Examples and Exercises

Transcription

1 Examples and Exercises Guerino Mazzola January 6, 00 Example of A Rigorous Proof Claim: Let a, b, c be sets. Then we have c (a b) = (c a) (c b). Proof. By definition of equality of sets, we have to prove both inclusions, and.. means: x c (a b) IMPLIES x (c a) (c b). Let us give the proof stepwise:. IF x c (a b) IMPLIES x c AND x a b (by definition of the difference of sets).. x a b IMPLIES x a AND x b (by definition of the union a b) 3. Insert implication. into the implication.: x c (a b) IMPLIES (x c AND x a AND x b) (such logical operations are allowed without altering the truth value). 4. But x c (a b) IMPLIES (x c AND x a AND x b) has the same truth value as x c (a b) IMPLIES ((x c AND x a) AND (x c AND x b)) 5. The latter means, by definition of set difference and intersection that x c (a b) IMPLIES x (c a) (c b). 6. The converse inclusion can also be shown with this rigor, and we have our claim,.

2 Allen Forte s set complexes Let T P, where P is the pitch class set P = {C, C, D, D, E, F, F, G, G, A, A, B}. Forte defines a subset K(T ) P, which he calls the set complex of T, by the attribute S K(T ) iff. S T OR. S T OR In Forte s notation, the symbol X Y stands for X Y OR Y X. Prove that this is equivalent to the following conditions in terms of Boolean algebra operations. S T = S OR. S T = T OR 3. S + T = (S T ) OR 4. S T = Hint for the third condition: S T implies S T = P and therefore S + T = (S T ). On the other hand, if S T is false, then there is an element not in S and not in T, so S T P, therefore S + T (S T ). Therefore S T is equivalent to S + T = (S T ). 3 Relations Take the set a = {r, s, t, u, v} with the mutually different elements r, s, t, u, v. Consider these relations on a:. R = {(r, r), (u, u), (t, t), (v, v), (s, s), (r, s)}. R = {(s, t), (r, s), (t, u), (u, v)} 3. R 3 = a a 4. R 4 = {(r, s), (s, t), (t, u), (u, v), (r, t), (r, u), (r, v), (s, t), (s, u), (s, v), (t, u), (t, v)} 5. R 5 = {(r, s), (t, u)} a {(r, s), (t, u)} Which of these relations is reflexive, transitive, symmetric, linear, an equivalence relation? Take the partition a = {r, v} {s, t} {u} of the above set. Exhibit the equivalence relation on a giving rise to this partition. For any two relations R X, S Y, one defines the Cartesian product relation R S (X Y ) by the condition that ((x, y), (x, y )) R S iff (x, x ) R AND

3 (y, y ) S. Example: Take this Cartesian product ordering on the Cartesian product N N of the natural number set N = {0,,, 3,...} with itself, and with the natural ordering among natural numbers. Prove that this is a partial ordering, but not a linear ordering. Give an example of the violation of the latter condition. Take alternatively the lexicographic ordering on N N (view it as the set N, and = {0, } being ordered naturally by 0 < ). We know that the lexicographic ordering is linear. So exhibit the order relation for that same example. 4 Natural Numbers and Recursion We want to make some examples of the proof by induction technique enabled by the fifth Peano axiom. (We know from the last lecture that addition is associative and shall use this fact now.) Show that for all natural numbers a, b, a + b = b + a. To do so, first, show that a + = + a by induction on a. For a = 0, we have 0 + = = + 0, and if a + = + a holds, then for the successor of a, we have a + + = (a + ) + = ( + a) + = + (a + ) = + a +, and we are done. We then show that a + 0 = 0 + a = a for all a by induction on a. This is evidently true for a = 0, check! And if it holds for a it also holds for its successor. In fact, a = a + = a + = (0 + a) + = 0 + (a + ) = 0 + a +. To prove the statement a + b = b + a, we make induction on a. For a = 0, we have a + b = 0 + b = b + 0 from the above. If a + b = b + a holds, then for the successor of a, we have a + + b = (a + ) + b = a + ( + b) = a + (b + ) = (a + b) + = (b + a) + = b + (a + ), and we are done. Next, we give a proof by induction of the famous formula n n(n + ) i = i=0 for the sum of all natural numbers up to n. This formula was found by yound Carl Friedrich Gauss (the inventor of the complex numbers) while he as a pupil and always annoyed the teacher by solving math problems in no time. So the teacher would ask Gauss to calculate the sum to keep him quiet for a while. But Gauss immediately found the solution by adding + 00, + 99, , etc., So it is 50 0 = We prove the formula by induction on n. It is evidently true for n = 0. Then, suppose it is true for n. We want to prove it for n + = n +. In fact, n+ ( n ) i = i + (n + ) = i=0 i=0 n(n + ) + (n + ) = (n + )(n + ) = (n + )((n + ) + ). (Here we are supposing that the well-known laws of rational number arithmetic have been proved.) 3

4 Next, we want to prove the formula for counting the bijections f : a a of a finite set a onto itself. Such bijecions are also called permutations of the set a. Let us denote the set of bijections by Sym(a), so this is a subset of Set(a, a). The formula to be proved is card(sym(a)) = card(a)! For example, if card(a) = 4, then card(sym(a)) = 4! = 3 4 = 4. The proof of this formula goes by induction on card(a). For the starting value card(a) = 0, i.e., a =, we have Sym(a) = {Id a }, so in fact card(sym( )) = = 0! is the correct formula. To prove the induction step from card(a) = n to card(a) = n +, we do the following. We first fix a bijection u : a n + such that every element x a correponds one-to-one to an element q(x) n +. This means that we index the a-elements by the n + natural numbers from 0 to n. Evidently, we may now as well calculate the permutations of the set n + instead of a. The idea of proof is this one: Each such permutation f : n + n + sends the maximal element n to a number f(n). If we have f(n) = n, f leaves n fixed, which means that f is essentially a permutation of the set n of the first n elements of n +. So there are n! such permutations, by the induction hypothesis. Now, if f(n) = k < n, we may append to f another very simple permutation of n +, namely the exchange of k and n. This one is denoted by (k, n) and is called the transposition of k and n. But then, the composed permutation (k, n) f leaves n fixed, and therefore is one of the already counted n! permutations. Hence we get n! new permutations for all permutations senting n to that k. Now, this k may be any of the n elements of n. We therefore get n such cases, and, counting also the very first case, where n was left fixed, this adds up to n + cases, each of which having n! permutations, all in all n! (n + ) = (n + )!, and we are done. We now want to give a roof by induction of the formula card( a ) = card(a) for the cardinality of the powerset of a finite set a. We proceed by induction on the cardinality of a. The formula evidently holds for card(a) = 0 card(a) =. As before, for the general induction step n to n +, we might just fix a bijection of a with the number n + = card(a) and transfer our discussion to n +. To count the subsets of n +, distinguish those subsets which are already contained in the subset n n +. Their number is, by induction hypothesis, n. The other subsets all contain the element n and are arbitrary besides that condition. So such a subset can be written as x = (x n) {n}. Therefore it is characterized by its intersection x n, so again there are n such subsets. All in all we now have n + n = n = n+ subsets as required. Let us do some examples of the division theorem a = q b+r. For the following numbers a, b, all written in decimal representation, do exhibit the numbers q, r, with 0 r < b:. a = 3, b = 3 4

5 . a = 45, b = 3. a = 55, b = 6 5 Classical Number Domains Consider the 5-adic (!!!) representation of a real number x = , where 43 means that these two entries are repeated until infinity to the right, i.e., x = We know that this must be a rational number u v. Calculate this fraction! Use our method as discussed in class: multiply x by a power 5 m of the basis 5 and subtract x from this power product 5 m x in order to eliminate the periodic repetition And if you do not recall the discussion in class, here is the solution: You first have to multiply x by a power 5 m such that the resulting product 5 m x has the period at the same place as the original number x. Obviously, we may take m = since the period has length, so we get 5 x = Then we subtract x from 5 x: 5 x = x = (5 )x = = So, since 5 = 44, we have(always in 5-adic notation!) x = Prove that the real number 0.9 in the decimal representation equals.0. Proceed as follows: Use first that 0.9 is represented by the Cauchy sequence x 0 = 0, x = 0.90, x = 0.990, x 3 = ,... x n = (the nth number having 9 until the nth position to the right of the dot and 0 after that). Show that the difference sequence.0 x 0,.0 x,.0 x,.0 x 3,....0 x n,... is a zero sequence. This implies that 0.9 =.0. Consider the chromatic octave set χ c = {c, d, d,...b} from c in just tuning (built upon the primes,3,5), as represented by points on the Euler space Q 3,i.e., c = (0, 0, 0), g = (,, 0), etc., see the graphic which was distributed in class. Consider the corresponding octave set χ g built by the same interval distances from g χ c. What is the intersection χ c χ g? Here is the solution, which you may draw from the image which I distributed in class. The intersection has three elements, shown in circles: 5

6 Use the representation of complex numbers x = Re(x)+i Im(x) to prove Re(x) = x+x and Im(x) = x x i. Prove that for complex numbers x, y, we have x y = x y by use of the definition x = x x, and by use of the fact that for any two non-negative real numbers u, v, we have u v = u v. Calculate the inverse x for x = + i, calculate x y for x = 3 i, y = 3 + i 4. 6 Graphs and Nerves Consider the following digraph: Find a cycle of length in this digraph. Write the cycle as sequence x x... x of its arrows with the shown names, the cycle starting at x and ending at x. Which of the following graphs (they are the complete graphs with, 3, 4, and 5 vertexes) are having an Euler cycle? Use the criterium, which we have discussed in class (degree of a vertex). And also draw such a cycle, if it exists. We have seen that the nerve of the triadic covering C (3) = {I, II,..., V II} of the major scale C = {c, d, e, f, g, a, b} yields a Moebius strip, as shown to the left in the following figure. Show that if we replace the triad I by the seventh triad I 7+ in this covering, the Moebius strip is enriched by a tetrahedron, as shown in red color in figure??. How does the nerve of the covering C (4) = {I 7+, II 7,..., V II 7 } by the tetradic (i.e., seventh) degrees look like? Consider the two sets A = {a, b, c}, B = {x, y} of cardinality 3 and, respectively. How many digraph morphisms f : Loop(A) Loop(B) are there? How many morphisms f : Loop(A) [] into the directed -chain [] are there? How many f : [] Loop(A)? Give explicit lists of such morphisms. 7 Algebra and Matrixes Let us consider the so-called affine functions on the integers Z. They are determined by the selection of two integers t, m and denoted by T t m. By definition, such a function T t m : Z Z evaluates to T t m(x) = t + mx. These functions are central in music theory. In fact: If we interpret Z as the set of pitches (keys or MIDI pitches) T t is just the transposition of a pitch by t semitones, verify this! What is the musical meaning of the affine function T if 60 denotes middle c? Make an image! Denote by Aff(Z) the set of all affine functions on the integers. Show that this set defines a submonoid of the monoid Set(Z, Z) of all set functions f : Z Z, together with the monoid operation =, i.e., the compositions of set functions. First check the defining properties of a monoid for (Set(Z, Z),, Id Z ) and then also for this set Aff(Z). Show in particular, that the neutral element of Set(Z, Z) is an affine function. And then exhibit 6

7 the shape of the composition T t m T s n! To calculate this shape, take an argument x Z and calculate the resulting value (T t m T s n)(x). Next, do calculate the subgroup Aff(Z) of invertible elements of Aff(Z). Describe this group in terms of musical operations if we interpret Z as the set of pitches as above. Find out whether this group Aff(Z) is commutative or not. Consider the symmetric group S 8, i.e., the group of permutations of the set {,, 3,... 8} of natural numbers. Consider the permutations p = q = Calculate the product q p (attention: do not forget that the permutation to the right comes first, the one to the left second!). Consider the set p = {Id, p, p, p 3,...} of all powers of p. Show that this is a subgroup. What is its order (= cardinality?) Cycle decomposition of permutations: To do so, start with the cycle starting at. Then take the smallest number not contained in this cycle, an so forth, until all cycles are calculated. As a preliminary exercise, we calculate the cycle decomposition of q. The cycle of is C = (, 4,, 6, 5). The smallest number not in this first cycle is 3. The cycle of 3 is C = (3, 8, 7), so the total decomposition is q = C C. Now calculate the cycle decomposition of the permutation p. Consider the symmetric group S 4 and its element 3 4 r = 4 3 Write r as a product of the transpositions (, ),, 3), (, 4) (we know from the last lesson that this is possible). Hint: first multiply r by a transposition such that the product leaves 4 fixed. This can be done by prepending the transposition (, 4) to r, i.e., 3 4 (, 4) r = (, 4) = = (, ) 3 4 Then multiply that product by a transposition such that 3 remains fixed, etc. Then use the fact that transpositions are their own inverses. How? In our case, we have the even simpler situation that 7

8 (, 4) r = (, ) Therefore, multiplying the equation by (, 4) from the left we get i.e., since (, 4) (, 4) = Id, (, 4) (, 4) r = (, 4) (, ) r = (, 4) (, ) and we are done. Consider the set G = {x, y, z, u, v, w} of cardinality 6, and a multiplication : G G G given by the table X Y x y z u v w x u v w x y z y v w x y z u z w x y z u v u x y z u v w v y z u v w x w z u v w x y Is commutative? Is it associative? Is there a neutral element? Is this a monoid? Is it even a group? Exhibit inverses of its elements if possible. 7. Generators of Groups and Cyclic Groups. If we are given any subset X G of a group G, there is a unique smallest subgroup H G such that X H. There are two proofs of this fact. The first is quite abstract, but quite immediate: Take the set Σ of all subgroups K G such that X K. For example, G Σ. So Σ is not the empty set. Then consider the intersection S = Σ of all these subgroups. Evidently, S contains X as a subset since every element of Σ does so. But S is also a subgroup. Verify this! Then this intersection is the smallest subgroup of G that contains X. The second proof is more constructive and concrete. If this group exists, it must contain not only all elements x X, but also all elements of the following types:. the neutral element e G of the group G,. all products x x... x k, where the x i are either elements of X or inverses thereof. But these two types of elements are already a subgroup of G. Prove this! So this is the smallest subgroup of G containing X as a subset. This subgroup is denoted by X and called the subgroup of G generated by X. If the set is a singleton, i.e., contains just one element x: X = {x}, then we also write x instead of {x} and call this group a cyclic group. If x is an finite group, its order of order( x ) (recall, that this is the cardinality of the group) is called the order of x and denoted by order(x). If x is an infinite group, the order of x is also called infinite. 8

9 Example. Take the additive group of integers G = Z. Show that we have Z =, i.e., Z is cyclic. Can you find another generator x such that Z = x? Example. Show that every subgroup H of Z is cyclic. To do so, procede as follows: First, look at the trivial subgroup H = {0} = 0Z. Check that this is cyclic (which is its generator?). Then, take a non-trivial subgroup H Z. This one contains at least one positive integer h. Why? (Think of all possibilities and use the fact that H is a subgroup and therefore contains the inverse element for each of its elements.) Then, take the smallest positive integer m H. Now look at any element x H. By the division theorem we know that we may write x = a m + r, with a remainder 0 r < m. But then, since x and a m are both elements of H, we must have r H. As m is the minimal positive number in H, we must have r = 0. This shows that all elements of H are (positive or negative) multiples of m, i.e., we have H = m, the subgroup which we usually denote by mz. Example 3. The above results allow us to completely classify the cyclic groups, i.e., to give a complete description of such groups. In fact, suppose that a subgroup H of any group G is cyclic. This means that we have an element x G such that H = x. Consider now the group homomorphism f : Z G, which is defined by sending z Z to x z, more precisely: f(0) = e if z > 0, f(z) = x z (= x x x... x, z times) if z < 0, f(z) = (x ) z (= x x x... x, z times) For example, f(3) = x 3 = x x x and f( ) = x = (x ) = x x. Verify that this is a group homomorphism! By the theorem about kernels and images of group homomorphisms, which we have discussed in class, we know that we have a canonical isomorphism of groups Z/Ker(f) Im(f). But evidently, Im(f) = x. Therefore we know that Z/Ker(f) x. Using the classifiation of all subgroups of Z, we know that Ker(f) = mz for m = 0 or a positive m Z. This means that there is a (unique) integer m 0 such that Z m = Z/mZ x If m = 0, no power of x is the neutral element e G, and we have Z 0 = Z/0Z = Z x If m > 0, then x m = e is the smallest positive power of x that yiels e. Let us make a number of examples of cyclic groups! Take the element p = (, 3, 5, 6) S 6. What is the group p? Calculate the order of p and then describe the group generated by p as a quotient group of the integers! 9

10 Take the element i C (imaginary unit of the complex numbers) and describe the group i generated by i in the unit circle U = {z z = } (group operation = product of complex numbers). Show that a cyclic group is always commutative. conclude from this that the symmetric group S 3 is not cyclic. Use the Lagrange formula (for a finite group G and any subgroup H G, we have order(g) = (G : H) order(h)) to answer the following question: Is there an element p S 4 or order 7? Use the Lagrange formula to show that any group of prime order is cyclic. The group of pitch classes Z is evidently cyclic (why?). Find all generators x such that Z = x. Show that the soubgroup of the affine group GL(Z) = Aff(Z) generated by an inversion T t.( ) is isomorphic to Z. Consider the normal subgroup T Z of translations in GL(Z). Show that the quotient group GL(Z)/T Z is isomorphic to Z. In the set Z Z of notes (x, y), whose first coordinates are time (integer values), and whose second coordinates are pitch (integer values), consider the group C of permutations of the set Z Z generated by the inversion at zero I, i.e., I(x, y) = (x, y) and the retrograde R, i.e., R(x, y) = ( x, y). Show that C Z Z, the latter being the so-called Kleinian 4 group. Hint: Define a bijection C Z Z, which is a group isomorphism. 7. Matrixes and Affine Maps We have discussed matrixes and their operations. Let us first check by examples, whether we have understood these operations. Take the following six matrixes: K =, L = 5 3, M = , N = , t = 3, P = 4 Calculate the following matrixes (calcuate first, then compare to our results!). K + L. K L. 0

11 3. K 4. M N 5. N M 6. L 7. K 8. (affine maps!) T t.k(p ) And here are the results: 0. K + L = 6. K L = 3. K = 4. M N = 8 4 ( 4 ) 0 6 ( 5.5 ) N M = L does not exist, since det(l) = 0 7. K exists since det(k) = is different from zero. We have K = 3 det(k) = a b =. (Recall that for a matrix K = with non-vanishing 5 5 c d ( d b determinant, its inverse is K = det(k) 8. T t.k(p ) = t + K P = 4 c a ).)

12 x Next, we would like to know which vector Z = of the plane R y is mapped to the vector P = under the bijection of the plane defined by the matrix S =. So the problem is to find P such that P = S Z. Since S is invertible (check!), we can solve this equation by multiplying both sieds by S : S P = S S Z. This means Therefore we have Z = S P = Z. = 0. Given the -matrix R = ( ), we may define a map (denoted by the same x symbol R) R : R R via R(x, y) = R. Use the rules of matrix arithmetic to verify y that this is a group homomorphism of the additive group R onto the additive group R. Describe the image group Im(R) and the kernel Ker(R)! f Consider the digraph = a b, and the diagram D : GL(Z ) of affine g maps on Z, defined by D(f) = T.5, D(g) = T.3. Calculate all networks of this diagram. Recall for this that a network is a pair (x a, x b ) of elements of Z (one for each vertex of ) such that we have x b = D(f)(x a ) and x a = D(g)(x b ). Check that these are exactly the two networks (, 0) and (8, 6), and as we know from the class, this number of networks must be a divisor of.