Conference rusure::math

> Back to basics on this one: using the Pythagorean Thm. or any branch of 
> math that rests on it (analytic geometry, trig, etc.) begs the question.

    Re .-1:  Can I use Euclid's Fifth Postulate?
    
    Does Euclid's Fifth Postulate necessarily follow from the existence
    of a triangle with sides a,b,c such that a^2 + b^2 = c^2?
    
    Hint:  I don't know the answer to this.
    
    Dan

    I am pretty sure that the Parallel Postulate, being a postulate,
    is not derived from Pythagoras' theorem.

    The Parallel Postulate is "equivalent" to such statements as
    
        - the sum of the angles of a triangle is 180 degrees
        - there exist similar but non-congruent triangles
        - [I think] there exist triangles of arbitrarily large area
    
    Many of the early "proofs" of the Parallel Postulate used one of
    these other statements.  What I was asking was if the existence
    of an "a^2 + b^2 = c^2" triangle was yet another equivalent to
    the above.
    
    Dan

I don't recall seeing any proof of PyThm. that makes any reference to the 
Parallel Postulate. Maybe it's buried in some intermediate constructions,
e.g. of a square with given side, or something.

    Using "the sum of the angles of a triangle is 180 degrees" with
    the following diagram, we see that the center is a square.  Letting
    the inner square have side c, we see that (a+b)^2 = 2ab + c^2, so
    that a^2 + b^2 = c^2, proving the Pythagorean Theorem.

	      b       a
	   +-------+----+
	   |     /  \   |
	 a |   /     \  | b
	   | /        \ |
	   +           \|
	   |\           +
	   | \        / |
	 b |  \     /   | a
	   |   \  /     |
	   +----+-------+
	     a      b

    Now to prove te reverse.  Suppose we have a triangle with sides
    a, b, and c, such that a^2 + b^2 = c^2, but the angle opposite c
    is not a right angle.  Construct a triangle with sides a and b,
    with a right angle between those sides.  Then the third side must
    equal c (by the Pythagorean Theorem, above), and so it is similar
    to the first triangle.  Hence (by the law of similarity of triangles?),
    the first triangle must also have a right angle opposite side c.

    Re: .2, .4
    
    The Pythagorean theorem is equivalent to the parallel postulate.
    
    In the classic non-Euclidean geometries, the analogues to the
    Pythagorean Theorem are a^2 + b^2 > c^2 and a^2 + b^2 < c^2 .

    Hence to prove the converse you do in fact need the 5th postulate
    or one of its many equivalents.

    I think a proof of .0 not using the PthM (as .6 did) could be
    constructed by starting out with

		P
		+
    		|\
    		| \ c		P, R, and S are points,
    	      b	|  \ 		a, b, and c are lengths
    		|   \
    		+----+
    		R a  S
    
    and constructing point Q so that PR || SQ and SQ is of length b.
    	
		P    Q
		+    +
    		|\   |
    		| \ c|		P, Q, R, and S are points,
    	      b	|  \ | b	a, b, and c are lengths
    		|   \|
    		+----+
    		R a  S
    
    Therefore RQ (not drawn) = PS = c, since RQ^2 = a^2 + b^2

    Therefore triangle PRS is congruent to triangles RSQ, SQP, and QPR
    (by Side-Side-Side), hence all four angles of PQRS are equal.  Since
    the angles of a quadrilateral sum to 4 right angles and all are
    equal, well you get the idea...

      John

    It seems to me that both .6 and .7 are assuming the Pyth. Thm.
    to start.  Why do you say (in .7) RQ^2 = a^2 + b^2.  Also, assuming
    that the angles of a quadrilateral sum to 4 right angles means
    assuming that the angles of a triangle sum to 2 right angles (draw
    a diagonal), which is the same as assuming the Pythagorean Theorem.
    
    Shouldn't it be possible to prove this, instead of assuming it,
    given a triangle with sides a,b,c such that a^ + b^2 = c^2.
    
    Dan

.8>    ... Why do you say (in .7) RQ^2 = a^2 + b^2.  ...
    
    Oops, you're right.  We are given that a^2 + b^2 = c^2, but haven't
    established that PRS is congruent to RSQ, so at that point in the
    proof can't assume that angle R is the same as S, which is what
    you need to sail thru the rest.
    
    But you will have to "take the fifth" (nice phrase) sooner or later.
    I thought it was rather cute to use the 5th right at the start where
    SQ is constructed parallel to PR.
    
      John

    Re .6: I think I read somewhere that this proof of Pyth Thm was
    discovered by the Chinese some 1000 years before Pythagoras. Anybody
    know of any evidence of this? I wish I'd come across this proof
    before I did my 'O' level maths - I could never remember the usual
    proof for more than a few days.

    Any proof as a reply here that states "construct the parallel to ..."
    should be met with a loud
    
                    "WHICH PARALLEL???"
    
    or a
    
                    "BUT THERE AREN'T ANY!!!"
    
    (-:
    
    Dan

I'd like to defend Gilbert's proof in .6.  He did *not* use the
Pythagorean theorem, he first proved it.  For those that persist and
say "yeah, but then he used it, right?", merely redo .6 and
call it something other than Pythagorean.  After all, anything you
prove along the way you can name anything you want.

The only thing I might ask for in .6 is that the "sum of angles
of triangle equals 180" might need proof.

/Eric

I agree with .-1; what I was seeking is a direct constructive proof, much 
like the one Gilbert cited, but without putting the bucket of water on the 
floor.


In case my reference was cryptic, there is the following old chestnut:

Problem 1:
Given: a bucket of water on the floor, and a stove with burner lighted;
to boil the water. 
Solution: pick up the bucket of water and put it on top of the burner.

Problem 2: 
Given: a bucket of water on a table, and a stove with burner lighted;
to boil the water. 
Solution: pick up the bucket of water and put it on the floor. See problem 1.

    One part of .6 was a hand-wave.  That is, if the corresponding sides
    of two triangles are equal, then the corresponding angles are equal.
    Does this also depend on the parallel postulate?

    Re .15
    
>>    One part of .6 was a hand-wave.  That is, if the corresponding sides
>>    of two triangles are equal, then the corresponding angles are equal.
>>    Does this also depend on the parallel postulate?
    
    I am sure that it doesn't.
    
    Dan

>>    One part of .6 was a hand-wave.  That is, if the corresponding sides
>>    of two triangles are equal, then the corresponding angles are equal.
>>    Does this also depend on the parallel postulate?

    		   . P
    	 	.     .
    	   Q  .          . R		If you draw a kite and construct
    	       .        .		the chord QR, you then have two
       		.      .		triangles QPR and QSR that share
    		 .    .			side QR.  Since QR = QR does it
    		  .  .			then follow that angle P = angle S?
    		    . S

    You've got to have more restrictions than listed above, and depend
    on stronger theorems, some if not all of which are equivalent to
    the parallel postulate.

      John

    Aha.  I have finally convinced myself that some form of the
    parallel postulate is needed for this proof.  Consider a triangle
    on the surface of a sphere.  Put one vertex at the north pole, the
    other two on the equator a quarter circumference apart.  This triangle
    will have three right angles.  Now let a and b be the "longitudinal"
    sides, from the north pole to the equator.  Side c will be the edge
    that runs along the equator.  Increase the length of side c by
    increasing the angle at the north pole between sides a and b.  You
    can increase the length of side c until c^2 = a^2 + b^2 = 2 a^2 = 2 b^2.
    This triangle will not have a right angle opposite side c.
    
    So, all objections about proofs using something equivalent to the
    parallel postulate are withdrawn.  It is needed here.
    
    Dan

    Re .6, .17
    
>>   >>    One part of .6 was a hand-wave.  That is, if the corresponding sides
>>   >>    of two triangles are equal, then the corresponding angles are equal.
>>   >>    Does this also depend on the parallel postulate?
>>
>>   You've got to have more restrictions than listed above,
    
    I took "corresponding sides of two triangles are equal" to mean
    all three sides in turn are equal.  Your example uses only one of the
    three sides.  I haven't re-examined .6 yet to see which he meant,
    though.
    
    Dan

    Yes, solving the problem in .0 will require the fifth postulate.
    The pythagorean theorem is not necessarily true in non-euclidean
    geometries.
    
    And yes, that is equivalent to saying that the sum of angles in
    a triangle is equal to two right angles.
    
    And it is equivalent to saying that if all three sides of two triangles
    are proportional then the corresponding angles are equal.
    
    But it is not equivalent to the statement that if all three
    corresponding sides of two triangles are equal, then the corresponding 
    angles are also equal.  This is true in Euclidean geometry and the two
    non-euclidean geometries I know about.  I seem to remember proving
    that in my youth, before we started using the fifth postulate, but
    that was long ago.

    Re .-1
    
    I am also sure that I have heard that it requires Euclidean geometry
    (i.e., the parallel postulate) to prove that there can be "similar
    triangles" that are not "congruent".
    
    Now what do the words I quoted mean?
    
         similar = the same angles?
         similar = proportional sides?
         similar = the same angles and proportional sides?
    
         congruent = the same sides?
         congruent = the same sides and the same angles?
    
    Do any of these possible definitions imply a later one on the list?
    
    Anyway, in non-Euclidean geometries I think the area of a triangle is
    a function of the difference between the sum of its angles and 180
    degrees.  So if you "scale" the triangle, changing the sides but all
    by the same factor, the resulting "proportional" triangle has a
    different area and so a different sum of angles.
    
    Dan

    You cannot define what an 'angle' is without first defining a metric.

    So in a sense, .0 takes the modern view of geometry, where the Pythagorean
    theorem (defining the euclidean metric) is reduced to a postulate, and
    then the ramifications of it are deduced.  The advantages seem to be
    that it generalizes naturally to arbitrary differentiable manifolds and
    lets one define 'intrinsic' properties of them without the clumsy
    requirement of imbedding them in another space.  The first example
    of this was Gauss' studies of surfaces (I believe this is the case.)
    This intrinsic-ness of space was philosophically very important for the
    development of the theory of relativity...

    - Jim

Congruence and similarity I remember; the rest I need to think about a 
bit.  Euclid said that two triangles are similar if their sides are in
proportion.  Two triangles are congruent if a one-to-one coorespondence
can be constructed between their vertices such that corresponding sides
are congruent and corresponding angles are congruent.  Congruence of line
segments and angles were accepted by Euclid as "common notions".

The line segment congruence axiom says, informally that you can always "move"
one line segment onto a ray and mark off a second line segment congruent to
the first.  The congruent angle axiom similarly says you can "lay off" a
congruent angle on a line segment from an existing angle.

The last congruence axioms is the old side-angle-side axiom for congruent
triangles.

I believe it is the Sacheri (Saccheri?)-Legendre theorem which shows
that the best you can do without the parallel postulate is to say that
the sum of the angles in a triangle is <= 2 right triangles.  You need
the parallel postulate to prove that the sum = 2 right triangles.

re: < Note 846.21 by ZFC::DERAMO "Daniel V. D'Eramo" >
                -< I wish I could remember more of this stuff. >-

>    I am also sure that I have heard that it requires Euclidean geometry
>    (i.e., the parallel postulate) to prove that there can be "similar
>    triangles" that are not "congruent".
    
    Yes, this is my favorite statement of the fifth postulate: similar,
    non-congruent triangles exist.
    
>         similar = the same angles?
>         similar = proportional sides?
>         similar = the same angles and proportional sides?
    
    Yes, in Euclidean geometry all these are equivalent.  I believe
    that the second statement defines similar, and the fifth postulate
    is used to prove the first.
    
>         congruent = the same sides?
>         congruent = the same sides and the same angles?

    As I remember it, the second statement is the definition of congruent,
    and the first four postulates (and probably a few things taken for
    granted) are used to show that the first is equivalent, that is,
    equal sides imply equal angles.  

re: < Note 846.23 by COOKIE::WAHL "Dave Wahl: Database Systems AD, CX01-2/N22" >

> I believe it is the Sacheri (Saccheri?)-Legendre theorem which shows
> that the best you can do without the parallel postulate is to say that
> the sum of the angles in a triangle is <= 2 right triangles.  
        
    This theorem must refer to a special case.  On the surface of a
    sphere, the sum of angles in a triangle > 2 right angles.

> You need
> the parallel postulate to prove that the sum = 2 right triangles.

    the parallel postulate or one of its many equivalents

>> I believe it is the Sacheri (Saccheri?)-Legendre theorem which shows
>> that the best you can do without the parallel postulate is to say that
>> the sum of the angles in a triangle is <= 2 right triangles.  
>        
>    This theorem must refer to a special case.  On the surface of a
>    sphere, the sum of angles in a triangle > 2 right angles.

    The Saccheri stuff holds in hyperbolic geometry, i.e., more than
    one nonintersecting line.  The reverse is true in elliptic geometry,
    i.e, two lines always intersect iff the sum of angles in a triangle
    is greater than 2 right angles.