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90 = 100: a proof

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A (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 meetC.
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 clearit'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 CBEBEC 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 ABCDEC 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 onhow 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 ofinputsto the constructionthe length of CD is arbitrary; no matter how long or short the proof should apply equally well.Similarly, the angle 100seems 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.

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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) typesthe difference between angles and their measures, in this case.
  • Sometimes a line's justification is glibly given asby 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-levelsteps, once you appreciate that they are being omitted.

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Source:  OpenStax, Intro to logic. OpenStax CNX. Jan 29, 2008 Download for free at http://cnx.org/content/col10154/1.20
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