# More Fallacies, Flaws, and Flimflam

Edward J. Barbeau
Series: Spectrum
Edition: 1
Pages: 188
https://www.jstor.org/stable/10.4169/j.ctt13x0n8w

1. Front Matter
(pp. i-viii)
(pp. ix-xiv)
3. Preface
(pp. xv-xvi)
Ed Barbeau
4. CHAPTER 1 Arithmetic
(pp. 1-18)

Many mathematical missteps in the public sphere involve percentages. It is a good habit to note explicitly the quantity to which a percentage is applied, as a common error is to add or subtract percentages applied to different quantities.

For example, suppose a woman earned a salary of $40,000 last year and got a raise of 10% this year, but expects a salary cut of 10% next year. This will not leave her where she started. The first 10% is applied to$40,000, so her earnings this year are $44,000. It is this amount to which the 10% reduction is... 5. CHAPTER 2 School Algebra (pp. 19-56) For some light relief, we begin with this item which has been circulating in the ether as the scan of a pupil’s script for many years. It appears to be genuine. The title of this item was written by the teacher, along with an “X,” beside the following solution: Problem Expand (a+b)n.${(a + b)^n} = {(a + b)^n} = {(a + b)^n} = {(a + b)^n} = (a + $Li Zhou sends the following problem, which appeared on a sample group examination for the Florida Math Olympics: Problem If$x > 0$and$x + \frac{1}{x} = 1$, then evaluate${x^2} + \frac{1}{{{x^2}}}$. This is a mathematics contest for Florida community college students. It is usually held in April at the University... 6. CHAPTER 3 Geometry (pp. 57-72) With inadequate care in the use of the double-angle formula, we can conclude that the standard 3-4-5 right triangle does not exist. These arguments are due to Jeff Suzuki. Argument 1. Ifθis the angle opposite the side of length 4, we have$\sin 2\theta = 2\sin \theta \cos \theta = 2\sin \theta \sqrt {1 - {{\sin }^2}\theta } $. Plugging this formula into the calculator with$\sin \theta = 0.8$we find that$\sin 2\theta \approx 0.96$,$\sin 4\theta \approx 0.5376$,$\sin 8\theta \approx 0.9066$, and so forth, so that all ofθ, 2θ, 4θ, and so on, must be between 0° and 180°. We conclude that θ = 0. Argument 2. Start with$\cos \theta = 3/5$and use$\cos 2\theta = 2{\cos ^2}\theta - 1$. Then$\cos \theta = 0.6$,$\cos 2\theta = - 0.28$,$\cos 4\theta \approx - 0.8432$,$\cos 8\theta \approx 0.4219$. Since the equation... 7. CHAPTER 4 Limits, Sequences and Series (pp. 73-82) Theorem.The cardinality of the class of subsets of the closed unit interval does not exceed the cardinality of the closed unit interval itself. Proof.LetSbe a subset of [0, 1]. We assign toSan element of [0, 1], written in binary form as follows. IfSis void, assign toSthe number 0.00000 . . . . Otherwise, the first binary digit after the decimal point of the corresponding number is 1. IfShas nonvoid intersection with the interval [0, 1/2], let the second binary digit of the corresponding number be 1; otherwise, assign... 8. CHAPTER 5 Differential Calculus (pp. 83-94) TheoremIf the real-valued function f defined onRis continuous at a, then there exists an interval I around a on which f is continuous. Proof fis continuous ata. Therefore, for all$ \in > 0$, there exists$\delta > 0$such that,$|x - a| < \delta \Rightarrow |f(x) - f(a)| < \in /2$Choosebin the interval$I = (a - \delta /4,a + \delta /4)$. Clearly$|b - a| < \delta $, and so, for anyxinI,$|f(x) - f(b)| = |f(x) - f(a) + f(a) - f(b)| \le |f(x) - f(a) + f(a) - f(b)| < \frac{ \in }{2} + \frac{ \in }{2} = \in $. It follows thatfis continuous atb. Sincebwas arbitrary inI,fis continuous onI. Contributed by Russ Euler and Jawad Sadek. CommentOne way to get a handle on the situation is to... 9. CHAPTER 6 Integral Calculus (pp. 95-114) Having students perform integration by hand is one way to help them become familiar with the properties of standard functions and to develop judgment as to the best way to proceed. However, different approaches can lead to results that might appear at variance with each other. It is a nice exercise for students to reconcile them. Here are some examples submitted by readers of theCollege Mathematics Journalover the years. Example 1 Integrate$\int {\frac{x}{{{{(x - 1)}^2}}}} dy$. Solution 1An integration by parts withu=xanddv(x– 1)–2dxyields the answer$\int {\frac{x}{{{{(x - 1)}^2}}}} dx = \frac{{ - x}}{{x - 1}} + \int {\frac{1}{{(x - 1)}}dx} = \frac{{ - x}}{{x - 1}} + \ln (x - 1) + C$. Solution 2The substitutionx= 1... 10. CHAPTER 7 Combinatorics (pp. 115-124) Joseph G. R. Martinez describes how students in a teaching methods course were frustrated with a word problem that appeared on a competency examination that no one had been able to solve: Mrs. Amos has 28 students in her class. 17 students have brown hair and hazel eyes. 15 students have brown hair. 10 students have hazel eyes. How many students have neither brown hair nor hazel eyes? The multiple choice answers were: 5, 6, 7, 10. Martinez writes, “The problem seemed simple enough, but none of the students’ attempted solutions made sense, and none yielded answers to match the... 11. CHAPTER 8 Probability and Statistics (pp. 125-136) ProblemAandBare standing at the respective points (0, 0) and (2, 1) in a square grid. At the same instant, with the same speed, each walks towards the position of the other.Amoves only to the right and up, whileBmoves only to the left and down; both must travel on grid lines (at least one of the coordinates is an integer). What is the probability that they will meet? Solution 1Each person has three possible paths, according as the route goes along the segments,x= 0 ,x= 1 andx... 12. CHAPTER 9 Complex Analysis (pp. 137-142) Teik-Cheng Lim provides the following development to show that$\sqrt { - 1} $is real. Start with$\sqrt { - 1} = 1 + (\sqrt { - 1} - 1)$and manipulate to obtain$\sqrt { - 1} = 1 + \frac{{(\sqrt { - 1} - 1)(\sqrt { - 1} + 1)}}{{(\sqrt { - 1} + 1)}} = 1 - \frac{2}{{(1 + \sqrt { - 1} )}} = 1 - \frac{2}{{(1 + \sqrt { - 1} - 1)}}$. Iterating this process leads to the beautiful continued fraction$\sqrt { - 1} = 1 - \frac{2}{{2 - \frac{2}{{2 - \frac{2}{{2 - \frac{2}{{2 - \frac{2}{{2 - \frac{2}{{2 - \cdots }}}}}}}}}}}}$. A very bright student calculated the square of the modulus of a complex number$z = a + b\sqrt { - 1} $thus:$|z{|^2} = z\bar z = (a + b\sqrt { - 1} )(a - b\sqrt { - 1} ) = {a^2} - {b^2}\sqrt { - 1} \sqrt { - 1} = {a^2} - {b^2}\sqrt {( - 1)( - 1)}  = {a^2} - {b^2}\sqrt 1 = {a^2} - {b^2}$, so that when$b > a > 0$, then$|z{|^2}$is negative. Peter M. Jarvis and Paul S. Shuette, the contributors of this item, emphasize the care needed in defining roots of negative numbers and warn about the pitfalls of “defining”ias$\sqrt { - 1} \$

Mark Lynch provides a conundrum that causes his beginning students...

13. CHAPTER 10 Linear and Modern Algebra
(pp. 143-154)

Ayoub B. Ayoub asked his students to check whether the vector (2,4,1,5) belonged to the span of {(1,1,1,1), (1,0,1,–2), (1,–3,1,1)} in R⁴. A student who noticed that the three vectors are pairwise orthogonal, used the dot product to calculate the scalar factorsx,y,zin the linear combination

(2,4,1,5) =x(1,1,1,1) +y(1,0,1,–2) +z(1,–3, 1,1).

Taking the dot product of both sides with (1,1,1,1) yieldsx= 3. Similarly, he goty= –7/6 andz= –1/3. Then the student claimed that the vector belonged to the subspace.

However, this is not...

14. CHAPTER 11 Miscellaneous
(pp. 155-168)

Sung Soo Kim recalls his puzzlement when he first learned of the countability of the rationals and the uncountability of the irrationals. In view of the density of both rationals and irrationals in the reals, he wondered how an uncountable number of gaps could be produced from only a countable number of points while only a countable number of gaps could be generated from an uncountable number of separated points.

The front page of the June 15, 2004, edition of theNational Postin Toronto carried the account of a formula to measure the effectiveness of jokes. According to the...

15. Index
(pp. 169-170)