90 = 100: A Proof1.82005/03/14 11:12:28 US/Central2007/04/07 18:09:10.781 GMT-5IanBarlandibarland@radford.eduMosheVardivardi@cs.rice.eduPhokionKolaitiskolaitis@cse.ucsc.eduMatthiasFelleisenmatthias@ccs.neu.eduJohnGreinergreiner@cs.rice.eduIanBarlandibarland@radford.eduJohnGreinergreiner@cs.rice.edu0 = 190 = 100falsegeometryparadoxproofA (purported) geometric proof that 90=100.
Construct a four-sided figure ABED as follows:
|∠ABE| = 90°
|∠DEB| = 100°
|AB| = |ED|
Using that as a starting point, we now tinker a bit to show that 90=100:
Draw the perpendicular bisectors to BE and AD;
call the point where they meet “C”.
Actually, we must prove that those two perpendicular bisectors really
do meet at all (i.e., that the point C even exists).
In this case, it turns out to be pretty clear —
it's not hard to argue that lines AD and BE aren't parallel,
and therefore their perpendicular bisectors aren't parallel,
and so they must intersect (in Euclidean geometry).
Still, be alert for people making glib assertions in proofs.
A construction to help prove that 90=100
Looking at this figure, some warning flags should be going up:
How do we know C lies below BD?
Might it lie above BD? Or exactly on BD?
It turns out that the argument below is the same in all of these cases,
though you'll certainly want to verify this to yourself later.
1|AB| = |ED|By construction.2|BC| = |EC|
C is on the perpendicular bisector of BE
(thus △BEC is isosceles).
3∠CBE≅∠BEC
Base angles of isosceles triangle BEC are congruent.
4|∠CBE| = |∠BEC|
Congruent angles have equal measures; line 3.
5|AC| = |DC|
C is on the perpendicular bisector of AD
(thus △ADC is isosceles).
6△ABC≅△DEC (!!)
Triangles with three congruent sides are congruent
(Euclid's Side-Side-Side congruence theorem); lines 1,2,5.
7(From here, it's just routine steps to conclude 90=100:)8∠ABC≅∠DEC
Corresponding parts of congruent triangles are congruent; line 6.
9|∠ABC| = |∠DEC|
Congruent angles have equal measures; line 8.
10|∠ABC| = |∠ABE| + |∠CBE|
By construction.
11|∠DEC| = |∠DEB| + |∠BEC|
By construction.
12|∠DEC| = |∠DEB| + |∠CBE|
Substituting equals with equals; lines 11 and 4.
13|∠ABC| = |∠DEB| + |∠CBE|
Substituting equals with equals; lines 12 and 9.
14|∠ABE| + |∠CBE| = |∠DEB| + |∠CBE|
Substituting equals with equals; lines 13 and 10.
15|∠ABE| = |∠DEB|
Subtracting equals from equals remains equal.
1690 = |∠DEB|
By construction, and substituting equals with equals; line 15.
1790 = 100
By construction, and substituting equals with equals; line 16.
If you feel this result is incorrect,
then the challenge for you is to find the first line which is false.
The flaw is extremely hard to find.
We won't actually give the solution, but here's a hint on
how to go about attacking the puzzle:
Note that finding the bug in the proof is the
same skill as debugging a program.
A good approach is to try various degenerate inputs.
In this case, there are a couple of “inputs” to
the construction—the length of CD is arbitrary; no matter
how long or short the proof should apply equally well.
Similarly, the angle 100° seems arbitrary;
fiddling with inputs like these (making them very small or very large)
might give you some clues as to where the bug is.
A very careful drawing will clear things up.
You may have noticed that the proof given here has some very minuscule
steps—e.g. “Congruent angles have equal measure.”
Usually such simple steps can be omitted, since they are obvious to any reader.
We include them for a few reasons:
As a careful thinker, you should recognize that such small steps
really are part of the complete reasoning,
even if they're not worth mentioning continually.
If a computer is checking a proof, it needs to actually include
those steps.
Programmers do need to be concerned with distinctions about (abstract)
types—the difference
between angles and their measures, in this case.
Sometimes a line's justification is glibly given
as “by construction”, when
that may not even be correct !-).
In this course, we'll spend a few weeks working with proofs which
do include all the small, pedantic steps,
to instill a mental framework for what a rigorous proof is.
But after that, you can relax your proofs to leave out
such low-level
steps, once you appreciate that they are being omitted.