Inside Collection (Course): Intro to Logic
Summary: A (purported) geometric proof that 90=100.
Construct a four-sided figure ABED as follows:
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. |
16 | 90 = |∠DEB| | By construction, and substituting equals with equals; line 15. |
17 | 90 = 100 | By construction, and substituting equals with equals; line 16. |
A useful corollary: 0=1.
1 | 90 = 100 | Previous theorem. |
2 | 0 = 10 | Subtracting equals (90) from equals remains equal. |
3 | 0 = 1 | Dividing equals by non-zero equals (10) remains equal. |
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:
!-)
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