Euclid's postulates and proof of the Pythagorean theorem are summarized. The document discusses Euclid's first five postulates, including his fifth postulate which many tried but failed to prove. It then summarizes Euclid's proof of the Pythagorean theorem using squares constructed on the sides of a right triangle and properties of parallel lines.
3. First postulate -
Euclid’s first postulate state that two points determine a line segment. In other words,
we can say a straight line may be drawn from anyone point to any other point.
This postulate tells us that at least one straight line passes through two distinct points,
but it does not say that there cannot be more than one such line. However, in his
work, Euclid has frequently assumed, without mentioning, that there is a unique line
joining two distinct points.
How many lines passing through P also pass through
Q? Only one, that is, the line PQ. How many lines
passing through Q also pass through P? Only one,
that is, the line PQ. Thus, the postulate is self-evident.
4. Second postulate -
Euclid’s second postulate states that a line segment can be extended indefinitely
along a line.
What we call a line segment now-a-days is what Euclid called a terminated line.
We have a line segment AB, we
have two endpoints A and B on
this line, and the line segment can
be produced from both sides to
form a straight line.
5. Third postulate -
In Euclid’s third postulate, he states that a circle can be drawn with a center and any radius.
Fourth postulate -
Euclid’s fourth postulate states that all right angles are congruent. In other words, we can
say that all right angles are equal to one another.
For example, we have two right angles P and Q. Angle P is equal to angle Q since both are right angles.
Let us arrange the angles in such a way that the orientation of the angle has changed. You can observe that after
changing the orientation, angle P remains equal to angle Q.
6. Fifth postulate -
Euclid’s fifth postulates states: If a straight line falling on two straight lines makes the
interior angles on the same side of it taken together less than two right angles, then the
two straight lines, if produced indefinitely, meet on that side on which the sum of angles is
less than two right angles.
For example, the line PQ in falls on lines AB and
CD such that the sum of the interior angles 1
and 2 is less than 180° on the left side of PQ.
Therefore, the lines AB and CD will eventually
intersect on the left side of PQ.
Many mathematicians, including Euclid himself, were convinced that the fifth postulate
is actually a theorem that can be proved using just the first four postulates and other
axioms. However, all attempts to prove the fifth postulate as a theorem have failed.
7. Attempts made to prove Euclid’s fifth postulate -
It's hard to add to the fame and glory of Euclid who managed to write an all-time
bestseller, a classic book read and scrutinized for the last 23 centuries. However
insignificant the following point might be, I'd like to give him additional credit for
just stating the Fifth Postulate without trying to prove it. For attempts to prove it
were many and all had failed.
By the end of the last century, it was also shown that the fifth postulate
is independent of the remaining axioms, i.e., all the attempts at proving it had been
doomed from the outset.
Did Euclid sense that the task was
impossible?
8. The earliest source of information on attempts to prove the fifth
postulate is the commentary of Proclus on Euclid's Elements.
Proclus mentions Ptolemy's (2nd century) attempts to prove the
postulate and demonstrates that Ptolemy had unwittingly
assumed what in later years became known as Playfair's Axiom.
Proclus left a proof of his own, but the latter rests on the
assumption that parallel lines are always a bounded distance
apart, and this assumption can be shown to be equivalent to the
fifth postulate.
9. al-Gauhary (9th century) deduced the fifth postulate from the
proposition that through any point interior to an angle it is
possible to draw a line that intersects both sides of the angle.
He deduced the proposition from an implicit
assumption that if the alternating angles
determined by a line cutting two other lines are
equal, then the same will be true for all lines cutting
the given two.
The proposition was implicitly used by A.M.
Legendre (1800) in his proof of the fifth postulate.
10. Al-Haytham made an attempt to prove the postulate in the
tenth century. His method was criticized by Omar Khayyam
in the eleventh century whose own proof was published
for the first time in 1936.
Later in thirteen century Nasir ad-Din at-Tusi analyzed the work
of Al-Haytham, Omar Khayyam and Al-Gauhary.
In one of his own attempts, at-Tusi tried to prove the postulate
by a reductio ad absurdum. This appears to be the first attempt
to prove the postulate by deriving a contradiction from the
assumption that the fifth postulate is wrong.
11. John Wallis had been inspired by the work of at-Tusi
and delivered a lecture at Oxford on July 11, 1663.
To prove the postulate he made an explicit assumption that for
every figure there is a similar one of arbitrary size. Unlike many
(even later) mathematicians, John Wallis realized that his proof
was based on an assumption (more natural in his view but still)
equivalent to the postulate.
So as we can see, many tried to prove this amazing postulate of
Euclid’s. But all these attempts failed. It is indeed a postulate and not a
theorem.
12. Euclid’s proof for the Pythagoras Theorem -
Let ACBACB be a right-angled triangle with right angle CAB.
On each of the sides BC, AB, and CA, squares are drawn: CBDE, BAGF, and ACIH, in
that order. The construction of squares requires the immediately preceding theorems in
Euclid and depends upon the parallel postulate.
13. From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L,
respectively. Join CF to AD , forming the triangles BCF and BDA.
Angles CAB and BAG are both right angles; therefore C, A, and G are collinear. Similarly for B, A,
and H. Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since
both are the sum of a right angle and angle ABC.
Since AB is equal to FB and BD is equal to BC, triangle ABD must be congruent to triangle FBC.
Since A-K-L is a straight line parallel to BD, rectangle BDLK has twice the area of
triangle ABD because they share the base BD and have the same altitude BK, i.e. a line
normal to their common base, connecting the parallel lines BD and AL.
14. Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC. Therefore,
rectangle BDLK must have the same area as square BAGF, which is AB^2.
Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH, which
is AC^2. Adding these two results, AB^2 + AC^2 = BD X BK + KL X KC.
Since BD=KL, BD × BK + KL × KC = BD(BK + KC) = BD × BC.
Therefore, AB^2 + AC^2 = BC^2 since CBDE is a square.
Hence Proved.
