Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Vincent's theorem wikipedia , lookup
Georg Cantor's first set theory article wikipedia , lookup
Large numbers wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Fundamental theorem of algebra wikipedia , lookup
Series (mathematics) wikipedia , lookup
Location arithmetic wikipedia , lookup
Elementary mathematics wikipedia , lookup
Four color theorem wikipedia , lookup
“Puzzles and Pythagoras in the Classroom” Presented by Professor Spencer Hurd School of Science and Mathematics The Citadel, Charleston, SC Sigma Xi Talk – The Citadel, September 2008 Necessary Background - about that of an entering freshman Prerequisites - High School Algebra and a Sense of Humor A little puzzle to start with: Label the edges AB, BC, CD, and DA with reciprocals of positive whole numbers. Say AB 1 , w BC 1 , x CD 1 , y DA 1 . z And, do this so that, for each vertex A, B, C, and D, the sum of the two associated reciprocals is also a reciprocal. This means there must be positive integers a, b, c, and d so that 1 1 1 , a w z 1 1 1 , b x w 1 1 1 , c x y 1 1 1 . d y z The ancient Egyptians had crude number symbols, essentially base ten, using repetitions: │=1 ││ = 2 │││ = 3. ∩ = 10 ∩∩ = 20 ∩∩∩ = 30 And │││ ∩∩∩ = 33, and so on. Fractions were indicated by putting a bar (or eye) over the integer symbols. In effect, all fractions were “unit fractions” or reciprocals. In our notation, all fractions were of the form 1 . n If they wanted to write 2/5 they had to use 1 1 . 3 15 This has led to a huge literature in puzzles and theoretical problems of great difficulty. How many ways are there to write might as 2 as the Egyptians 35 2 1 1 for different positive integers x, y. 35 x y What is the fewest number of distinct reciprocals so that they add to 3 . (Ans: it can be done with 3 reciprocals.) 7 Can you find an Egyptian labeling for a square? An Egyptian graph is defined to be a connected graph whose edges have different positive Egyptian fraction labels and whose vertices have different labels also, and such that, for each vertex, the sum of the labels of the edges incident with a vertex is the label for that vertex. A graph is Egyptian, then, if such a labeling is known to exist. Before going further, can you construct an edge labeling of, say, a square so that the labeling is Egyptian. Table 1: Egyptian Labels for the Square 1 2 3 4 5 6 7 8 9 10 AB 6 9 10 12 12 8 9 9 10 10 BC 12 18 15 24 24 24 18 18 15 40 CD 60 36 60 40 72 168 90 144 60 60 DA 30 72 40 60 36 56 72 72 90 90 A 5 8 8 10 9 7 8 8 9 9 B 4 6 6 8 8 6 6 6 6 8 (All entries represent reciprocals.) C 10 12 12 15 18 21 15 16 12 24 D 20 24 24 24 24 42 40 48 36 36 Pythagoras Pythagoras of Samos (about 500-600 BCE) was the sage-philosopher who founded a cult based on his unwritten teachings. He founded a commune on the coast of Italy which became successful in the sense that the cult lasted for 200 years, at least. Their philosophy was based on the notion that “all is number.” He is credited with discovering (or naming) prime and composite numbers, odd and even numbers, figurate numbers, polygonal numbers amicable numbers and perfect numbers. Figurate numbers are fore-runners of today’s labeled graphs. But that’s another talk. His eponymous theorem is a theorem about areas – the area of the two squares erected on the legs of a right triangle is equal to the area of the square erected on the hypotenuse. We call a positive integer triple (a, b, c), which satisfies a 2 b 2 c 2 , a Pythagorean Triple or PT. If gcd(a, b, c) = 1, then the triple is called primitive. A Pythagorean Quadruple, or PQ, is a 4-tuple of integers (a, b, c, x) such that a 2 b 2 c 2 x 2 . Teachers for 2500 years have used side lengths with whole number values to illustrate Pythagoras’s theorem. Pythagoras himself is credited with discovering an infinite sequence of triples in which the hypotenuse and the larger leg differ by unity. a 3 5 7 9 11 c = b+1 B 4 12 24 40 ? c 5 13 25 41 ? “Let’s find one example.” Students find all the examples by trial and error. Then they look for patterns. Observe that Also So b + c = a2 . c = b + 1. a2 = 2b + 1 Algorithm: select any odd positive integer for a. Then Find two consecutive positive integers which sum to a2. So, b + b +1 = 112 and b = 60, c = 61. Conclusion: Arrange for a parameter N. N = 1 corresponds to (3, 4, 5) N = 2 corresponds to (5, 12, 13) Solution: take a = 2N + 1 Then b (a 2 1) / 2 2 N 2 2 N And c (a 2 1) / 2 2 N 2 2 N 1 Student Check: (2 N 1)2 (2 N 2 2 N )2 (2 N 2 2 N 1)2 (i.e., that a 2 b 2 c 2 ) Euclid proved in The Elements (about 300 BCE): If (a, b, c) is a PPT, then there are two numbers m and n (one odd and the other even and gcd(n, m) = 1) so that a = 2mn, b = n2 – m2, and c = n2 + m2. It is easy to have students check that: (2mn)2 (n2 m2 )2 (n2 m2 )2 and, thus, such (a, b, c) is always a PT. The hard part was to show that the existence of the m and n. The philosopher Plato is credited with discovering a second infinite sequence of triples in which the hypotenuse and the larger leg differ by two. a 4 8 12 16 11 c = b+2 b 3 15 ? ? ? c 5 17 ? ? ? A set of Discovery lessons A Pythagorean Quadruple, or PQ, is a 4-tuple of integers (a, b, c, x) such that a 2 b 2 c 2 x 2 . 1. For simplicity, let a = 1. Can someone find a PQ with a = 1?” Someone will find (1, 2, 2, 3). Ask her to put it on the board. “Do you think (2, 4, 4, 6) is a PQ?” For PTs, multiples of a PT corresponded to similar right triangles and gave new PTs, is this true for PQs as well? HMWK Prove or disprove: (1) There is a PQ given by (1, 1, b, c) for some integers b and c. (2) (a, b, c, d) is a PQ if and only if (ka, kb, kc, kd) is a PQ. 2. Remember that 32 + 42 = 52 and 52 + 122 = 132. Put these facts together as 13 2 12 2 5 2 12 2 4 2 3 2 . HMWK: string together known PTs to make at least a dozen new PQs. Put them in a chart or organized list. Are there any patterns? 3. Going back to a = 1, the table of sums of squares gives values of k so that k = m2 + n2. Look for values of k so that k itself is one less than a square. That is, look for k = x2 – 1. The table shows 80 = 82 + 42 and 80 = 81 – 1 = 92 – 1. Put together, this gives 4 2 8 2 80 9 2 1 . And re-arranging, we get 12 4 2 8 2 81 9 2 . HMWK: Find as many of these as you can, put them in a chart, and see if you can find any patterns. If yes, state and prove an identity (or 2). 4. Using the PQ (1, 2, 2, 3), one might suppose that there are others in the form (1, 2n, 2n, x). If so, they satisfy 12 + (2n)2 + (2n)2 = x2, from which we get x 1 8n 2 . Use a spreadsheet to search for such quadruples. In column A, beginning with cell A2=0, put the integers from 1 to 100. (Put =1+A2 in cell A3.) In column B, in B3, put the formula =1+8*A3^2. In cell C3 put =sqrt(B3). [If the entry in cell C is an integer, we have found a PQ.] Select A3, B3 and C3 and FILL down to n = 100 at least. Make a list of the solutions found, if any. Can you find anything? Note: (1, 2, 2, 3) (1, 12, 12, 17) (1, 70, 70, 99) and (1, 2378, 2378, 3363) are in this pattern, a rather sparse sequence. Are there more? 5. Maybe a = 2 is promising. In this case x 2 – 4 = b2 + c2. Lets look in the table for k = x2 – 4. Adam , you check x = 1,..., 5 and see if x2 – 2 is in the table, Betty, you check x = 6,..., 10. Cathy, you check x = 11,..., 15. Continue until each student has some to check. A spreadsheet that extends to 40^2 + 40^2 would get more, but your students should find (2, 3, 6, 7) and (2, 5, 14, 15). Can somebody find some more like (2, 7, c, c+1)? Perhaps they can find these (2, 7, 26, 27), (2, 9, 42, 43), (2, 11, 62, 63). Analysis: 22 + b2 + c2 = (c+1)2 gives, on removing parentheses and simplifying, c = (b2+3)/2. This shows that b can be any odd number. Put another way, if b = 2n+1, then c = 2n2 + 2n + 2 and x = c+1 = 2n2 + 2n + 3. We get identity. 2 2 (2n 1) 2 (2n 2 2n 2) 2 (2n 2 2n 3) 2 HMWK: Put this in a table headed by n, a, b, c, and x. Then prove the identity. 6. The method above can be used over and over. Let a be any fixed number. Let x = c+1. For example, a = 3, and we get c = (8+b2)/2. So b has to be even. We get the sequence (3, 2, 6, 7), (3, 4, 12, 13), (3, 6, 22, 23), and so on. In this sequence, a = 3 and b = 2n. Express c and x in terms of n. One gets, 32 (2n) 2 c 2 (c 1) 2 . Express c and x in terms of n, make a chart for n = 1,...,5. n a=3 b = 2n c = 4 + 2n2 x = 5 + 2n2 1 3 2 6 7 2 3 4 12 13 3 3 6 22 23 4 3 5 3 HMWH: Prove the identity: 32 (2n) 2 (4 2n 2 ) 2 (5 2n 2 ) 2 HMWK: Let a = 5 and x = c+1 and find a sequence of PQs. Make a chart for those you find. Determine an identity in terms of a single parameter n and prove the identity. 7. Let 2n and 2n+1 represent consecutive numbers. Note (2n+1)2 – (2n)2 = 4n2 + 4n + 1 – 4n2 = 4n + 1. This can be viewed backwards – that is, any number of the form 4n + 1 is a difference of two squares! Let F1 denote the set {4n+1| n = 1, 2, 3, ...}. Let F3 denote the set {4n+3| n = 0, 1, 2, ...}. It can be shown that no member of F3 can be written as a sum of two squares, but many members of F1 can be so written. [Had you noticed that EVERY odd number in the list of sums of two squares is from F1 ? NONE are from F3.] Pick any odd number from the list. (It is already a sum of two squares, that is how it got in the list.) Write it as a difference of squares – pick 53 for example; 53 = 27 + 26 so also 53 = 272 – 262. But, from the list, 53 = 72 + 22. Put together 2 2 7 2 27 2 26 2 which means 2 2 7 2 26 2 27 2 . Thus, every odd number in the list can be used to create a PQ. 8. For the PQ (a,b,c,x), suppose a and b are consecutive. Are there any examples like that so far that you have seen? Make a list of them. HW: find a pattern and express a,b,c,x in terms of a paramenter n. 9. Pick any algebra formula involving squares. Say (n+1)2 = 1 + 2n + n2. If 2n is a square, say, 2n = m2, then (n+1)2 = 12 + m2 + n2, a new PQ. HMWK: express m in terms of n and get a one-parameter family of PQs. HMWK: Choose any other identity involving squaring and create a new family of PQs.. 10. Exploration: let x = c + 2 for a change. What can you find. 11. Exploration: Are there infinitely many PQs with a “leg” equal to 6? (Are there infinitely many primitive PTs with a leg of 6?) 12. What is the length of the diagonal of a casket (rectangular box) if the dimensions of the casket are 2-by-3-by-6 ? Graph Labeling Problems: for which graphs or classes of graphs are such labelings possible? 1. Label the vertices with colors so that no two vertices which share an edge also share a color. Use the minimum number of colors. The minimum number is the chromatic number of the graph. How is the chromatic number of a graph related to other numbers associated with the graph. 2. Label the edges with plus (+) or minus (-) so that all cycles are “positive” in the sense that they have an even number of minuses. 3. Label the nodes with distinct non-negative integers. Then label each edge with the absolute value of the difference of it corresponding vertex labels. If the graph edge labels then run from 1, 2, …, e, the graph is graceful or gracefully numbered. 4. Do any chains have graceful labelings? 5. An edge labeling which uses the consecutive integers 1, 2, …, e so that the vertex sums are consecutive e+1, e+2,…, e+v is super magic. (A vertex sum is the sum of the labels of the edges incident with the vertex. ) 6. Review hydrocarbon chains in basic chemistry and redescribe them in graph-theoretic terms.