name: due date: the history of...

NAME: Due date: The History of Mathematics

Weekender #1

1: Textbook Problems.

1a. page 71 problem 1;

1b. page 125 problem 1;

1c. page 126 problem 7.

2: Babylonian Square Roots. The Babylonian approxima-tion of

√a2 + h is given by

a+h

2a,

where 0 < h < a2.

2a. Take a = 43

and h = 29

to approximate√

2. (Thisis an approximation used by the Babylonians.)

2b. Take a = 1712

and h = − 1144

to approximate√

2.

2c. Take a = 2 and h = 1 to approximate√

5.

2d. This approximation was still used in the 20th cen-tury as a quick estimate for square roots. Usually,a2 would be the largest perfect square less thana2 + h. For example, to estimate

√70, we would

take a2 = 64, which implies h = 70−64 = 6. Then

√70 =

√

82 + 6 ≈ 8 +6

2 · 8= 8

3

8.

Interestingly, 8 38

as a decimal is 8.375, and my TI-

84 tells me that√

70 ≈ 8.367. So this method ofapproximation good enough for a rough estimate.

Determine similar estimates using this “Babylo-nian” method for

√56 and

√90.

2e. A better approximation formula (used only onceby the Babylonians) is

√

a2 + h ≈ a+h

2a− h2

8a3.

Repeat parts 2a, 2c, and 2d using this better ap-proximation.

3: It’s All Greek To Me. In the alphabetic Greek nu-meral system the numbers 1000, 2000, . . . , 9000 wereoften represented by priming the symbols for 1, 2, . . . ,9. Thus 1000 might appear as α′. The number 10,000,or myriad, was denoted by M . The multiplicative prin-ciple was used for multiples of 10,000. Thus 20,000,300,000, and 4,000,000 appeared as βM , λM , and νM .For example, here’s how the number 345,720 would bewritten. Since

300, 000 is λM,

40, 000 is δM,

5, 000 is η′,

700 is ψ, and

20 is κ,

then 345,720 is λMδMη′ψκ.

3a. Write, in alphabetic Greek, the numbers 5,780,72,803, 450,082, and 3,257,888.

3b. In order to write numbers less than 1000, howmany different symbols must one memorize in thealphabetic Greek numeral system?

4: False Position Must Be True. Solve the following bythe method of false position. Both are from the Rhind

Papyrus.

4a. A quantity and its 17

added together become 19.What is the quantity?

4b. A quantity, its 23, its 1

2, and its 1

7added together

become 33. What is the quantity?

5: Then Double False Position Must Be False. One of theoldest methods for approximating the real roots of anequation is the method of double false position. Thismethod seems to have originated in India and was usedby the Islamic mathematicians; however, there is evi-dence that the method was used in China first. In brief,the method (in modern notation) is this: Let x1 and x2

be two numbers lying close to and on each side of a rootx of the equation f(x) = 0. Then the intersection withthe x-axis of the chord joining the points (x1, f(x1))and (x2, f(x2)) gives an approximation x3 to the soughtroot. Symbolically,

x3 =x2f(x1) − x1f(x2)

f(x1) − f(x2).

The process can now be repeated with the appropriatepair x1, x3 or x2, x3.

For example, to solve f(x) = x2 − 3x− 15 = 0, we firstpick two numbers which make f(x) as close to zero aspossible, where one number gives a positive value andthe other gives a negative value. We pick x = 5 andx = 6 since f(5) = −5 and f(6) = 3. Next, we find thechord (line) between the two points (5,−5) and (6, 3)to be y = 8x − 45. This line intersects the x-axis atx = 45

8= 5.625. Notice that the “formula” above gives

this number also, where we use (5,−5) as (x1, f(x1))and (6, 3) as (x2, f(x2)):

6f(5) − 5f(6)

f(5) − f(6)=

6 · −5 − 5 · 3

−5 − 3=

−45

−8= 5.625.

This number, x = 5.625, is our first approximationto the root of x2 − 3x − 15. Then, since f(5.625) =−0.234375, which is negative, we can repeat this pro-cess with f(5.625) and f(6) (since f(6) is positive).

Your task: Use double false position to approximate theroot of x3 − 36x+ 72 = 0 which lies between 2 and 3 totwo decimal places.

6: Finger Numbers. Finger numbers were widely usedfor many centuries; from this use, finger processes weredeveloped for some simple computations. One of theseprocesses, by giving the product of two numbers, eachbetween 5 and 10, served to reduce the memory workconnected with the multiplication tables. For example,to multiply 7 by 9, raise 7 − 5 = 2 fingers on one handand 9 − 5 = 4 fingers on the other hand. Now add theraised fingers, 2+4 = 6, for the tens digit of the product,and multiply the closed fingers, 3 × 1 = 3, for the units

c© Dr. Charles R. Garner, Jr., 2004, 2010, 2013, 2016


digit of the product, giving the result 63. This processis still used by some European peasants. Why does themethod gives correct results? Bonus: Prove that themethod gives correct results.

7: Babylonians Got 3 33322221 Problems But Base 60 Ain’t 1.Solve the following problems found on a Babylonian claytablet dating from 1800 bc. The numerical values inthe problem are in Babylonian base-60, where the com-mas separate powers of 60 and a semi-colon acts as the“decimal” point. For example, 2, 1; 50 is translated as2(60) + 1 + 50

60, which simplifies to 121 5

6.

7a. “An area consisting of the sum of two squares is16, 40. The side of one square is 10 less than 0; 40of the side of the other square. What are the sidesof the squares?” (Your answer does not need to bein Babylonian numerals!)

7b. Find the area of a trapezoid with bases 14 and 50and sides 30. (Notice that the numerals used hereare Babylonian, but does that really matter?)

8: Circles and Sagittas and Chords, Oh My.

8a. Interpret mathematically the following, found ona Babylonian tablet dating from 2600 bc:

“60 is the circumference, 2 is the sagitta,find the chord.”“Thou, double 2 and get 4, dost thounot see? Take 4 from 20, thou gettest16. Square 20, thou gettest 400. Square16, thou gettest 256. Take 256 from 400,thou gettest 144. Find the square rootof 144. 12, the square root, is the chord.Such is the procedure.”

8b. Using part 8a as a guide, derive a correct formulafor the length of a chord in terms of the radius rand sagitta s.

9: Egyptians Got 222222222||||||||| Problems But Hiero-glyphics Ain’t |. Solve these problems which are takenfrom the Moscow Papyrus without permission. (Hope-fully some ancient Egyptian mummy won’t come afterme for stealing math problems.)

9a. The area of a rectangle is 12 and the width is 34

the length. What are the dimensions?

9b. One leg of a right triangle is 2 12

times the otherand the area is 20. What are the dimensions?

10: Egyptian Pi. In the Moscow Papyrus, the area of a cir-cle is repeatedly taken as equal to the area of a squarewhose side length is 8

9of the diameter of the circle. This

implies that the Egyptians used what value for π?

11: Empirical Mathematics. The idea of averaging is im-portant in empirical work. Thus, we find in the Rhind

Papyrus the area of a quadrilateral having successivesides a, b, c, and d given by

K =(

a+ c

2

)(

b+ d

2

)

.

11a. How does this formula demonstrate averaging?(Don’t overthink this.)

11b. This formula gives too large a result for all non-rectangular quadrilaterals. For instance, considera parallelogram with sides 20, 5, 20, and 5 anda height of 4 between the sides of length 20. Thearea is known to be (base)×(height) = 4×20 = 80.Calculate the incorrect area given by the Egyptianformula, and determine the percentage error.1

11c. Now assume that the Egyptian formula is correct.Under this assumption, show that the area of a tri-angle would be given by half the sum of two sidesmultiplied by half the third side.2

12: Babylonian Quadratics. A problem from Babylon asksfor the side of a square if the area of the square dimin-ished by the side of the square is 870. The solution isdescribed as follows: “Take half of 1, which is 1

2; mul-

tiply 12

by 12, which is 1

4; add the 1

4to 870, to obtain

870 14. This last is the square of 29 1

2. Now add 1

2to 29 1

2;

the result is 30, which is the side of the square.”

12a. Show that the Babylonian solution above is ex-actly equivalent to solving the quadratic equationx2 − px = q by the formula

x =

√

(

p

2

)2

+ q +p

2.

(Demonstrate this by using the quadratic formulaon x2 − px− q = 0, and rewriting the result.)

Another Babylonian text solves the equation 11x2 +7x = 6 1

4by first multiplying through by 11 to obtain

(11x)2 + 7(11x) = 68 34, which, by setting y = 11x, has

the “normal” form y2 + py = q. This is solved by theformula

y =

√

(

p

2

)2

+ q − p

2;

and, finally, x = y

11.

12b. Solve 7x2 + 5x = 2 by this method.

1Percentage error is |estimated value − actual value| × 100 ÷ actual value.2This incorrect formula for the area of a triangle is found in a property deed from Egypt, dating 1500 years after the Rhind Papyrus.



Weekender #2



1b. page 145–146 problem 4.

2: The Seqt of a Pyramid. The Egyptians measured thesteepness of a face of a pyramid by the ratio of the “run”to the “rise”—that is, by giving the reciprocal of whatwe consider the slope of the face of the pyramid. Thevertical unit (the “rise”) was the cubit and the horizon-tal unit (the “run”) was the hand; there were 7 hands ina cubit. (So that means the “rise” and “run” are differ-ent units.) With these units, the measure of steepnesswas called the seqt of a pyramid.

2a. Solve Problem 56 of the Rhind Papyrus which asks:What is the seqt of a pyramid 250 cubits high andwith a square base 360 cubits on a side? (Don’tforget to convert the horizontal distance to hands.)

2b. The great pyramid of Cheops has a square base440 cubits on a side and a height of 280 cubits.What is the seqt of this pyramid?

3: Geometrical Experiments—No Lab Equipment Needed.For the following problems, solve the problems by ex-periment and intuition. (There are no wrong answers—except an unjustified one!)

3a. Let F , V , and E denote the number of faces, ver-tices, and edges of a polyhedron. For the tetra-hedron, cube, triangular prism, pentagonal prism,square pyramid, pentagonal pyramid, cube withone corner cut off, and cube with a square pyra-mid erected on one face, we find

V − E + F = 2.

Do you feel that this formula holds for all polyhe-dra? Why?

3b. There are convex polyhedra all faces of which aretriangles (i.e., a tetrahedron3), all faces of whichare quadrilaterals (i.e., a cube), all faces of whichare pentagons (i.e., a regular dodecahedron). Doyou think the list can be continued? Why?

3c. Consider an ellipse with semiaxes a and b. If a = bthe ellipse becomes a circle and the two expressions

P1 = π(a+ b) and P2 = 2π√ab

each become 2πa, which gives the perimeter (cir-cumference) of the circle. This suggests that P1 orP2 may give the perimeter of any ellipse. Whichone is the right one? Why?

3d. If the inside of a race track is a noncircular ellipse,and the track is of constant width, is the outsideof the track also an ellipse? How could you justifyyour answer? (Notice I am not asking you to proveyour answer is the right one, but simply asking youto think about how you go about justifying it.)

4: The Thales Puzzle. There are two versions of howThales, when in Egypt, evoked admiration by calculat-ing the height of a pyramid by shadows. The earlier ac-count given by Hieronymous, a pupil of Aristotle, saysthat Thales determined the height of the pyramid bymeasuring the shadow it cast at the moment a man’sshadow was equal to his height. The later version, givenby Plutarch, says that he set up a stick and then madeuse of similar triangles. Both versions fail to mentionthe very real difficulty, in either case, of obtaining thelength of the shadow of the pyramid—that is, the dis-tance from the shadow of the top of the pyramid to the

center of the base of the pyramid.

This unaccounted-for difficulty is called the Thales puz-zle: Devise a method, based on shadow observationsand similar triangles and independent of latitude andspecific time of day or year, for determining the heightof the pyramid. (Hint: Use two shadow observationsspaced a few hours apart.)

pyramid

shadow

5: The Pythagorean Theorem Over and Over and Over andOver and. . . . Certainly no mathematical propositionboasts a greater number of different proofs than the the-orem of Pythagoras, for which well over 400 different ar-guments can be found in E.S. Loomis’ The Pythagorean

Proposition, published in 1968. These may seem to ap-proach mathematical overkill; on the other hand, theyshould satisfy even the most hard-boiled sceptic.

For each proof outlined below, we begin with a right tri-angle BAC with right angle at A (thus the hypotenuseis labeled a and the legs are b and c). As you workthrough the details of these proofs, ask yourself wherewe use the assumption that we have a right triangle.

5a. This is the proof that we may attribute to Pythag-oras himself. We begin with identical squares ofside length b+ c, decomposed as shown below.

c b

b

c

bc

c

b

c b

c

b

cb

c

b

a1. Show that the area of the left-hand square is2bc+ b2 + c2.

a2. Prove that the inner figure on the right is itselfa square.

a3. Show that the area of the right-hand sidesquare is 2bc+ a2.

a4. Now prove the Pythagorean Theorem fromthese preliminaries.

3Look at my poster above the bulletin board for a picture.



5b. This proof is due to the 12th century Indian math-ematician Bhaskara. Assemble four copies of theoriginal triangle as shown below.

cb c

b

cbc

b

a

a

a

a

b1. Prove that the large quadrilateral is a square.

b2. Prove that the inner quadrilateral is a square.

b3. Equate the area of the large square with thetotal areas of the small square and the fourtriangles to prove the Pythagorean Theorem.

5c. This proof is usually attributed to the 17th cen-tury British mathematician John Wallis, althoughit surely had been discovered prior to him. It isregarded as the shortest proof of all.

B C

A

D a

c b

c1. From A draw altitude AD to the hypotenuseand prove ADC ∼ BAC ∼ BDA.

c2. Conclude that b/CD = a/b and that c/DB =a/c. From this, complete the proof.

5d. Here is another similarity proof. In right triangleBAC, mark off BD = BA, then bisect 6 ABC byline BE, where E is on side AC. Also, draw EDand call its length x.

B C

A

E

D

x

d1. Prove BAE ∼= BDE.

d2. Show EDC ∼ BAC.

d3. Set up the resulting proportions from part d2and use these to eliminate x and thereby de-rive the theorem.

5e. Here is a clever “inscribed circle” proof. Again be-gin with BAC. Inscribe within a circle havingcenter O and radius r. Draw OD, OE, and OF ,as shown. You may use one key fact from Euclidabout inscribed circles: (Prop. IV.4) The point O– the center of the inscribed circle – is the inter-section of the three bisectors of the angles of thetriangle.

A C

B

E

OD

F

r

r r

e1. Carefully prove that BOD ∼= BOF andCOE ∼= COF .

e2. Prove quadrilateral AEOD is a square.

e3. Explain why b+ c− 2r = a and solve this forr.

e4. Why is the area of BAC equal to 12bc?

e5. Decomposing BAC into the square and thetwo pairs of congruent triangles, prove thatthe area of BAC is equal to r(b+ c− r).

e6. Finally, equate the two expressions for thearea of BAC from parts e4 and e5, substi-tute for r from part e3, and do some morealgebra to derive the theorem.

5f. The last proof is due to Congressman (later Presi-dent) James A. Garfield of Ohio, who published itin the New England Journal of Education in 1876.This proof depends on calculating the area of thetrapezoid in the figure below in two different ways:by the formula for the area of the trapezoid andby the sum of the areas of the triangles. Carry outthis proof in detail.

c b

c

ba

a



Weekender #3


1a. page 167 problem 4(a);

1b. page 168 problem 5.

2: Two From Nine. The most famous ancient Chinesemathematical work is the Jiuzhang suanshu, or Nine

Chapters on the Mathematical Art. Compiled around200 bc by one of ancient China’s most renowned math-ematicians, Liu Hui, but probably written 900 years ear-lier, this work formed the basis of mathematics educa-tion in China for nearly a millennium. Solve the follow-ing two problems from the Nine Chapters.

2a. The height of a wall is 10 ch’ih.4 A pole of un-known length leans against the wall so that its topis even with the top of the wall. If the bottom ofthe pole is moved 1 ch’ih farther from the wall, thepole will fall to the ground. What is the length ofthe pole?

2b. A square pond has side 10 ch’ih with a reed grow-ing in the center whose top is 1 ch’ih out of thewater.5 If the reed is pulled to the shore, the topjust reaches the shore. What is the depth of thewater and the length of the reed?

3: Pythagorean Triples Are Never Identical. An or-dered triple (a, b, c)—called a Pythagorean Triple—isan ordered triple of numbers that are also sides of aright triangle. For example, (3, 4, 5) and (5, 12, 13) arePythagorean triples; in fact, they are called primitive

Pythagorean triples since the greatest common factorof the three numbers is 1. For instance, (6, 8, 10) isa Pythagorean triple, but it is not primitive since thegreatest common factor of 6, 8, and 10 is 2.

3a. Using algebra, show that, for any positive integern, the three numbers 2n, n2 −1, and n2 +1 consti-tute a Pythagorean triple. It is believed that thePythagoreans knew this.

3b. Let integers u and v have greatest common factor1, let one be even and one odd, and let u > v.Then every primitive Pythagorean triple (a, b, c) isgiven by

a = u2 − v2, b = 2uv, c = u2 + v2.

For example, if u = 2 and v = 1, then a = 3, b = 4,and c = 5. Find the 16 primitive Pythagoreantriples for which c < 100. Suggestion: Make atable with column headings u, v, a, b, and c toorganize your work.

4: Come On Columbus, the Greeks Already Knew the EarthWas Round. The Greek philosopher Eratosthenes (276-194 bc), in 240 bc, made a measurement of the earth.He observed at Syene, at noon on the summer solstice,that a vertical stick cast no shadow, while at Alexandria(which he believed to be on the same meridian as Syene)the sun’s rays were inclined to the vertical 1

50of a com-

plete circle. He then calculated the circumference of theearth from the known distance of 5000 stades betweenAlexandria and Syene. Obtain Eratosthenes’ result of250,000 stades for the circumference of the earth. Thereis reason to believe that a stade was equal to about 559feet. Assuming this, calculate from the above result thediameter of the earth in miles.6

5: Pushing Past Geometrical Limits.

5a. Knowing how to measure lengths of straight-linesegments, how might one define the length of thecircumference of a circle?

5b. Knowing how to measure areas of polygons, howmight one define the area of a circle?

5c. Let Pn and An denote the perimeter and the areaof a regular n-gon circumscribed about a circle ofunit diameter. Find

limn→∞

Pn and limn→∞

An.

5d. Knowing the volume of a pyramid is given by one-third the area of its base times its altitude, howmight one arrive at the formula for the volume ofa circular cone?

6: The Usefulness of Definitions. The definition of a tech-nical term (beyond primitive ones) of a logical discourseserves merely as an abbreviation for a complex com-bination of terms already present. Thus a new termintroduced by definition is really arbitrary, though con-venient, and may be entirely ignored; but then the dis-course would immediately increase in complexity. Con-sider, for example, the following definitions taken from ageometry text: The diagonals of a quadrilateral are thetwo straight line segments joining the two pairs of op-posite vertices of the quadrilateral; parallels are straightlines lying in the same plane and never meeting, howeverfar they are extended in either direction; a parallelogram

is a quadrilateral having its opposite sides parallel.

6a. Without using any of the italicized words above,restate the proposition: “The diagonals of a par-allelogram bisect each other.”

6b. By means of appropriate definitions, reduce thethe following sentence to one containing no morethan five words: “The movable seats with four legswere restored to a sound state by the person whotakes care of the building.”

4The ch’ih is an ancient Chinese unit of length.5A reed grows from the bottom of a pond.6The actual diameter of the earth to the nearest mile is 7900 miles.



7: The Quadrature of the Lune. Hippocrates of Chios wasan ancient Greek mathematician and astronomer wholived in the 5th century bc.7 He is credited with writ-ing the first book on geometry. Below is the figure Hip-pocrates used in his geometric proof of the area of thelune. The lune is the shaded area in the figure. In whatfollows, you will prove his result algebraically (which is,of course, totally alien to Greek mathematics).

A O B

C

D

F

E

r r

r

7a. Letting r be the radius of semicircle ACB, find thearea of triangle AOC in terms of r.

7b. Now determine the area of circular segmentAFCD in terms of r. (Use the post-Hippocrateanfact that the area of a circle is πr2.)

7c. Next, find the area of semicircle AECD in termsof r.

7d. Using parts 7b and 7c, find the area of lune AECFin terms of r.

7e. Finally, put all of this together to explain why Hip-pocrates’ lune is quadrable. (That is, why is itpossible to construct a square whose area is equalto that of the lune?)

7Not to be confused with the other Hippocrates, who lived around the same time, who is credited as the father of Western medicine,

and for whom the physician’s “Hippocratic Oath” is named.



Weekender #4




2: Time To Get Mean. If a and b are two real numbers,the following means have been found useful:

i. arithmetic mean:a+ b

2

ii. geometric mean:√ab

iii. harmonic mean:2ab

a+ b

iv. root-mean-square:

√

a2 + b2

2

Here is one way to explain what these means represent.Let a and b, with a > b, denote the lengths of the lowerand upper bases of a trapezoid. Then any line segmentparallel to the bases with endpoints on the other sidesis some mean of a and b:

i. arithmetic: bisects the sides of the trapezoid

ii. geometric: divides the trapezoid into two similartrapezoids

iii. harmonic: passes through the intersection of thediagonals

iv. root-mean-square: bisects the area of the trapezoid

2a. Find each of the four means of 8 and 16.

2b. A trapezoid has bases of lengths 6 and 12. What isthe length of the line that cuts the trapezoid intotwo trapezoids of equal area? What is the lengthof the line that cuts the trapezoid into two similartrapezoids?

2c. Let s be the side of a square inscribed in a righttriangle and having one angle coinciding with theright angle of the triangle. Show that s is half theharmonic mean of the legs of the triangle.

2d. A car travels at the rate r1 mph from A to B,and then returns at the rate r2 mph from B to A.Show that the average rate for the round trip isthe harmonic mean of r1 and r2.

Equally used in ancient times were the “progression” in-terpretations of the arithmetic, geometric, and harmonicmeans. We say that three numbers p, q, r (in that order)are in arithmetic progression if

q − p = r − q;

they are in geometric progression if

q

p=r

q;

and they are in harmonic progression if

q − p

p=r − q

r.

2e. If a, b, and c are in harmonic progression, thenprove that their reciprocals 1

c, 1

b, and 1

a, are in

arithmetic progression.

2f. If a2, b2, and c2 are in arithmetic progression, thenprove that b+ c, c+ a, and a+ b are in harmonicprogression.

2g. Since 8 is the harmonic mean of 12 and 6, Philo-laus, a Pythagorean of about 425 bc, called thecube a “geometrical harmony.” Why a cube andnot some other solid?

3: Geometric Algebra. Let r and s denote the roots of theequation

x2 − px+ q2 = 0

where p and q are positive numbers.

3a. Show that r + s = p and rs = q2.

3b. Show that if q ≤ p

2then r and s are both positive.

3c. To solve the equation x2 − px + q2 = 0 geomet-rically, we must find segments r and s from givensegments p and q. The figure below is the con-struction made, from segments p and q, to find rand s. Explain the construction and explain whyit works.

p

q

r s

3d. Now consider the equation

x2 − px− q2 = 0

where p and q are positive numbers. Here, theroots are real, but one, say s, is negative and theother, r, is positive. To solve this equation geo-metrically, we again must find segments r and sfrom given segments p and q. The figure below isthe construction made, from segments p and q, tofind r and s. Explain why the construction works.

p

q

r

s

With these two examples of geometric algebra, it is easyto see why ancient cultures considered each equation acompletely different problem.

4: The Euclidean Algorithm. There is non-geometricalmaterial in the Elements, such as the Euclidean algo-rithm, a process for finding the greatest common divisor(gcd) of two numbers. This is found at the beginningof Book VII, but was presumably known before Euclid’stime. Stated as a rule, the process is this:



Divide the larger of the two numbers by thesmaller one. Then divide the divisor by theremainder. Continue this process, of dividingthe last divisor by the last remainder, untilthere is no remainder. The final divisor is thegcd.

For example, let us find the gcd of 126 and 210. Wedivide 210 by 126 to get 1 with remainder 84. Next, wedivide the divisor, 126, by the remainder, 84, to get 1with new remainder 42. Now divide the previous divi-sor, 84, by the previous remainder, 42, to get 2 with noremainder. Now that we have no remainder, we knowthat the gcd was the last divisor we used: 42. So thegcd of 126 and 210 is 42.

4a. Find the gcd of 481 and 851.

4b. Find the gcd of 5913 and 7592.

4c. Find the gcd of 1827 and 3248.

5: The Platonic Solids—They Swear They Are Just Friends.The final book of the Elements is devoted to the con-struction of the regular polyhedra known as the PlatonicSolids: the tetrahedron (all 4 faces are equilateral trian-gles), cube (6 squares), octahedron (8 equilateral trian-gles), dodecahedron (12 regular pentagons), and icosa-hedron (20 equilateral triangles).8 The first three solidswere known before the Greeks, and there is an extantbronze dodecahedron dating from the seventh centurybc. The icosahedron was apparently first studied byTheaeteus (417-369 bc), who also proved that these arethe only regular polyhedra. Euclid constructed theseby inscribing them in spheres; then compared the exactside lengths of each solid to the diameter of the sphere.If we let the diameter of the sphere be 1 unit, these arethe side lengths found by Euclid:

Tetrahedron:√

23

Cube:√

13

Octahedron:√

12

Dodecahedron:

√15 −

√3

6

Icosahedron:

√

50 − 10√

5

10

What is truly remarkable is that, with the lack of a goodnumber system and the complete absence of algebraicnotation, Euclid accomplished this task! Your task isto find decimal approximations to these measurements,then to find the (approximate) surface area of each solid.

6: The Logical Role of the Parallel Postulate. None of thepropositions in Book I prior to Proposition I.29 uses theParallel Postulate in its proof, whereas all the later re-sults in Book I depend on the Parallel Postulate, with

a single exception. Find it. Speculate as to why Eucliddidn’t put it before I.29.9

7: The Pythagorean Theorem. . . One More Time. Theprevious six proofs of the Pythagorean Theorem fromWeekender #2 give an array of possible ways of estab-lishing the theorem, but it should be noted that Euclidused none of them in the Elements. Read the proofcontained in the Elements (Prop. I.47). Which of theprevious six could he have used as Prop. I.47? Do yougive Euclid high marks for the proof he actually devisedfor the Elements, or did he miss an easy proof?

8: There are Infinitely Many Primes. From Euclid’s Prop.IX.20, we know that we’ll never run out of primes.10 Inaddition, since all primes but 2 are odd, we can dividethe odd primes into two categories—those that are onemore than a multiple of 4 (e.g., 5, 13, 17, 29, . . . ) andthose that are three more than a multiple of 4 (e.g., 3,7, 11, 19, . . . ). Obviously, at least one of these two cat-egories of primes must be infinite. In what follows, we’llmodify Euclid’s proof of IX.20 to show that there areinfinitely many primes of the form p = 4n+ 3.

8a. Prove that the product of two numbers, each ofwhich is one more than a multiple of 4, is itselfone more than a multiple of 4. In other words, ifa = 4m + 1 and b = 4n + 1, then ab also has thisform. (Of course, all numbers here are positiveintegers.)

8b. Now suppose that a, b, c, d, . . . , w is a finite col-lection of primes (in Euclid’s words, an “assignedmultitude”), each having the form 4n + 3. Weintroduce a new number M = 4(abcd · · ·w) − 1.Mimic Euclid’s two cases from IX.20 to show thatthere must be a prime of the form 4n+3 not amongthe original multitude. Conclude that there are in-finitely many primes of the form 4n+3. (Hint: Usethe property that positive integers have a uniqueprime factorization and part 8a.)

8c. Are there infinitely many composites of the form4n+ 3? Explain.

9: A Smattering of Results.

9a. Read the proof of Prop. II.1. Translate this into asimple algebraic identity.

9b. Read the proof of Prop. II.13. What is the famousresult from trigonometry that you should recognizehere?

9c. Read the proof of Prop. III.1. Use a compassand a straightedge to follow along with the proof,and construct the center of a given circle yourself.What method of proof did Euclid use?

8Look at my poster above the bulletin board for a picture.9For this and the next three problems, you may find it helpful to consult the home of the amazing on-line version of the Elements at

http://aleph0.clarku.edu/~djoyce/java/elements/elements.html. Your browser will need a Java plug-in for the interactive diagrams.10If you haven’t read IX.20 yet, you really should before starting this problem.



Weekender #5




2: The Quadrature of the Parabola. One of Archimedes’most remarkable accomplishments was in determiningthe area of a parabolic segment. Let AV C be a parabolicsegment with vertex V and base AC. Archimedesthen proceeds to inscribe triangles. The first trian-gle is AV C. In the two portions left over, he in-scribes two triangles, ABV and CDV ; in the foursmaller portions left over, he inscribes four triangles,and so on—completely filling the parabola with trian-gles. Archimedes then calculated that the total areas ofthe triangles at each stage was 1

4the area of the trian-

gles from the previous stage. Thus, after n stages, thesum of the areas of all inscribed triangles is equal to

(

1 +1

4+ · · · +

1

4n

)

AV C.

As n increases indefinitely, all triangles equal theparabolic segment. Hence, since the sum of the “in-finitely many” fractions is an infinite geometric series,we have that the area of the parabolic segment is equalto

1

1 − 14

AV C =4

3AV C.

A

V

C

DB

Thus, the area of any parabolic segment is 43

the area ofthe inscribed triangle!

For example, to find the area of the segment formed byy = 2 −x2 and the x-axis, we inscribe a triangle so thatthe base lies on the x-axis and one vertex is on the y-axis(like in the picture above—imagine AC is the x-axis andthe y-axis goes through V ). Then the base of the trian-gle is the distance between the x-intercepts of y = 2−x2;the x-intercepts are −

√2 and

√2, so the base has length√

2 − (−√

2) = 2√

2. The height of the triangle is they-intercept of y = 2 − x2; this is 2. Therefore the areaof the inscribed triangle is 1

2· 2 · 2

√2 = 2

√2. Thus, the

area of the parabolic region is 43

· 2√

2 = 83

√2.

Without using calculus, find the area of the region be-tween:

2a. y = 4 − x2 and the x-axis;

2b. y = 10 − x2 and the x-axis;

2c. y = 16 − x2 and the x-axis.

3: Archimedes, an Approximating βαδασσβαδασσβαδασσ. In the midstof his approximation for the value of π, Archimedesneeded a value for

√3 and he used

265

153<

√3 <

1351

780.

How good is this as a decimal?

4: You, an Approximating βαδασσβαδασσβαδασσ. An interesting ques-tion raised in the previous problem is how one wouldget such a sharp approximation for square roots with-out benefit of a calculator. A very nice algorithm is thefollowing.

To approximate√A, begin with an initial approxima-

tion (obtained by “eyeballing it”) of x0. Then let thenext approximation be

x1 =x2

0 +A

2x0,

then the next one is

x2 =x2

1 +A

2x1

and generally, use the recursive definition

xn+1 =x2

n +A

2xn

. (1)

4a. Assuming that the sequence of successive approxi-mations converges to a limit L, show that L =

√A.

(Hint: Take limits of both sides of Eq. (1) asn → ∞.)

4b. Now suppose we want to approximate√

3 and westart with the rational number x0 = 5

3(this is

reasonable since ( 53)2 = 25

9≈ 27

9= 3). Apply the

recursion formula twice to approximate this squareroot.

4c. Notice anything? Do you think Archimedes wason to something?

5: Archimedean “Formulas” in Geometric Disguise. In hismasterpiece On the Sphere and Cylinder, Archimedesstated his results about spherical volume and area bycomparing his figures with such better-understood fig-ures as cylinders and cones. Assuming we know themodern formulas for the key properties of cones andcylinders, translate the following statements into famil-iar, modern-day formulas.

5a. “Any sphere is equal (by volume) to four times thecone which has its base equal to the greatest circlein the sphere and its height equal to the radius ofthe sphere.”

5b. “Every cylinder whose base is the greatest circle ina sphere and whose height is equal to the diameterof the sphere is half again as large as the sphere.”

5c. “Every cylinder whose base is the greatest circlein a sphere and whose height is equal to the di-ameter of the sphere has surface (together with itsbases) that is half again as large as the surface ofthe sphere.”



6: The Quadrature of the Arbelos. Let A, C, and B bethree collinear points, with C between A and B. Semi-circles having AC, BC, and AB as diameters are drawnon the same side of the line. The figure bounded by thethree semicircles is called an arbelos.

The term “arbelos” means shoemaker’s knife in Greek,and this term is applied to the figure which resembles theblade of a knife used by ancient cobblers. Once again, itwas Archimedes who investigated the fascinating prop-erties of this figure.

A C B

6a. At C, erect a perpendicular to AB that intersectssemicircle AB at G. Let the common externaltangent to the two smaller semicircles touch thesesemicircles at T and W . You will prove that GCand TW are equal.

a1. Draw TW and connect the radii of the twosmaller circles to TW . Then construct a seg-ment parallel to TW from the center of thesmallest semicircle. This creates a right tri-angle. Find the sides of the right triangle interms of AC and BC.

a2. Show that TW is equal to the length of thelongest leg of the right triangle, and find asimplified expression for TW in terms of ACand BC.

a3. Find an expression for CG in terms of AC andBC, and conclude that TW = GC. (Hint:

Use Problem 3 from Weekender #4.)

6b. Prove that the area of the arbelos equals the areaof the circle that has GC as a diameter.

7: Ptolemy’s Corollary. Prove the following usingPtolemy’s Theorem: If P lies on the arc AB of the cir-cumcircle of an equilateral triangle ABC, then PC =PA+ PB.

8: Approximating Polygons with Heron. Heron (10-75 ad)was the first Greek geometer to be concerned, almost ex-clusively, with practical problems and procedures (for-mulas) for finding areas and volumes of various shapes.Heron determined all kinds of area relationships. Al-though the areas of some regular polygons are relativelyeasy to find (square and octagon, for instance) Heronapproximated the area of many other regular polygons.

For example, if we let s be the side length and An thearea of the regular n-gon, Heron gave

A3 =13

30s2, A5 =

5

3s2,

A7 =43

12s2, A9 =

51

8s2.

8a. Using the modern formula

An =1

4ns2 cot

(

π

n

)

determine the percentage error11 in Heron’s ap-proximations.

8b. What approximation for√

3 is implied by Heron’svalue for A3?

9: Having Fun with Heron.

9a. We own a four-sided piece of land, whose areawe must determine for tax purposes. Pacing offthe sides, we get lengths of 100 yards, 170 yards,250 yards, and 240 yards (as shown), and then wemarch along the diagonal and find its length to be260 yards. How many square yards of property dowe own?

260

100 240

250170

9b. An equilateral triangle has each side 4x units long,where x is some positive real number. Find its areaboth by the standard formula (a = 1

2bh) and by

Heron’s formula and verify that the results agree.

9c. Consider the triangle below.

B C

A

D

85

132

157

c1. Find its area via Heron’s formula.

c2. Find its altitude, AD, using part c1.

c3. What do you notice about ABC from partc2?

c4. Suppose a circle is inscribed in this triangle.Determine the area that is left if we removethe inscribed circle from ABC. (Hint: Usea result from Problem 5e on Weekender #2.)

11Percentage error is |estimated value − actual value| × 100 ÷ actual value.



Weekender #6




1c. page 119 problem 2;

1d. page 119 problem 4.

2: Battle of the Millenia Round One: Pythagoras and HeronSquare Off. Suppose we have a right triangle BAC withright angle at A with sides a, b, and c. We extend CAto D so that AD = AC = b and then we draw BD, asshown below.

CA

B

b

a c

D

2a. Show that BAC ∼= BAD.

2b. Explain why the semiperimeter of DBC is s =a+ b.

2c. Apply Heron’s formula to DBC to deduce thatthe area of DBC is equal to

√

b2 (a2 − b2).

2d. Apply the simple formula for triangular area (A =12bh) to prove that the area of DBC is bc.

2e. Finally equate the area expression from parts 2cand 2d, simplify algebraically, and deduce thePythagorean Theorem.

3: Battle of the Millenia Round Two: Pythagoras and HeronSquare Off. Beginning with a right triangle of sides a, b,and c, determine its area in two ways: by the standardformula (A = 1

2bh) and by Heron’s formula. Equate

these and manipulate the resulting, dreadful equationuntil you have derived the Pythagorean Theorem.

4: The Referee’s Decision in the Battle of the Millenia.

4a. Which proof of “Heron implies Pythagoras” do youlike better: Problem 2 or 3?

4b. Do either of these proofs in Problems 2 and 3 pro-vide valid proofs of the Pythagorean Theorem?That is, do they contain circular reasoning?

5: Prismatoids. A polyhedron all of whose vertices lie intwo parallel planes is called a prismatoid. The sectionparallel to the bases and midway between them is calledthe midsection of the prismatoid. The figure below showa prismatoid with rectangular bases. If we denote theareas of the upper base, lower base, and midsection byU , L, and M , we have the modern formula for the vol-ume of any prismatoid of height h:

V =h(U + L+ 4M)

6.

ab

c

d

L

U

M

In Book II of Heron’s Metrica, Heron gives the volume ofa prismatoid with rectangular bases. Denote the lowerlength and width by a and b, and the upper by c and d.Then Heron gives this formula:

V = h

[

(a+ c)(b+ d)

4+

(a− c)(b− d)

12

]

.

5a. Write U , L, and M in terms of a, b, c, and d.

5b. Show that Heron’s formula is equivalent to themodern one.

5c. Show that the Egyptian formula for a truncatedpyramid is a special case of the prismatoid for-mula.

6: Perfect and Amicable Numbers Have Such Nice Person-alities.

6a. Show that 1184 and 1210 are amicable by findingthe factors of each and then adding the factors ofeach.

6b. Thabit ibn Qurra invented the following rule forfinding amicable numbers: If

p = 3(2n) − 1, q = 3(2n−1) − 1,

r = 9(22n−1) − 1

are three odd primes, then 2npq and 2nr are apair of amicable numbers. Verify this for n = 2and n = 4.

6c. Why does ibn Qurra’s rule fail for n = 3?

6d. Why must n be prime in Euclid’s propositionabout perfect numbers?

6e. What is the fourth perfect number furnished byEuclid’s formula?

7: Figurate Numbers Have Such Lovely Shapes.

7a. List the first four pentagonal numbers and the firstfour hexagonal numbers.

7b. Prove, using algebra, that 8 times any triangularnumber plus 1 is a square number.

7c. Show that the nth pentagonal number equals thenth square number plus the (n − 1)th triangularnumber.



8: Nicomachus’ Amazing Formula of Formulas. Nico-machus (60-120 ad) was the first to treat arithmeticseparately from geometry. He wrote a comprehensiveset of school texts based on the quadrivium: arithmetic,harmonics (music), geometry, and astronomy. In thebook on arithmetic, he gives the following formula,

12(m− 2)n2 + 1

2(4 −m)n,

for finding the nth m-gonal number. For instance, tofind the 8th pentagonal number, let n = 8 and m = 5;the result is 92. We may also use the formula to findmore formulas; Let m = 3 and we have 1

2n2 − 1

2n as

the formula for triangular numbers. Find formulas forpentagonal, hexagonal, and heptagonal numbers.

9: Diophantus Got ϙθ Problems But Algebra Ain’t α. Thefollowing problems are all from the Arithmetica.

9a. A common technique employed by Diophantus tosolve a system of equations is the following. Con-sider the system

x+ y = a

x2 + y2 = b.

Diophantus sets x = a2

+ z and y = a2

− z. (Notethat this ensures that x+ y = a.) He then substi-tutes these expressions in the second equation; thisresults in a single quadratic in the new variable z,whose solution is straightforward. Having foundz, it is then easy to produce the values of x andy. This technique was employed by many culturesbefore the Greeks; this leads some to believe thatthere was transmission of ideas between societies.Solve the following two systems in this manner:

x+ y = 20

x2 + y2 = 208and

x+ y = 22

x2 + y2 = 274

9b. Diophantus also used a similar technique to theone above for systems of the form

x+ y = a

x3 + y3 = c.

In this instance, with x = a2

+ z and y = a2

− z, wehave that

z =

√

c− 2( a2

)3

3a

so that x and y are easily found. Use this techniqueto solve

x+ y = 20

x3 + y3 = 2240.

9c. Solve Problem 17 of Arithmetica Book I: Find fournumbers, the sum of every arrangement three at atime being 22, 24, 27, and 20.

9d. All we know of Diophantus’ personal life is foundin the summary of an epitaph given in Pappus’Collection, about 350 ad:

“This tomb holds Diophantus. Ah, howgreat a marvel! The tomb tells mathe-matically the measure of his life. Godgranted him to be a boy for one-sixth ofhis life, and adding one-twelfth to this,he grew a beard; he was married afteranother one-seventh, and five years afterhis marriage God granted him a son—Alas!; after living half his father’s life,chill Fate took him. After consoling hisgrief by mathematics for four years Dio-phantus ended his life. Tell me the num-ber of years Diophantus lived.”

9e. Bonus: If m is any positive integer and

x = m2, y = (m+ 1)2, z = 2(x+ y + 1),

prove that the three numbers

xy + z, yz + x, zx+ y

are all squares.

10: Here’s Why You Should Be Happy with Algebraic Sym-bols.

10a. Write 2x4 − 21x3 + 12x2 − 7x+ 33 in the notationof Diophantus.

10b. Write 3xy + 2x+ 2y +√

13 − 8 in the notation ofBrahmagupta.

10c. Write the following in Bombelli’s notation:

√

3

√√68 + 2 − 3

√√68 − 2.

10d. Bombelli indicated√

−11 by di m R q 11. Write,in modern notation, the following expression whichappears in Bombelli’s work:

R c ⌊ 4 p di m R q 11 ⌋p R c ⌊ 4 m di m R q 11 ⌋ .

10e. Viète qualified the coefficients of a polynomialequation so as to render geometric meaning, andhe used our present + and − signs, but he hadno symbol for equality. He would have written theequation 5BA2 − 2CA+A3 = D as

B 5 in A quad − C plano 2 in A

+A cub aequatur D solido.

Note how the coefficients C and D are qualifiedso as to make each term of the equation three-dimensional. Write, in Viète’s notation, A3 −3BA2 + 4CA = 2D.



Weekender #7





1d. page 77 problem 5;

1e. page 131 problem 1.

2: There’s No Such Thing as New Algebra Problems.

2a. From the Greek Anthology, assembled byMetrodorus in 500 ad, solve the following: Makea crown of gold, copper, tin, and iron weighing 60minæ. Gold and copper shall be two thirds of it;gold and tin three fourths of it; and gold and ironthree fifths of it. Find the weights of gold, copper,iron, and tin required.

2b. Solve this problem, from Chuquet’s 1484 book,Three parts in the science of numbers: Find twonumbers in the proportion 5 : 7 such that thesquare of the smaller multiplied by the larger gives40.

2c. Solve this problem, also from Chuquet: A mer-chant visited three fairs. At the first, he doubledhis money and spent 30; at the second he tripledhis money and spent 54; at the third he quadru-pled his money and spent 72; and then he had 48left. How much had he at the start?

2d. Solve the following, from Rudolff’s book Coss of1525: I am owed 3240 florins. The debtor pays me1 florin the first day, 2 the second day, 3 the thirdday, and so on. How many days does it take topay off the debt?

2e. From Recorde’s Whetstone of Witte of 1557, wehave the following challenge: A gentleman, willingto prove the cunning of a bragging Arithmetician,said thus—I have in both hands 8 crowns. But ifI count the sum of each hand by itself and add toit the squares and the cubes of the both, it willmake in number 194. Now tell me, what is in eachhand? (This is an easy problem, but the way it’swritten makes it more difficult than it appears.)

2f. Bonus: From Pacioli’s Summa de arithmetica of1494, solve this problem: The radius of the in-scribed circle of a triangle is 4, and the segmentsinto which one side is divided by the point of con-tact are 6 and 8. Determine the other two sides.

3: Gelosia. Multiply 80,342 and 7318 by the gelosiamethod. Make sure you demonstrate the method!

4: Casting Out Nines. The divisibility rule for 9 has beenknown for hundreds of years: a number is divisible by9 if the sum of its digits is divisible by 9. This factwas used to determine the following two theorems. Theremainder when a number is divided by 9 is called theexcess for that number.

I. The excess for a sum is equal to sum of the excessesfor each addend.

II. The excess for a product of two numbers is equalto the product of the excesses for each factor.

These facts furnish a method for checking arithmeticcalled casting out 9s. For example, to check the calcu-lation 87305 + 93295 = 180600, we find the excess ofeach number: excess(87305) = 5, excess(93295) = 1,excess(180600) = 6. Since the excesses add correctly(5 + 1 = 6), the arithmetic must be correct as well.

There is another theorem based on divisibility by 9.

III. If the digits of any integer are rearranged to forma new number, then the difference between the oldand new numbers is divisible by 9.

Rearrange 93295 to get 29953. Subtract: 93295 −29953 = 63342. Excess(63342) = 0, so 63342 is divisibleby 9.

4a. By hand, add 478 and 993, and then, also by hand,multiply 478 and 993. Show that you checked youranswers by casting out 9s.

4b. Why does Theorem III work?

Theorem III is still used today. It is called thebookkeeper’s check: if the sums of debit entries andcredit entries do not balance, and the differencebetween the two sums is divisible by 9, then it isquite likely that the error is due to accidentally re-versing (transposing) digits when writing down adebit or credit.

5: The Myth of Gematria. Since many of the ancientnumeral systems were also alphabetical, it was naturalto substitute number values for the letters in a name.This led to the pseudo-science called arithmography, orgematria, which was very popular among the ancients,and was revived during the Middle Ages.

Michael Stifel (1487-1567) was an ordained priest whoused gematria to predict the world would end on Oc-tober 18, 1533. This, in itself, was not a punishableoffense, but he had persuaded his congregation to giveall their wordly goods to him for “safe-keeping” in thehereafter. He was tried for heresy, and turned his ob-vious mathematical skill to better use by writing booksusing the new algebraic symbols of Rudolff.

An extreme example of Stifel’s gematria is his “proof”that Pope Leo X was the “beast” mentioned in the Book

of Revelation. From the Latin LEO DECIMVS he re-tained the letters L, D, C, I, M, and V since these havevalues in the Roman numeral system. He then addedX, both for Leo X and because Leo decimus contains 10letters, and omitted M because it stands for mysterium.A rearrangement of the letters gave DCLXVI, or 666,the “number of the beast.”

Some years later, Napier, the inventor of Logarithms,showed (by similar methods) that 666 stands for thePope of Rome, and a Jesuit priest, Father Bongus, de-clared that 666 stands for Martin Luther. Bongus’ rea-soning went as follows. (The Latin alphabet is just like



the English, except there is no J and no W, and in uppercase, U appears as a V.)

Bongus let the letters A to I represent 1 to 9, K to Srepresent 10 to 90 (by tens), and T to Z represent 100to 500 (by hundreds). Then we have, in Latin,

M A R T I N30 1 80 100 9 40

L V T E R A20 200 100 5 80 1

whose sum is 666.

It has been shown that if you express 666 in the nu-meric letter symbols of the Aramaic language in whichthe Book of Revelation was originally written, 666 spellsNero.

During World War I, gematria was used to show that666 must be Kaiser Wilhelm, and later it was shown torepresent Hitler. By using values assigned to Englishletters, it was “proven” that, of the three, Rooseveltwas a “greater” leader than Churchill or Stalin. Morerecently, former President Ronald Wilson Reagan hasbeen “proven” to be the beast since each of his threenames has six letters.

5a. “Prove,” by Stifel’s method, that each of theserepresent 666: LVDOVICVS (Ludovicus — KingLouis XIV) and SILVESTER SECVNDVS (Sil-vester Secundus — Pope Sylvester II)

5b. What about your own name? Can you assign val-ues to the English alphabet (similar to the way inwhich Bongus assigned values to the Latin alpha-bet) to prove how “great” (or how beastly!) youare?

6: The Incredible Al-Karajî. Abu Bekr ibn Muham-mad ibn al-Husayn Al-Karajî (953-1029) was an Islamicmathematician who can be regarded as the first personto free algebra from geometrical operations and replacethem with the type of operations which are at the core ofalgebra today. Around the year 1020, Al-Karajî wrotea book on algebra called the al-Fakhrî. (The book isnamed after his patron, Fakhr al-Mulk, the grand vizierof Baghdad at the time.) In it, one is asked to find tworational numbers such that the sum of their cubes is asquare. In other words, find rational numbers x, y, andz such that x3 +y3 = z2. Al-Karajî takes as the solution

x =n2

1 +m3, y = mx, and z = nx,

where m and n are any rational numbers.

6a. Find x, y, and z when m = 2 and n = 3. Use thesevalues to verify that x3 + y3 equals z2.

6b. What is the relationship between x and y whenm = 1 and n is even? Why does n have to be evenin this case?

6c. Find the exact values (in fractions—no decmials!)of x, y, and z when m = 1

4and n = 1

8.

7: The Fantastic Abu Kamıl. Abu Kamıl (850-930) wasan Islamic mathematician best known for writing a com-mentary on Al-Khwarizmı’s Algebra, in addition to writ-ing a number of clever algebra problems. This is one ofthose clever problems: “The number 50 is divided by acertain other number. If the divisor is increased by 3,then the quotient decreases by 3 3

4. What is the divisor?”

Solve this problem.

8: The Mathematical Pope. Gerbert (950-1003) was bornin France and from an early age, he showed amazingabilities. He was one of the first Christians to studyin the Moslem schools in Spain, and may have broughtback the Hindu-Arabic numerals to Christian Europe.He is said to have constructed abaci, terrestrial and ce-lestial globes, a clock, and reportedly an organ. Suchaccomplishments led some to believe that he had soldhis soul to the devil. Nevertheless, he rose steadily inthe Catholic church and was elected Pope as Sylvester IIin 999. He was considered a profound scholar, althoughhe contributed very little that was original to mathe-matics. He did write several textbooks on the subject,however. The following problem is from Gerbert’s Ge-

ometry, and was considered quite difficult at the time;solve it.

Determine the legs of a right triangle whose hypotenuseis 5 and area is 3 9

25.

9: Fibonacci Got 99 Problems But Rabbits Ain’t One.Solve the following problems, which appear in the Liber

abaci.

9a. If A gets from B 7 denarii, then A’s sum is fivefoldB’s; if B gets from A 5 denarii, then B’s sum issevenfold A’s. How much has each?

9b. A certain king sent 30 men into his orchard to planttrees. If they could set out 1000 trees in 9 days, inhow many days would 36 men set out 4400 trees?

9c. There are two numbers that differ by 5. If thegreater is multiplied by

√8 and the lesser by

√10,

the results are equal. Find the numbers.

9d. A man left to his oldest son one bezant and a sev-enth of what was left; then, from the remainder,to his next son he left two bezants and a seventhof what was left; then, from the new remainder, tohis third son he left three bezants and a seventh ofwhat was left. He continued this way, giving eachson one more bezant than the previous son and aseventh of what remained. By this division it de-veloped that the last son received all that was leftand all the sons shared equally. How many sonswere there and how large was the man’s estate?

9e. A man entered an orchard through seven gates,and there took a certain number of apples. Whenhe left the orchard he gave the first guard half theapples that he had and one apple more. To the sec-ond guard he gave half his apples that he had andone apple more. He did the same to each of theremaining five guards, and left the orchard withone apple. How many apples did he gather in theorchard?



Weekender #8





1d. page 205 problem 4(a).

2: The Broken Bamboo. As a small example of how simi-lar problems occur in different cultures, solve the follow-ing two “broken bamboo” problems. The first is fromBrahmagupta’s Correct Astronomical System of Brahma

(628); the second is from Yang Hui’s updated versionof the Chinese work Nine Chapters, called Arithmetic

Rules in Nine Sections (1261).

2a. A bamboo 18 cubits high was broken by the wind.Its top touched the ground 6 cubits from the root.Tell the lengths of the segments of bamboo.

2b. There is a bamboo 10 ch’ih high, the upper endof which being broken reaches the ground 3 ch’ih

from the stem. Find the height of the break.

3: The Spectacular Brahmagupta. Brahmagupta was aHindu astronomer born in 598 ad in northwestern India.Most of his mathematical discoveries were embedded inhis great astronomic work Brahmasphut.asiddhanta, orCorrect Astronomical System of Brahma. Brahmaguptaproved amazing results concerning cyclic quadrilaterals.Consider the following three theorems that he discov-ered.

I. The area of a cyclic quadrilateral with sides a, b,c, and d is

√

(s− a)(s− b)(s− c)(s− d)

where s is the semiperimeter.

II. The diagonals m and n of a cyclic quadrilateralwith consecutive sides a, b, c, and d are

m =

√

(ac+ bd)(ab+ cd)

ad+ bc

and

n =

√

(ac+ bd)(ad+ bc)

ab+ cd.

III. Let x, y, z, X, Y , and Z be positive integers suchthat x2 + y2 = z2 and X2 + Y 2 = Z2. (In otherwords they are Pythagorean triples.) If a = xZ,b = zY , c = yZ, and d = zX are consecutivesides of a cyclic quadrilateral, then the diagonalsare perpendicular and the diameter of the circle isgiven by

D =

√

(ad+ bc)(ab+ cd)

ac+ bd.

Moreover, the area and diagonals are rational num-bers. A cyclic quadrilateral with these propertiesis called a Brahmagupta trapezium.

Find the sides, diagonals, diameter, and area of a Brah-magupta trapezium determined by the Pythagoreantriples (3, 4, 5) and (5, 12, 13). Also find the area of thecircle.

4: Ten Into Two Parts. There is a wide variety of prob-lems found in ancient and medieval mathematics texts,so any similarities stand out easily. One of the peculiarsimilarities are the problems that we call “ten into twoparts” problems. The idea is that 10 is split into twounknown parts, say x and 10 − x, then some other re-lationship is defined between the parts that allows oneto find the numerical values of the parts. Solve the fol-lowing problems. The first two are from the works ofAl-Khwârizmî (780-850); the next one is from the worksof Abu Kamıl (850-930); the last is from the work of oneof the first Italian mathematicians, Antonio de’ Mazz-inghi (1353-1383).

4a. “I have divided 10 into two parts, and having mul-tiplied each part by itself, I have put them to-gether, and have added to them the difference ofthe two parts previous to their multiplication, andthe amount of all this is 54. Find the two parts.”

4b. “I have divided 10 into two parts; I multiply theone by 10 and the other by itself, and the productswere the same. Find the two parts.”

4c. “One says that 10 is divided into two parts, eachof which is divided by the other, and when each ofthe quotients is multiplied by itself and the smalleris subtracted from the larger, then there remains2. Find the two parts.”

4d. “Divide 10 into two parts such that if one squaresthe first, subtracts it from 97, and takes its squareroot, then squares the second, subtracts it from100, and takes its square root, the sum of the tworoots is 17. Find the two parts.”

5: A Mighty Word Problem. Typical of Hindu andmuch Islamic mathematical writing, the following prob-lem from the works of Mahavıra (circa 850) illustratesthe verbal nature of mathematics before symbolism.Find the smallest possible solution to this indeterminateproblem:

“Into the bright and refreshing outskirts of aforest, which was full of numerous trees withtheir branches bent down with the weightof flowers and fruits, trees such as jambutrees, lime trees, plantains, areco palms, jacktrees, date palms, hintala trees, palmyras,punnago trees, and mango trees—outskirts,the various quarters whereof were filled withmany sounds of crowds of parrots and cuck-oos found near springs containing lotuseswith bees roaming about them—into such aforest outskirts a number of weary travelersentered with joy. There were 63 numericallyequal heaps of plantain fruit put together andcombined with 7 more of those same fruits,and these were equally distributed among 23



travelers so as to have no remainder. Youwill tell me now the numerical measure of aheap of plantains.”

6: A “Cardanesque” Approach to Quadratics. Herewe adapt the geometric/algebraic ideas of Cardano’sArs Magna to an old friend—the quadratic equationax2 + bx+ c = 0.

6a. Subdivide a square of side length t to prove geo-metrically that

(t− u)2 + 2u(t− u) = t2 − u2.

u t− u6b. Now convert the original quadratic equation into

x2 +b

ax = − c

a

and introduce the auxiliary variables t and u,where x = t − u. Then, using part 6a, find t andu (and ultimately x) in terms of a, b, and c.

6c. Does your solution for x ring a bell?

7: Using Cardano’s Formula.

7a. Solve x3 + 3x = 10 by Cardano’s Formula.

7b. Solve x3 + 63x = 316 by Cardano’s Formula.

7c. Solve x3 − 63x = 162 by Cardano’s Formula.

7d. By expanding and simplifying, show that

(

−3 + 2√

−3)3

= 81 + 30√

−3

and(

−3 − 2√

−3)3

= 81 − 30√

−3,

thereby proving that the root given in part 7c is−6 in disguise.

7e. Solve 9x3 − 9x = 4 by Cardano’s Formula, andsimplify your solution as much as you can. (Thisdoes not mean obtain a decimal approximation!)

8: A Cubic Transformation. The transformation

x = z − b

3a

converts the general cubic

ax3 + bx2 + cx+ d = 0

into one of the form

z3 +Hz +G = 0.

8a. Find H and G in terms of a, b, c, and d.

8b. Transform x3−9x2+21x−5 = 0; solve the reducedequation; solve the given equation.

9: Using Viète’s Method.

9a. Solve x3 + 63x = 316 by Viète’s method.

9b. By Viète’s method, obtain the cubic associatedwith the quartic equation

x4 + 6x2 + 36 = 60x.

10: Viète Does Cardno One Better. It was Rafael Bombelliwhose 1572 Algebra took the first successful steps towardresolving the troubling “irreducible case of the cubic.”This case arose from the depressed cubic x3 + mx = nin the situation where 1

4n2 + 1

27m3 < 0. Using Car-

dano’s Formula introduces the square root of a negativenumber. Bombelli realized that, in such a case, the real

solutions of a real cubic require us to detour into therealm of the complex numbers. While we begin and endwith real numbers, we seem to have to journey throughthe strange uncharted world of complex numbers. How-ever, Viète’s approach to the “irreducible” case avoidedcomplex numbers; alas, it moved beyond algebra and in-troduced trigonometry. We present Viète’s ideas in thefollowing.

10a. Prove the trigonometric identity

cos3(θ) − 3

4cos(θ) =

1

4cos(3θ).

Hint: Use cos(α+β) = cosα cosβ−sinα sin β withα = 2θ and β = θ.

10b. Now, to solve the depressed cubic x3 + mx = n,Viète made the substitution x = b cos(α), where αand b are unknowns to be determined. Use part10a to show

α =1

3cos−1

(

n/2√

−m3/27

)

and

b =

√

−4m

3.

10c. Apply Viète’s technique to get one real solution ofx3 − 63x = 162. You may resort to decimals andcalculators here. Does your result agree with thatobtained in Problem 7d? What does that tell you?

10d. Apply Viète’s technique to get one real solution ofx3 − 6x = 4. This time, express your answer as asimplified radical—no decimals and no trigonome-try! Hint: You may need to use the not-so-famousfact that

cos(

π

12

)

=

√

1

2+

√3

4.

11: A Quartic by Cardano. Cardano solved the quarticequation

13x2 = x4 + 2x3 + 2x+ 1

by adding 3x2 to both sides. Do this and solve the equa-tion for all four roots.



Weekender #9





1d. page 214 project 3.

2: A Couple of Quintics.

2a. The equation

x5 − x4 − x+ 1 = 0

can be completely solved by factoring. Do this andfind all the roots.

2b. Bonus: I claim that the only real root of the equa-tion

x5 − 5x4 − 10x3 − 10x2 − 5x− 1 = 0

is the number

1 +5√

2 +5√

4 +5√

8 +5√

16.

Either verify my claim or disprove it. (Convertingthis number to decimals and graphing the equationis neither verification nor disproof.) Hint: Expand−(x+ 1)5 and compare.

3: Da Vinci’s “Dynamic Dissection of the Lune”. LeonardoDa Vinci is most famously known as an artist, but hewas an intellectual who sought any and all knowledge.He was an inventor, an artist, a philosopher, and atechnological innovator. He is widely recognized as oneof the most diversely talented individuals ever to havelived. However, it still comes as a surprise to many thathe was also interested in mathematics. Here is an alge-braic version of Da Vinci’s proof of the quadrature ofthe lune.

3a. Begin with a square of side length 2r and uponeach side contruct a semicircle. Simlutaneouslycircumscribe a circle about the square, as shownbelow. Find the combined area of the four semi-circles in terms of r.

3b. Now find the area of the circumscribed circle interms of r. Notice anything?

3c. According to parts 3a and 3b, we get the samearea if we remove the circumscribed circle fromthe figure above as we get if we instead remove thefour semicircles from the figure above. Use thisto explain why a single lune from the picture isquadrable.

4: Aristotle’s Wheel. Explain the following paradox, orig-inating with Aristotle, but here considered by Galileoin his The Two New Sciences of 1638: Suppose thelarge circle has made one revolution in rolling along thestraight line from A to B, so that AB is equal to thecircumference of the large circle. Then the small circle,fixed to the large one, has also made one revolution, sothat CD is equal to the circumference of the small circle.It follows that the two circles have equal circumferences.Clearly, this conclusion is incorrect! Why?

A

C

B

D

5: Kepler’s Third Law.

5a. Check Kepler’s Third Law using the followingmodern figures. (AU stands for astronomical unit,the length of the semimajor axis of Earth’s orbit.)

Planet Years Semimajor Axis

Mercury 0.241 0.387 AUVenus 0.615 0.723 AUEarth 1.000 1.000 AUMars 1.881 1.524 AUJupiter 11.862 5.202 AUSaturn 29.457 9.539 AU

5b. How many years would it take a planet to orbitthe sun if the semimajor axis is 100 AU?

5c. What would be the semimajor axis of a planetwhose orbit takes 125 years?

6: Galileo’s Strange Remark. Consider the remark inGalileo’s The Two New Sciences that

“neither is the number of squares less thanthe totality of all numbers, nor the lattergreater than the former.”

What does he mean?

7: You Cannot Put Descartes Before De Horse. Solve thefollowing problems from the works of Descartes.

7a. Prove, using analytic geometry, that the midpointsof the sides of a planar quadrilateral are the ver-tices of a parallelogram. (Begin by labeling thevertices of a quadrilateral (a, b), (c, d), etc., thenfinding the midpoints. Joining the four midpointsis clearly another quadrilateral, now prove that thequadrilateral must be a parallelogram! Hint: Re-call that parallelograms have two pairs of parallel



sides—in analytic geometry, how do we know iftwo lines are parallel?)

7b. Solve the equation

x3 −√

3x2 +26

27x− 8

27√

3= 0

by first multiplying through by(

3√

3)3

, then bysubstituting

z =(

3√

3)

x

to get an equation with integer coefficients. Solvethe resulting equation, then undo the substitutionto get the solution to the original equation.

8: Pascal’s Square Triangle. Establish the following rela-tions, all of which were developed by Pascal, involvingthe numbers of the triangle. Pascal’s organized his tri-angle as a square array (like the one below), and it is inthis context that the relations are stated.

1 1 1 1 1 1 · · ·1 2 3 4 5 6 · · ·1 3 6 10 51 21 · · ·1 4 10 20 35 56 · · ·1 5 15 35 70 126 · · ·1 6 21 56 126 252 · · ·...

......

......

.... . .

8a. The mth element in the nth row is(

m+ n− 2

n− 1

)

.

8b. The element in the mth row and nth column isequal to the element in the nth row and mth col-umn.

8c. The sum of elements along any diagonal is twicethe sum of the elements along the receding diago-nal.

8d. The sum of the elements along the nth diagonal is2n−1.

9: Always Bet on Fermat. Find the division of stakes ina game of chance between two equally skilled players Aand B where

9a. A needs 1 more point to win and B needs 4 morepoints to win;

9b. A needs 3 more points to win and B needs 4 morepoints to win;

9c. A needs 3 more points to win and B needs 5 morepoints to win.

10: Three Dice Means Triple the Fun. The following prob-lem was considered in a paper by Galileo: For a roll ofthree dice, show that both a 9 and a 10 can be achievedin six different ways. Nevertheless, show that the prob-ability of rolling a 10 is higher than that of rolling a9.



Weekender #10

1: Tangent Methods. Find the slope of the tangent at thepoint (3, 4) on the circle x2 + y2 = 25 by

1a. Barrow’s method;

1b. Newton’s method of fluxions;

1c. the modern method.

2: Newton Had No Limits—But He Had Bounds. Oneof the early uses for calculus was finding roots of poly-nomials. Associated with this is a method for approx-imating an upper bound for the real roots of a poly-nomial. The procedure is as follows: Let n be thedegree of the polynomial. Take the smallest integerthat will make f (n−1)(x) positive. Substitute this in-teger in f (n−2)(x). If the result is negative, increasethe integer successively by ones until an integer is foundthat makes f (n−2)(x) positive. Now proceed with thenew integer as before. Continue in this manner un-til an integer is found that makes all the functionsf (n−1)(x), f (n−2)(x), . . . , f ′(x), f(x) positive.

For example, consider

f(x) = x3 + 2x2 − 13x+ 11.

Since n = 3, we find f ′′(x) = 6x + 4. The smallest in-teger that makes 6x+ 4 positive is 0. Now substitute 0into f ′(x) = 3x2 + 4x− 13 to obtain −13. Since this isnegative, we try substituting 1: f ′(1) = −6; still neg-ative, so we try f ′(2) = 7. This is positive, so now wemove to f(x). Substitute 2 to obtain f(2) = 1. Thisresult is positive; thus, all real roots are less than 2, theupper bound.

Find an upper bound for the roots of

x4 − 3x3 − 4x2 − 2x+ 9.

3: Bashing Newton. The following is a paraphrasing fromNewton’s Quadrature of Curves (1704).

“In the same time that x, by growing, be-comes x+ o, the power x3 becomes (x+ o)3,or x3 + 3x2o + 3xo2 + o3, and the growths,or increments, o and 3x2o+ 3xo2 + o3 are toeach other as 1 to 3x2 +3xo+o2. Now let theincrements vanish, and their last proportionwill be 1 to 3x2, whence the rate of changeof x3 with respect to x is 3x2.”

3a. George Berkeley (1685-1753) was an Irish bishopand philosopher whose best known contribution tomathematics is his attack on the logical foundationof the calculus as developed by Newton. ExplainBishop Berkley’s sarcastic description of deriva-tives as “ghosts of departed quantities.”

3b. Johann Bernoulli, an 18th century world-renownedmathematician and proponent of calculus, madethe following postulate to allow operations likethat illustrated above: “A quantity which is in-creased or decreased by an infinitely small quan-tity is neither increased nor decreased.” Commentintelligently (yes, intelligently!) on the merits ofthis postulate.

4: L’Hôpital Sees His First Patient. The following limit isthe first illustration l’Hôpital gives of his rule (in 1696):

limx→a

√2a3x− x4 − a

3√a2x

a− 4√ax3

.

4a. Verify that if x = a, both numerator and denomi-nator are zero.

4b. Use l’Hôpital’s Rule to determine the limit.

5: Fee Phi Fo Fer, I See a Problem of Mr. Euler. Givenany positive integer n, Euler defined the function φ(n)as the quantity of positive integers less than and rela-tively prime to n. For example, φ(12) = 4 since thereare 4 positive integers less than and relatively prime 12;namely, 1, 5, 7, and 11.

5a. Evaluate φ(24) and φ(30).

5b. Show that φ(16) · φ(9) = φ(16 · 9).

5c. It is generally not true that, for positive integersm and n, φ(m) · φ(n) = φ(m · n). So why did part5b work?

5d. Show that if p is prime, then φ(p) = p− 1.

5e. Euler discovered and proved the following easy wayto calculate φ(n) for any n: If n has the prime fac-torization

n = pa1

1 pa2

2 · · · pak

k

then

φ(n) = n

(

1 − 1

p1

)

· · ·(

1 − 1

pk

)

.

Compute φ(1000) with this formula.

6: Fee Phi Fo Fum, I See That Euler Has a Theorem. Fer-mat’s Little Theorem is a special case of the followingtheorem discovered and proved by Euler: If a and mare relatively prime, then aφ(m) divided by m leaves aremainder of 1.

6a. What happens in Euler’s Theorem when m isprime? (This is why Fermat’s Theorem is a specialcase of Euler’s.)

6b. Use Euler’s Theorem to find the remainder when:76 is divided by 18; 716 is divided by 18; when 542

is divided by 24.

7: All Aboard the Euler Line. Euler made the quite re-markable discovery that is now called the Euler line:The orthocenter, centroid, and circumcenter of a trian-gle all lie, in that order, on the same line. Moreover, thecentroid splits the line into the ratio 2 : 1.

Euler did much with analytic geometry as well. Let a, b,and c be the sides of triangle ABC, with A at the origin,B on the x-axis, and with area K. Euler then discov-ered that the coordinates of the orthocenter, centroid,and circumcenter are, respectively,

(

b2 + c2 − a2

2c,

8K2 + c2(c2 − a2 − b2)

4cK

)

,



(

3c2 + b2 − a2

6c,

2K

3c

)

,

(

c

2,c(a2 + b2 − c2)

8K

)

.

Find the coordinates of the orthocenter, centroid, andcircumcenter for the triangle with vertices A(0, 0),B(6, 0), and C(2, 4), verify that they are collinear, thenfind the equation of the Euler line.

8: Euler’s Constant.

8a. Use your calculator or a computer to determineexactly how many terms we must sum in the har-monic series before it totals 4.5 or more. That is,find the smallest n so that

n∑

k=1

1

k> 4.5.

8b. Euler’s constant γ is defined by

γ = limn→∞

(

n∑

k=1

1

k− ln(n)

)

.

This constant is approximately equal to 0.577215.(It remains one of the great unsolved problems ofmathematics to determine whether this constantis rational or irrational.) Use this value of γ to es-timate the number of terms it takes the harmonicseries to grow to 10 or more.

8c. How many terms does it take the harmonic seriesto grow to 15 or more?

9: Euler’s Deranged Rearrangement. Euler was posed thefollowing problem, for which he found three differentmethods of solution:

“Given any series of n letters a, b, c, d, etc., to find howmany ways they can be arranged so that none returnsto the position it initially occupied.”

Euler called this unique rearrangement a derangement.He empirically discovered the first few values: If Π(n) isthe number of derangements of n objects, then

Π(1) = 0, Π(2) = 1, Π(3) = 2,

Π(4) = 9, Π(5) = 44.

After much thought, he discovered a general formula:

Π(n) = n!

(

1

2!− 1

3!+ · · · +

(−1)n

n!

)

9a. Find Π(6) and Π(7).

9b. Given any series of 7 letters, find the probabilitythat none returns to its original position.

10: The Platonic Solids—They Are Still Just Friends. Eu-clid determined the regular polyhedra by examining theangle sums at each vertex (Book XIII). There is still

quite a different approach to establishing that five andonly five regular solids exist, and it involves Euler’s For-mula for Polyhedra. This says that if we have a polyhe-dral solid with F faces, V vertices, and E edges, then

V + F = E + 2.

This result was probably known to Archimedes, wasmearly discovered by Descartes, but was first publishedby Euler in 1752.

10a. Show that Euler’s Formula holds for the GreatPyramid of Cheops.

10b. Now suppose that we have a regular solid with Vvertices, F faces, and E edges, where each face ofthe solid is a regular n-gon. Suppose further thatthe number of regular solids meeting at each vertexis k. Note immediately that n ≥ 3 and k ≥ 3.

b1. Initially, one might conclude that, since eachof the F faces is a regular polygon with nedges, there should be nF edges in the reg-ular solid. Explain why the correct result isE = nF/2.

b2. Explain why V = nF/k.

b3. Now subtitute parts b1 and b2 into Euler’sFormula to get

F

4k(2n+ 2k − kn) = 1,

and conclude that 2n+ 2k − kn > 0.

b4. The inequality in part b3 provides the key tothe analysis of the regular solids. Begin withthe minimal case n = 3. Show that this im-plies 3 ≤ k < 6 and thus yields three regularsolids.

b5. If n = 4, show the inequality yields only oneregular solid.

b6. What happens if n = 5?; n = 6?; n = 7?; etcetera.

10c. Create a table of data about the regular solids,combining the results of Euclid with the resultsof his problem. Your table should have seven en-tries for each of the five solids. Those seven entriesshould be headed

• Number of Faces (F ),

• Number of Vertices (V ),

• Number of Edges (E),

• Type of Regular Polygon at Each Face (n),

• Number of Faces at Each Vertex (k),

• Number of Degrees in Each Face Angle,

• Number of Degrees in Each Polyhedral Angle.

You might look for interesting symmetries hiddenamong the entries in your table—you can be surethe Greeks would have.



Weekender #11




1c. page 183 problem 6(a, b, c, d);

1d. page 199 problem 2.

2: The Goldbach Conjecture. This is one of the mosttroubling unsolved problems in mathematics, for it is sosimple to state: any even number greater than 2 can bewritten as the sum of two primes. Alas, no one fromGoldbach to Euler to Newton to Gauss to Erdős hasproved it.

2a. Verify the Goldbach Conjecture for 38, 538, and2016.

2b. Is the conjecture true if we replace “even” by“odd”?

2c. Is the conjecture true if we replace “sum” by “prod-uct”?

2d. Suppose that tomorrow someone proved the Gold-bach Conjecture. Show that it would then followthat any integer (even or odd) greater than 5 canbe written as the sum of three primes.

3: Quadratic Reciprocity. Gauss’ Quadratic ReciprocityTheorem only applies to odd primes. However, thefollowing theorem enables us to determine quadraticresidues of 2 and an odd prime:

If p is an odd prime, then(

2

p

)

=

1 p ≡ ±1 mod 8

−1 p ≡ ±3 mod 8.

3a. Determine whether or not 2 is a quadratic residuemodulo 17.

3b. Determine whether or not 2 is a quadratic residuemodulo 379. (379 is prime.)

3c. Determine whether or not 41 is a quadratic residuemodulo 53.

3d. Show that if p and q are distinct odd primes con-gruent to 3 mod 4, then

(

q

p

)

= −(

p

q

)

.

4: Gaussian Integers. Which of the following real primesare primes as Gaussian integers: 2, 3, 5, 7, 11, 13, 17,19, 23, 29, 31, 37? Demonstrate a factorization of thosethat are composite Gaussian integers.

5: The Prime Number Theorem. Suppose we let π(n)denote the number of primes less than n. The PrimeNumber Theorem states that

limn→∞

π(n) =n

lnn.

In other words, n/(lnn), as n increases, is a better andbetter approximation to π(n). This remarkable and im-portant theorem was first conjectured by Gauss afterexamining a large table of primes. The theorem was fi-nally proved in 1896 by two men working independently:Jacques Hadamard (1865-1963) and Charles de la ValléePoussin (1866-1962).

Approximate π(n) for n = 500, 105, 108, 109.

6: Twin Primes–They Are Neither Fraternal Nor Paternal.Recall that if p and p+ 2 are both prime, we call them“twins.”

6a. Find the first ten pairs of twin primes.12

6b. How many pairs of twins are there between 100and 199? Between 1500 and 1599? Between 6000and 6099?

7: Constructible Numbers and Trisections. In an 1837paper, Pierre Wantzel (1814-1848) proved the followingkey result:

If a cubic equation with integer coefficientsx3 + ax2 + bx + c = 0 has no rational solu-tions, then it has no constructible solutions.

Wantzel proved this result by contradiction. He beganby assuming that, although this cubic had no rationalsolutions, it nonetheless did have a constructible solu-tion of the form r = p + q

√β, where this solution was

“minimal” in the sense that it used as few or fewer squareroots in its formula than any other constructible solu-tions. He next showed that s = p − q

√β will also be a

root of the cubic above.

7a. Explain why this implies that −a − 2p is also aroot of the cubic. Hint: Use the fact that −a isthe sum of the roots of the cubic.

7b. What did this contradict and why?

7c. Now, carefully indicate exactly where in this ar-gument we used the fact that the cubic had norational roots. That is, show precisely how the de-sired contradiction falls apart if rational roots arepermitted.

7d. With this key result, Wantzel proved that the 60

angle could not be trisected with compass andstraightedge. Use this fact to prove the following.

d1. It is impossible to construct a regular nonagon(9-gon) with Euclidean tools.

d2. It is impossible to construct a regular 27-gonwith Euclidean tools.

d3. It is impossible to construct an 85 angle withEuclidean tools.

12You may find the list of primes found at https://primes.utm.edu/lists/small/1000.txt helpful.



Weekender #12

1: De Polignac’s Conjecture. In 1848, French mathemati-cian Alphonse de Polignac (1826-1863) asserted: “Everyodd number can be expressed as the sum of a power of2 and a prime.”

1a. Verify this for odd numbers 3, 5, 7, 9, 11, 13, 15,17, and 19. (That is, write each of these num-bers as the sum of a power of 2 and a prime; i.e.,11 = 22 + 7.)

1b. Verify this for 61 and 119.

1c. Try to verify this for 509. What does this implyabout de Polignac’s Conjecture?

2: DeMorgan’s Little Puzzle. Augustus DeMorgan, 19thcentury mathematician, was fond of saying: “I was xyears old in the year x2.” In what year was DeMorganborn?

3: A Model of Lobachevskian Geometry. The followingis a model of Lobachevskian geometry using a suitableinterpretation of Euclidean postulates: Take a fixed cir-cle, Λ, and interpret plane as the interior of Λ; point

as a point in Λ; and, line as that part of the Euclideanline contained within Λ. (Note the italics: “line” is Eu-clidean; “line” is Lobachevskian.) Verify the followingstatements.

3a. Two distinct points determine exactly one line.

3b. Two distinct lines intersect in at most one point.

3c. Through a point P not on line m, there can bepassed infinitely many lines not intersecting m.

3d. Let a Euclidean line determined by the points

P and Q intersect Λ in S and T , in the orderS, P,Q, T . Interpret Lobachevskian distance fromP to Q as

log

(

QS · PTPS ·QT

)

.

If P , Q, and R are three points on a line, thenshow that distance PQ + distance QR = distance

PR.

Notice that this interpretation of distance implies thefollowing: Let point P be fixed and let point Q movealong a fixed line through P towards T . Then it can beshown that

limQ→T

distance PQ = ∞.

This model was devised by Felix Klein (1849-1929).With the interpretation above, and with a suitable in-terpretation of angle between two lines, it can be shownthat all the assumptions necessary for Euclidean geom-etry, except the parallel postulate, are true propositionsin the model.

4: Non-Euclidean Physics—Because Regular Physics Is NotCrazy Enough. Because of the apparent inextricableentanglement of space and matter it may be impossibleto determine by astronomical methods whether physical

space is Euclidean or non-Euclidean. Since all measure-ments involve both physical and geometrical assump-tions, an observed result can be explained in many dif-ferent ways by making a suitable change in our assumedqualities of space and matter: an observation may beexplained by holding to laws of, say, optics, and usingnon-Euclidean geometry; or by holding to Euclidean ge-ometry and changing some law of optics. Henri Poincaré(1854-1912) maintained that asking which geometry isthe “true” geometry is therefore irrelevant. To clarifythis view, Poincaré devised an imaginary universe Ω oc-cupying the interior of a sphere of radius R in which heassumed the following physical laws to hold.

L1. At any point P of Ω the absolute temperature Tis given by T = R2 − r2, where r is the distance ofP from the center of Ω.

L2. The linear dimensions of a body vary directly withthe absolute temperature of its location.

L3. All bodies in Ω immediately assume the tempera-tures of their locations.

4a. What happens to the temperature of a body mov-ing away from the center of Ω? What happens tothe body’s size?

4b. Would an inhabitant of this universe be aware ofthe laws governing it? Explain.

4c. Would an inhabitant of this universe believe thatit’s universe is infinite or finite? Explain.

In this universe, it can be shown that the shortest dis-tance between points is an arc that always bends towardthe center, except if the two points lie on a diameter, inwhich case the shortest distance is a line. If we extendeach arc so that the endpoints lie on the sphere, we canalso show that the arc is perpendicular to the surface ofthe sphere. We now impose a fourth law.

L4. Light travels along the arcs of shortest distance.

4d. When observing what to us is an arc, would aninhabitant of this universe see an arc or a straightline?

4e. Demonstrate that the Lobachevskian parallel pos-tulate holds in this universe, so that an inhabitantwould believe it’s universe is non-Euclidean.

Here, we finally see Poincaré’s point: we have a piece ofordinary and supposedly Euclidean space, but because of

different physical laws, it appears to be non-Euclidean!

5: Snarks and Trones. Consider the following set of pos-tulates about certain objects called trones, and certaincollections of trones called snarks.

P1: Every snark is a collection of trones.

P2: There exist at least two trones.

P3: If p and q are two trones, then there exists exactlyone snark containing both p and q.

P4: If M is a snark, then there exists a trone not inM .



P5: If M is a snark, and p is a trone not in M , thenthere exists exactly one snark containing p and notcontaining any trone that is in M .

5a. Interpret snark as “committee” and trone as “com-mittee member.” Are the postulates consistent?

5b. Interpret trone as any one of the three lettersa, b, c, and snark as any one of the three pairsab, ac, bc. Are the postulates consistent?

5c. Consider the postulates under the following inter-pretation: a trone is a point and a snark is a line.

6: Quaternions: Vectors’ Best Pals. Let q = (1, 0,−2, 3)and r = (1, 1, 2,−2) be two quaternions.

6a. Find q + r.

6b. Find qr and rq.

7: The First of a Pair of Ordered Pairs. Consider the set ofall ordered real number pairs and define the following.

1: (a, b) = (c, d) if and only if a = c and b = d.

2: (a, b) + (c, d) = (a+ c, b+ d).

3: (a, b)(c, d) = (0, ac).

4: k(a, b) = (ka, kb).

7a. Show that multiplication is commutative, associa-tive, and distributive over addition.

7b. Show that the product of three or more factors isalways equal to (0, 0).

7c. Construct a multiplication table for the units u =(1, 0) and v = (0, 1).

8: Lie Algebra. Sophus Lie (1842-1899, pronounced“Lee”) made major advances in the theory of fields anddifferential equations. He created an algebra that laterproved useful to describe certain aspects of physics. Thebasic elements of Lie algebras are square matrices withequality and addition defined as usual for matrices, butwith multiplication defined as

A B = AB −BA

where AB is the usual matrix (Cayley) product. Let

A =

[

1 0−1 0

]

, B =

[

1 1−1 1

]

,

and C =

[

1 10 1

]

.

8a. Calculate A B.

8b. Calculate B A.

8c. Show that A B = −B A for all A and B.

8d. Use the definition of to simplify A (B C) and(A B) C algebraically.

8e. Apparently, which fundamental property does aLie algebra lack?

9: Jordan Algebra. Pascual Jordan (1902-1980) was oneof the founders and foremost proponents of quantumphysics, as well as the author of numerous books and es-says on theoretical physics, biophysics, and astrophysics,in which he sought to popularize the new theories. Asan aid to describe certain quantum processes, he createdwhat is now called a Jordan algebra. This algebra hassquare matrices for elements, with equality and additiondefined as usual for matrices, but with multiplicationdefined as

A ⋆ B =AB +BA

2,

where AB is the usual matrix (Cayley) product. Let A,B, and C, be defined as in the previous problem.

9a. Calculate A ⋆ B.

9b. Calculate B ⋆ A.

9c. Use the definition of ⋆ to simplify A ⋆ (B ⋆C) and(A ⋆ B) ⋆ C algebraically.

9d. Apparently, which fundamental property does aJordan algebra lack?



Weekender #13


1a. page 153 problem 2(a, b);

1b. page 153 problem 3(a).

2: The Second of a Pair of Ordered Pairs.

2a. Any integer, be it positive, zero, or negative, canbe represented as the difference m− n of two nat-ural numbers m and n. If m > n, then m− n is apositive integer; if m = n, then m − n is the zerointeger; if m < n, then m − n is a negative inte-ger. This suggests the idea of logically introduc-ing the integers as ordered pairs (m,n) of naturalnumbers, where by (m,n) we actually mean thedifference m − n. With this interpretation of or-dered pairs of natural numbers, how should we de-fine equality, addition, and multiplication of theseordered pairs?

2b. Any rational number, be it positive, zero, or neg-ative, can be represented as the quotient m/n oftwo integers m and n, with n 6= 0; in fact, the wordrational has its origin in this fact. This suggeststhe idea of logically introducing the rational num-bers as ordered pairs (m,n) of integers, where by(m,n) we actually mean the quotient m/n. Withthis interpretation of ordered pairs of integers, howshould we define equality, addition, and multipli-cation of these ordered pairs?

3: Puzzling Paradoxes.

3a. Most students will agree to the following: “If twofractions are equal and have equal numerators,then the denominators are equal.” Now considerthe following problem. We want to solve the equa-tion

x+ 5

x− 7− 5 =

4x− 40

13 − x.

Combining the terms on the left, we have

4x− 40

7 − x=

4x− 40

13 − x.

But by the statement above, it must be that7 − x = 13 − x, but this implies 7 = 13. Whatwent wrong?

3b. Explain the following: Clearly 3 > 2. Multiplyingboth sides by log( 1

2), we have

3 log(

1

2

)

> 2 log(

1

2

)

,

or

log(

1

2

)3

> log(

1

2

)2

.

Thus, ( 12)3 > ( 1

2)2, or 1

8> 1

4.

3c. By standard procedure, we find∫ 1

−1

dx

x2= − 1

x

∣

∣

∣

1

−1= −1 − 1 = −2.

But the function y = 1/x2 is never negative; hencethe answer above cannot be correct. Explain.

4: Counting on Denumerable Sets. Prove the followingstatements.

4a. The union of a finite number of denumerable setsis a denumerable set.

4b. The union of a denumerable number of denumer-able sets is a denumerable set.

4c. The set of all irrational numbers is nondenumer-able.

4d. The set of all transcendental numbers is nondenu-merable.

5: Counting on Nondenumerable Sets. Prove or disprovethe following statements.

5a. The difference of two transcendental numbers istranscendental.

5b. The product of two transcendental numbers istranscendental.

6: Counting on Algebraic Numbers, Part ℵ0.

6a. Show that 1 is the only polynomial of height 1.

6b. Show that x and 2 are the only polynomials ofheight 2.

6c. Show that x2, 2x, x+ 1, x− 1, and 3 are the onlypolynomials of height 3, and they give the alge-braic numbers 0, 1, and −1.

6d. Form all possible polynomials of height 4 and showthat the only new real algebraic numbers con-tributed are −2, − 1

2, 1

2, and 2.

6e. Show that polynomials of height 5 contribute 12more real algebraic numbers.

7: Counting on Algebraic Numbers, Part c.

7a. Show that 3√

3 +√

2 is algebraic by determining aspecific polynomial with integer coefficients whichthis number satisfies. (Hint: Let x = 3

√3 +

√2;

then x −√

2 = 3√

3, and then cube both sides, etcetera. . . )

7b. Likewise, find a polynomial that guarantees that

13√

3 +√

2

is an algebraic number.

8: Intuition vs. Abstraction: Live on Pay Per View. Onepurpose of formal axiomatics, with its abstraction andsymbolism, is to furnish a barrier against the use of in-tuition. Answer the following questions intuitively, andthen check your answers by calculation.

8a. A car travels from P to Q at 40 mph and thenreturns from Q to P at 60 miles per hour. Whatis the average rate for the entire trip?



8b. A man sells half his apples at 3 for 17 cents andthen sells the other half at 5 for 17 cents. At whatrate should he sell all his apples in order to makethe same income?

8c. If a ball of yarn 4 inches in diameter costs 20 cents,how much should you pay for a ball of yarn 6 inchesin diameter?

8d. A clock strikes six in 5 seconds. How long will ittake to strike twelve?

8e. A bottle and a cork together cost $1.10. If the bot-tle costs a dollar more than the cork, how muchdoes the cork cost?

8f. Two jobs have a starting salary of $60,000 per yearwith salaries paid every six months. One job offersan annual raise of $8000 and the other job offersa semiannual raise of $2000. Which is the betterpaying job?

8g. Suppose a large sheet of paper one one-thousandthof an inch thick is torn in half and the 2 pieces puttogether, one on top of the other. These are thentorn in half, and the 4 pieces put together in a pile.If this process of tearing in half and piling is done50 times, will the final pile of paper be more or lessthan a mile high?

8h. Four-fourths exceeds three-fourths by what frac-tional part?

8i. A student wants the arithmetical average of 8grades. He averages the first 4 grades, then thelast 4 grades, and then finds the average of theseaverages. Is this correct?

9: A Blast From the Past—Egyptian Unit Fractions. TheBritish mathematician J. J. Sylvester (1814-1897) re-fined the procedure for uniquely expressing any rationalfraction between 0 and 1 as a sum of unit fractions.

i. Find the largest unit fraction less than the givenfraction.

ii. Subtract this unit fraction from the given fraction.

iii. Find the largest unit fraction less than the result-ing difference.

iv. Subtract again, and continue the process.

(To find the largest unit fraction less than the givenfraction, divide the denominator of the given fraction bythe numerator and take the next integer greater than thequotient as the denominator of the unit fraction sought.)

For example, let the given fraction be 711

. Division ofdenominator by numerator (11 divided by 7) results ina quotient of 1; take the next integer greater: 2. Thenthe largest unit fraction less than 7

11is 1

2. Subtract:

711

− 12

= 322

. Now repeat the process with the frac-tion 3

22: 22 divided by 3 gives 7; take 8. Then subtract:

322

− 18

= 188

. Thus, we have represented a fraction asthe sum of unit fractions:

7

11=

1

2+

1

8+

1

88.

9a. Express 27

as the sum of unit fractions.

9b. Express 297

as the sum of unit fractions.(The representations you obtain in parts a and bare the same as given in the Rhind Papyrus.)

9c. Express 4577

as the sum of unit fractions.

9d. Show that for any integer n, the fraction 2n

can berepresented by the sum

1

n+

1

2n+

1

3n+

1

6n.

(In the Rhind Papyrus, only 2101

is represented inthis manner.)

10: Conclusions Are Not Enough, They Must Also Be Valid.In each of the following, is the given conclusion a validdeduction from the given premises?

10a. If today is Saturday, then tomorrow will beSunday.

But tomorrow will be Sunday.Therefore, today is Saturday.

10b. Germans are heavy drinkers.Germans are Europeans.Therefore, Europeans are heavy drinkers.

10c. If a is b, then c is d.But c is d.Therefore, a is b.

10d. All a’s are b’s.All a’s are c’s.Therefore, all c’s are b’s.

This problem illustrates how a person may allow themeanings which are associated with words to dominatethe logical analysis.

11: Counting on Gödel Numbers.

11a. Find the Gödel number of the formula (fu(fv)),where u and v are variables of types 1 and 2, pos-sessing Gödel numbers 19 and 192.

11b. Find the formula corresponding to the Gödel num-ber

25311537191113.

11c. Find the Gödel indices of − 1260

and 936

.

11d. Find the rational number whose Gödel index is2250.


name: due date: the history of...

Documents