Mathematical Fallacies, Flaws and Flimflam by Edward J. Barbeau

By Edward J. Barbeau

Via demanding adventure, mathematicians have discovered to topic even the main “evident” assertions to rigorous scrutiny, as instinct and facile reasoning can frequently lead them off beam. even though, the impossibility and impracticality of thoroughly watertight arguments ensure that mistakes to slide by means of the main watchful eye. they can be refined and hard of detection. while stumbled on, they could educate us much and will current a true problem to straighten out.

Presenting scholars with defective arguments to troubleshoot will be a good way of aiding them seriously comprehend fabric, and it truly is accordingly that i started to bring together fallacies and post them first within the Notes of the Canadian Mathematical Society and later within the university arithmetic magazine within the Fallacies, Flaws and Flimflam part. I was hoping to problem and amuse readers, in addition to to supply them with fabric appropriate for instructing and pupil assignments. This ebook collects the goods from the 1st 11 years of publishing within the CMJ.

One resource of such mistakes is the paintings of scholars. sometimes, a textual content publication will weigh in with a specious outcome or resolution. Nonprofessional assets, equivalent to newspapers, are chargeable for a goodly variety of mishaps, really in mathematics (especially probabilities) and likelihood; their use in study rooms might help scholars develop into severe readers and listeners of the media. various goods come from expert mathematicians. The reader will locate during this e-book a few goods that aren't inaccurate yet appear to be. those want a fuller research to explain the location. the entire goods are awarded in your leisure and use.

The mathematical issues lined contain algebra, trigonometry, geometry, likelihood, calculus, linear algebra, and smooth algebra.

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Surprising symmetry David Wells, in his book You are a mathematician (John Wiley & Sons, 1995, p. 88), makes the interesting observation that the nonsymmetric condition a = b + c leads to the symmetric result a4 + b4 + c4 = 2b2 c2 + 2c2 a2 + 2a2 b2 . Does this mean that a = b + c is equivalent to b = c + a and c = a + b? ♣ The nature of the symmetry becomes evident when we factor: (2b2 c2 + 2c2 a2 + 2a2 b2 ) − (a4 + b4 + c4 ) 22 Mathematical Fallacies, Flaws, and Flimflam = (a + b + c)(a + b − c)(b + c − a)(c + a − b).

1 intersections of the segments F D and GD with the arc QV . Then the rays M A and N A yield the desired trisection of angle BAC. First proof. Let H be the midpoint of arc QV . Consider the \triangles" DM H and DF E, one side of each being a circular arc. Since the arcs RW and QV are parallel, ∠DF E = ∠DM H and ∠DEF = ∠DHM , so that triangle DM H is similar to triangle DF E. Since 2DH = DE, it follows that arc M N = 2 arc M H = arc F E. Now, arc QV = (3/4) arc RW = 3 arc F E and arc QM = arc N V .

1 P I 44 Mathematical Fallacies, Flaws, and Flimflam angle OAB is√isosceles. √ Triangle ON B is a right isosceles triangle, hence BN = OB/ 2 = 1/(4 2). Thus, the √ area of triangle OAB is equal to /2 = 1/(32 2). Finally, the area of the octagon is (base)(height)/2 = OABN√ 8 times this, namely 1/(4 2). Second solution. Area(octagon) = Area(square F HBD) + 4Area(triangle BHA). We first note that JP is the median from vertex J to side LI of triangle JLI, and thus F H = HR. By a similar argument, F H = QF , and since QF +F H +HR = 1, they must each be 1/3.

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