Solutions to Quiz 1 (April 17, 2008) 1. P, Q, R (P Q) R (P R) (Q R). P Q R (P Q) R (P R) (Q R) X T T T T T T T T T T T T T T T F T T F T T T F F T F F

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1 Quiz 1 Due at 9:00 p.m. on Thursday, April 17, 2008 Division: ID#: Name: 1. P, Q, R (P Q) R (P R) (Q R). P Q R (P Q) R (P R) (Q R) X T T T F T T F F T F T T T F F F F T T F F T F T F F T F F F F F (1) (7) p.44 (1)-(4) P Q ( P ) Q 3. (P R) (Q R) ( ),, 4. X P, Q, R,, Message HP

2 Solutions to Quiz 1 (April 17, 2008) 1. P, Q, R (P Q) R (P R) (Q R). P Q R (P Q) R (P R) (Q R) X T T T T T T T T T T T T T T T F T T F T T T F F T F F F T F F F T F T T T F T T T T T T F T T T T F F T T F F F T F F F F T F F F T T F T T T T F T T T T T T F F T F F T T F F F T F F T F F T F F T F F F T T F T T T F T T F F F F F F F T F F T F T F T F F (1) (7) p.44 (1)-(4) P Q ( P ) Q. 1.1 P Q R (7) (6) (P Q) R ( P Q) R (3)(5) ( P R) ( Q R) (7) (P R) (Q R). 3. (P R) (Q R) ( ),,. 1.1 (2), (6) S T ( (S T )) ( S T ) (P R) (Q R) ( P Q) R ( ( P Q) R). (P R) (Q R) ( P R) ( Q R) (P R) (Q R). 4. X P, Q, R,,. P, Q, R T, F, T T F P Q R F, T, F T F P Q R T X X (P Q R) ( P Q R)

3 Quiz 2 (Due at 9:00 p.m. on Thursday, April 24, 2008) Division: ID#: Name: A, B, C, D 1. (A B) C A (B C) Venn 2. (A B) C A (B C) A, B. C 3. (A B) C A (B C) Venn 4. A A B C D B D A Message HP

4 Solutions to Quiz 2 (April 24, 2008) A, B, C, D 1. (A B) C A (B C) Venn. 2. (A B) C A (B C) A, B. C. A = C = {1} B = (A B) C = = {1} = A (B C). Venn 3. (A B) C A (B C) Venn. A, B, C A B C X C = X C A C A (A B) C = (A B) C = (A C) (B C) A (B C).. x (A B) C(= {y y A B, y C}) x A (B C) X Y (A B) C A (B C) x A B x C x A B x A x B x A x A A (B C) x B x C x B C x B C A (B C) x A (B C) (A B) C A (B C) A C = A A C A C = A C = A (A B) C = A (B C) 4. A A B C D B D A. A a A b B A B C D (a, b) A B C D = {(c, d) c C, d D} b D b B B D A = B = {1}, A = C, D = = A B C D {1} = B D = a A A =

5 Quiz 3 (Due at 9:00 p.m. on Thursday, May 1, 2008) Division: ID#: Name: R Z a, b R b a Z a b a R [a] = {x R x a} 1. a b R For all a, b, c R, (i) a a, (ii) a b b a, (iii) a b b c a c. 2. a, b, c, d R a b c d a + c b + d 3. a, b, c, d R a b c d ac bd 4. a, b R 0 < b a < 1 [a] [b] = 5. S = {x R 0 x < 1} R = [a]. a S Message ICU HP

6 Solutions to Quiz 3 (May 1, 2008) R Z a, b R b a Z a b a R [a] = {x R a x} 1. a b R For all a, b, c R, (i) a a, (ii) a b b a, (iii) a b b c a c. (i) a a = 0 Z a a. (ii) a b b a Z a b = (b a) Z b a. (iii) a b b c b a Z c b Z c a = (b a) + (c b) Z a c. 2. a, b, c, d R a b c d a + c b + d. a b c d b a Z d c Z (b + d) (a + c) = (b a) + (d c) Z a + c b + d 3. a, b, c, d R a b c d ac bd. 0 1, < 2 < = 2 Z 4. a, b R 0 < b a < 1 [a] [b] =. x [a] [b] x a Z x b Z b a = (x a) (x b) Z 0 < b a < S = {x R 0 x < 1} R = [a].. a S [a] R R [a] a S x R n n x < n x n < 1 a = x n S x a = n Z a x x [a] = {y R a y} a S x [a] R [a] a S a S R = [a] a S 4 S R/Z = {[a] a S} S 1.5, 1.6 R/Z 2 [a] + [b] = [a + b] R/Z Z n = Z/nZ a S

7 Quiz 4 (Due at 9:00 p.m. on Thursday, May 15, 2008) Division: ID#: Name: 1. X, Y, Z f : X Y, g : Y Z f g g f : X Z (x g(f(x))) (a) f (b) g 2. R f : R R f(1) = 1, x 1 f(x) = x+1 x 1 (a) f f : R R (b) f (c) f Message HP

8 Solutions to Quiz 4 (May 15, 2008) 1. X, Y, Z f : X Y, g : Y Z f g g f : X Z (x g(f(x))) (a) f. h = g f x 1, x 2 X f(x 1 ) = f(x 2 ) x 1 = x 2 f(x 1 ) = f(x 2 ) h(x 1 ) = (g f)(x 1 ) = g(f(x 1 )) = g(f(x 2 )) = (g f)(x 2 ) = h(x 2 ) h = g f x 1 = x 2 f (b) g. X = Z = {1}, Y = {1, 2}, f(1) = 1, g(1) = g(2) = 1 h = g f h(1) = 1 X Z g 1 2 g(1) = g(2) X = Z g f g g 6 2. R f : R R f(1) = 1, x 1 f(x) = x+1 x 1 (a) f f : R R. (f f)(1) = f(f(1)) = 1, x 1 (f f)(x) = f( x + 1 x+1 x 1 ) = x x+1 x 1 1 = x x 1 x + 1 x + 1 = 2x 2 = x. x = 1, x 1 (f f)(x) = x f f = id R id R x R id R (x) = x (b) f. 1(a) f [ ] x 1 f(x) = x+1 x+1 x 1 1 x 1 = 1 x + 1 = x 1 2 = 0 x, y 1 f(x) f(y) f(x) = f(y) x+1 x 1 = y+1 y 1 (x + 1)(y 1) = (x 1)(y + 1) 2x = 2y x = y (c) f. y R f(f(y)) = y f(y) = x x = f(y) R f [ ] f(1) = 1 1 f y 1 x = y+1 y 1 x 1 f(x) = y (2(a) ) f (b)(c) f f (f f)(x) = x (c) (a)

9 Quiz 5 Due at 9:00 p.m. on Thursday, May 22, 2008 Division: ID#: Name: 1. f : R R f(xy) = xf(y) + yf(x) ( x, y R) (a) f(1) = 0 (b) u R, n N f(u n ) = nu n 1 f(u) 2. g : N R g(1) = g(2) = 1 g(n) = 1 2 (g(n 1) + 2/g(n 2)) ( n 3) 1 g(n) 2 Message HP

10 Solutions to Quiz 5 (May 22, 2008) 1. f : R R f(xy) = xf(y) + yf(x) ( x, y R) (a) f(1) = = 1 0 = f(1) f(1) = f(1 1) 1 f(1) = 1 f(1) + 1 f(1) 1 f(1) = f(1). f(1) = 0 (b) u R, n N f(u n ) = nu n 1 f(u). n = 1 f(u) = 1u 0 f(u) n = k 1 f(u k ) = ku k 1 f(u) f(u k+1 ) = f(u k u) = u f(u k ) + u k f(u) = u ku k 1 f(u) + u k f(u) = (k + 1)u k f(u). n f(u n ) = nu n 1 f(u). (a) x, y f(1) = 0 x, y f f(x+y) = f(x)+f(y) f(xy) = xf(y) + yf(x) R (derivation) R = R {0} L : R R (x f(x)/x) L(xy) = L(x) + L(y) L(1) = 0 L(x n ) = nl(x) log 2. g : N R g(1) = g(2) = 1 g(n) = 1 2 (g(n 1) + 2/g(n 2)) ( n 3) 1 g(n) 2. g(1) = g(2) = 1 n = 1, 2 1 g(n) 2 n 3 1 k < n 1 g(k) 2 1 g(k 1) g(k 2) 2 1 = 1 2 (1 + 1) 1 2 ( ) 2 g(k 1) + g(k 2) 1 (2 + 2) = 2. 2 n 1 g(n) 2. 1 n = k n = k + 1 k 1 k 2 k n = 1, 2 n = 3 g(n) = a n a 1 = a 2 = 1, a n = 1 2 (a n a n 2 ) 1 a n 2 2 a 1 = a 2 = 1

11 Quiz 6 (Due at 9:00 p.m. on Thursday, May 29, 2008) Division: ID#: Name: Z 11 = {[0], [1],..., [10]} n 1. n 10 n = a k a k 1 a 1 a 0 = a k 10 k + a k 1 10 k a a 0, 0 a i < 10 [n] = [( 1) k a k + ( 1) k 1 a k ( 1) 1 a 1 + ( 1) 0 a 0 ] 2. [n] [0] m [m][n] = [1] 3. d = gcd{526, 11} d = 526m + 11l m l 4. [n] [0] [n 10 ] = [1] 5. [ ] = [x] x (0 x 10) Message HP

12 Solutions to Quiz 6 (May 29, 2008) Z 11 = {[0], [1],..., [10]} n 1. n 10 n = a k a k 1 a 1 a 0 = a k 10 k + a k 1 10 k a a 0, 0 a i < 10 [n] = [( 1) k a k + ( 1) k 1 a k ( 1) 1 a 1 + ( 1) 0 a 0 ]. [x + y] = [x] + [y], [xy] = [x][y] [10] = [ 1] [n] = [a k 10 k + a k 1 10 k a a 0 ] = [a k ][10] k + [a k 1 ][10] k [a 1 ][10] + [a 0 ] = [a k ][ 1] k + [a k 1 ][ 1] k [a 1 ][ 1] 1 + [a 0 ] = [a k ][( 1) k ] + [a k 1 ][( 1) k 1 ] + + [a 1 ][( 1) 1 ] + [a 0 ][( 1) 0 ] = [( 1) k a k + ( 1) k 1 a k ( 1) 1 a 1 + ( 1) 0 a 0 ]. [3492] 11 = [ ] 11 = [5] , [n] [0] m [m][n] = [1]. [n] [0] n 11 gcd{n, 11} = = mn + l11 l, m [m][n] = [mn] = [1 11l] = [1] [m] [n] 1 [1] 1 = [1], [2] 1 = [6], [3] 1 = [4], [4] 1 = [3], [5] 1 = [9], [6] 1 = [2], [7] 1 = [8], [8] 1 = [7], [9] 1 = [5], [10] 1 = [10]. [ a] 1 = [ 1][a] 1 3. d = gcd{526, 11} d = 526m + 11l m l. 526 = , 11 = , 9 = gcd{526, 11} = 1 2 = 11 9, 9 = = = 9 4 (11 9) = (4 + 1) = 5 ( ) 4 11 = ( ) 11 = m = 5, l = [526] = [ ] = [ 2] = [9] [5][526] = [5][9] = [1] = = 1 [5] [9] 1 = [5] l, m 1 = 526m + 11l = 525m + 11l 526(m m ) = 11(l l) m m 526 l l l = l s m = m 11s l, m 4. [n] [0] [n 10 ] = [1]. [2] 2 = [4], [2] 3 = [4][2] = [8], [2] 4 = [8][2] = [5], [2] 5 = [5][2] = [ 1], [2] 6 = [ 1][2] = [ 2], [2] 7 = [ 2][2] = [ 4], [2] 8 = [ 4][2] = [ 8], [2] 9 = [ 8][2] = [ 5], [2] 10 = [ 5][2] = [1]. Z 11 [0] [n] = [2] m [n] 10 = ([2] m ) 10 = ([2] 10 ) m = [1] m = [1]. 5. [ ] = [x] x (0 x 10) = [3 10 ] = [1] x = 1 [ ] = [3 10 ] 6720 [3 5 ] = [3 5 ] = [9][9][3] = [ 2][ 2][3] = [1].

13 Quiz 7 (Due at 9:00 p.m. on Thursday, June 5, 2008) Division: ID#: Name: a, b R (a < b) (a, b) {x R a < x < b} 1. f : R (0, 1) (x 1 e x + 1 ) 2. a, b R (a < b) R (a, b) R (a, b) 3. Q + = {r Q r > 0} g : Q + N Q + N Q + N Message HP

14 Solutions to Quiz 7 (June 5, 2008) a, b R (a < b) (a, b) {x R a < x < b} 1. f : R (0, 1) (x 1 e x + 1 ). e x f x e x 0 f(x) 1 x e x f(x) 0 0 < y < 1 x 1 < x 2 f(x 1 ) > y f(x 2 ) < y f f(x) = y x f R (0, 1). f (x) = ex (e x + 1) 2 < 0. g(y) y = 1/(e g(y) + 1) e g(y) = 1/y 1 = (1 y)/y g : (0, 1) R (y log(1 y) log(y)) f g = id (0,1) f g f = id R f 2. a, b R (a < b) R (a, b) R (a, b). h h : (0, 1) (a, b) (x (b a)x + a) h h f : R (a, b) (x b a e x a). e x e x arctan(x) 3. Q + = {r Q r > 0} g : Q + N Q + N Q + N. r Q + r = p(r)/q(r), p(r), q(r) N gcd(p(r), q(r)) = 1 g : Q + N (r 2 p(r) 1 (2q(r) 1)) g 1 : Q + N N (r (p(r), q(r))) g 2 : N N N ((m, n) 2 m 1 (2n 1)) g = g 2 g 1 g 1 g 2 g g 1 g 3 : N N Q + ((m, n) m/n) g 3 g 1 = id Q + N Q (1) Q + n I(n) N Q + N. g 1 N N N

15 Quiz 8 (Due at 9:00 p.m. on Thursday, June 12, 2008) Division: ID#: Name: A = {x R 0 x 1} 1. X Y X = Y X Y 2. R A (Hint: Quiz 7) 3. R = A 4. A A = R R Message ICU ICU

16 Solutions to Quiz 8 (June 12, 2008) A = {x R 0 x 1} 1. X Y X = Y X Y. X Y f : X Y X = Y X Y g : X Y X Y X = Y X Y 2. R A (Hint: Quiz 7). Quiz 7 f : R (0, 1) (x 1/(e x + 1)) g : R [0, 1] = A (x 1/(e x + 1))) R A 3. R = A. h : A R (x x) A R A R X, Y X Y Y X X = Y 2 A = R 4. A A = R R. 3 A = R A R φ : A R Φ : A A R R ((a, b) (φ(a), φ(b))) φ(a) R, φ(b) R Φ A A R R Φ(a, b) = Φ(a, b ) (φ(a), φ(b)) = Φ(a, b) = Φ(a, b ) = (φ(a ), φ(b )) φ(a) = φ(a ) φ(b) = φ(b ) φ a = a b = b (a, b) = (a, b ) Φ (x, y) R R x, y R φ : A R φ(a) = x, φ(b) = y a, b A Φ(a, b) = (φ(a), φ(b)) = (x, y) Φ Φ : A A R R A A = R R

Solutions to Quiz 1 (April 20, 2007) 1. P, Q, R (P Q) R Q (P R) P Q R (P Q) R Q (P R) X T T T T T T T T T T F T F F F T T F T F T T T T T F F F T T F

Solutions to Quiz 1 (April 20, 2007) 1. P, Q, R (P Q) R Q (P R) P Q R (P Q) R Q (P R) X T T T T T T T T T T F T F F F T T F T F T T T T T F F F T T F Quiz 1 Due at 10:00 a.m. on April 20, 2007 Division: ID#: Name: 1. P, Q, R (P Q) R Q (P R) P Q R (P Q) R Q (P R) X T T T T T T F T T F T T T F F T F T T T F T F T F F T T F F F T 2. 1.1 (1) (7) p.44 (1)-(4)