unstated assumption).
This idea gives a valid proof, of course, but it again raises a
question: What in the Postulates says we can move any
figure? Common Notion 4 seems to be hinting at this, but that
is pretty ambiguous. 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.
Outline of proof of Proposition 5
Diagram is constructed by extending AC to D
(Postulate 2), extending AB to E with BE = CD
(Proposition 3), then joining AE and BD (Postulate 1)
First, ΔDCB ≅ ΔECA (Proposition 4)
Therefore, ΔAEB ≅ ΔBDA (Proposition 4 again – note
the angle at D is the same as the angle at E by
Common Notion 2.) This shows ᐸ DAB = ᐸ EBA
Then by the two triangle congruences,
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.
Line Segment Bisection
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Proposition 10. To bisect a given line segment.
Construction: Let AB be the given line segment
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.
“Erecting” a perpendicular
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Proposition 11. To construct a line at right angles to
a given line from a point on the line.
Construction is closely related to Proposition 10:
Given point A on the line, use Postulate 3 to
construct two other points on the line B, C with AB =
AC.
Construct an equilateral triangle ΔBCD (Proposition
1)
Then DA is perpendicular to the line at A.
“Dropping” a perpendicular
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Proposition 12. To drop a perpendicular to a given
line from a point not on the line.
Construction: Given point A not on the line and P
on the other side of the line, use Postulate 3 to
construct a circle with radius AP and center A – it
intersects the line in points B, C with AB = AC.
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Let D be the midpoint of BC (Proposition 10)
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Then DA is perpendicular to the line at D.
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Proof: ΔADB and ΔADC are congruent by
Proposition 8 (“SSS”). Hence <ADB = <ADC are right
angles. QEF
Propositions about angles
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Proposition 13. If from a point on a line a ray is drawn, then this
ray forms with the line two angles whose sum is the same as
two right angles.
Proposition 14. If two angles have a side in common, and if the
noncommon sides are on different sides of the common side,
and if the angles are together equal to two right angles, then the
noncommon sides lie along the same straight line.
This is a converse of Proposition 13. The reasoning is similar in
that it is based just on the Common Notions.
Note: Euclid did not consider 180˚ (“straight”) angles as angles.
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Propositions about angles, continued
Proposition 15. Vertical angles are equal.
Proof: <BPC + <CPD is the same as two right angles by
Proposition 13. Similarly for <APB + <BPC. Hence <CPD + <BPC
= <APB + < BPC by Postulate 4. Therefore, <CPD = <APB by
Common Notion 3. QED
A group of propositions about angles,
continued
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Proposition 16. In a triangle, an exterior angle is greater than
either of the nonadjacent interior angles.
The statement is that <DCB is greater than <CAB, <CBA:
Proof of Proposition 16
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Euclid's proof is clever! To show <DCA is greater than <BAC:
Construct the midpoint E of AC (Proposition 10) and extend BE
to BF with BE = EF (Postulate 2 and Proposition 3). Construct
CF (Postulate 1). Note that <AEB = <FEC by Proposition 15.
Proof of Proposition 16, concluded
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Therefore ΔAEB and ΔCEF are congruent (Proposition 4 –
“SAS”). Hence <BAE = <ECF.
But <ECF is a part of the exterior angle <DCA. So the exterior
angle is larger (Common Notion 5). QED
A similar argument shows <DCA is larger than < ABC.
Proposition 17
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Proposition 17. In any triangle, the sum of any two angles is
less than two right angles.
This follows pretty immediately from Proposition 16 and
Proposition 13 (which says that the interior and exterior angles
sum to two right angles).
Note that Euclid has not yet proved that the sum of all the
angles in a triangle is equal to two right angles (or 180˚). So he
cannot use any facts related to that yet.
The angle sum theorem is coming later, in Proposition 32, after
facts about parallel lines have been established.
We will skip lightly over the next group of propositions –
important for geometry, but off our main track here(!)
Sides in triangles
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Proposition 18. In a triangle, if one side is greater than another
side, then the angle opposite the larger side is larger than the
angle opposite the smaller side.
Proposition 19. In a triangle, if one angle is greater than another
angle, the side opposite the greater angle is larger than the side
opposite the smaller angle.
Proposition 20. In a triangle, the sum of any two sides is
greater than the third side.
We will omit Propositions 21, 24 entirely.
Constructing triangles and angles
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Proposition 22. To construct a triangle if the three sides are
given.
The idea should be clear – given one side, find the third corner
by intersecting two circles (Postulate 3). This only works if the
statement of Proposition 20 holds.
Proposition 23. To construct with a given ray as a side an angle
that is congruent to a given angle
This is based on finding a triangle with the given angle
(connecting suitable points using Postulate 2), then applying
Proposition 22.
Additional triangle congruences
Proposition 26. Two triangles are congruent if
a) One side and the two adjacent angles of one triangle are equal
to one side and the two adjacent angles of the other triangle
b) One side, one adjacent angle, and the opposite angle of one
triangle are equal to one side, one adjacent angle, and the
opposite angle of the other triangle.
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Both statements here are cases of the “AAS” congruence
criterion as usually taught today in high school geometry.
Euclid's proof here does not use motion in the same way that
his proof of the SAS criterion (Proposition 4) did.
Theory of parallels
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Proposition 27. If two lines are intersected by a third line so
that the alternate interior angles are congruent, then the two
lines are parallel.
As for us, parallel lines for Euclid are lines that, even if
produced indefinitely, never intersect
Say the two lines are AB and CD and the third line is EF as in
the following diagram
Proposition 27, continued
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The claim is that if <AEF = <DFE, then the lines AB and CD,
even if extended indefinitely, never intersect.
Proof: Suppose they did intersect at some point G
Proposition 27, concluded
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Then the exterior angle <AEF is equal to the opposite interior
angle <EFG in the triangle ᐃEFG.
But that contradicts Proposition 16. Therefore there can be
no such point G. QED
Parallel criteria
Proposition 28. If two lines AB and CD are cut by third line EF,
then AB and CD are parallel if either
a)
Two corresponding angles are congruent, or
b) Two of the interior angles on the same side of the transversal
sum to two right angles.
Parallel criteria
Proof: (a) Suppose for instance that <GEB = <GFD. By
Proposition 15, <GEB = <AEH. So <GFD = <AEH (Common
Notion 1). Hence AB and CD are parallel by Proposition 27.
Parallel criteria
Proof: (b) Now suppose for instance that <HEB + <GFD = 2 right
angles. We also have <HEB + < HEA = 2 right angles by
Proposition 13. Hence <GFD = <HEA (Common Notion 3).
Therefore AB and CD are parallel by Proposition 27. QED
Familiar facts about parallels
Proposition 29. If two parallel lines are cut by a third line, then (a)
the alternate interior angles are congruent, (b) corresponding
angles are congruent, (c) the sum of two interior angles on
the same side is equal to 2 right angles.
Familiar facts about parallels
Proof of (c): The claim is that, for instance, if AB and CD are
parallel, then <BEH+<DFG = 2 right angles. Suppose not. If the
sum is less, then Postulate 5 implies the lines meet on that
side of GH. But this is impossible since AB and CD are
parallel. If the sum is greater, then since <BEH = <GEA and <
DFG = <HFC (Proposition 15), while <GEA + <AEH = 2 right
angles = <HFC + CFG (Proposition 13), then <AEH + <CFG is
less than 2 right angles, and Postulate 5 implies the lines
meet on the other side. QED
Comments about Proposition 29
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This is the first use of Postulate 5 in Book I of the Elements
It is almost as if Euclid wanted to delay using it as long as
possible (maybe to see if it was necessary??)
Recall how much less intuitive and “obvious” the statement is
– there was a long tradition of commentary that ideally
Postulate 5 should be a Proposition with a proof derived from
the other 4 Postulates and the Common Notions – for
instance in quotation from Proclus on page 27 of McCleary's
book.
The other parts of Proposition 29 are proved similarly.
Further properties of parallels
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Proposition 30. If two lines are parallel to the same line then
they are parallel to one another.
That is, in modern language, parallelism is a transitive relation
on lines(!)
The proof Euclid gives depends on intuitive properties of
parallels that are “obvious” from a diagram, but that do not
follow directly from the other Postulates and previously
proved theorems.
In this case, though, the gap can be filled with additional
reasoning – will appear on a future problem set(!).
Construction of parallels
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Proposition 31. To construct a line parallel to a given line and
passing through a given point not on that line.
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Construction: Say AB is the line and F is the given point.
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Pick any point E on AB and construct EF (Postulate 1)
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Construct FG so that <GFE = <BEF (Proposition 23)
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Proof: Then FG and AB, produced indefinitely, are parallel lines
(Proposition 27). QED
Note: in some modern geometry textbooks, the statement that
there is exactly one such parallel line is used as a substitute for
Euclid's Postulate 5. Not hard to see they are equivalent
statements – we'll return to this(!)
The angle sum theorem
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Proposition 32. In any triangle, (a) each exterior angle is
equal to the sum of the two opposite interior angles and (b)
the sum of the interior angles equals 2 right angles.
Construction: Given ᐃABC, extend AB to D and construct
BE parallel to AC (Proposition 31).
The angle sum theorem, proof
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Proof: (a) The exterior angle <DBC is equal to <DBE + <EBC.
But <DBE = <BAC and <EBC = < ACB by Proposition 29 parts
(a) and (b).
(b) <ABC + <DBC = 2 right angles by Proposition 13.
Therefore, using part (a), the sum of the three angles in the
triangle equals 2 right angles. QED
Parallelograms
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Proposition 33. If two opposite sides of a quadrilateral are
equal and parallel, then the other pair of opposite sides are
also equal and parallel.
Let AB and CD be the given parallel sides and construct CB
(Postulate 1)
Parallelograms
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Proof: We have <BCA = <DCB by Proposition 29 (a). Hence
ᐃABC and ᐃDCB are congruent (Proposition 4 – SAS).
Therefore BD = AC and <BCA = <DBC. But then BD and AC
are also parallel by Proposition 27. QED
More on parallelograms
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Proposition 34. In a parallelogram, the opposite sides are
congruent and the opposite angles are congruent. Moreover
a diagonal divides the parallelogram into two congruent
triangles.
This follows from Propositions 33 and 29.
Comparing areas
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Proposition 35. Two parallelograms with the same base and
lying between the same parallel lines are equal in area.
Comment
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This depends on knowing that the entire segment EF lies on
one side of CD.
There are other possible arrangements too! Euclid does not
address this, but it is not too difficult to adjust the argument
to handle the case where the upper sides of the
parallelograms overlap too (to appear on a future problem
set!)
A corollary
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Proposition 36. Two parallelograms with congruent bases
and lying between the same two parallel lines are equal in
area.
This follows from Proposition 35 and Common Notion 1 – say
AB = GH. Then areas ABDC = ABFE = GHFE.
Corresponding facts for triangles
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Proposition 37. Two triangles with equal bases and lying
between the same two parallel lines are equal in area.
ᐃABC and ᐃABD below have the same area if CD is parallel
to AB:
Corresponding facts for triangles
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Proof comes by constructing parallelograms with diagonals BC
and AD, then applying Propositions 34 and 35.
Corresponding facts for triangles
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Proposition 38. Two triangles with congruent bases and lying
between the same two parallel lines are equal in area.
Follows from Proposition 37 by the same construction used to
deduce Proposition 36 from Proposition 35.
The next two Propositions – 39 and 40 – are corollaries of 37
and 38. Not needed for our purposes, so omitted.
Proposition 41. If a parallelogram and a triangle have the
same base and lie between two parallel lines, then the
parallelogram has double the area of the triangle.
This follows from the previous statements and Proposition 34.
Construction of a square
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Propositions 42 through 45 are also not needed for our
purposes.
Proposition 46. To construct a square on a given line
segment.
Construction: Let AB be the given line segment. Erect a
perpendicular AC at A (Proposition 11) and find D on AC
with AC = AB (Proposition 3 or just use Postulate 3)
Construct a parallel line to AB passing through D (Proposition
31)
Construct a parallel line to AB passing through D (Proposition
31)
Let E be the intersection of the parallels. Then ABED is a
square.
Where we started last Friday
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Proposition 47. In a right triangle, the square on the
hypotenuse is equal to [the sum of] the squares on the two
other sides.
The “Theorem of Pythagoras,” but stated in terms of areas
(not as an algebraic identity!)
The proof Euclid gives has to rank as one of the masterpieces
of all of mathematics, although it is far from the simplest
possible proof (as we have seen already or will discuss
shortly).
The thing that is truly remarkable is the way the proof uses just
what has been developed so far in Book I of the Elements.
Euclid's proof, construction
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Let the right triangle be ᐃABC with right angle at A
Construct the squares on the three sides (Proposition 46) and
draw a line through A parallel to BD (Proposition 31)
Euclid's proof, step 1
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<BAG + <BAC = 2 right angles, so G,A,C are all on one line
(Proposition 14), and that line is parallel to FB (Proposition 28)
Consider ᐃFBG
Euclid's proof, step 2
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Proposition 37 implies areas of ᐃFBG and ᐃFBC are equal
Euclid's proof, step 3
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AB = FB and BC = BD since ABFG and BCED are squares
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<ABD = <FBC since each is a right angle plus <ABC
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Hence ᐃBFC and ᐃABD are congruent (Proposition 4 – SAS)
Euclid's proof, step 4
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Proposition 37 again implies ᐃBDA and ᐃBDM have the
same area.
Hence square ABFG and rectangle BLKD have the same
area (twice the corresponding triangles – Proposition 41)
Euclid's proof, conclusion
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A similar argument “on the other side” shows ACKH and
CLME have the same area
Therefore BCDE = ABFG + ACKH. QED